Q
stringlengths 4
3.96k
| A
stringlengths 1
3k
| Result
stringclasses 4
values |
|---|---|---|
Lemma 2. Suppose \( f\left( z\right) \) is holomorphic in a circle \( \left| {z - {z}_{0}}\right| \leqq R \), and has at least \( n \) zeros in the circle \( \left| {z - {z}_{0}}\right| \leqq r < R \) (counting multiplicities). Assume \( f\left( {z}_{0}\right) \neq 0 \) . Then\n\n\[{\left\lbrack \left( R - r\right) /r\right\rbrack }^{n} \leqq B/\left| {f\left( {z}_{0}\right) }\right|\]\n\nwhere \( B \) is the maximum of \( \left| {f\left( z\right) }\right| \) on the larger circle. | Proof. We may assume \( {z}_{0} = 0 \) . Let\n\n\[f\left( z\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {z - {a}_{i}}\right) \varphi \left( z\right)\]\n\nwhere \( {a}_{i} \) are the zeros of \( f \) in the small circle. Then obviously on the large circle, we have\n\n\[\left| {\varphi \left( z\right) }\right| \leqq \left| {f\left( z\right) }\right| /{\left( R - r\right) }^{n} \leqq B/{\left( R - r\right) }^{n}.\]\n\nSince \( \left| {a}_{i}\right| \leqq r \) for each \( i \), we get our inequality by looking at \( \varphi \left( 0\right) \) . | Yes |
Proposition 3. Let a be a number \( > 0 \) . Then \( {L}^{\prime }/L\left( s\right) \) is bounded for \( \operatorname{Re}\left( s\right) = 1 + a \), and\n\n\[{\Gamma }^{\prime }/\Gamma \left( s\right) = \log s + O\left( {1/{\left| s\right| }^{2}}\right)\]\n\n\[\\text{for}\\operatorname{Re}\left( s\right) = 1 + a\\text{and}\\left| s\right| \\rightarrow \\infty \\text{.} \] | Proof. The first assertion follows at once from the product expansion of \( L \), and the second follows from Stirling’s formula (differentiating inside the integral giving the error term). | No |
Lemma 3. On the space of almost BCL functions, the preceding limit exists. | Proof. In a neighborhood of 0 , the denominator\n\n\[ \n\\left| {{e}^{x/2} - {e}^{-x/2}}\\right| \n\]\n\nbehaves like \( \\left| x\\right| \\left( {\\;\\operatorname{mod}\\{x}^{2}\\right) \) . It will then suffice to prove our assertion when we replace this denominator by \( \\left| x\\right| \) and \( \\beta \) by a function which tends rapidly to 0 at infinity.\n\nUsing linearity, consider first the case of a characteristic function. If the origin does not lie in its interval, then the limit clearly exists. If the origin lies in the interval, then we are led to consider an integral of type\n\n\[ \n\\varphi \\left( \\lambda \\right) = {\\int }_{0}^{b}\\frac{\\left( 1 - {e}^{-{\\lambda x}}\\right) }{x}{dx},\\;b > 0,\\;\\lambda \\geqq 1. \n\]\n\nWe can differentiate under the integral sign, and taking \( b = 1 \) for simplicity, we get\n\n\[ \n{\\varphi }^{\\prime }\\left( \\lambda \\right) = \\frac{1}{\\lambda } - \\frac{{e}^{-\\lambda }}{\\lambda } \n\]\n\nHence \( \\varphi \\left( \\lambda \\right) = \\log \\lambda + \) an integral which converges for \( \\lambda \\rightarrow \\infty \) . From this it is clear that the term \( {2\\beta }\\left( 0\\right) \\log \\lambda \) will cancel, and leave an integral whose limit exists as \( \\lambda \\rightarrow \\infty \) .\n\nNext, suppose that \( \\beta \) is BCL and tends rapidly to 0 at infinity. Decomposing \( \\beta \) into its odd and even parts, and subtracting a characteristic function we may assume that \( \\beta \) is even and \( \\beta \\left( 0\\right) = 0 \) . In that case, \( \\beta \\left( x\\right) /\\left| x\\right| \) is bounded in some neighborhood of the origin, and hence the integral\n\n\[ \n{\\int }_{-\\infty }^{\\infty }\\left( {1 - {e}^{-\\lambda \\left| x\\right| }}\\right) \\frac{\\beta \\left( x\\right) }{\\left| x\\right| }{dx} \n\]\n\nhas a limit as \( \\lambda \\rightarrow \\infty \) . The term involving \( {2\\beta }\\left( 0\\right) \\log \\lambda \) is 0, and so our assertion is proved in that case.\n\nLet \( \\beta \) be a BCL function. If we write\n\n\[ \n\\beta \\left( x\\right) = \\beta \\left( x\\right) - \\beta \\left( 0\\right) + \\beta \\left( 0\\right) \n\]\nwe obtain by linearity\n\n\[ \nW\\left( \\beta \\right) = \\beta \\left( 0\\right) W\\left( 1\\right) + {\\int }_{-\\infty }^{+\\infty }\\frac{\\beta \\left( x\\right) - \\beta \\left( 0\\right) }{\\left| {e}^{x/2} - {e}^{-x/2}\\right| }{dx}. \n\]\n\nThere is no convergence problem about this last integral, and thus we find: | Yes |
Lemma 5. Let \( \\left\\{ {\\beta }_{n}\\right\\} \) be a sequence of \( {BCL} \) functions, converging to a \( {BCL} \) function \( \\beta \) . Assume also that the functions \( \\left\\{ {\\beta }_{n}\\right\\} \) are uniformly bounded, that the convergence is uniform on every compact set, and that the Lipshitz constants \( \\operatorname{lip}{\\beta }_{n} \) are bounded on every compact set. Then \( W\\left( {\\beta }_{n}\\right) \) converges to \( W\\left( \\beta \\right) \) . | Proof. We write for each \( n \) ,\n\n\[ \n{\\beta }_{n}\\left( x\\right) = {\\beta }_{n}\\left( x\\right) - {\\beta }_{n}\\left( 0\\right) + {\\beta }_{n}\\left( 0\\right) .\n\] \n\nThen \( {\\beta }_{n}\\left( 0\\right) \) converges to \( \\beta \\left( 0\\right) \) . This reduces the proof to considering the sum of the integrals \n\n\[ \n\\int \\frac{{\\beta }_{n}\\left( x\\right) - {\\beta }_{n}\\left( 0\\right) }{\\left| {e}^{x/2} - {e}^{-x/2}\\right| }{dx} \n\] \n\nover intervals \n\n\[ \nA \\leqq \\left| x\\right| \n\] \n\n\[ \n\\epsilon \\leqq \\left| x\\right| \\leqq A \n\] \n\n\[ \n\\left| x\\right| \\leqq \\epsilon \n\] \n\ntaking \( \\epsilon > 0 \) small and \( A \) large. For \( A \) large, the exponential function in the denominator makes the integral small. For \( \\epsilon \) small, the last integral has a small value in view of the uniform bound for the Lipshitz constants. The integral in the middle range is then close to the corresponding integral for \( \\beta \\left( x\\right) - \\beta \\left( 0\\right) \) . Thus our lemma is clear. | Yes |
Lemma 7. The functional \( W \) is a distribution. Let \( \beta \) be a BCL function. The convolution of \( W \) with \( {T}_{\beta } \) (the distribution represented by \( \beta \) ) is represented by the function whose value at \( x \) is \( W\left( {{\beta }^{ - }x}\right) \) . Symbolically, \[ \left( {W * {T}_{\beta }}\right) \left( x\right) = W\left( {\beta }^{ - }\right) . \] This function is continuous. | Proof. If \( T \) is a distribution, which is represented outside some compact set by a function tending exponentially to 0 at infinity, and \( \alpha \) is a \( {C}^{\infty } \) - function which is bounded, then by the theory of distributions, one knows that \[ T * {T}_{\alpha } \] is represented by the function \( T\left( {{\alpha }^{ - }{}_{x}}\right) \), which has a meaning in this case. We can apply this result to the functions \( \beta * {\rho }_{n} \), and hence our lemma follows. [Cf. TD, Theorem XI of Chapter VI, §4 and formula (VI, 1; 2).] | No |
Lemma 8. Let \( x \) be a characteristic function of an interval which does not hove 0 as its endpoints. Then the distribution \( W * {T}_{\chi } \) is represented by a \( {C}^{\infty } \) -function locally at every point other than the endpoints of the interval, and its value at such a point \( x \) is the value \( W\left( {{x}^{ - }x}\right) \) . | Proof. This follows from the general properties of convolutions of distributions, e.g. TD, Chapter VI, Theorem III of §3 and Theorem XI of §4. | No |
Lemma 9. The convergence of this limit is uniform on every compact set. | Proof. The first assertion is clear. As to the second, observe that the sum in the expression for \( {\mathrm{q}}_{M} \) is bounded from below by\n\n\[ \frac{n + \frac{1}{2}}{{\left( n + \frac{1}{2}\right) }^{2} + {t}^{2}} \leqq \frac{1}{n + \frac{1}{2}}. \]\n\nSay \( t > 0 \) . For \( M \leqq t \) the expression for \( {\mathrm{Y}}_{M} \) is bounded by \( \log M \leqq \log t \) . For \( M \geqq t \), we observe:\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{M} \geqq \mathop{\sum }\limits_{{n = t}}^{M} \geqq \log M - \log t - \text{ constant,}\]\n\nwhich gives us again what we want. | No |
Lemma 10. As a distribution, \[ {\widehat{T}}_{\mathrm{q}} = \widehat{\mathrm{q}} = - {\pi W} \] | Proof. The boundedness condition of Lemma 9 insures that the limit of the Fourier transforms is the Fourier transform of the limit. (Use TD, Example 3 of Chapter VII, §7.) | No |
Proposition 4. Let \( F \) be a function satisfying properties (A) and (B). Let\n\n\[ \n\psi = {\widehat{F}}^{ - }\text{. }\n\]\n\nThen \( \langle \psi ,\Psi \rangle \) exists, and\n\n\[ \n\langle \psi ,\mathrm{T}\rangle = \mathop{\lim }\limits_{{M \rightarrow \infty }}\left\langle {F,{\widehat{\mathrm{T}}}_{M}}\right\rangle\n\]\n\nor by what we have just seen, it is equal to \( - {\pi W}\left( F\right) \) . | Proof. Writing \( F \) as a sum of an even function and an odd function, i.e.\n\n\[ \nF\left( x\right) = \frac{F\left( x\right) + F\left( {-x}\right) }{2} + \frac{F\left( x\right) - F\left( {-x}\right) }{2},\n\]\n\nhas the same effect on the Fourier transform. Thus it suffices to prove our proposition for even and odd functions separately. But \( \mathbf{Y}\left( t\right) \) is even. Thus for odd functions \( F \), the integral \( \langle \psi ,\Psi \rangle \) is 0 . The limit on the right is also equal to 0 (each term being 0) and so our assertion is trivial.\n\nWe may therefore assume from now on that \( F \) is even. In particular, \( F \) is continuous at the origin.\n\nWe shall now prove the assertion of Proposition 4 for a more general type of function. Namely, we assume that \( F \) satisfies the following conditions:\n\n(i) \( F \) is almost BCL, and is even.\n\n(ii) \( F \) is in \( {\mathfrak{L}}_{2} \), it is differentiable except at its points of discontinuity, and its derivative is in \( {\mathfrak{L}}_{2} \) .\n\nAny even function satisfying conditions (A) and (B) also satisfies conditions (i) and (ii). | Yes |
Denoting as usual by \( {T}_{f} \) the distribution represented by a function \( f \), we have\n\n\[ {\widehat{T}}_{\psi \mathrm{q}} = {\widehat{T}}_{\mathrm{q}} * {T}_{F} \] | Proof. This follows from the theory of distributions, and hypotheses (i) and (ii), because \( W \) decreases rapidly and \( {T}_{f} \) is tempered \( ({TD} \) , Theorem XV of Chapter VII, §8). | No |
Lemma 12. Let \( f \) be the function \( \psi \mathrm{T} \) . Except at \( \pm {a}_{\nu },{\widehat{T}}_{f} \) is represented by the continuous function given by the integral\n\n\[{\int }_{-\infty }^{+\infty }f\left( t\right) {e}^{-{itx}}{dt}\] | Proof. Let \( A \) be a compact interval, \( - T \leqq t \leqq T \), and let \( {f}_{A} \) be the function \( f \) multiplied by the characteristic function of \( A \) . Our statement is true if we replace \( f \) by \( {f}_{A} \) . Hence \( {T}_{fA} \) approaches \( {T}_{f} \) as a tempered distribution, and consequently, \( {\widehat{T}}_{fA} \) approaches \( {\widehat{T}}_{f} \) also as a tempered distribution, hence as a distribution locally at each \( x \neq \pm {a}_{\nu } \) . But the integral expressing \( {\widehat{T}}_{fA} \) converges to\n\n\[{\int }_{-\infty }^{+\infty }f\left( t\right) {e}^{-{itx}}{dt}\]\n\nuniformly (hence as a distribution) on every compact set not containing the \( \pm {a}_{\nu } \) . Hence \( {\widehat{T}}_{f} \) is represented by this integral in the desired range. | Yes |
