Abstract
Full Text
P. E. SOBOLEVSKII
ON SECOND-ORDER DIFFERENTIAL EQUATIONS IN A BANACH SPACE
(Presented by Academician I. G. Petrovskii on 28 III 1962)
We consider the problem
\[ v''+A(t)v'+B(t)v=f(t)\quad (0\leq t\leq T);\qquad v(0)=v_0,\quad v'(0)=v'_0 . \tag{1} \]
in a Banach space \(E\). We shall call a function \(v(t)\) a solution of problem (1) if, for all \(t\) in \([0,T]\), it satisfies equation (1) and the functions \(v''(t)\), \(A(t)v'(t)\), and \(B(t)v(t)\) are continuous on \([0,T]\).
In this note we give a theorem on the existence of a solution of problem (1), as well as of some nonlinear problems. An equation with a small parameter \(\varepsilon>0\) at the derivative \(v''(t)\) is considered, and it is shown that the solution of this problem tends to the solution of a degenerate first-order equation. We note that the stimulus for writing this paper was the work \((^1)\), in which problem (1) with a small parameter and constant operators \(A\) and \(B\) was considered. The method presented below for investigating problem (1), etc., differs from the method developed in \((^1)\). The latter method, apparently, does not generalize to equations with variable operators.
- In the work \((^2)\) the problem
\[ v'+A(t)v=0\quad (0\leq s\leq t\leq T);\qquad v(s)=v_0 \tag{2} \]
was studied. The solution of this problem is written in the form \(U(t,s)v_0\). The operator \(U(t,s)\) makes it possible to determine the solution of the problem
\[ v'+A(t)v=f(t)\quad (0\leq t\leq T);\qquad v(0)=v_0 \tag{3} \]
by the formula
\[ v(t)=U(t,0)v_0+\int_0^t U(t,s)f(s)\,ds . \tag{4} \]
Let now \(v(t)\) be a solution of problem (1). Integrating (1) from \(0\) to \(t\), performing integration by parts and using formula (4), we arrive at the relation
\[ v(t)=U(t,0)v_0+\int_0^t U(t,s)\left\{v'_0+A(0)v_0+\int_0^s f(\tau)d\tau+\int_0^s [A'(\tau)-B(\tau)]v(\tau)d\tau\right\}ds . \tag{5} \]
Finally, making the substitution \(x(t)=A(t)v(t)\) and applying formula (2.45) from \((^3)\), we obtain
\[ \begin{aligned} x(t)=&\left[I+\int_0^t W(t,\tau)A'(\tau)A^{-1}(\tau)d\tau\right]A(0)v_0 +\left[I-W(t,0)+\int_0^t W(t,\tau)A'(\tau)A^{-1}(\tau)d\tau\right]v'_0 \\ &+\int_0^t \left[I-W(t,\tau)+\int_\tau^t W(t,s)A'(s)A^{-1}(s)ds\right]f(\tau)d\tau \\ &+\int_0^t \left[I-W(t,\tau)+\int_0^t W(t,s)A'(s)A^{-1}(s)ds\right] [A'(\tau)-B(\tau)]A^{-1}(\tau)x(\tau)d\tau , \end{aligned} \tag{6} \]
where \(W(t,\tau)=A(t)U(t,\tau)A^{-1}(\tau)\).
Thus it is proved:
Theorem 1. Suppose that the operator \(-A(t)\), for each \(t\in[0,T]\), is the infinitesimal generator of a strongly continuous semigroup \(\exp\{-\tau A(t)\}\)
\((\tau \geqslant 0)\), whose norm satisfies the inequality
\[ \left\|\exp\{-\tau A(t)\}\right\|\leqslant \exp\{-\delta\tau\}\quad(\delta>0). \tag{7} \]
Suppose that the operator \(A(t)\) has a domain of definition \(D\) independent of \(t\), and that the operator-function \(A(t)A^{-1}(0)\) is twice strongly continuously differentiable. Suppose that the operator-function \(B(t)A^{-1}(0)\) is once strongly continuously differentiable. Suppose that the function \(f(t)\) is continuously differentiable. Finally, suppose that \(v_0\) and \(v'_0\) belong to \(D\). Then problem (1) has one and only one solution.