Theorem 2.1 If \( {R}_{0} \leq 1 \), then \( {E}_{f} \) is globally asymptotically stable. | Proof Let us consider the Lyapunov function\n\n\[ \n{U}_{f}\left( {T, I, V}\right) = \frac{1}{1 + a{T}_{0}}\left( {T - {T}_{0} - {T}_{0}\ln \frac{T}{{T}_{0}}}\right) + I + \frac{p}{k}V + \frac{m}{h}C. \]\n\nIt is easily seen that \( {U}_{f}\left( {T, I, V, C}\right) \geq 0 \) and \( {U}_{f}\left( {T, I, V, C}\right) = 0 \) if and only if \( T = {T}_{0}, I = V = C = 0. \) We now compute the time derivative of \( {U}_{f} \) along the solutions of (0.2). One then has\n\n\[ \n{\dot{U}}_{f}\left( t\right) = \frac{1}{1 + a{T}_{0}}\dot{T} - \frac{1}{1 + a{T}_{0}}\frac{{T}_{0}}{T}\dot{T} + \dot{I} + \frac{p}{k}\dot{V} + \frac{m}{h}\dot{C} = \]\n\n\[ \n\frac{1}{1 + a{T}_{0}}\left( {\lambda - {dT} - \frac{\beta TV}{(1 + {aT})(1 + {bV})}}\right) - \frac{1}{1 + a{T}_{0}}\frac{{T}_{0}}{T}\left( {\lambda - {dT} - \frac{\beta TV}{(1 + {aT})(1 + {bV})}}\right) + \]\n\n\[ \n\frac{\beta TV}{\left( {1 + {aT}}\right) \left( {1 + {bV}}\right) } - \frac{p\mu }{k}V - \frac{mn}{h}C = \]\n\n\n\[ \n\frac{1}{1 + a{T}_{0}}\left\lbrack {-d\left( {T - {T}_{0}}\right) + d\frac{{T}_{0}}{T}\left( {T - {T}_{0}}\right) }\right\rbrack + \frac{1}{1 + a{T}_{0}}\frac{\beta {T}_{0}V}{\left( {1 + {aT}}\right) \left( {1 + {bV}}\right) } + \frac{\beta TV}{\left( {1 + {aT}}\right) \left( {1 + {bV}}\right) } - \]\n\n\[ \n\frac{1}{1 + a{T}_{0}}\frac{\beta TV}{\left( {1 + {aT}}\right) \left( {1 + {bV}}\right) } - \frac{p\mu }{k}V - \frac{mn}{h}C = \frac{-d}{T\left( {1 + a{T}_{0}}\right) }{\left( T - {T}_{0}\right) }^{2} + \]\n\n\[ \n\frac{p\mu }{k}\frac{V}{(1 + {bV})}\left\lbrack {\frac{k}{p\mu }\left( {\frac{ - 1}{1 + a{T}_{0}}\frac{\beta T}{1 + {aT}} + \frac{\beta {T}_{0}}{1 + a{T}_{0}}\frac{1}{1 + {aT}} + \frac{\beta T}{1 + {aT}}}\right) - (1 + {bV})}\right\rbrack - \frac{mn}{h}C = \]\n\n\[ \n- \left\lbrack {\frac{d}{T\left( {1 + a{T}_{0}}\right) }{\left( T - {T}_{0}\right) }^{2} + \frac{p\mu b}{k\left( {1 + {bV}}\right) }{V}^{2}}\right\rbrack + \frac{p\mu }{k}\frac{V}{\left( 1 + bV\right) }\left( {{R}_{0} - 1}\right) - \frac{mn}{h}C. \]\n\nIt is obvious that \( {R}_{0} \leq 1 \) ensure \( {\dot{U}}_{f} \leq 0 \) for all \( T, I, V, C > 0 \) . Hence, the uninfected steady \( {E}_{f} \) is stable. And \( {\dot{U}}_{f} = 0 \), when \( T = {T}_{0} \) and \( V = C = 0 \) . Let \( {\sum }_{0} \) be the largest invariant set in set\n\n\[ \n\sum = \{ T, I, V, C)\left| {{\dot{U}}_{f}\left( {T, I, V, C}\right) = 0\} = \{ T, I, V, C)}\right| T = {T}_{0}, I \geq 0, V = C = 0\} . \]\n\nWe have from the third equation of (0.2) that \( {\sum }_{0} = \left\{ {E}_{f}\right\} \) . By the Lyapunov-LaSalle invariance \( {\text{principle}}^{\left\lbrack {15},{16}\right\rbrack },{E}_{f} \) is global asymptotically stable. | Yes |
Theorem 2.2 Assume that \( f : \left\lbrack {0, T}\right\rbrack \times R \rightarrow R \) is continuous and satisfies\n\n\[ \mathop{\lim }\limits_{{u \rightarrow 0}}\mathop{\sup }\limits_{{t \in \left\lbrack {0, T}\right\rbrack }}\left| \frac{f\left( {t, u}\right) }{u}\right| = m \]\n\nwhere \( m \) is a positive constant.\n\nSuppose also that \( \rho ,\lambda \) and \( m \) satisfy\n\n\[ \frac{\left( {{e}^{\varrho T} - 1}\right) \lambda \left( {m + 1}\right) {T}^{2}}{\rho \left( {{e}^{\varrho T} + 1}\right) } < 4 \]\n\nThen the anti-periodic boundary value problem for the third-order differential equation(0.2) has at least one solution. | In order to use Leray-Schauder nonlinear alternative, we set \( r < \delta \) . Using the same method in the proof of Theorem 2.1, we can prove \( A\overline{{K}_{r}} \subseteq \overline{{K}_{r}} \) is a completely continuous operator. Taking \( u \in \partial {K}_{r} \) such that \( u = {\mu Au},0 < \mu < 1 \) . By the definition of \( {K}_{r} \) and the operator \( A \), we have\n\n\[ \parallel u\parallel = \mu \parallel {Au}\parallel < \parallel {Au}\parallel .\n\nOn the other hand, according to the proof of Theorem 2.1 and the assumption of Theorem 2.2, we\n\nhave\n\[ \left| {{Au}\left( t\right) }\right| \leq \frac{\left( {{e}^{\rho T} - 1}\right) \lambda \left( {m + \varepsilon }\right) {T}^{2}}{{4\rho }\left( {{e}^{\rho T} + 1}\right) }\parallel u\parallel < \frac{\left( {{e}^{\rho T} - 1}\right) \lambda \left( {m + 1}\right) {T}^{2}}{{4\rho }\left( {{e}^{\rho T} + 1}\right) }\parallel u\parallel \leq r.\n\nHence, we have\n\n\[ \parallel u\parallel < \parallel {Au}\parallel = \mathop{\max }\limits_{{t \in \left\lbrack {0, T}\right\rbrack }}\left| {{Au}\left( t\right) }\right| \leq r.\n\nIt is a contradiction to \( u \in \partial K \) . By Lemma 1.3, the operator \( A \) has at least one fixed point, which implies the anti-periodic boundary value problem for the third-order differential equation (0.2) has at least one solution. The Theorem 2.2 is proved. | Yes |
Consider the following anti-periodic boundary value problem for the third-order differential equation\n\[ \left\{ \begin{array}{l} {u}^{\prime \prime \prime } - {u}^{\prime } + \left( {t + 1}\right) {u}^{2} = 0,\;t \in J = \left\lbrack {0,2}\right\rbrack , \\ {u}^{\prime }\left( 0\right) + {u}^{\prime }\left( 2\right) = {u}^{\prime \prime }\left( 0\right) + {u}^{\prime \prime }\left( 2\right) = 0, \\ u\left( 0\right) + u\left( 2\right) = 0. \end{array}\right. \] | Here, \( \rho = \lambda = 1, T = 2, f\left( {t, u}\right) = \left( {t + 1}\right) {u}^{2} \) is continuous and\n\[ \mathop{\lim }\limits_{{u \rightarrow 0}}\mathop{\sup }\limits_{{t \in \left\lbrack {0,2}\right\rbrack }}\left| \frac{f\left( {t, u}\right) }{u}\right| = 0,\;\frac{\left( {{e}^{\varrho T} - 1}\right) \lambda {T}^{2}}{\rho \left( {{e}^{\varrho T} + 1}\right) } = \frac{4\left( {{e}^{2} - 1}\right) }{\left( {e}^{2} + 1}\right) < 4. \]\nObviously. The anti-periodic boundary value problem for the third-order differential equation(3.1) satisfies all assumptions of Theorem 2.1. Hence, it has at least one solution. | Yes |
Consider the following anti-periodic boundary value problem for the third-order differential equation\n\[ \n\\left\\{ \\begin{array}{l} {u}^{\\prime \\prime \\prime } - 4{u}^{\\prime } + \\frac{1}{2}\\sin \\left( {u{e}^{ut}}\\right) = 0,\\;t \\in J = \\left\\lbrack {0,\\pi }\\right\\rbrack , \\\\ {u}^{\\prime }\\left( 0\\right) + {u}^{\\prime }\\left( \\pi \\right) = {u}^{\\prime \\prime }\\left( 0\\right) + {u}^{\\prime \\prime }\\left( \\pi \\right) = 0, \\\\ u\\left( 0\\right) + u\\left( \\pi \\right) = 0. \\end{array}\\right.\n\] | Obviously. The anti-periodic boundary value problem for the third-order differential equation(3.2) satisfies all assumptions of Theorem 2.2. Hence, it has at least one solution. | No |
Theorem 1.1 \\( \\; \\) Let \\( \\left( {{u}_{1}\\left( {x, t}\\right) ,{u}_{2}\\left( {x, t}\\right) ,{u}_{3}\\left( {x, t}\\right) ,{u}_{4}\\left( {x, t}\\right) }\\right) \\in \\lbrack C\\left( {\\overline{\\Omega }\\times \\lbrack 0, T}\\right) )\\xrightarrow[]{\\cap {C}^{2,1}}\\left( {\\Omega \\times \\left( {0, T}\\right) }\\right) {\\rbrack }^{3}\\;(\\forall T \\) \\( > 0) \\) be the solution of (0.1) with the initial values \\( {u}_{10},{u}_{20},{u}_{30},{u}_{40} \\geq 0\\left( { \\neq 0}\\right) \\), then system (0.1) has a unique nonnegative solution, and satisfy\n\n\\[ \n0 \\leq {u}_{i}\\left( {x, t}\\right) \\leq {M}_{i},\\;i = 1,2,3.\n\\]\n\nwhere \\( {M}_{1}{M}_{2},{M}_{3} \\) and \\( {M}_{4} \\) are a positive constant depending on the coefficient of \\( \\left( {0.1}\\right) \\) and the initial values \\( {u}_{i0}\\left( {i = 1,2,3,4}\\right) \\) . | Proof Since the reaction functions of (0.1) are smooth in \\( {\\Re }_{ + }^{3} \\), from the standarded theory of PDE shows that has a unique local solution. By the strong maximum principle (see [6]), we know that if \\( {u}_{i0}\\left( x\\right) \\neq 0\\left( {i = 1,2,3,4}\\right) \\), then \\( {u}_{i}o\\left( {x, t}\\right) > 0 \\) on \\( \\overline{\\Omega } \\times \\left( {0, T}\\right) \\) for all \\( t > 0 \\) .\n\nLet \\( z = {u}_{1} + {u}_{2} + {u}_{3} + {u}_{4} \\), from partial integral formula and Young inequality we have\n\n\\[ \n\\frac{d}{dt}{\\int }_{\\Omega }{zdx} = {\\int }_{\\Omega } - {r}_{1}{u}_{1} - {r}_{2}{u}_{3} + {b}_{1}{u}_{2}\\left( t\\right) - {a}_{11}\\left( t\\right) {u}_{2}^{2} + {b}_{2}{u}_{4} - {a}_{22}{u}_{4}^{2}\\left( t\\right) {dx} \\leq\n\\]\n\n\\[ \n- f{\\int }_{\\Omega }{zdx} + \\left| \\Omega \\right| {\\left( {b}_{1} + f\\right) }^{2}/{4f}{a}_{11} + \\Omega \\mid {\\left( {b}_{2} + f\\right) }^{2}/{4f}{a}_{22}.\n\\]\n\nThus \\( \\| z\\left( {x, t}\\right) {\\| }_{1} \\leq {M}_{1}\\overset{ = }{ = }{M}_{2}\\overset{ = }{ = }\\max \\{ \\mid \\Omega \\mid {\\left( a + f\\right) }^{2}/{4a\\rho f},\\| {u}_{10}{\\| }_{\\infty } + \\| {u}_{20}{\\| }_{\\infty } + \\| {u}_{30}{\\| }_{\\infty }\\overset{ + }{ = }\\| {u}_{40}{\\| }_{\\infty }\\} ,\\;\\forall t > \\) \\( 0 \\), where \\( f = \\min \\left( {{r}_{1},{r}_{2}}\\right) \\), so \\( {\\begin{Vmatrix}{u}_{i}\\left( t\\right) \\end{Vmatrix}}_{\\infty } \\leq {M}_{i}\\left( {i = 1,2,3,4}\\right) \\) . From [7, Section 3.5] we know that there exists a positive constant \\( {M}_{i}\\left( {i = 1,2,3,4}\\right) \\) which depends only on the coefficients of (0.1), \\( \\Omega \\) and \\( {\\begin{Vmatrix}{u}_{i}\\left( 0\\right) \\end{Vmatrix}}_{\\infty }\\left( {i = 1,2,3,4}\\right) \\), but not depend on t, such that so the proof of Theorem1.1 is competed. | Yes |
Theorem 1.1 For the model (1.3), if \( g \leq k \leq g + m,{v}_{1} \leq h \), then \( 0 \leq {c}_{i0}\left( t\right) \leq 1,0 \leq \) \( {c}_{e}\left( t\right) \leq 1 \), for any \( t \in \left\lbrack {0, T}\right\rbrack \) . | Proof The proof is completed by the standard argument [5], we omit it here. | No |
Theorem 2.2 If \( \left( {{u}^{ * },{v}^{ * }}\right) \) is an optimal control and \( \left( {{p}^{ * },{c}_{0}^{ * },{c}_{e}^{ * }}\right) \) is the corresponding optimal state, then\n\n\[ \n{u}_{i}^{ * }\left( {a, t}\right) = {\mathcal{L}}_{i}\left( \frac{\left\lbrack {{w}_{i}\left( {a, t}\right) - {q}_{i}\left( {a, t}\right) }\right\rbrack {p}_{i}^{ * }\left( {a, t}\right) }{{C}_{i}}\right) \;i = 1,2,3,\text{ a.e. in }Q, \n\]\n\n(2.11)\n\n\[ \n{v}^{ * }\left( t\right) = {\mathcal{L}}_{4}\left( \frac{{q}_{7}\left( t\right) }{{C}_{4}}\right) \;\text{ a.e. in }\left( {0, T}\right) , \n\]\n\nwhere\n\n\[ \n{\mathcal{L}}_{j}\left( x\right) = \left\{ \begin{array}{ll} 0, & x < 0 \\ x, & 0 \leq x \leq {H}_{j} \\ {H}_{j} & x > {H}_{j} \end{array}\right. ,\;j = 1,2,3,4. \n\]\n\n\( \left( {2.12}\right) \)\n\nand \( \left( {{q}_{1},{q}_{2},\cdots ,{q}_{7}}\right) \) is the solution of following adjoint system corresponding to \( \left( {{u}^{ * },{v}^{ * }}\right) \)\n\n\[ \n\begin{cases} \frac{\partial {q}_{i}}{\partial a} + \frac{\partial {q}_{i}}{\partial t} & = \left\lbrack {{\mu }_{i}\left( {a,{c}_{i0}^{ * }\left( t\right) }\right) + {\lambda }_{ik}\left( {a, t}\right) {P}_{k}^{ * }\left( t\right) + {u}_{i}^{ * }\left( {a, t}\right) }\right\rbrack {q}_{i}\left( {a, t}\right) + \\ & \left\lbrack {{k}_{1}{c}_{e}^{ * }\left( t\right) - {g}_{1}{c}_{i0}^{ * }\left( t\right) }\right\rbrack {q}_{7}\left( t\right) - {q}_{i}\left( {0, t}\right) {\beta }_{i}\left( {a,{c}_{i0}^{ * }\left( t\right) }\right) + {w}_{i}\left( {a, t}\right) {u}_{i}^{ * }\left( {a, t}\right) , \\ \frac{\mathrm{d}{q}_{3 + i}}{\mathrm{\;d}t} & = {\int }_{0}^{{a}_{ + }}\frac{\partial {\mu }_{i}\left( {a,{c}_{i0}^{ * }\left( t\right) }\right) }{\partial {c}_{i0}}{p}_{i}^{ * }\left( {a, t}\right) {q}_{i}\left( {a, t}\right) \mathrm{d}a + \left( {g + m}\right) {q}_{3 + i}\left( t\right) - {g}_{1}{P}_{i}^{ * }\left( t\right) {q}_{7}\left( t\right) - \\ & {q}_{i}\left( {0, t}\right) {\int }_{0}^{{a}_{ + }}\frac{\partial {\beta }_{i}\left( {a,{c}_{i0}^{ * }\left( t\right) }\right) }{\partial {c}_{10}}{p}_{i}^{ * }\left( {a, t}\right) \mathrm{d}a, \\ \frac{\mathrm{d}{q}_{7}}{\mathrm{\;d}t} & = - k\mathop{\sum }\limits_{{i = 1}}^{3}{q}_{4}\left( t\right) + {k}_{1}{P}^{ * }\left( t\right) {q}_{7}\left( t\right) + h{q}_{7}\left( t\right) , \\ {q}_{i}\left( {a, T}\right) & = {q}_{i}\left( {a +, t}\right) = 0,\;{q}_{i}\left( T\right) = 0,\;i = 1,2,3, j = 4,\cdots ,7. \end{cases} \n\]\n\n\( \left( {2.13}\right) \) | Proof The existence of a unique bound solution to the system (2.13) can be treated in a same manner as that for (1.3). For any given \( \left( {{\nu }_{1},{\nu }_{2}}\right) \in {\mathcal{F}}_{\mathcal{U}}\left( {{u}^{ * },{v}^{ * }}\right) \) (the tangent cone of at \( \left( {{u}^{ * },{v}^{ * }}\right) ,\;{u}^{ * } = \left( {{u}_{1}^{ * },\cdots ,{u}_{3}^{ * }}\right) ,\;{\nu }_{1} = \left( {{\nu }_{11},\cdots ,{\nu }_{31}}\right) ),\;\left( {{u}^{ * } + \varepsilon {\nu }_{1},{v}^{ * } + \varepsilon {\nu }_{2}}\right) \in \mathcal{U} \) provided that \( \varepsilon \) small enough[14, p21]. Then from \( J\left( {{u}^{ * } + \varepsilon {\nu }_{1},{v}^{ * } + \varepsilon {\nu }_{2}}\right) \leq J\left( {{u}^{ * },{v}^{ * }}\right) \) it is derived that\n\n\[ \n\mathop{\sum }\limits_{{i = 1}}^{3}{\int }_{0}^{T}{\int }_{0}^{{a}_{ + }}{w}_{i}\left( {{u}_{i}^{ * } + \varepsilon {\nu }_{i1}}\right) {p}_{i}^{\varepsilon }\mathrm{d}a\mathrm{d}t - \frac{1}{2}\mathop{\sum }\limits_{{i = 1}}^{3}{\int }_{0}^{T}\left\lbrack {{\int }_{0}^{{a}_{ + }}{C}_{i}{\left( {u}_{i}^{ * } + \varepsilon {\nu }_{i1}\right) }^{2}\mathrm{\;d}a - {C}_{4}{\left( {v}^{ * } + \varepsilon {\nu }_{2}\right) }^{2}}\right\rbrack \mathrm{d}t \leq \n\]\n\n\[ \n\mathop{\sum }\limits_{{i = 1}}^{3}{\int }_{0}^{T}{\int }_{0}^{{a}_{ + }}{w}_{i}{u}_{i}^{ * }\left( {a, t}\right) {p}_{i}^{ * }\mathrm{\;d}a\mathrm{\;d}t - \frac{1}{2}\mathop{\sum }\limits_{{i = 1}}^{3}{\int }_{0}^{T}\left\lbrack {{\int }_{0}^{{a}_{ + }}{C}_{i}{u}^{*2}\left( {a, t}\right) | Yes |