In proving the smoothness of solutions of equation (8), the following is used.
Lemma 1. Suppose the inequality
\[ \max_{0\leqslant t\leqslant T}\left\|A'(t)A^{-2}(t)\right\|<1 \tag{8} \]
holds. Then the identity
\[ \begin{aligned} \int_0^t W(t,s)f(s)\,ds ={}& A^{-1}(t)\left[I-A'(t)A^{-2}(t)\right]^{-1}f(t) \\ &- W(t,0)A^{-1}(0)\left[I-A'(0)A^{-2}(0)\right]^{-1}f(0) \\ &+ \int_0^t W(t,s)A^{-1}(s)A'(s)A^{-1}(s) \left[I-A'(s)A^{-2}(s)\right]^{-1}f(s)\,ds \\ &- \int_0^t W(t,s)A^{-1}(s) \left(\left[I-A'(s)A^{-2}(s)\right]^{-1}\right)' f(s)\,ds \\ &- \int_0^t W(t,s)A^{-1}(s) \left[I-A'(s)A^{-2}(s)\right]^{-1}f'(s)\,ds . \end{aligned} \tag{9} \]
The requirement (8) imposes essentially no additional restrictions on the problem, since by means of the substitution \(t=kt\) one can always ensure that it is satisfied.
- Consider the nonlinear problem
\[ v''+A(t)v'=f(t,v,v')\quad(0\leqslant t\leqslant T);\qquad v(0)=v_0,\quad v'(0)=v'_0 . \tag{10} \]
Theorem 2. Suppose that the operator \(A(t)\) and the elements \(v_0\) and \(v'_0\) satisfy the conditions of Theorem 1. Suppose that the operator \(f(t,A^{-1}(0)v,w)\) is differentiable with respect to \(t\in[0,T]\) and \(v,w\in E\) (the derivatives with respect to \(v,w\) are Fréchet derivatives). Suppose that the derivatives
\(f_t(t,A^{-1}(0)v,w)\), \(f_v(t,A^{-1}(0)v,w)\), and \(f_w(t,A^{-1}(0)v,w)\) are continuous in the aggregate of the variables and bounded when \(v\) and \(w\) belong to a bounded set in \(E\). Then problem (10) has one and only one solution, defined on some segment \([0,t_0]\subset[0,T]\).
- Consider the problem with a small parameter
\[ \varepsilon v''+A(t)v'=f(t,v)\quad(0\leqslant t\leqslant T);\qquad v(0)=v_0,\quad v'(0)=v'_0\quad(0<\varepsilon\leqslant 1) \tag{11} \]
and the corresponding degenerate problem \((\varepsilon=0)\)
\[ A(t)v'=f(t,v)\quad(0\leqslant t\leqslant T);\qquad v(0)=v_0 . \tag{12} \]
Theorem 3. Suppose the conditions of Theorem 2 are fulfilled. Then problem (11), for every \(\varepsilon\in(0,1]\), has a solution \(v_\varepsilon(t)\), defined on some segment \([0,t_0]\subset[0,T]\) independent of \(\varepsilon\). As \(\varepsilon\to0\), the function \(A(t)v_\varepsilon(t)\) converges uniformly on \([0,T]\) to the function \(A(t)v_0(t)\), where \(v_0(t)\) is the solution of problem (12). The function \(A(t)v'_\varepsilon(t)\) is uniformly bounded with respect to \(\varepsilon\) on \([0,T]\), and for \(t>0\) it converges to the function \(A(t)v'_0(t)\) as \(\varepsilon\to0\). Finally, the residual \(A(t)v'_\varepsilon(t)-f[t,v'_\varepsilon(t)]\) is uniformly bounded with respect to \(\varepsilon\) on \([0,T]\), and for \(t>0\) tends to zero as \(\varepsilon\to0\).
The proof of Theorem 3 is based on the following lemmas.