Consider the delay integral equation\n\n\[ \nx\left( t\right) = {\int }_{t - \sigma }^{t}\frac{a\left( s\right) }{1 + {x}^{2}\left( s\right) }{ds},\;t \in \mathbb{R}.\n\]\n\nwhich is a model arising in an epidemic problem. If \( a\left( t\right) \in {WPAA}\left( {\mathbb{R},{\mathbb{R}}^{ + },{\rho }_{1},{\rho }_{2}}\right) \) with \( {\rho }_{1} = {e}^{t} \) , \( {\rho }_{2} = 1 + {t}^{2} \), and assume \( \mathop{\inf }\limits_{{t \in \mathbb{R}}}a\left( t\right) = {a}_{0} > 0,\mathop{\sup }\limits_{{t \in \mathbb{R}}}a\left( t\right) = {a}_{1} < 1/\sigma \), then \( f\left( {t, x}\right) = \frac{a\left( t\right) }{1 + {x}^{2}} \in \) \( {WPAA}\left( {\mathbb{R} \times {\mathbb{R}}^{ + },{\mathbb{R}}^{ + },{\rho }_{1},{\rho }_{2}}\right) \) and \( f\left( {t, \cdot }\right) \) is nonincreasing in \( {\mathbb{R}}^{ + } \), i.e., \( \left( {H}_{2}\right) \) and \( \left( {H}_{4}\right) \) holds. It is easy to see that \( \left( {H}_{5}^{\prime }\right) \) holds since \( {a}_{0} > 0 \) . | Let \( \psi \left( \alpha \right) = \frac{1 + {\alpha }^{2}}{2},\alpha \in \left( {0,1}\right) \), it is easy to see that \( \psi \in \Lambda \) . For all \( t \in \mathbb{R},\alpha \in \left( {0,1}\right) \) and \( x \in \left\lbrack {0,\alpha }\right\rbrack , \) one has \( \left( {1\widetilde{ - }{\alpha }^{2}}\right) \left( {{\alpha }^{2}\widetilde{ - }{x}^{2}}\right) \geq 0 \), then \( {\alpha }^{2}\widetilde{ - }{\alpha }^{2}{x}^{2} \geq {\alpha }^{4}\widetilde{ + }{x}^{2} \), so \( 2{\alpha }^{2}\left( {1\widetilde{ + }{x}^{2}}\right) \geq \left( {1\widetilde{ + }{\alpha }^{2}}\right) \left( {{\alpha }^{2}\widetilde{ + }{x}^{2}}\right) , \) by this, one has\n\n\[ \n\frac{f\left( {t,{\alpha }^{-1}x}\right) }{f\left( {t, x}\right) } = \frac{1 + {x}^{2}}{1 + {\alpha }^{-2}{x}^{2}} = \frac{{\alpha }^{2}\left( {1 + {x}^{2}}\right) }{1 + {x}^{2}} \geq \frac{1 + {\alpha }^{2}}{2} = \psi \left( \alpha \right) ,\n\]\n\nwhich means that \( {M}_{\psi }^{f} \geq 1 \), so \( \left( {H}_{6}^{\prime }\right) \) holds by \( {a}_{1} < 1/\sigma \) . Hence,(8) has a unique WPAA solution with positive infinimum by Corollary 2.1. | Yes |
Consider the following Lasota-Wazewska model:\n\n\[ \n{x}^{\prime }\left( t\right) = - \lambda \left( t\right) x\left( t\right) + \mu \left( t\right) {e}^{-{\gamma x}\left( {t - \tau }\right) },\;t \in \mathbb{R}, \n\] \n\nwhich is a model for the survival of red blood cells in an animal. Assume that \( \gamma ,\tau > 0 \) are two fixed constants, and \( \lambda \in {AA}\left( {\mathbb{R},{\mathbb{R}}^{ + }}\right) ,\mu \in {WPAA}\left( {\mathbb{R},{\mathbb{R}}^{ + },{\rho }_{1},{\rho }_{2}}\right) ,{\rho }_{1},{\rho }_{2} \in {U}_{T} \) satisfying \( \mathop{\inf }\limits_{{t \in \mathbb{R}}}\lambda \left( t\right) > 0,\mathop{\inf }\limits_{{t \in \mathbb{R}}}\mu \left( t\right) > 0 \) . Then \( x\left( t\right) \) is a bounded continuous solution to \( \left( 9\right) \) if and only if \( x\left( t\right) \) is a bounded continuous solution to the following integral equation [19]:\n\n\[ \nx\left( t\right) = {\int }_{-\infty }^{t}{e}^{-{\int }_{s}^{t}\lambda \left( u\right) {du}}\mu \left( s\right) {e}^{-{\gamma x}\left( {s - \tau }\right) }{ds},\;t \in \mathbb{R}, \n\] \n\ni.e., \n\n\[ \nx\left( t\right) = {\int }_{-\infty }^{t}{e}^{-{\int }_{s + \tau }^{t}\lambda \left( u\right) {du}}{\mathbf{1}}_{\left\lbrack \tau , + \infty \right\rbrack }\left( {t - s}\right) \mu \left( {s + \tau }\right) {e}^{-{\gamma x}\left( s\right) }{ds},\;t \in \mathbb{R}. \n\] | Let \( \alpha \left( t\right) = h\left( t\right) = 0,\beta \left( t\right) = 1, f\left( {t, x}\right) = \mu \left( {t + \tau }\right) {e}^{-{\gamma x}} \) and \( a\left( {t, s}\right) = {e}^{-{\int }_{t - s + \tau }^{t}\lambda \left( u\right) {du}}{\mathbf{1}}_{\left\lbrack \tau , + \infty \right\rbrack }\left( s\right) \) , then (9) can be rewritten as the form (4). It is not difficult to see that \( \left( {H}_{1}\right) - \left( {H}_{5}\right) \) hold. Let \( \psi \left( \alpha \right) = {e}^{\alpha - 1},\alpha \in \left( {0,1}\right) \), so \( \psi \in \Lambda \) . For all \( t \in \mathbb{R},\alpha \in \left( {0,1}\right), x \in \left\lbrack {0,\alpha /\gamma }\right\rbrack \), one has \n\n\[ \n\frac{f\left( {t,{\alpha }^{-1}x}\right) }{f\left( {t, x}\right) } = {e}^{{\gamma x} - \gamma {\alpha }^{-1}x} \geq \psi \left( \alpha \right) \n\] \n\nwhich means that \( {M}_{\psi }^{f} \geq \frac{1}{\gamma } \) . If \( \mathop{\sup }\limits_{{t \in \mathbb{R}}}\mu \left( t\right) \leq \mathop{\inf }\limits_{{t \in \mathbb{R}}}\lambda \left( t\right) \), then \( \left( {H}_{6}\right) \) hold. Whence, by Theorem 2.1,(9) has a unique WPAA solution with positive infinimum. | Yes |
In (4), let \( f\left( {t, x}\right) = \frac{v\left( t\right) }{\sqrt{1 + x}}, v\left( t\right) \mathrel{\text{:=}} m\sin \frac{1}{{\cos }^{2}t + {\cos }^{2}{\pi t} + 2} + m{\left( 1 + {t}^{2}\right) }^{-1} \) , where \( 0 < m < \frac{1}{4 + \pi } \), then \( v\left( t\right) \in {WPAA}\left( {\mathbb{R},{\mathbb{R}}^{ + },{\rho }_{1},{\rho }_{2}}\right) \) with \( {\rho }_{1} = {e}^{t},{\rho }_{2} = 1 + {t}^{2} \), and assume that \( \alpha \left( t\right) = h\left( t\right) = v\left( t\right) ,\beta \left( t\right) = 1, a\left( {t, s}\right) = \frac{1}{1 + {s}^{2}} \), hence \( \left( {H}_{1}\right) - \left( {H}_{4}\right) \) hold and it is not difficult to see that \( \left( {H}_{5}\right) \) holds. | Let \( \psi \left( \alpha \right) = \sqrt{\alpha },\alpha \in \left( {0,1}\right) \), then \( \psi \in \Lambda \) . For all \( t \in \mathbb{R},\alpha \in \left( {0,1}\right) \) and \( x \in \left\lbrack {0,\alpha }\right\rbrack \), one has\n\n\[ \frac{f\left( {t,{\alpha }^{-1}x}\right) }{f\left( {t, x}\right) } = \frac{\sqrt{1 + x}}{\sqrt{1 + {\alpha }^{-1}x}} = \frac{\sqrt{\alpha }\sqrt{1 + x}}{\sqrt{x + \alpha }} \geq \sqrt{\alpha } = \psi \left( \alpha \right) ,\]\n\nso \( {M}_{\psi }^{f} \geq 1 \) . Since \( m < \frac{1}{4 + \pi } \), then \( \frac{1}{1 - \parallel \alpha \parallel }\mathop{\sup }\limits_{{t \in \mathbb{R}}}\left( {\beta \left( t\right) {\int }_{-\infty }^{t}a\left( {t, t - s}\right) f\left( {s,0}\right) {ds} + h\left( t\right) }\right) < 1 \), so \( \left( {H}_{6}\right) \) holds. By Theorem 2.1, the following delay integral equation:\n\n\[ x\left( t\right) = v\left( t\right) x\left( {t - \tau }\right) + {\int }_{-\infty }^{t}a\left( {t, t - s}\right) f\left( {s, x\left( s\right) }\right) {ds} + v\left( t\right) ,\;t \in \mathbb{R},\]\n\nhas a unique WPAA solution with positive infinimum. | Yes |
Lemma 1.4 \( {}^{\left\lbrack 4\right\rbrack } \) Let \( \alpha > 0 \), and \( f;g : \left\lbrack {a;b}\right\rbrack \rightarrow R \) be \( {C}^{n} \) functions. Then,\n\n\[ \n{\int }_{a}^{b}g{\left( t\right) }^{C}{D}_{{a}^{ + }}^{\alpha }f\left( t\right) {dt} = {\int }_{a}^{b}f\left( t\right) {D}_{{b}^{ - }}^{\alpha }g\left( t\right) {dt} + \mathop{\sum }\limits_{{j = 0}}^{{n - 1}}{D}_{{b}^{ - }}^{\alpha + j - n}g{\left( t\right) }^{n - \alpha - 1}{D}_{{b}^{ - }}^{n - 1 - j}f\left( t\right) {\left. \right| }_{a}^{b}, \n\] \n\nand \n\n\[ \n{\int }_{a}^{b}g{\left( t\right) }^{C}{D}_{{b}^{ - }}^{\alpha }f\left( t\right) {dt} = {\int }_{a}^{b}f\left( t\right) {D}_{{a}^{ + }}^{\alpha }g\left( t\right) {dt} + \mathop{\sum }\limits_{{j = 0}}^{{n - 1}}{\left( -1\right) }^{n + j}{D}_{{a}^{ + }}^{\alpha + j - n}g{\left( t\right) }^{n - \alpha - 1}{D}_{{a}^{ + }}^{n - 1 - j}f\left( t\right) {\left. \right| }_{a}^{b}. \n\] | In the particular case when \( 0 < \alpha < 1 \), by Lemma1.4. we have \n\n\[ \n{\int }_{a}^{b}g{\left( t\right) }^{C}{D}_{{a}^{ + }}^{\alpha }f\left( t\right) {dt} = {\int }_{a}^{b}f\left( t\right) {D}_{{b}^{ - }}^{\alpha }g\left( t\right) {dt} + {\left. f\left( t\right) {I}_{{b}^{ - }}^{1 - \alpha }g\left( t\right) \right| }_{a}^{b}, \n\] \n\nand \n\n\[ \n{\int }_{a}^{b}g{\left( t\right) }^{C}{D}_{{b}^{ - }}^{\alpha }f\left( t\right) {dt} = {\int }_{a}^{b}f\left( t\right) {D}_{{b}^{ - }}^{\alpha }g\left( t\right) {dt} - {\left. f\left( t\right) {I}_{{a}^{ + }}^{1 - \alpha }g\left( t\right) \right| }_{a}^{b}. \n\] \n\nIn addition, if \( f \) is such that \( f\left( a\right) = f\left( b\right) = 0 \) ; then \n\n\[ \n{\int }_{a}^{b}g{\left( t\right) }^{C}{D}_{{a}^{ + }}^{\alpha }f\left( t\right) {dt} = {\int }_{a}^{b}f\left( t\right) {D}_{{b}^{ - }}^{\alpha }g\left( t\right) {dt} \n\] \n\nand \n\n\[ \n{\int }_{a}^{b}g{\left( t\right) }^{C}{D}_{{b}^{ - }}^{\alpha }f\left( t\right) {dt} = {\int }_{a}^{b}f\left( t\right) {D}_{{b}^{ - }}^{\alpha }g\left( t\right) {dt}. \n\] | Yes |
Theorem 3.2 Let \( L\left( {t;y;z;u}\right) \) be jointly convex (concave) in \( \left( {y, z, u}\right) \) . If \( {y}_{0} \) satisfies conditions (3), then \( {y}_{0} \) is a global minimizer (maximizer) to problems (1)-(2). | Proof We will give the proof for only the convex case (and similarly we can prove it for the concave case). Since \( L \) is jointly convex in \( \left( {y, z, u}\right) \) for any admissible function \( {y}_{0} + \eta \left( t\right) ,\eta \left( t\right) \) is an admissible function such that \( \eta \left( a\right) = 0,\eta \left( b\right) = 0 \), we have\n\n\[ J\left( {{y}_{0} + \eta }\right) - J\left( {y}_{0}\right) = \]\n\n\[ {\int }_{a}^{b}\left\lbrack {L\left( {t,\left( {{y}_{0} + \eta }\right) {,}^{C}{D}_{{a}^{ + }}^{\alpha }\left( {{y}_{0} + \eta }\right) {,}^{C}{D}_{{b}^{ - }}^{\alpha }\left( {{y}_{0} + \eta }\right) }\right) - L\left( {t,{y}_{0}{,}^{C}{D}_{{a}^{ + }}^{\alpha }{y}_{0}{,}^{C}{D}_{{b}^{ - }}^{\alpha }{y}_{0}}\right) }\right\rbrack {dt} \geq \]\n\n\[ {\int }_{a}^{b}{\int }_{a}^{b}\left( {\eta \frac{\partial L}{\partial y} + {}^{C}{D}_{{b}^{ - }}^{\alpha }\eta \frac{\partial L}{\partial {}^{C}{D}_{{a}^{ + }}^{\alpha }y} + {}^{C}{D}_{{a}^{ + }}^{\alpha }\eta \frac{\partial L}{\partial {}^{C}{D}_{{b}^{ - }}^{\alpha }y}}\right) {dt}. \]\n\nBy using integration by parts,\n\n\[ {\int }_{a}^{b}{}^{C}{D}_{{b}^{ - }}^{\alpha }\eta \frac{\partial L}{\partial {}^{C}{D}_{{a}^{ + }}^{\alpha }y}{dt} = {\int }_{a}^{b}\eta {D}_{{a}^{ + }}^{\alpha }\frac{\partial L}{\partial {}^{C}{D}_{{a}^{ + }}^{\alpha }y}{dt} + {\left. \eta {I}_{{a}^{ + }}^{1 - \alpha }\frac{\partial L}{\partial {}^{C}{D}_{{a}^{ + }}^{\alpha }y}\right| }_{a}^{b} = {\int }_{a}^{b}\eta {D}_{{a}^{ + }}^{\alpha }\frac{\partial L}{\partial {}^{C}{D}_{{a}^{ + }}^{\alpha }y}{dt}, \]\n\nand\n\n\[ {\int }_{a}^{b}{}^{C}{D}_{{a}^{ + }}^{\alpha }\eta \frac{\partial L}{\partial {}^{C}{D}_{{b}^{ - }}^{\alpha }y}{dt} = {\int }_{a}^{b}\eta {D}_{{b}^{ - }}^{\alpha }\frac{\partial L}{\partial {}^{C}{D}_{{b}^{ - }}^{\alpha }y}{dt} + \eta {I}_{{b}^{ - }}^{1 - \alpha }\frac{\partial L}{\partial {}^{C}{D}_{{b}^{ - }}^{\alpha }y}{\left. \right| }_{a}^{b} = {\int }_{a}^{b}\eta {D}_{{b}^{ - }}^{\alpha }\frac{\partial L}{\partial {}^{C}{D}_{{b}^{ - }}^{\alpha }y}{dt}. \]\n\nwe get\n\n\[ J\left( {{y}_{0} + \eta }\right) - J\left( {y}_{0}\right) \geq \]\n\n\[ {\int }_{a}^{b}{\int }_{a}^{b}\left( {\eta \frac{\partial L}{\partial y} + {}^{C}{D}_{{b}^{ - }}^{\alpha }\eta \frac{\partial L}{\partial {}^{C}{D}_{{a}^{ + }}^{\alpha }y} + {}^{C}{D}_{{a}^{ + }}^{\alpha }\eta \frac{\partial L}{\partial {}^{C}{D}_{{b}^{ - }}^{\alpha }y}}\right) {dt} = \]\n\n\[ {\int }_{a}^{b}\left( {\frac{\partial L}{\partial y} + {D}_{{a}^{ + }}^{\alpha }\frac{\partial L}{\partial {}^{C}{D}_{{a}^{ + }}^{\alpha }y} + {D}_{{b}^{ - }}^{\alpha }\frac{\partial L}{\partial {}^{C}{D}_{{b}^{ - }}^{\alpha }y}}\right) {\eta dt} = 0. \]\n\nSince \( {y}_{0} \) satisfies conditions (3), we obtain \( J\left( {{y}_{0} + \eta }\right) - J\left( {y}_{0}\right) \geq 0 \), which completes the proof. | Yes |
Theorem 2.1 Suppose that assumptions \( \left( {A}_{1}\right) \left( {A}_{2}\right) \) and \( \left( {A}_{3}\right) \) hold. Then the prey species \( \left( {{x}_{1}\left( t\right) ,{x}_{2}\left( t\right) }\right) \) of system (0.1) is permanent. | Here, we prove Theorem 2.1 under assumptions \( \left( {A}_{1}\right) \left( {A}_{2}\right) \) and \( \left( {A}_{3}\right) . We will use the Proposition 2.1 and Proposition 2.2 to complete the proof of Theorem 2.1. | No |
Theorem 2.2 Suppose that \( \left( {A}_{1}\right) \left( {A}_{2}\right) \) and \( \left( {A}_{3}\right) \) hold. If\n\n\[ \n{A}_{\omega }\left\lbrack {-g\left( t\right) + \frac{h\left( t\right) {x}_{1}^{ * }\left( t\right) }{k\left( t\right) + {\left( {x}_{1}^{ * }\left( t\right) \right) }^{2}}}\right\rbrack > 0.\n\]\n\n(2.1)\n\nthen predator species in system (0.1) is permanent, where \( \left( {{x}_{1}^{ * }\left( t\right) ,{x}_{2}^{ * }\left( t\right) }\right) \) is the positive \( \omega \) -periodic solution of system (1.2) given by Lemma 1.1. | Here, we prove Theorem 2.2 under assumptions \( \left( {A}_{1}\right) \left( {A}_{2}\right) \) and \( \left( {A}_{3}\right) \) . We will use the Proposition 2.3 and Proposition 2.4 to complete the proof of Theorem 2.2. | No |
Lemma 1.2 Assume \( \left( {H}_{1}\right) - \left( {H}_{6}\right) \) hold. Then the solutions of Eq. (5) and (6) are defined on \( \lbrack - \bar{\tau },\infty ) \), positive and bounded on \( \lbrack 0,\infty ) \) . | Proof By Lemma 1.1, we only need to prove that the solutions of (7) and (8) are defined on \( \lbrack - \bar{\tau },\infty ) \) and are positive on \( \lbrack 0,\infty ) \) . From (7) and (8) we have that for any\n\n\[ \varphi \in L\left( {\left\lbrack {-\widetilde{\tau },0}\right\rbrack ,\lbrack 0,\infty }\right) ),\;\varphi \left( 0\right) > 0,\;t > 0 \]\n\n\[ y\left( t\right) = \varphi \left( 0\right) {e}^{-{\int }_{0}^{t}a\left( s\right) {ds}} + {\int }_{0}^{t}\lambda \left( s\right) g\left( {s, y\left( {s - {\tau }_{0}\left( s\right) }\right), y\left( {s - {\tau }_{1}\left( s\right) }\right) ,\cdots, y\left( {s - {\tau }_{n}\left( s\right) }\right) }\right) {ds}. \]\n\nHence, \( y\left( t\right) \) is defined on \( \lbrack - \bar{\tau },\infty ) \) and are positive on \( \lbrack 0,\infty ) \) . Now, we prove that every solution of Eq.(7) is bounded. Otherwise, there exists an unbounded solution \( y\left( t\right) \) .\n\nFrom Eq.(7) we obtain\n\n\[ \begin{aligned} {y}^{\prime }\left( t\right) & = - a\left( t\right) x\left( t\right) + \lambda \left( t\right) g\left( {t, y\left( {t - {\tau }_{0}\left( t\right) }\right), y\left( {t - {\tau }_{1}\left( t\right) }\right) ,\cdots, y\left( {t - {\tau }_{n}\left( t\right) }\right) }\right) \\ & \leq - {ay}\left( t\right) + \bar{\lambda }L. \end{aligned} \]\n\n(9)\n\nThus there exist \( 0 < {t}_{1} < {t}_{2} \) such that \( y\left( {t}_{1}\right) < y\left( {t}_{2}\right) \) and \( - {ay}\left( {t}_{2}\right) + \bar{\lambda }L < 0 \) .\n\nHence, \( {y}^{\prime }\left( {t}_{2}\right) < 0 \) . Let \( y\left( {t}_{3}\right) = \mathop{\max }\limits_{{{t}_{1} \leq t \leq {t}_{2}}}\{ y\left( t\right) \} \) . It is easy to see that \( {t}_{3} \neq {t}_{1} \) and \( {t}_{3} \neq {t}_{2} \) . So \( {y}^{\prime }\left( {t}_{3}\right) = 0 \) . From (9) we have\n\n\[ {y}^{\prime }\left( {t}_{3}\right) \leq - {ay}\left( {t}_{3}\right) + \bar{\lambda }L \leq - {ay}\left( {t}_{2}\right) + \bar{\lambda }L < 0 \]\n\nWhich is a contradiction. Consequently, \( y\left( t\right) \) is bounded. The proof is complete. | Yes |
Example 2.1 Consider the impulsive delay differential equation\n\n\[ \left\{ \begin{array}{ll} {x}^{\prime }\left( t\right) = - \left( {\log 2}\right) \left| {\sin t}\right| x\left( t\right) + \frac{3}{{8\pi }\widetilde{b}}\left\lbrack {{\left( x\left( t - \tau \right) + 1\right) }^{2}{e}^{-x\left( {t - \tau }\right) } + 1}\right\rbrack & \text{ a.e. }{}_{t} > 0,\;t \neq {t}_{k}; \\ x\left( {t}_{k}^{ + }\right) - x\left( {t}_{k}\right) = {b}_{k}x\left( {t}_{k}\right) , & k = 1,2,\cdots \end{array}\right. \] | where \( \tau > 0 \) is a fixed constant, \( {b}_{k} \in \left( {-1,0}\right) \) . Set \( a\left( t\right) = - \left( {\log 2}\right) \left| {\sin t}\right| ,\omega = \pi, f\left( {t, u}\right) = \) \( \frac{3}{{8\pi }\bar{b}}\left\lbrack {{\left( u + 1\right) }^{2}{e}^{-u} + 1}\right\rbrack \) ; then we have \( \bar{b} < 1,{\int }_{0}^{\omega }a\left( t\right) {dt} = 2\left( {\log 2}\right), A = \frac{1}{\exp \left( {{\int }_{0}^{\omega }a\left( \xi \right) {d\xi }}\right) - 1} = \) \( \frac{1}{3} \), Again \( f\left( {t, u}\right) \) is a monotonic decreasing function for \( u > 0, A\bar{b}\bar{p}{\omega L} = \frac{1}{3}\bar{b}\pi \frac{3}{{8\pi }\bar{b}}2 = \frac{1}{4} < \) \( \widetilde{1}, A{\bar{b}}^{2}\bar{p}{\omega L} < 1 \)\n\n\[ {e}^{-\frac{1}{4}}\frac{3}{{8\pi }\bar{b}}2 \leq \frac{3}{{8\pi }\bar{b}} \leq f\left( {t, u}\right) \leq \frac{3}{{8\pi }\bar{b}}2 \]\n\nAnd for\n\n\[ 0 < \lambda < 1, \]\n\n\[ f\left( {t,{\lambda u}}\right) = \frac{3}{{8\pi }\bar{b}}\left\lbrack {{\left( \lambda u + 1\right) }^{2}{e}^{-{\lambda u}} + 1}\right\rbrack = \frac{3}{{8\pi }\bar{b}}\left\lbrack {{\left( \lambda u + 1\right) }^{2}{e}^{\left( {1 - \lambda }\right) u} \cdot {e}^{-u} + {e}^{\left( {1 - \lambda }\right) u} \cdot {e}^{\left( {\lambda - 1}\right) u}}\right\rbrack = \]\n\n\[ {e}^{\left( {1 - \lambda }\right) u}\frac{3}{{8\pi }\bar{b}}\left\lbrack {{\left( \lambda u + 1\right) }^{2} \cdot {e}^{-u} + {e}^{\left( {\lambda - 1}\right) u}}\right\rbrack \leq {e}^{\left( {1 - \lambda }\right) \parallel u\parallel }\frac{3}{{8\pi }\bar{b}}\left\lbrack {{\left( u + 1\right) }^{2} \cdot {e}^{-u} + 1}\right\rbrack = \]\n\n\[ {e}^{\left( {1 - \lambda }\right) \parallel u\parallel }f\left( {t, u}\right) \]\n\nBy Theorem 2.1 and Theorem 2.2, we see that the above equation has a unique positive \( \omega \) -periodic solution. | Yes |
Theorem 0.1 Suppose that \( \left( {H}_{1}\right) \) and \( \left( {H}_{2}\right) \) hold. Then, for any \( \tau > 0 \) sufficiently small, there exist speeds \( c \) such that the Eq. \( \left( {0.1}\right) \) with \( \left( {0.6}\right) \) has a traveling wave solution \( u\left( {x, t}\right) = U\left( z\right) \) connecting \( {u}_{0} \) to \( {u}_{e} \) (that is \( \mathop{\lim }\limits_{{z \rightarrow - \infty }}U\left( z\right) = {u}_{0},\mathop{\lim }\limits_{{z \rightarrow + \infty }}U\left( z\right) = {u}_{e} \) ), where \( z = x + {ct} \) is the wave variable and \( c \geq 0 \) is the wave speed. | ## 1 The Proof of Theorem 0.1\n\nTo prove Theorem 0.1, we first introduce the following result on invariant manifolds which is due to Fenichel \( {}^{\left\lbrack 9\right\rbrack } \) . For convenience, we use a version of this theorem due to Jones \( {}^{\left\lbrack {10}\right\rbrack } \) .\n\nLemma 1.1 (Geometric Si | No |
Lemma 1.1 (Geometric Singular Perturbation Theorem). Given a \( {C}^{\infty } \) vector field of the form \( {x}^{\prime } = f\left( {x, y,\varepsilon }\right) ,\;{y}^{\prime } = {\varepsilon g}\left( {x, y,\varepsilon }\right) \) such that when \( \varepsilon = 0 \), the system has a compact, normally hyperbolic manifold of critical points \( {\mathcal{M}}_{0} \) which is given as the graph of a \( {C}^{\infty } \) function \( {h}^{0}\left( y\right) \) , then for every \( r > 0 \), there exists an \( {\varepsilon }_{0} > 0 \), such that if \( \left| \varepsilon \right| < \left| {\varepsilon }_{0}\right| \), there exists a manifold \( {\mathcal{M}}_{\varepsilon } \) : | (I) which is locally invariant under the flow of the system;\n\n(II) which is \( {C}^{r} \) in \( x, y \) and \( \varepsilon \) ;\n\n(III) for which \( {\mathcal{M}}_{\varepsilon } = \left\{ {\left( {x, y}\right) \mid x = {h}^{\varepsilon }\left( y\right) }\right\} \) for some \( {C}^{r} \) function \( {h}^{\varepsilon } \) and \( y \) in some compact set \( K \) ;\n\n(IV) for which there are locally stable and unstable manifolds, \( {W}^{s}\left( {\mathcal{M}}_{\varepsilon }\right) \) and \( {W}^{u}\left( {\mathcal{M}}_{\varepsilon }\right) \), that lie within \( \mathcal{O}\left( \varepsilon \right) \) of and are diffeomorphic to \( {W}^{s}\left( {\mathcal{M}}_{0}\right) \) and \( {W}^{u}\left( {\mathcal{M}}_{0}\right) \) . | Yes |
Lemma 1.3 System (1.1) has a positive periodic solution \( {S}^{ * }\left( t\right) \) and for every solution \( S\left( t\right) \) of (1.1), \( \left| {S\left( t\right) - {S}^{ * }\left( t\right) }\right| \rightarrow 0 \) as \( t \rightarrow \infty \), where \( {S}^{ * }\left( t\right) = \frac{q{e}^{-D\left( {t - {nT}}\right) }}{1 - {e}^{-{DT}}}, t \in ({nT},\left( {n + 1}\right) T\rbrack \) . | Therefore, system (0.1) has a microorganism-free periodic solution\n\n\[ \left( {{S}^{ * }\left( t\right) ,0,0}\right) = \left( {\frac{q{e}^{-D\left( {t - {nT}}\right) }}{1 - {e}^{-{DT}}},0,0}\right) ,\;t \in ({nT},\left( {n + 1}\right) T\rbrack . \] | Yes |
Theorem 3.1 For each positive solution \( \left( {S\left( t\right) ,{x}_{1}\left( t\right) ,{x}_{2}\left( t\right) }\right) \) of system (0.1), there exists a constant \( M > 0 \) such that \( S\left( t\right) \leq M,{x}_{i}\left( t\right) \leq M, i = 1,2 \) for \( t \) large enough. | Proof Define a function\n\n\[ V\left( t\right) = S\left( t\right) + \frac{{e}^{D{\tau }_{1}}{x}_{1}\left( {t + {\tau }_{1}}\right) }{\alpha } + \frac{{e}^{D{\tau }_{2}}{x}_{2}\left( {t + {\tau }_{2}}\right) }{\beta }, \]\n\nthen the upper right derivative of \( V\left( t\right) \) along a solution of (0.1) is described as\n\n\[ {D}^{ + }V\left( t\right) = - {DV}\left( t\right) \]\n\nwhen \( t = {nT} \), we obtain \( V\left( {n{T}^{ + }}\right) = V\left( {nT}\right) + q \) . According to Lemma 1.4, we have\n\n\[ V\left( t\right) \leq V\left( {0}^{ + }\right) {e}^{-{Dt}} + \frac{q{e}^{DT}}{{e}^{DT} - 1} + \frac{q{e}^{-D\left( {t - T}\right) }}{{e}^{DT} - 1} \rightarrow \frac{q{e}^{DT}}{{e}^{DT} - 1}, t \rightarrow \infty . \]\n\nTherefore, \( V\left( t\right) \) is ultimately bounded and there exists a constant \( M > 0 \), such that \( {x}_{i}\left( t\right) \leq M, i = \) \( 1,2 \) and \( S\left( t\right) \leq M \) for each positive solution \( \left( {S\left( t\right) ,{x}_{1}\left( t\right) ,{x}_{2}\left( t\right) }\right) \) of system (0.1) for \( t \) large enough. The proof is complete. | Yes |
Lemma 2.1 The system (2.1) is uniformly bounded. | The Proof is referenced in [10]. | No |
Theorem 2.1 Assume that \( {a}_{11}^{iL} > 0,{a}_{22}^{jL} > 0\left( {i = 0,1,\ldots ,{2m};j = 0,1,\ldots, m}\right) ,{b}_{k}^{L} > \) \( 0,{r}_{k}^{L} > 0,{d}_{k}^{L} > 0,{f}_{k}^{L} > 0,{e}_{k}^{L} > 0\left( {k = 1,2}\right) \), then system (0.3) is permanent. | Proof Combined with Lemma 1.6 and Lemma 1.7, the conclusion is obvious. | No |
Lemma 2.2 Assume that conditions \( {\left( H1\right) }^{\prime } \) and \( {\left( H2\right) }^{\prime } \) hold, then\n\n\[ \operatorname{Re}\left( {{d\lambda }\left( \mu \right) /{d\mu }\left| {}_{\mu = {\mu }_{1}^{ + }}\right. }\right) > 0,\operatorname{Re}\left( {{d\lambda }\left( \mu \right) /{\left. d\mu \right| }_{\mu = {\mu }_{1}^{ - }}}\right) < 0. \] | Proof Substituting \( \lambda \left( \mu \right) \) into (2.3), and by derivative for both sides of it we have\n\n\[ {d\lambda }\left( \mu \right) /{d\mu } = 2{g}^{\prime }\left( 0\right) {e}^{-{\lambda \tau }}/\left( {{2\tau \mu }{g}^{\prime }\left( 0\right) {e}^{-{\lambda \tau }} - 1}\right) . \]\n\nBy \( \mu = {\mu }_{1}^{ + } \), we will get that \( {d\lambda }\left( \mu \right) /{d\mu } = - \left( {{m}_{1} + i{\omega }_{j}^{ + }}\right) /\left( {\mu }_{j}^{ + }\right) \left( {\tau {m}_{1} + 1 + {i\tau }{\omega }_{j}^{ + }}\right) . \)\n\nTherefore,\n\n\[ \operatorname{Re}\left( {{d\lambda }\left( \mu \right) /{d\mu }\left| {}_{\mu = {\mu }_{1}^{ + }}\right. }\right) = \left( {{m}_{1} + \tau {m}_{1}^{2} + \tau {\left( {\omega }_{j}^{ + }\right) }^{2}}\right) /{D}_{1}{\mu }_{j}^{ + } > 0, \]\n\nwhere \( {m}_{1} = \left( {{a}_{1} - {\delta }_{1} - {\delta }_{11} - 2{a}_{11}{f}^{\prime }\left( 0\right) }\right) ,{D}_{1} = {\left( 1 + \tau \right) }^{2}{\left( {a}_{1} - {\delta }_{1} - {\delta }_{11} - 2{a}_{11}{f}^{\prime }\left( 0\right) \right) }^{2} + \tau {\left( {\omega }_{j}^{ + }\right) }^{2} \) . Similar, we easy get \( \operatorname{Re}\left( {{d\lambda }\left( \mu \right) /{\left. d\mu \right| }_{\mu = {\mu }_{1}^{ - }}}\right) < 0 \) . | Yes |