Lemma 2. Let the operator-function \(A(t)A^{-1}(0)\) be once strongly continuously differentiable. Then
\[ \|U_\varepsilon(t,s)\|\leq \exp\left\{-\frac{t-s}{\varepsilon}\delta\right\}, \tag{13} \]
\[ \|W_\varepsilon(t,s)\|\leq C\exp\left\{-\frac{t-s}{\varepsilon}\delta\right\}. \tag{14} \]
Let the operator-function \(A(t)A^{-1}(0)\) be twice strongly continuously differentiable. Then
\[ \|A(t)W_\varepsilon(t,s)A^{-1}(s)\|\leq C\exp\left\{-\frac{t-s}{\varepsilon}\delta\right\}. \tag{15} \]
Here \(U_\varepsilon(t,s)\) and \(W_\varepsilon(t,s)\) are operator-functions constructed from the operator-function \(\frac{1}{\varepsilon}A(t)\) in the same way as the operator-functions \(U(t,s)\) and \(W(t,s)\) are constructed from the operator-function \(A(t)\).
Lemma 3. Let the operator-function \(A(t)A^{-1}(0)\) be once strongly continuously differentiable, and let the function \(f(t)\) be strongly continuously differentiable. Then
\[ \lim_{\varepsilon\to 0}\frac{1}{\varepsilon}A(t)\int_0^t U_\varepsilon(t,s)f(s)\,ds=f(t). \tag{16} \]
If, moreover, \(A(t)A^{-1}(0)\) is twice strongly continuously differentiable, then
\[ \lim_{\varepsilon\to 0}\frac{1}{\varepsilon}A(t)\int_0^t W_\varepsilon(t,s)f(s)\,ds=f(t). \tag{17} \]
4. Problem (1) was investigated by us under the assumption that the operator \(B\) is subordinate to the operator \(A\). We shall show that in some cases this requirement can be weakened. Let \(v(t)\) be a solution of problem (1). Then it is easy to see that the identity
\[ \left[A(t)\int_0^t v'\,ds\right]' + A_1(t)\left[A(t)\int_0^t v'\,ds\right] = f(t)-B(t)v_0-v'', \tag{18} \]
holds, where \(A_1(t)=B(t)A^{-1}(t)-A'(t)A^{-1}(t)\). Let \(U_1(t,s)\) be the operator-function with whose aid the solution of the problem \(v' + A_1(t)v=0\) \((0\leq s\leq t\leq T)\), \(v(s)=v_0\), is written in the form \(U_1(t,s)v_0\). Then from (19) it follows that
\[ A(t)\int_0^t v'\,ds = \int_0^t U_1(t,s)[f(s)-B(s)v_0]\,ds - \int_0^t U_1(t,s)v''\,ds. \tag{19} \]
Integrating by parts in the last integral, we find:
\[ \left[\int_0^t v'\,ds\right]' + A(t)\left[\int_0^t v'\,ds\right] = \]
\[ = U_1(t,0)v'_0 +\int_0^t U_1(t,s)A_1(s)v'\,ds +\int_0^t U_1(t,s)[f(s)-B(s)v_0]\,ds. \tag{20} \]
Hence it follows that
\[ \int_0^t v'\,ds = \int_0^t U(t,\tau)\left\{ U_1(\tau,0)v'_0 + \int_0^\tau U_1(\tau,s)[f(s)-B(s)v_0]\,ds \right\}d\tau + \]
\[ + \int_0^t U(t,\tau)\left\{ \int_0^\tau U_1(\tau,s)A_1(s)v'\,ds \right\}d\tau. \tag{21} \]
Using formula (2.46) from \({}^{(3)}\), we obtain:
\[ \begin{aligned} v'(t)=&\Bigg[W(t,0)-\int_{0}^{t} W(t,s)A'(s)A^{-1}(s)U_1(s,0)\,ds \\ &+\int_{0}^{t}W(t,s)\frac{\partial}{\partial s}U_1(s,0)\,ds\Bigg]v'_0 -\int_{0}^{t}W(t,\tau)A'(\tau)A^{-1}(\tau) \left\{\int_{0}^{\tau}U_1(\tau,s)[f(s)-\right.\\ &\left.-B(s)v_0]\,ds\right\}d\tau +\int_{0}^{t}\Bigg[W\left(t,\frac{t+s}{2}\right)U_1\left(\frac{t+s}{2},s\right) +\int_{\frac{t+s}{2}}^{t}W(t,\tau)\frac{\partial}{\partial \tau}U_1(\tau,s)\,d\tau-\\ &-\int_{s}^{\frac{t+s}{2}}\frac{\partial}{\partial \tau}W(t,\tau)U_1(\tau,s)\,d\tau\Bigg][f(s)-B(s)v_0]\,ds -\int_{0}^{t}W(t,\tau)A'(\tau)A^{-1}(\tau)\times\\ &\times\left\{\int_{0}^{\tau}U_1(\tau,s)A_1(s)v'(s)\,ds\right\}d\tau +\int_{0}^{t}\Bigg[W\left(t,\frac{t+s}{2}\right)U_1\left(\frac{t+s}{2},s\right)+\\ &+\int_{\frac{t+s}{2}}^{t}W(t,\tau)\frac{\partial}{\partial \tau}U_1(\tau,s)\,d\tau -\int_{s}^{\frac{t+s}{2}}\frac{\partial}{\partial \tau}W(t,\tau)U_1(\tau,s)\,d\tau\Bigg]A_1(s)v'(s)\,ds. \end{aligned} \tag{22} \]
Thus, for example, the following is proved:
Theorem 4. Let, for every \(t\in[0,T]\) and any \(\lambda\) with \(\operatorname{Re}\lambda\geq 0\), the operator \(A(t)+\lambda I\) have a bounded inverse, whose norm satisfies the inequality\(^*\)
\[ \left\|[A(t)+\lambda I]^{-1}\right\|\leq C[|\lambda|+1]^{-1}. \tag{23} \]
Suppose the operator \(A(t)\) has a dense domain of definition \(D\) in \(E\), independent of \(t\); suppose the operator-function \(A(t)A^{-1}(0)\) is continuously differentiable and its derivative \(A'(t)A^{-1}(0)\) satisfies some Hölder condition. Suppose the operator \(A_1(t)\) satisfies condition (23), has a domain of definition \(D_1\) independent of \(t\), and suppose the operator-function \(A_1(t)A_1^{-1}(0)\) satisfies some Hölder condition. Suppose the operator-function \(A_1(t)A^{-\rho}(0)\) satisfies a Hölder condition for some \(\rho\in[0,1)\). Suppose the functions \(B(t)v_0\) and \(f(t)\) satisfy some Hölder condition. Finally, suppose \(v_0\in D[A^{\rho_1}(0)]\) for some \(\rho_1>\rho\). Then problem (1) has one and only one solution.
By a solution here is meant a function \(v(t)\), continuously differentiable once on \([0,T]\), satisfying the initial conditions (1), and, for \(t>0\), equation (1), such that the functions \(v''(t)\) and \(A(t)v'(t)\) are continuous for \(t>0\), while the functions \(\|v''(t)\|\) and \(\|A(t)v'(t)\|\) are summable on \([0,T]\).
Similarly, one can consider the nonlinear problem
\[ v''+A(t)v'+B(t,v,v')v=f(t,v,v');\qquad v(0)=v_0,\quad v'(0)=v'_0 \tag{24} \]
and the problem with a small parameter at the highest derivative.
Voronezh Agricultural Institute
Received
27 III 1962
REFERENCES
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\({}^{2}\) T. Kato, J. Math. Soc. Japan, No. 5 (1953).
\({}^{3}\) P. E. Sobolevskii, Tr. Mosk. matem. obshch., 10 (1961).
\({}^{4}\) M. A. Krasnosel’skii, P. E. Sobolevskii, DAN, 129, No. 3 (1959).
\(^*\) For operators satisfying (23), all fractional powers are defined \({}^{(4)}\).