Theorem 2.3\n\n(1) when \( \mu \in \left( {{\mu }_{1}^{ - },{\mu }_{1}^{ + }}\right) \), the trivial equilibrium \( \left( {0,0}\right) \) of equation (2.1) is asymptotic stable.\n\n(2) when \( \mu \in R \smallsetminus \left\lbrack {{\mu }_{1}^{ - },{\mu }_{1}^{ + }}\right\rbrack \), the trivial equilibrium \( \left( {0,0}\right) \) of (2.1) is not stable.\n\n(3) \( {\mu }_{j} \in \left\{ {\mu }_{j}^{ \pm }\right\}, j = 1,2,\cdots \), are Hopf bifurcation of system (2.1). | The conclusion of \( \left( 1\right) \) and \( \left( 2\right) \) hold obviously. In fact, notice that from lemma 2.2, we can obtain that as \( \mu < {\mu }_{1}^{ - } \) and \( \mu > {\mu }_{1}^{ + } \), the equation (2.3) has at least a positive real root, and get the conclusion (2). | No |
Lemma 3.2 \( \; \) If the \( - {b}_{i}\left( t\right) + \mathop{\sum }\limits_{{j = 1}}^{n}\left( {{a}_{ij}\left( t\right) + {a}_{ij}^{T}\left( t\right) }\right) \neq 0,\; \) then the system (3.1) \( \; \) at most there exists a equilibrium solution on the same sub-domain. | By definition 3.1, similar lemma1 of Theorem in [17]. Lemma 3.2 can be proved. Omit these details. | No |
Theorem 3.5 Assume that (3.3), (3.6) hold, if there exists a positively definite diagram \( \begin{matrix} \text{matrix}\;D\left( t\right) = \text{diag}\left( {{d}_{1}\left( t\right) ,\cdots ,{d}_{n}\left( t\right) }\right) , \\ \min {d}_{i}\left( t\right) = {d}_{i}^{\prime } > 0, i = 1,\cdots, n,\;\text{such}\;\text{that}\;\text{the}\;\text{diagonal} \end{matrix} \) matrix \[ \bar{H} = \left( \begin{matrix} - 2{B}^{l}\left( t\right) + 2{A}^{M}\Phi \left( {x}^{ * }\right) + {C}_{1}^{M} + {C}_{2}^{M} + {D}^{M}{\left( {A}^{M}\right) }^{T}\Phi \left( {x}^{ * }\right) \\ {\left( {\left( {A}^{M}\right) }^{T}\Phi \left( {x}^{ * }\left( t\right) \right) \right) }^{T}\Phi - {D}^{l} \end{matrix}\right) \] is negative defined, then the zero solution \( z = 0 \) of this system (3.7) is almost sure exponential stability, where \( - \lambda = {\lambda }_{\max }\left( \bar{H}\right) ,\lambda > 0. \) | Proof For the \( z, y \in {\Omega }_{0} \subset {R}^{n} \), select the Lyapunov function \( V\left( {z, t}\right) = {z}^{2} \), then by system (3.7) to yield the following that operator for \[ {LV}\left( {z, y}\right) = 2{z}^{T}\left( {-B\left( t\right) + A\left( t\right) \Phi \left( {{x}^{ * }\left( t\right) }\right) + {A}^{T}\left( t\right) \Phi \left( {{x}^{ * }\left( t\right) }\right) y}\right) + {trace}\left( {\bar{\sigma }{\left( z, y\right) }^{T}\bar{\sigma }\left( {z, y}\right) }\right) . \] From (3.6), we have \[ \begin{matrix} {LV}\left( {z, y}\right) \leq 2\left( {-B\left( t\right) + A\left( t\right) \Phi \left( {{x}^{ * }\left( t\right) }\right) }\right) z + {z}^{T}{A}^{T}\left( t\right) \Phi \left( {{x}^{ * }\left( t\right) }\right) y + {y}^{T}\Phi {\left( {x}^{ * }\left( t\right) \right) }^{T}{\left( {A}^{T}\left( t\right) \right) }^{T}z + \end{matrix} \] \[ {\left\lbrack \begin{matrix} {z}^{T}{C}_{1}\left( t\right) z + {y}^{T}{C}_{2}\left( t\right) y + {y}^{T}{\Phi }^{T}\left( {{x}^{ * }\left( t\right) }\right) {\left( {A}^{T}\left( t\right) \right) }^{T}z + {y}^{T}\left( {-D\left( t\right) }\right) y \end{matrix}\right\rbrack }^{ - } \] \[ {z}^{T}\left( {{C}_{2}\left( t\right) + D\left( t\right) }\right) z + {y}^{T}{C}_{2}\left( t\right) y + {y}^{T}D\left( t\right) = \] \[ \left( {{z}^{T},{y}^{T}}\right) \bar{H}\left( \begin{array}{l} z \\ y \end{array}\right) - {z}^{T}\left( {{C}_{2}\left( t\right) + D\left( t\right) }\right) z + {y}^{T}\left( {{C}_{2}\left( t\right) + D\left( t\right) }\right) y \leq \] \[ - \mathop{\sum }\limits_{{i = 1}}^{n}\left( {\lambda + {\delta }_{i}^{l} + {d}_{i}^{l}}\right) {z}_{i}^{2} + \mathop{\sum }\limits_{{i = 1}}^{n}\left( {\lambda + {\delta }_{i}^{l} + {d}_{i}^{l}}\right) {y}_{i}^{2} \] By the construction matrix of \( \bar{H} \), we easy obtained \( \lambda \leq {d}_{i}, i = 1,2,\cdots, n \) . \[ {\lambda }_{1} = \mathop{\min }\limits_{{1 \leq i \leq n}}\left\{ {\lambda + {\delta }_{i} + {d}_{i}}\right\} ,\;{\lambda }_{2} = \mathop{\min }\limits_{{1 \leq i \leq n}}\left\{ {\left( {{\delta }_{i} + \left( {{d}_{i}^{L} - \lambda }\right) }\right) /\left( {\lambda + {\delta }_{i}^{l} + {d}_{i}^{l}}\right) }\right\} \] Let \( \Phi \in C\left( {{\mathbb{R}}^{n};\mathbb{R}}\right) \) and \( {\varphi }_{i} \in C\left( {\mathbb{R};{\mathbb{R}}_{ + }}\right) \) as bellow \[ \mu \left( z\right) = {\lambda }_{1}^{-1}\mathop{\sum }\limits_{{i = 1}}^{n}\left( {\lambda + {\delta }_{i}^{l} + {d}_{i}^{l}}\right) {z}_{i}^{2},\mu \left( {y}_{i}\right) = {\lambda }_{1}^{-1}\mathop{\sum }\limits_{{i = 1}}^{n}\left( {\lambda + {\delta }_{i}^{l} + {d}_{i}^{l}}\right) {y}_{i}^{2}. \] Obviously, we have (3.1),(3.2)and (3.3) are hold of the Lemma 3.4. Then the zero solution \( z = 0 \) of this system (3.7) is almost sure exponential stability by lemma 3.2. | Yes |
Lemma 1.3 The equation \(\Delta \left( {\lambda, c}\right) = 0\) has no such root with the form \(\lambda = {\lambda }_{1}\left( c\right) + {i\beta },\beta \neq 0.\) | Proof Assume, for a contradiction, that the equation \(\Delta \left( {\lambda, c}\right) = 0\) has such root \(\lambda = {\lambda }_{1}\left( c\right) + {i\beta }.\) Thus we have\n\n\[- c{\lambda }_{1} + D\left\lbrack {{e}^{{\lambda }_{1}}\cos \beta + {e}^{-{\lambda }_{1}}\cos \beta - 2}\right\rbrack - d +\n\]\n\[ \frac{{b}^{\prime }\left( 0\right) }{2\pi }{\int }_{0}^{\infty }f\left( a\right) {e}^{-{da}}\mathop{\sum }\limits_{{l = - \infty }}^{\infty }\beta \left( {a, l}\right) {e}^{{\lambda }_{1}\left( {l - {ca}}\right) }\cos \left( {l - {ca}}\right) \mathrm{d}a = 0,\]\n\n(1.4)\n\nand\n\n\[- {c\beta } + D\left\lbrack {{e}^{{\lambda }_{1}}\sin \beta + {e}^{-{\lambda }_{1}}\sin \beta }\right\rbrack +\n\]\n\[ \frac{{b}^{\prime }\left( 0\right) }{2\pi }{\int }_{0}^{\infty }f\left( a\right) {e}^{-{da}}\mathop{\sum }\limits_{{l = - \infty }}^{\infty }\beta \left( {a, l}\right) {e}^{{\lambda }_{1}\left( {l - {ca}}\right) }\sin \left( {l - {ca}}\right) \mathrm{d}a = 0.\]\n\n(1.5)\n\nUsing \(\Delta \left( {{\lambda }_{1}, c}\right) = 0\), we change 1.4 and 1.5 into\n\n\[D\left\lbrack {{e}^{{\lambda }_{1}} + {e}^{-{\lambda }_{1}}}\right\rbrack \cos \beta = D\left\lbrack {{e}^{{\lambda }_{1}} + {e}^{-{\lambda }_{1}}}\right\rbrack +\n\]\n\[ \frac{{b}^{\prime }\left( 0\right) }{2\pi }{\int }_{0}^{\infty }f\left( a\right) {e}^{-{da}}\mathop{\sum }\limits_{{l = - \infty }}^{\infty }\beta \left( {a, l}\right) {e}^{{\lambda }_{1}\left( {l - {ca}}\right) }\left( {1 - \cos \left( {l - {ca}}\right) }\right) \mathrm{d}a \mathrel{\text{:=}}\]\n\n(1.6)\n\n\[D\left\lbrack {{e}^{{\lambda }_{1}} + {e}^{-{\lambda }_{1}}}\right\rbrack + {\Delta }_{1}\]\n\nand\n\n\[D\left\lbrack {{e}^{{\lambda }_{1}} + {e}^{-{\lambda }_{1}}}\right\rbrack \sin \beta = {c\beta } - \frac{{b}^{\prime }\left( 0\right) }{2\pi }{\int }_{0}^{\infty }f\left( a\right) {e}^{-{da}}\mathop{\sum }\limits_{{l = - \infty }}^{\infty }\beta \left( {a, l}\right) {e}^{{\lambda }_{1}\left( {l - {ca}}\right) }\sin \left( {l - {ca}}\right) \mathrm{d}a \mathrel{\text{:=}}\]\n\n(1.7)\n\n\[{c\beta } - {\Delta }_{2}\]\n\nSquare (1.6)-(1.7) respectively and plus them together, then we have\n\n\[2{D}^{2}\left( {{\cos }^{2}\beta - {\sin }^{2}\beta - 1}\right) = {2D}\left( {{e}^{{\lambda }_{1}} + {e}^{-{\lambda }_{1}}}\right) {\Delta }_{1} + {\Delta }_{1}^{2} + {\left( c\beta - {\Delta }_{2}\right) }^{2}.\]\n\n(1.8)\n\nSince the left side of (1.8) is negative and another side is positive, that leads to a contradiction. | Yes |
Theorem 2.2 Assume \( \left( {H}_{1}\right) - \left( {H}_{4}\right) \) hold. \( {c}_{ * },{\lambda }_{ * } \) and \( {\lambda }_{1} \) are defined in Lemma 1.1. Let \( \widetilde{\phi }\left( {x + {ct}}\right) \) is a traveling wavefront of (0.2) with speed \( c > {c}_{ * } \) satisfying (1.9). Then \( \bar{\phi }\left( \xi \right) = \phi \left( {\xi + {\xi }_{0}}\right) \), where \( {\xi }_{0} \in \mathbb{R} \) is a constant, and \( \phi \left( {x + {ct}}\right) \) is a traveling wavefront with speed \( c \) which is determined by Theorem 2.1. | Proof From Theorem 2.1, we have\n\n\[ \mathop{\lim }\limits_{{\xi \rightarrow - \infty }}\bar{\phi }\left( \xi \right) {e}^{-{\lambda }_{1}\xi } = \bar{\rho }\;\text{ and }\;\mathop{\lim }\limits_{{\xi \rightarrow - \infty }}\phi \left( \xi \right) {e}^{-{\lambda }_{1}\xi } = \rho . \]\n\nLet \( {\xi }_{0} = \frac{1}{{\lambda }_{1}}\ln \frac{\bar{\rho }}{\rho } \) and \( v\left( x\right) = \left\lbrack {\bar{\phi }\left( \xi \right) - \phi \left( {\xi + {\xi }_{0}}\right) }\right\rbrack {e}^{-{\lambda }_{1}\xi } \) . Then we have \( v\left( {\pm \infty }\right) = 0 \) . Furthermore, \( \mathop{\max }\limits_{{x \in \mathbb{R}}}v\left( x\right) \) and \( \mathop{\min }\limits_{{x \in \mathbb{R}}}v\left( x\right) \) exist. Without generality, we assume that \( \mathop{\max }\limits_{{x \in \mathbb{R}}}v\left( x\right) \geq \left| {\mathop{\min }\limits_{{x \in \mathbb{R}}}v\left( x\right) }\right| \) . Then there has \( \bar{\xi } \) satisfying\n\n\[ v\left( \bar{\xi }\right) = \mathop{\max }\limits_{{x \in \mathbb{R}}}v\left( x\right) \geq 0\;\text{ and }\;{v}^{\prime }\left( \bar{\xi }\right) = 0. \]\n\nWe claim that \( v\left( {\bar{\xi } \pm 1}\right) = v\left( \bar{\xi }\right) \) . If that is not true, we consider two cases: \( v\left( {\bar{\xi } + 1}\right) < v\left( \bar{\xi }\right) \) and \( v\left( {\bar{\xi } - 1}\right) < v\left( \bar{\xi }\right) \)\n\nFrom \( \left( {\mathrm{H}}_{1}\right) \) and (1.1), we have\n\n\[ 0 = c{v}^{\prime }\left( \bar{\xi }\right) = - c{\lambda }_{1}v\left( \bar{\xi }\right) + D\left\lbrack {v\left( {\bar{\xi } + 1}\right) {e}^{{\lambda }_{1}} + v\left( {\bar{\xi } - 1}\right) {e}^{-{\lambda }_{1}} - {2v}\left( \bar{\xi }\right) }\right\rbrack - {dv}\left( \bar{\xi }\right) + \]\n\n\[ \frac{1}{2\pi }{\int }_{0}^{\infty }f\left( a\right) {e}^{-{da}}\mathop{\sum }\limits_{{l = - \infty }}^{\infty }\beta \left( {a, l}\right) \left\lbrack {b(\bar{\phi }\left( {\bar{\xi } + l - {ca}}\right) - b(\phi \left( {\bar{\xi } + {\xi }_{0} + l - {ca}}\right) \rbrack {e}^{{\lambda }_{1}\bar{\xi }}\mathrm{d}a \leq } \]\n\n\[ - c{\lambda }_{1}v\left( \bar{\xi }\right) + D\left\lbrack {v\left( {\bar{\xi } + 1}\right) {e}^{{\lambda }_{1}} + v\left( {\bar{\xi } - 1}\right) {e}^{-{\lambda }_{1}} - {2v}\left( \bar{\xi }\right) }\right\rbrack - {dv}\left( \bar{\xi }\right) + \] | Yes |
Lemma 1 If \( S\left( 0\right) > 0, D\left( 0\right) > 0, R\left( 0\right) > 0, Q\left( 0\right) > 0 \), the solution \( S\left( t\right), D\left( t\right), R\left( t\right), Q\left( t\right) \) of system (1) is positive for all \( t > 0 \) . | Proof If the conclusion does not hold, then at least one of \( S\left( t\right), D\left( t\right), R\left( t\right), Q\left( t\right) \) is not positive. Thus, we have the following four cases.\n\n1) there exists a first time \( {t}_{1} \) such that\n\n\[ S\left( {t}_{1}\right) = 0,{S}^{\prime }\left( {t}_{1}\right) < 0, D\left( t\right) > 0, R\left( t\right) > 0, Q\left( t\right) > 0,\;0 < t < {t}_{1}, \]\n\n(2)\n\n2) there exists a first time \( {t}_{2} \) such that\n\n\[ D\left( {t}_{2}\right) = 0,{D}^{\prime }\left( {t}_{2}\right) < 0, S\left( t\right) > 0, R\left( t\right) > 0, Q\left( t\right) > 0,\;0 < t < {t}_{2}, \]\n\n(3)\n\n3) there exists a first time \( {t}_{3} \) such that\n\n\[ R\left( {t}_{3}\right) = 0,{R}^{\prime }\left( {t}_{3}\right) < 0, S\left( t\right) > 0, D\left( t\right) > 0, Q\left( t\right) > 0,\;0 < t < {t}_{3}, \]\n\n(4)\n\n4) there exists a first time \( {t}_{4} \) such that\n\n\[ Q\left( {t}_{4}\right) = 0,{Q}^{\prime }\left( {t}_{4}\right) < 0, S\left( t\right) > 0, D\left( t\right) > 0, R\left( t\right) > 0,\;0 < t < {t}_{4}, \]\n\n(5)\n\nIn case 1), we have\n\n\[ {S}^{\prime }\left( {t}_{1}\right) = \Lambda > 0 \]\n\n(6)\n\nwhich is a contradiction with \( {S}^{\prime }\left( {t}_{1}\right) < 0 \) .\n\nIn case 2), we have\n\n\[ {D}^{\prime }\left( {t}_{2}\right) = {\rho R}\left( {t}_{2}\right) > 0 \]\n\n(7)\n\nwhich is a contradiction with \( {D}^{\prime }\left( {t}_{2}\right) < 0 \) .\n\nIn case 3), we have\n\n\[ {R}^{\prime }\left( {t}_{3}\right) = {\phi D}\left( {t}_{3}\right) > 0 \]\n\n(8)\n\nwhich is a contradiction with \( {R}^{\prime }\left( {t}_{3}\right) < 0 \) .\n\nIn case 4), we have\n\n\[ {Q}^{\prime }\left( {t}_{4}\right) = {\omega R}\left( {t}_{4}\right) > 0 \]\n\n(9)\n\nwhich is a contradiction with \( {Q}^{\prime }\left( {t}_{4}\right) < 0 \) .\n\nThus, the solution \( S\left( t\right), D\left( t\right), R\left( t\right), Q\left( t\right) \) of system (1) remains positive for all \( t > 0 \) . | Yes |
Lemma 2 All feasible solutions of the system (1) are bounded and enter the region\n\n\\[ \n\\Omega = \\left\\{ {\\left( {S, D, R, Q}\\right) \\in {R}_{ + }^{4} : S + D + R + Q \\leq \\frac{\\Lambda }{\\mu }}\\right\\} .\n\\] | Proof Let \\( \\left( {S, D, R, Q}\\right) \\in {R}_{ + }^{4} \\) be any solution with positive initial condition, adding the first four equations of (1), we have\n\n\\[ \n\\frac{d}{dt}\\left( {S + D + R + Q}\\right) = \\Lambda - {\\mu S} - {\\mu D} - {\\mu R} - {\\mu Q} - {d}_{1}D - {d}_{2}R - {d}_{3}Q =\n\\]\n\n\\[ \n\\Lambda - \\mu \\left( {S + D + R + Q}\\right) - \\left( {{d}_{1}D + {d}_{2}R + {d}_{3}Q}\\right) \\leq\n\\]\n\n\\[ \n\\Lambda - \\mu \\left( {S + D + R + Q}\\right) =\n\\]\n\n\\[ \n\\Lambda - {\\mu N}\\left( t\\right)\n\\]\n\nIt follows that\n\\[ \n0 \\leq N\\left( t\\right) \\leq \\frac{\\Lambda }{\\mu } + N\\left( 0\\right) {e}^{-{\\mu t}}\n\\]\n\nwhere \\( N\\left( 0\\right) \\) represents initial value of the total population. Thus \\( 0 \\leq N\\left( t\\right) \\leq \\frac{\\Lambda }{\\mu } \\), as \\( t \\rightarrow \\infty \\) .\n\nTherefore all feasible solutions of system (1) enter the region\n\n\\[ \n\\Omega = \\left\\{ {\\left( {S, D, R, Q}\\right) \\in {R}_{ + }^{4} : S + D + R + Q \\leq \\frac{\\Lambda }{\\mu }}\\right\\} .\n\\]\n\nHence, \\( \\Omega \\) is positively invariant and it is sufficient to consider solutions of system \\( \\left( 1\\right) \\) in \\( \\Omega \\) . Existence, uniqueness and continuation results for system (1) hold in this region. It can be shown that \\( N\\left( t\\right) \\) is bounded and all the solutions starting in \\( \\Omega \\) approach, enter or stay in \\( \\Omega \\) . | Yes |
Theorem 2.1 The drinking model (1) has two steady states as follows:\n\n(1) The drinking-free equilibrium \( {E}_{0}\left( {\frac{\Lambda }{\mu },0,0,0}\right) \) always exists for all parameter values.\n\n\( \left( 2\right) \; \) If \( \;{R}_{0} > 1,\;{the}\;{system}\;\left( 1\right) \;{has}\;a\;{unique}\;{positive}\;{drinking} \) - \( {present}\;{equilibrium}\;{E}^{ * }({S}^{ * },\;{D}^{ * }, \) \( \left. {{R}^{ * },{Q}^{ * }}\right) \) . And no drinking-present equilibrium if \( {R}_{0} \leq 1 \) . | Moveover, \( {E}^{ * }\left( {{S}^{ * },{D}^{ * },{R}^{ * },{Q}^{ * }}\right) \) satisfies the following equality:\n\n\[ \n{S}^{ * } = \frac{\left( {\mu + {d}_{1} + \phi - \frac{\rho \phi }{\rho + \mu + {d}_{2} + \omega }}\right) \left( {\frac{\phi }{\left( \rho + \mu + {d}_{2} + \omega \right) } + \frac{\phi \omega }{\left( {\rho + \mu + {d}_{2} + \omega }\right) \left( {\mu + {d}_{3}}\right) } + 1}\right) }{\beta \left( {\rho + \mu + {d}_{2} + \omega }\right) + {\rho \phi } - \left( {\mu + {d}_{1} + \phi }\right) \left( {\rho + \mu + {d}_{2} + \omega }\right) }{D}^{ * } > 0, \n\]\n\n\[ \n{R}^{ * } = \frac{\phi }{\rho + \mu + {d}_{2} + \omega }{D}^{ * }, \n\]\n\n\[ \n{Q}^{ * } = \frac{\phi \omega }{\left( {\rho + \mu + {d}_{2} + \omega }\right) \left( {\mu + {d}_{3}}\right) }{D}^{ * }, \n\]\n\nwhere \( {D}^{ * } \) is a unique positive solution of \( f\left( D\right) = 0 \), where\n\n\[ \nf\left( D\right) = {D}^{2}\left( {-\frac{{\beta }^{2}}{\mu {R}_{0}^{2}} + \frac{\beta }{{R}_{0}} + \frac{\beta \phi }{\left( {\rho + \mu + {d}_{2} + \omega }\right) {R}_{0}} + \frac{\beta \phi \omega }{\left( {\rho + \mu + {d}_{2} + \omega }\right) \left( {\mu + {d}_{3}}\right) {R}_{0}} + \frac{{\beta }^{2}}{\mu {R}_{0}}}\right) + \n\]\n\n\[ \nD\left( {\frac{\beta \Lambda }{\mu {R}_{0}} - \frac{\beta \Lambda }{\mu }}\right) \n\] | Yes |
Theorem 2.2 If \( {R}_{0} < 1 \), drinking-free equilibrium \( {E}_{0} \) is globally asymptotically stable. | Proof We introduce the following Lyapunov function:\n\n\[ V = \left( {\rho + \omega + {d}_{2} + \mu }\right) D + {\rho R}. \]\n\nThe derivative of \( V \) is given by\n\n\[ {V}^{\prime } = \left( {\rho + \omega + {d}_{2} + \mu }\right) {D}^{\prime } + \rho {R}^{\prime } = \]\n\n\[ \left( {\rho + \omega + {d}_{2} + \mu }\right) \left\lbrack {\frac{\beta DS}{N} + {\rho R} - \left( {\phi + {d}_{1} + \mu }\right) D}\right\rbrack + \rho \left\lbrack {{\phi D} - \left( {\rho + \omega + {d}_{2} + \mu }\right) R}\right\rbrack = \]\n\n\[ \frac{\beta DS}{N}\left( {\rho + \omega + {d}_{2} + \mu }\right) - \left( {\phi + {d}_{1} + \mu }\right) \left( {\rho + \omega + {d}_{2} + \mu }\right) D + {\rho \phi D} = \]\n\n(15)\n\n\[ \left\{ {\frac{\beta S}{N}\left( {\rho + \omega + {d}_{2} + \mu }\right) - \left\lbrack {\left( {\phi + {d}_{1} + \mu }\right) \left( {\rho + \omega + {d}_{2} + \mu }\right) - {\rho \phi }}\right\rbrack }\right\} D. \]\n\nWe note that \( S \leq N \), then\n\n\[ \left. {{V}^{{}^{\prime }} \leq \left\{ {\beta \left( {\rho + \omega + {d}_{2} + \mu }\right) - \left\lbrack {\rho \left( {\mu + {d}_{1}}\right) + \left( {\mu + {d}_{1} + \phi }\right) \left( {\mu + {d}_{2} + \omega }\right) }\right\rbrack }\right. }\right\} D = \]\n\n\[ \left\lbrack {\rho \left( {\mu + {d}_{1}}\right) + \left( {\mu + {d}_{1} + \phi }\right) \left( {\mu + {d}_{2} + \omega }\right) }\right\rbrack \left\lbrack {\frac{\beta \left( {\rho + \mu + {d}_{2} + \omega }\right) }{\rho \left( {\mu + {d}_{1}}\right) + \left( {\mu + {d}_{1} + \phi }\right) \left( {\mu + {d}_{2} + \omega }\right) } - 1}\right\rbrack D = \]\n\n\[ \left\lbrack {\rho \left( {\mu + {d}_{1}}\right) + \left( {\mu + {d}_{1} + \phi }\right) \left( {\mu + {d}_{2} + \omega }\right) }\right\rbrack \left( {{R}_{0} - 1}\right) D. \]\n\nHence, if \( {R}_{0} < 1 \), we have \( {V}^{\prime } < 0 \) .\n\nThus, for system (1), the drinking-free equilibrium \( {E}_{0} \) is globally asymptotically stable if \( {R}_{0} < 1 \) . | Yes |
Lemma 1.3 If \( x\left( t\right) \in P \cap \left( {{\bar{\Omega }}_{2} \smallsetminus {\Omega }_{1}}\right) ,{H}_{2} \) hold, then \( {\left( \Phi x\right) }_{i}\left( t\right) > 0, i = 1,2,3 \) . | Proof Since \( x\left( t\right) \in P \cap \left( {{\bar{\Omega }}_{2} \smallsetminus {\Omega }_{1}}\right) \), from(1.9),(1.10) and \( {H}_{2} \) we have\n\n\[ \n{\left( \Phi x\right) }_{1}\left( t\right) = {\int }_{t}^{t + \omega }{G}_{1}\left( {t,\sigma }\right) {x}_{1}\left( \sigma \right) \left\lbrack {-{a}_{11}\left( \sigma \right) {x}_{1}\left( \sigma \right) + {a}_{12}\left( \sigma \right) {\int }_{-\infty }^{\sigma }{K}_{12}\left( {s - \sigma }\right) {x}_{2}\left( s\right) \mathrm{d}s + }\right. \n\]\n\n\[ \n\left. {{a}_{13}\left( \sigma \right) {\int }_{-\infty }^{\sigma }{K}_{13}\left( {s - \sigma }\right) {x}_{3}\left( s\right) \mathrm{d}s}\right\rbrack \mathrm{d}\sigma \geq \n\]\n\n\[ \n\frac{{A}_{1}{\eta }^{2}}{B}{\left| {x}_{1}\right| }_{0}\left( {{a}_{12}^{K}{x}_{2}^{ * } + {a}_{13}^{K}{x}_{3}^{ * } - \frac{B}{{\eta }^{5}A}{\bar{a}}_{11}{x}_{1}^{ * }}\right) > 0. \n\]\n\nObviously,\n\n\[ \n{\left( \Phi x\right) }_{2}\left( t\right) > 0,\;{\left( \Phi x\right) }_{3}\left( t\right) > 0. \n\]\nThe proof is complete. | Yes |
Theorem 2.1 In addition to the existence of positive periodic solutions, assume that\n\n\[ \mathop{\inf }\limits_{{t \in \lbrack 0, + \infty )}}\left\lbrack {{a}_{ii}\left( t\right) - \mathop{\sum }\limits_{{j = 1, j \neq i}}^{3}{\int }_{-\infty }^{0}{K}_{ji}\left( s\right) {a}_{ji}\left( {t - s}\right) \mathrm{d}s}\right\rbrack > 0,\;i = 1,2,3 \]\n\n(2.1)\n\n\( \textit{Then system}\;\left( {0.3}\right) \textit{ has a unique positive }\omega \) -periodic solution \( \;{x}^{ * }\left( t\right) = {\left( {x}_{1}^{ * }\left( t\right) ,{x}_{2}^{ * }\left( t\right) ,{x}_{3}^{ * }\left( t\right) \right) }^{T}\textit{ with} \) attracts all positive solutions \( x\left( t\right) = {\left( {x}_{1}\left( t\right) ,{x}_{2}\left( t\right) ,{x}_{3}\left( t\right) \right) }^{T} \) of system (0.3), that is\n\n\[ \mathop{\lim }\limits_{{t \rightarrow + \infty }}\left| {{x}_{i}^{ * }\left( t\right) - {x}_{i}\left( t\right) }\right| = 0,\;i = 1,2,3. \]\n\n(2.2) | Proof Let \( x\left( t\right) = {\left( {x}_{1}\left( t\right) ,{x}_{2}\left( t\right) ,{x}_{3}\left( t\right) \right) }^{T} \) be any positive solution of system (0.3). Consider a Lyapunov functional \( V\left( t\right) \) defined by\n\n\[ V\left( t\right) = {V}_{1}\left( t\right) + {V}_{2}\left( t\right) ,\;t \geq 0, \]\n\nwhere\n\n\[ {V}_{1}\left( t\right) = \mathop{\sum }\limits_{{i = 1}}^{3}\left| {\ln {x}_{i}^{ * }\left( t\right) - \ln {x}_{i}\left( t\right) }\right| \]\n\n\[ {V}_{2}\left( t\right) = \mathop{\sum }\limits_{{i = 1, j = 1, j \neq i}}^{3}{\int }_{-\infty }^{0}{K}_{ji}\left( s\right) \mathrm{d}s{\int }_{t + s}^{t}{a}_{ji}\left( {\theta - s}\right) \left| {{x}_{i}^{ * }\left( \theta \right) - {x}_{i}\left( \theta \right) }\right| \mathrm{d}\theta . \]\n\nFrom the definition of \( V\left( t\right) \), it is clear that \( V\left( t\right) \) is a positive definite function and\n\n\[ V\left( 0\right) < + \infty \]\n\n(2.3)\n\nand\n\n\[ V\left( t\right) \geq \mathop{\sum }\limits_{{i = 1}}^{3}\left| {\ln {x}_{i}^{ * }\left( t\right) - \ln {x}_{i}\left( t\right) }\right| ,\;t \geq 0. \]\n\n(2.4)\n\nCalculating the upper right derivative \( {D}^{ + }V\left( t\right) \) of \( V\left( t\right) \) at time \( t \), we obtain\n\n\[ {D}^{ + }V\left( t\right) \leq - K\mathop{\sum }\limits_{{i = 1}}^{3}\left| {{x}_{i}^{ * }\left( t\right) - {x}_{i}\left( t\right) }\right| \]\n\n(2.5)\n\nThen by (2.1), (2.3) and (2.5), one can obtain\n\n\[ \mathop{\lim }\limits_{{t \rightarrow + \infty }}\mathop{\sum }\limits_{{i = 1}}^{3}\left| {{x}_{i}^{ * }\left( t\right) - {x}_{i}\left( t\right) }\right| = 0 \]\n\nHence, we have\n\n\[ \mathop{\lim }\limits_{{t \rightarrow + \infty }}\left| {{x}_{i}^{ * }\left( t\right) - {x}_{i}\left( t\right) }\right| = 0,\;i = 1,2,3. \]\n\nThis completes the proof. | Yes |
Lemma 1 The solution of equation (3) satisfies \( \mathop{\lim }\limits_{{t \rightarrow + \infty }}u\left( t\right) = {u}_{1} \) if \( r > p \) . | Proof In equation (3), let\n\n\[ F\left( u\right) = {ru}\left( {1 - \frac{u}{K}}\right) - \frac{pu}{1 + {qu}}. \]\n\nThen equation (3) has three equilibrium points:\n\n\[ {u}_{0} = 0,\;{u}_{1} = \frac{-\bar{a} + \sqrt{{\bar{a}}^{2} - 4\bar{c}}}{2},\;{u}_{2} = \frac{-\bar{a} - \sqrt{{\bar{a}}^{2} - 4\bar{c}}}{2}. \]\n\nWhen \( r > p \), that is, \( \bar{c} < 0,{F}^{\prime }\left( {u}_{0}\right) = r - p \), consequently the equilibrium point \( {u}_{0} = 0 \) is unstable.\n\nIf \( r > p \), then \( {u}_{2} < 0 \), which is omitted. In order to see that the equilibrium point \( u = {u}_{1} \) is globally absorbing point and \( \mathop{\lim }\limits_{{t \rightarrow + \infty }}u\left( t\right) = {u}_{1} \), this paper discussed with two cases. Firstly, let\n\n\[ f\left( u\right) = r\left( {1 - \frac{u}{K}}\right) - \frac{p}{1 + {qu}}. \]\n\nThen we know that\n\n\[ f\left( {u}_{1}\right) = r\left( {1 - \frac{{u}_{1}}{K}}\right) - \frac{p}{1 + q{u}_{1}} = 0. \]\n\nCase 1. \( 0 < u < {u}_{1} \) . Notice that \( 1 + q{u}_{1} = {Kq} + \sqrt{{\left( 1 - Kq\right) }^{2} + \frac{4Kq}{r}\left( {r - p}\right) } > {Kq} \), and \( 1 + {qu} > 1 \) . So\n\n\[ f\left( u\right) - f\left( {u}_{1}\right) = \frac{r}{K}\left( {{u}_{1} - u}\right) - \left\lbrack {\frac{p}{1 + {qu}} - \frac{p}{1 + q{u}_{1}}}\right\rbrack = \]\n\n\[ \left( {{u}_{1} - u}\right) \left( {\frac{r}{K} - \frac{pq}{\left( {1 + {qu}}\right) \left( {1 + q{u}_{1}}\right) }}\right) > \]\n\n\[ \left( {{u}_{1} - u}\right) \left( {\frac{r}{K} - \frac{p}{K}}\right) = \]\n\n\[ \left( {{u}_{1} - u}\right) \frac{r - p}{K} > 0 \]\n\nCase 2. \( u > {u}_{1} \]\n\nNotice that \( 1 + q{u}_{1} = {Kq} + \sqrt{{\left( 1 - Kq\right) }^{2} + \frac{4Kq}{r}\left( {r - p}\right) } > {Kq} \), and \( 1 + {qu} > 1 \) . So\n\n\[ f\left( u\right) - f\left( {u}_{1}\right) = \frac{r}{K}\left( {u - {u}_{1}}\right) - \left\lbrack {\frac{p}{1 + q{u}_{1}} - \frac{p}{1 + {qu}}}\right\rbrack = \]\n\n\[ \left( {u - {u}_{1}}\right) \left( {\frac{pq}{\left( {1 + {qu}}\right) \left( {1 + q{u}_{1}}\right) } - \frac{r}{K}}\right) < \]\n\n\[ \left( {u - {u}_{1}}\right) \left( {\frac{p}{K} - \frac{r}{K}}\right) = \]\n\n\[ \left( {u - {u}_{1}}\right) \frac{p - r}{K} < 0 \]\n\nThe proof of the lemma is completed. | Yes |
Lemma 1.3 Consider the equation\n\n\\[ \n\\left\\{ \\begin{array}{l} {x}^{\\prime }\\left( t\\right) = \\alpha \\left( t\\right) x\\left( t\\right), t \\in J, t \\neq {t}_{k}, \\\\ x\\left( {t}_{k}^{ + }\\right) = x\\left( {t}_{k}\\right) + p, \\end{array}\\right.\n\\]\n\n(1.9)\n\nwhere \\( \\alpha \\left( t\\right) \\) is a continuous \\( \\omega \\) -periodic function, \\( p \\) is a positive constant and there is an integer \\( q > 0 \\) such that \\( {t}_{k + q} = {t}_{k} + \\omega \\) . Assume that \\( {\\int }_{0}^{\\omega }\\alpha \\left( t\\right) {dt} < 0 \\), then (1.9) has a unique positive, globally asymptotically stable \\( \\omega \\) -periodic solution. | Proof It is easy to show that the following function\n\n\\[ \n\\bar{x}\\left( t\\right) = \\left\\{ \\begin{array}{ll} {x}^{ * }\\left( t\\right) , & t \\in (0,\\omega \\rbrack , \\\\ x\\left( {t - {j\\omega }}\\right) , & t \\in ({j\\omega },\\left( {j + 1}\\right) \\omega \\rbrack, j \\in {Z}^{ + }, \\end{array}\\right.\n\\]\n\nwhere\n\n\\[ \n{x}^{ * }\\left( t\\right) = \\frac{p\\mathop{\\sum }\\limits_{{i = 1}}^{q}\\exp \\left( {{\\int }_{{t}_{i}}^{\\omega }\\alpha \\left( t\\right) {dt}}\\right) }{1 - \\exp \\left( {{\\int }_{0}^{\\omega }\\alpha \\left( t\\right) {dt}}\\right) }\\exp \\left( {{\\int }_{0}^{t}\\alpha \\left( s\\right) {ds}}\\right) + p\\mathop{\\sum }\\limits_{{0 < {t}_{i} < t}}\\exp \\left( {{\\int }_{{t}_{i}}^{t}\\alpha \\left( s\\right) {ds}}\\right)\n\\]\n\nis a \\( \\omega \\) -periodic solution of (1.9). | Yes |
Lemma 2.2 Let \( d\left( t\right) \) be a continuous \( \omega \) -periodic function. \( \omega > 0 \) and \( {\int }_{0}^{\omega }d\left( t\right) {dt} > 0 \), Then we have that\n\n\[ \n{e}^{c\left( {t - s}\right) } \leq {e}^{1 + {D\omega } + {\int }_{s}^{t}d\left( r\right) {dr}},\;\text{for}\;t \geq s, \n\] \n\n\( \textit{where}\;c : 0 < c \leq \min \{ {\int }_{0}^{\omega }d\left( t\right) {dt}/\omega ,\;1/\omega \} \;\textit{and}\;D = \max \{ \left| {d\left( t\right) }\right| : t \in \left\lbrack {0,\omega }\right\rbrack \} . \) | Proof Put \( t = s + {n\omega } + \mu \) where \( n \in \{ 0,1,2,\cdots \} \) and \( 0 \leq \mu < \omega \) . Then we have\n\n\[ \n{e}^{1 + {D\omega } + {\int }_{s}^{t}d\left( r\right) {dr}} = {e}^{1 + {D\omega } + {\int }_{s}^{s + {n\omega } + \mu }d\left( r\right) {dr}} \geq \n\] \n\n\[ \n{e}^{1 + n{\int }_{0}^{\omega }d\left( r\right) {dr}} \geq {e}^{1 + {nc\omega }} \geq \n\] \n\n\[ \n{e}^{c\left( {{n\omega } + \mu }\right) } = {e}^{c\left( {t - s}\right) }. \n\] \n\nThe proof is complete. | Yes |
Theorem 2.4 Suppose that \( {\int }_{0}^{\omega }d\left( t\right) {dt} > 0 \) and\n\n\[ \n{\int }_{0}^{\omega }a\left( s\right) {ds} + \ln \mathop{\prod }\limits_{{i = 1}}^{q}\left( {1 + {g}_{i}}\right) < 0 \n\]\n\n(2.8)\n\nhold, then the solution \( \left( {x\left( t\right), y\left( t\right) }\right) \) of system \( \left( {0.1}\right) \) with any positive initial value satisfies\n\n\[ \n\left( {x\left( t\right), y\left( t\right) }\right) \rightarrow \left( {0,{y}^{ * }\left( t\right) }\right) \text{ as }t \rightarrow \infty ,\n\]\n\nwhere \( {y}^{ * }\left( t\right) \) is unique positive \( \omega \) -periodic solution of the following equation\n\n\[ \n\left\{ \begin{array}{l} {y}^{\prime }\left( t\right) = - d\left( t\right) y\left( t\right), t \neq {t}_{k} \\ y\left( {t}_{k}^{ + }\right) = y\left( {t}_{k}\right) + p \end{array}\right.\n\]\n\n(2.9) | Proof From initial value condition (0.2), one can easily see that \( x\left( t\right) > 0, y\left( t\right) > 0 \) for \( t \geq 0 \) and \( \mathop{\lim }\limits_{{t \rightarrow \infty }}y\left( t\right) \geq {2}^{-1}\mathop{\min }\limits_{{t \in R}}{y}^{ * }\left( t\right) > 0 \) . Hence,\n\n\[ \n{x}^{\prime }\left( t\right) \leq a\left( t\right) x\left( t\right), t \neq {t}_{k} \n\]\n\nwhich implies that\n\n\[ \nx\left( t\right) \leq x\left( 0\right) \mathop{\prod }\limits_{{0 < {t}_{k} < t}}\left( {1 + {g}_{k}}\right) \exp \left( {{\int }_{0}^{t}a\left( s\right) {ds}}\right) \rightarrow 0,\text{ as }t \rightarrow \infty \n\]\n\nsince (2.8) holds. Therefore the system (0.1) follows that for any \( \varepsilon > 0 \), there exists \( {T}_{\varepsilon } > 0 \) such that\n\n\[ \n- d\left( t\right) y\left( t\right) \leq {y}^{\prime }\left( t\right) \leq \left( {-d\left( t\right) + \varepsilon }\right) y\left( t\right), t \neq {t}_{k}, t \geq {T}_{\varepsilon }. \n\]\n\nFrom Lemma 1.3, we obtain that\n\n\[ \n\left\{ \begin{array}{l} {u}^{\prime }\left( t\right) = \left( {-d\left( t\right) + {\varepsilon }^{ * }}\right) u\left( t\right), t \neq {t}_{k} \\ u\left( {t}_{k}^{ + }\right) = u\left( {t}_{k}\right) + p \end{array}\right. \n\]\n\n\( \left( {2.10}\right) \)\n\nhas positive globally asymptotically stable \( \omega \) -periodic solution \( {u}_{{\varepsilon }^{ * }} \) for \( 0 \leq {\varepsilon }^{ * } < {\int }_{0}^{\omega }d\left( t\right) {dt}/\omega \) and \( \mathop{\lim }\limits_{{{\varepsilon }^{ * } \rightarrow {0}^{ + }}}{u}_{{\varepsilon }^{ * }} = {u}_{0} \) . By comparison theorem of impulsive differential equations and the attractivity of \( {u}_{{\varepsilon }^{ * }} \), we can obtain that \( \mathop{\lim }\limits_{{t \rightarrow + \infty }}y\left( t\right) = {u}_{0}\left( t\right) = {y}^{ * }\left( t\right) \) . | Yes |
Theorem 1.1 If \( {R}_{0} < 1 \), the system (3) only have a disease-free equilibrium point \( {E}_{0} = \) \( \left( {\frac{\lambda }{d},0,0,0}\right) \) ; If \( {R}_{0}^{ * } \leq 1 < {R}_{0} \), the immune response does not play an important role, so we can ignore the immune response, this implies \( \frac{dz}{dt} \leq 0 \) and the equilibrium is \( {E}^{ * }\left( {{x}^{ * },{y}^{ * },{v}^{ * },0}\right) \) ; If \( {R}_{0} > {R}_{0}^{ * } > 1 \), the system exists immune respond equilibrium \( \bar{E}\left( {\bar{x},\bar{y},\bar{v},\bar{z}}\right) \) . | Proof When the system does not exist the immune respond and \( {R}_{0} > 1 \), means that the disease breaks out, and the immune system dosen't play a role in defending the disease. In view of mathematics, means \( {R}_{0}^{ * } \leq 1 < {R}_{0} \), then model (3) reduces to model (2).\n\nSetting the right hand side of model (2) to zero, we get the following biologically relevant equilibria.\n\n\[ \lambda - d{x}^{ * } - f\left( {{x}^{ * },{y}^{ * },{v}^{ * }}\right) {v}^{ * } = 0,\;f\left( {{x}^{ * },{y}^{ * },{v}^{ * }}\right) {v}^{ * }{e}^{-{m\tau }} - a{y}^{ * } = 0,\;k{y}^{ * } - u{v}^{ * } = 0. \]\n\nThen, we have\n\n\[ {v}^{ * } = \frac{k{y}^{ * }}{u},\;f\left( {{x}^{ * },\frac{\lambda - d{x}^{ * }}{a{e}^{m\tau }},\frac{k\left( {\lambda - d{x}^{ * }}\right) }{{au}{e}^{m\tau }}}\right) = \frac{{au}{e}^{m\tau }}{k},\;{y}^{ * } = \frac{\lambda - d{x}^{ * }}{{am}{e}^{m\tau }} \geq 0,\;{x}^{ * } \leq \frac{\lambda }{d}. \]\n\nDenote\n\[ g\left( {x}^{ * }\right) = f\left( {{x}^{ * },\frac{\lambda - d{x}^{ * }}{{am}{e}^{m\tau }},\frac{k\left( {\lambda - d{x}^{ * }}\right) }{{au}{e}^{m\tau }}}\right) - \frac{{au}{e}^{m\tau }}{k}, \]\n\nwe get\n\n\[ g\left( 0\right) < 0,\;g\left( \frac{\lambda }{d}\right) = \frac{{au}{e}^{m\tau }}{k}\left( {{R}_{0} - 1}\right) > 0,\;{g}^{\prime }\left( x\right) > 0, \]\n\nso the system (3) exists only one equilibrium point without immune response \( {E}^{ * }\left( {{x}^{ * },{y}^{ * },{v}^{ * },0}\right) \) . When \( {R}_{0} > {R}_{0}^{ * } > 1 \), means the disease break out and the immune system plays an important role in defending the disease, even so, the immune system can not stop the disease break out. So the system \( \left( 3\right) \) has an immune respond equilibrium point \( \bar{E}\left( {\bar{x},\bar{y},\bar{v},\bar{z}}\right) . \) | Yes |
Theorem 1.2 When \( {R}_{0} < 1 \), the disease-free equilibrium point \( {E}_{0} \) is globally asymptotically stable. | Proof When \( {R}_{0} < 1 \), model (3) reduces to model (2). Meanwhile the proof is given in the paper [9]. | No |
Lemma 1 (Positive invariant and bounded) \( \textit{System (1.1) is positively invariant and bounded} \) in \( {R}_{ + }^{3} \) . Moreover, we have\n\n\[ \mathop{\limsup }\limits_{{t \rightarrow + \infty }}x\left( t\right) \leq 1,\;\mathop{\limsup }\limits_{{t \rightarrow + \infty }}y\left( t\right) \leq \frac{{a}_{1} + {a}_{4}}{{a}_{2}},\;\mathop{\limsup }\limits_{{t \rightarrow + \infty }}z\left( t\right) \leq \frac{{a}_{2}{a}_{4}\left( {\mu + 1}\right) + \left( {\mu + {a}_{1}}\right) \left( {{a}_{1} + {a}_{4}}\right) }{{a}_{2}{a}_{4}\mu }.\] | Proof By the continuity argument, we can easily prove that system (1.1) is positively invariant in \( {R}_{ + }^{3} \) . Then from the first equation of (1.1) we have \( {x}^{\prime } \leq x\left( {1 - x}\right) \), thus we can conclude that\n\n\[ \mathop{\limsup }\limits_{{t \rightarrow + \infty }}x\left( t\right) \leq 1 \]\n\nFor time large enough, it follows from second equation of (1.1) that\n\n\[ {y}^{\prime }\left( t\right) = y\left( {{a}_{1} - {a}_{2}y - {a}_{3}z + {a}_{4}x}\right) \leq y\left( {{a}_{1} + {a}_{4} - {a}_{2}y}\right) \Rightarrow \]\n\n\[ \mathop{\limsup }\limits_{{t \rightarrow + \infty }}y\left( t\right) \leq \frac{{a}_{1} + {a}_{4}}{{a}_{2}} \]\n\nDefine \( w = x + {\theta }_{1}y + {\theta }_{2}z \), where \( 1 - {\theta }_{1}{a}_{4} > 0,{\theta }_{1}{a}_{3} - {\theta }_{2}{b}_{1} > 0,1 - {\theta }_{2}\beta > 0 \) . Then we have\n\n\[ {w}^{\prime } \leq x\left( {\mu + 1}\right) + {\theta }_{1}\left( {{a}_{1} + \mu }\right) y - {\mu w}. \]\n\nSince for any \( \varepsilon > 0 \), there exists a \( T > 0 \) such that \( x\left( t\right) \leq 1 + \varepsilon \) for all \( t > T \) . Therefore, for \( t > T \), we have\n\n\[ {w}^{\prime } \leq \left( {\mu + 1}\right) + \frac{\left( {\mu + {a}_{1}}\right) \left( {{a}_{1} + {a}_{4}}\right) }{{a}_{2}{a}_{4}} - {\mu w}. \]\n\nThus we can conclude that\n\n\[ \mathop{\limsup }\limits_{{t \rightarrow + \infty }}z\left( t\right) \leq \frac{{a}_{2}{a}_{4}\left( {\mu + 1}\right) + \left( {\mu + {a}_{1}}\right) \left( {{a}_{1} + {a}_{4}}\right) }{{a}_{2}{a}_{4}\mu }. \]\n\nTherefore, system (1.1) is bounded in \( {R}_{ + }^{3} \). | Yes |
Proposition 1 (Subsystem of species \( x \) and \( y \) ) The system (1.2) is globally stable at \( \left( {0,\frac{{a}_{1}}{{a}_{2}}}\right) \) if and only if \( {a}_{2} \leq {a}_{1} \) while it is globally stable at \( \left( {\frac{{a}_{2} - {a}_{1}}{{a}_{4} + {a}_{2}},\frac{{a}_{4} + {a}_{1}}{{a}_{4} + {a}_{2}}}\right) \) if and only if \( {a}_{2} > {a}_{1} \) . | Proof If \( {a}_{2} < {a}_{1} \), then (1.2) has only one locally asymptotically stable boundary equilibrium \( \left( {0,\frac{{a}_{1}}{{a}_{2}}}\right) \) and two unstable equilibria \( \left( {0,0}\right) \) and \( \left( {1,0}\right) \) . Since \( \mathop{\liminf }\limits_{{t \rightarrow \infty }}y\left( t\right) \geq \frac{{a}_{1}}{{a}_{2}} \) from system (1.2), we have the following inequality, if \( {a}_{2} < {a}_{1} \), when time \( t \) large enough\n\n\[ \n{x}^{\prime } = x\left( {1 - x - y}\right) \leq x\left( {1 - \frac{{a}_{1}}{{a}_{2}}}\right) \Rightarrow \mathop{\limsup }\limits_{{t \rightarrow \infty }}x\left( t\right) = 0.\n\]\n\nTherefore, we can obtain that \( \mathop{\lim }\limits_{{t \rightarrow \infty }}y\left( t\right) = \frac{{a}_{1}}{{a}_{2}} \) and \( \mathop{\lim }\limits_{{t \rightarrow \infty }}x\left( t\right) = 0 \) if \( {a}_{2} < {a}_{1} \) .\n\nIf \( {a}_{2} > {a}_{1} \), then (1.2) has an interior equilibrium \( \left( {\frac{{a}_{2} - {a}_{1}}{{a}_{4} + {a}_{2}},\frac{{a}_{4} + {a}_{1}}{{a}_{4} + {a}_{2}}}\right) \) which is locally asymptotically stable while \( \left( {0,0}\right) \) and \( \left( {1,0}\right) \) are unstable equilibria. By Poincare-Bendixson Therom \( {}^{\left\lbrack {13}\right\rbrack } \) , the omega limit set of system (1.2) is either a fixed point or a limit cycle. Now define a scalar function \( R\left( {x, y}\right) = \frac{1}{xy} \), and let\n\n\[ \nP\left( {x, y}\right) = x\left( {1 - x - y}\right) ,\;Q\left( {x, y}\right) = y\left( {{a}_{1} - {a}_{2}y + {a}_{4}x}\right) .\n\]\n\nThen we have\n\n\[ \n\frac{\partial \left( {R\left( {x, y}\right) P\left( {x, y}\right) }\right) }{\partial x} + \frac{\partial \left( {R\left( {x, y}\right) Q\left( {x, y}\right) }\right) }{\partial y} = - \frac{1}{y} - \frac{{a}_{2}}{x} < 0\;\text{ for all }\left( {x, y}\right) \in {R}_{ + }^{2}.\n\]\n\nTherefore, according to Dulac's criterion [13], we can conclude that the trajectory of any initial condition taken in \( {R}_{ + }^{2} \) convergies to an equilibrium point. Since the only locally stable equilibrium of (1.2) when \( {a}_{2} > {a}_{1} \) is the interior equilibrium point \( \left( {\frac{{a}_{2} - {a}_{1}}{{a}_{4} + {a}_{2}},\frac{{a}_{4} + {a}_{1}}{{a}_{4} + {a}_{2}}}\right) \), therefore, it must be global stable.\n\nIn the case that \( {a}_{1} = {a}_{2},\left( {1.2}\right) \) has equilibrium \( \left( {0,\frac{{a}_{1}}{{a}_{2}}}\right) \) and other two unstable equilibria. \( \left( {0,0}\right) \) and \( \left( {1,0}\right) \) . Then according to Poincare-Bendixson Therom \( {}^{\left\lbrack {13}\right\rbrack } \) again, we can conclude that all trajectories should converges to \( \left( {0,\frac{{a}_{1}}{{a}_{2}}}\right) \), thus \( \left( {0,\frac{{a}_{1}}{{a}_{2}}}\right) \) is global stable. The necessary conditions are easily derived from the local stability. | Yes |
Theorem 1 Species \( y \) is persistent in \( {R}_{ + }^{3} \) for system (1.1) if \( \mu > \beta + \frac{{b}_{1}\left( {{a}_{1} + {a}_{4}}\right) }{{a}_{2}} \) . In addition, System (1.1) has global stability at \( \left( {0,\frac{{a}_{1}}{{a}_{2}},0}\right) \) if \( \mu > \beta + \frac{{b}_{1}\left( {{a}_{1} + {a}_{4}}\right) }{{a}_{2}} \) and \( {a}_{1} > {a}_{2} \) . | Proof Assume that \( {a}_{1} > {a}_{2} \) for system (1.1). From Lemma 1, there is a \( T > 0 \) large enough such that for any \( t > T \), if \( \mu > \beta + \frac{{b}_{1}\left( {{a}_{1} + {a}_{4}}\right) }{{a}_{2}} \), we have\n\n\[ \n{z}^{\prime } = z\left( {{b}_{1}y + \frac{\beta x}{a + x} - \mu }\right) \leq z\left( {\frac{{b}_{1}\left( {{a}_{1} + {a}_{4}}\right) }{{a}_{2}} + \beta - \mu }\right) \Rightarrow \mathop{\limsup }\limits_{{t \rightarrow \infty }}z\left( t\right) = 0. \n\] \n\nHence, for any \( \epsilon > 0 \), there exists a \( T > 0 \) such that \( z\left( t\right) \leq \epsilon \) for all \( t > T \) . Therefore, for \( t > T \), we have \n\n\[ \n{y}^{\prime } = y\left( {{a}_{1} - {a}_{2}y - {a}_{3}z + {a}_{4}x}\right) \geq y\left( {{a}_{1} - {a}_{2}y - {a}_{3}\epsilon }\right) \Rightarrow \mathop{\liminf }\limits_{{t \rightarrow \infty }}y\left( t\right) \geq \frac{{a}_{1}}{{a}_{2}}. \n\] \n\nTherefore, by Lemma 1 we know that species \( y \) is persistent.\n\nFor any \( \varepsilon > 0 \) such that \( 1 + \varepsilon < \frac{{a}_{1}}{{a}_{2}} \) and time \( t \) large enough, we have that \n\n\[ \n{x}^{\prime } = x\left( {1 - x - y - \frac{z}{a + x}}\right) \leq x\left( {1 - y}\right) \leq x\left( {1 - \frac{{a}_{1}}{{a}_{2}} + \varepsilon }\right) \Rightarrow \n\] \n\n\[ \nx\left( t\right) < x\left( 0\right) {e}^{\left( {1 - \frac{{a}_{1}}{{a}_{2}} + \varepsilon }\right) t} \rightarrow 0\;\text{ as }\;t \rightarrow \infty . \n\] \n\nThis implies that the limiting system of (1.1) is \( {y}^{\prime } = y\left( {{a}_{1} - {a}_{2}y}\right) \), therefore, \( \mathop{\lim }\limits_{{t \rightarrow \infty }}y\left( t\right) = \frac{{a}_{1}}{{a}_{2}} \) . This indicates that system (1.1) is globally stable at \( \left( {0,\frac{{a}_{1}}{{a}_{2}},0}\right) \) if \( {a}_{1} > {a}_{2} \) and \( \mu > \beta + \frac{{b}_{1}\left( {{a}_{1} + {a}_{4}}\right) }{{a}_{2}} \) . | Yes |
Species \( x \) is persistent in \( {R}_{ + }^{3} \) for system (1.1) if \( \mu > \frac{{a}_{1}{b}_{1}}{{a}_{2}} \) and \( \frac{{a}_{1}}{{a}_{2}} < 1 \) or \( \min \left\{ {{a}_{1},{a}_{2}}\right\} > a{a}_{3} \) and \( \mu < \) \( \min \left\{ {\frac{a{a}_{3}{b}_{1} - {a}_{1}{b}_{1}}{a{a}_{3} - {a}_{2}},\frac{{a}_{1}{b}_{1}}{{a}_{2}}}\right\} \) . | Proof According to Lemma 1, we can restrict the dynamics of system (1.1) on the compact set \( C = \left\lbrack {0, M}\right\rbrack \times \left\lbrack {0, M}\right\rbrack \times \left\lbrack {0, M}\right\rbrack \), where \( M \) is positive constant. The omega limit set of the \( y - z \) subsystem restricted on \( C \) has two situations according to Proposition 2: If \( \mu > \frac{{a}_{1}{b}_{1}}{{a}_{2}} \), then the omega limit set of the \( y - z \) subsystem restricted on \( C \) for system (1.1) is \( \left( {0,\frac{{a}_{1}}{{a}_{2}},0}\right) \) while if \( \mu < \frac{{a}_{1}{b}_{1}}{{a}_{2}} \), the omega limit set is \( \left( {0,\frac{\mu }{{b}_{1}},\frac{{a}_{1}{b}_{1} - {a}_{2}\mu }{{a}_{3}{b}_{1}}}\right) \) . Therefore, according to Theorem 2.5 of \( {\text{Hutson}}^{\left\lbrack {14}\right\rbrack } \), species \( x \) is persistent in \( {R}_{ + }^{3} \) if\n\n\[ \mu > \frac{{a}_{1}{b}_{1}}{{a}_{2}},{\left. \;\frac{dx}{xdt}\right| }_{\left( 0,\frac{{a}_{1}}{{a}_{2}},0\right) } = 1 - \frac{{a}_{1}}{{a}_{2}} > 0 \Rightarrow {a}_{1} < {a}_{2}. \]\n\nor\n\n\[ \mu < \frac{{a}_{1}{b}_{1}}{{a}_{2}},{\left. \;\frac{dx}{xdt}\right| }_{\left( 0,\frac{\mu }{{b}_{1}},\frac{{a}_{1}{b}_{1} - {a}_{2}\mu }{{a}_{3}{b}_{1}}\right) } = 1 - \frac{\mu }{{b}_{1}} - \frac{1}{a}\frac{{a}_{1}{b}_{1} - {a}_{2}\mu }{{a}_{3}{b}_{1}} > 0 \Rightarrow \mu < \frac{a{a}_{3}{b}_{1} - {a}_{1}{b}_{1}}{a{a}_{3} - {a}_{2}}. \]\n\nif \( \mu > \beta + \frac{{b}_{1}\left( {{a}_{1} + {a}_{4}}\right) }{{a}_{2}} \), then from Lemma 1 and Theorem 1 we know that species \( y \) is persistence for system (1.1). Thus, if \( \mu > \beta + \frac{{b}_{1}\left( {{a}_{1} + {a}_{4}}\right) }{{a}_{2}} \), then sufficient conditions for the persistence of species \( x \) and \( y \) is equivalent to sufficient condition for the persistence of \( x \) . By Lemma 1 we have follows if time is large enough\n\n\[ {z}^{\prime } = z\left( {{b}_{1}y + \frac{\beta x}{a + x} - \mu }\right) \leq z\left( {\frac{{b}_{1}\left( {{a}_{1} + {a}_{4}}\right) }{{a}_{2}} + \beta - \mu }\right) . \]\n\nHence, species \( z \) of system (1.1) goes to extinct if \( \mu > \beta + \frac{{b}_{1}\left( {{a}_{1} + {a}_{4}}\right) }{{a}_{2}} \) . Therefore, the limiting set system of system (1.1) is the \( x - y \) subsystem. Then according to Proposition 1, system (1.1) has global stability at \( \left( {\frac{{a}_{2} - {a}_{1}}{{a}_{4} + {a}_{2}},\frac{{a}_{4} + {a}_{1}}{{a}_{4} + {a}_{2}},0}\right) \) if \( \mu > \beta + \frac{{b}_{1}\left( {{a}_{1} + {a}_{4}}\right) }{{a}_{2}} \) and \( \frac{{a}_{1}}{{a}_{2}} < 1 \) . | Yes |
Theorem 3 If \( {b}_{1} < \min \left( {\mu ,\frac{\beta {a}_{3}}{{a}_{4}}}\right) \), then equilibrium \( \left( {0,\frac{\mu }{{b}_{1}},\frac{{a}_{1}{b}_{1} - {a}_{2}\mu }{{a}_{3}{b}_{1}}}\right) = \left( {0,{y}^{ * },{z}^{ * }}\right) \) is globally asymptotically stable. | Proof Let\n\n\[ V = x + c\left( {y - {y}^{ * } - {y}^{ * }\ln \left( {y/{y}^{ * }}\right) }\right) + d\left( {z - {z}^{ * } - {z}^{ * }\ln \left( {z/{z}^{ * }}\right) }\right) ,\]\n\nwhere \( c \) and \( d \) are constants to be chosen later. \( V \) is continuous on the positive quadrant. It is also easy to verify that \( \mathrm{V} \) is positive for all values of \( x, y, z \) except for the positive boundary equilibrium \( \left( {0,{y}^{ * },{z}^{ * }}\right) \) where it is zero. The time derivative of V along a solution of system (1.1) is\n\n\[ \frac{dV}{dt} = \dot{x} + c\left( {y - {y}^{ * }}\right) \frac{\dot{x}}{y} + d\left( {z - {z}^{ * }}\right) \frac{\dot{z}}{z}. \]\n\n(1.4)\n\nOn substituting \( {a}_{1} - {a}_{2}{y}^{ * } - {a}_{3}{z}^{ * } = 0 \) and \( {b}_{1}{y}^{ * } - \mu = 0 \) into (1.4), we have\n\n\[ \frac{dV}{dt} = - {x}^{2} - {a}_{2}c{\left( y - {y}^{ * }\right) }^{2} + \left( {d{b}_{1} - {a}_{3}c}\right) \left( {y - {y}^{ * }}\right) \left( {z - {z}^{ * }}\right) + \left( {1 - {a}_{4}c{y}^{ * }}\right) x + \]\n\n\[ \left( {{a}_{4}c - 1}\right) {xy} + \frac{\left( {{d\beta } - 1}\right) {xz}}{a + x} - \frac{{d\beta x}{z}^{ * }}{a + x}. \]\n\nHere we pick \( d = \frac{1}{\beta } \), since \( x\left( t\right), y\left( t\right) \) and \( z\left( t\right) \) are positive, if \( {b}_{1}\max \left( {\frac{1}{{a}_{4}\mu },\frac{1}{\beta {a}_{3}}}\right) < c < \frac{1}{{a}_{4}} \) , then \( \frac{dV}{dt} \leq 0.\frac{dV}{dt} = 0 \) if and only if \( \left( {x, y, z}\right) = \left( {0,{y}^{ * },{z}^{ * }}\right) \) . Thus \( \left( {0,\frac{\mu }{{b}_{1}},\frac{{a}_{1}{b}_{1} - {a}_{2}\mu }{{a}_{3}{b}_{1}}}\right) \) is globally asymptotically stable. | Yes |
Lemma 1.1 \( \; \) If \( b = 0 \) , \( a{f}^{\prime }\left( 0\right) < 1 + c \), then the zero equilibrium \( {u}^{ * } \) of system (0.2) and (0.3) is local asymptotically stable. | Proof When \( b = 0 \), the characteristic equation (1.3) yields\n\n\[ \lambda + c{k}^{2} + 1 - a{f}^{\prime }\left( 0\right) = 0,\;k = 1,2,\ldots \]\n\n(1.4)\n\nIt is clear that all roots of (1.4) have negative real parts if \( a{f}^{\prime }\left( 0\right) < 1 + c \) . So the zero equilibrium \( {u}^{ * } \) of system (0.2) and (0.3) is local asymptotically stable. | Yes |
Lemma 1.2 Eq. (1.3) have purely imaginary roots if and only if \( b = {b}_{j}^{\left( k\right) }, k = 1,2,\ldots ;j = \) \( 0,1,2,\ldots \) Moreover, when \( b = {b}_{0}^{\left( 1\right) } \), all the roots of Equations (1.3), except \( \pm i{\omega }_{0}^{\left( 1\right) } \), have negative real parts. | Proof From the definition of \( {b}_{j}^{\left( k\right) } \), Equation (1.3) have purely imaginary roots if and only if \( b = {b}_{j}^{\left( k\right) } \) . In addition, we know that \( {b}_{0}^{\left( 1\right) } \) is the first value of \( b > 0 \) such that Equation (1.3) have roots appearing on the imaginary axis. Hence, all roots of all Equations (1.3) have negative real parts for \( b \in \left( {0,{b}_{0}^{\left( 1\right) }}\right) \), and all the roots of Equation (1.3), except \( \pm i{\omega }_{0}^{\left( 1\right) } \), have negative real parts when \( b > {b}_{0}^{\left( 1\right) } \) . | Yes |
Lemma 1.3 If \( a{f}^{\prime }\left( 0\right) < 1 + c \), then\n\n\[{\left. \frac{d\left( {\operatorname{Re}\lambda }\right) }{db}\right| }_{b = {b}_{j}^{\left( k\right) }} > 0.\] | Proof Differentiating Equation (1.3) with respect to \( b \), we obtain\n\n\[ \frac{d\lambda }{db} - {ab\tau }{f}^{\prime }\left( 0\right) {e}^{-{\lambda \tau }}\frac{d\lambda }{db} + a{f}^{\prime }\left( 0\right) {e}^{-{\lambda \tau }} = 0.\]\n\nThis gives\n\n\[ \frac{d\lambda }{db} = \frac{a{f}^{\prime }\left( 0\right) {e}^{-{\lambda \tau }}}{{ab\tau }{f}^{\prime }\left( 0\right) {e}^{-{\lambda \tau }} - 1} = \frac{{ab}{f}^{\prime }\left( 0\right) {e}^{-{\lambda \tau }}}{a{b}^{2}\tau {f}^{\prime }\left( 0\right) {e}^{-{\lambda \tau }} - b}\overset{\left( {1.3}\right) }{ = }\]\n\n\[ \frac{a{f}^{\prime }\left( 0\right) - 1 - c{k}^{2} - \lambda }{{b\tau }\left( {a{f}^{\prime }\left( 0\right) - 1 - c{k}^{2} - \lambda }\right) - b}.\]\n\nTherefore\n\n\[ {\left. \frac{d\left( {\operatorname{Re}\lambda }\right) }{db}\right| }_{b = {b}_{j}^{\left( k\right) }} = \operatorname{Re}\left\{ \frac{a{f}^{\prime }\left( 0\right) - 1 - c{k}^{2} - i{\omega }_{j}^{\left( k\right) }}{{b\tau }\left( {a{f}^{\prime }\left( 0\right) - 1 - c{k}^{2}}\right) - b - {b\tau i}{\omega }_{j}^{\left( k\right) }}\right\} = \]\n\n\[ \frac{{b\tau }\left( {a{f}^{\prime }\left( 0\right) - 1 - c{k}^{2}}\right) \left( {a{f}^{\prime }\left( 0\right) - 1 - c{\dot{k}}^{2} - 1}\right) }{{\left\lbrack b\tau \left( a{f}^{\prime }\left( 0\right) - 1 - c{k}^{2}\right) - b\right\rbrack }^{2} + {\left\lbrack b\tau {\omega }_{j}^{\left( k\right) }\right\rbrack }^{2}}.\]\n\nThus, if \( a{f}^{\prime }\left( 0\right) < 1 + c \), then we have\n\n\[ {\left. \frac{d\left( {\operatorname{Re}\lambda }\right) }{db}\right| }_{b = {b}_{j}^{\left( k\right) }} > 0 \] | Yes |
Example 4.1 Consider the following system\n\n\\[ \n\\begin{cases} {dx}\\left( t\\right) = & x\\left( t\\right) \\left\\lbrack {\\left( {{0.8} + {0.1}\\sin t}\\right) - \\left( {{0.6} + {0.1}\\sin t}\\right) x\\left( t\\right) - }\\right. \\\\ & \\left. \\frac{\\left( {{0.69} + {0.1}\\sin t}\\right) x\\left( t\\right) }{{0.26} + {0.01}\\sin t + {x}^{2}\\left( t\\right) }\\right\\rbrack {dt} + \\left( {{0.4} + {0.1}\\sin t}\\right) x\\left( t\\right) {dB}\\left( t\\right), t \\neq {t}_{k}, \\\\ x\\left( {t}_{k}^{ + }\\right) = & \\left( {1 + {I}_{k}}\\right) x\\left( {t}_{k}\\right) ,\\;t = {t}_{k}, k \\in \\mathbb{N}. \\end{cases} \n\\]\n\n(4.1) | Let \\( x\\left( 0\\right) = {1.0},{t}_{k} = k,{I}_{k} = {\\mathrm{e}}^{0.08} - 1 \\), then \\( {\\Phi }^{ * } = {0.94} > \\widehat{b}/2\\sqrt{\\widetilde{c}} = {0.7} \\) . It follows from Theorem 1.1 that species \\( x\\left( t\\right) \\) of system (4.1) is weakly persistent (see Figure 1). | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.