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Let $X$ be a solution to the boundary value problem $$ X^{\prime\prime}(s) = A(s)X(s), \quad X(0) = X_0, ~~X(t) = X_1,$$ where $0 < t \leq 1$, $A$ is some matrix-valued function, defined on $[0, 1]$, and $X_0$, $X_1$ are certain fixed matrices (suppose the problem has a unique solution for all $0<t \leq 1$.)

Does there exist a constant $C>0$ independent of $ t\in (0, 1]$ such that $|X(s)| \leq C$ for all $s \in [0, t]$? If yes, how does the constant depend on $A$?

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  • $\begingroup$ It seems that an answer to this problem requires, in particular, knowledge of all eigenvalues of Sturm-Liouville problems. Life would be nice if that were simple, but ... What kind of answer do you expect? $\endgroup$ Jan 11 '16 at 15:42
  • $\begingroup$ Honestly, I thought that this would be just true and quite elementary to show, even though I couldn't come up with a solution myself... $\endgroup$ Jan 11 '16 at 19:09
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If you mean $|X(s)|\leq C$ for $0\leq s\leq t$ and any $0<t\leq 1$, then the answer is yes; if you want $|X(s)|\leq C$ for $0\leq s\leq 1$ and any $0<t\leq 1$, then the answer is no in general.

Indeed, let $Y_j=Y_j(s)$ for $j=0,1$ be the solutions of the differential equation with $Y_0(0)=Y_1'(0)=I$ and $Y_0'(0)=Y_1(0)=0$, where $I$ stands for the identity matrix. Then the solution $X(\cdot,t)$ of the boundary value problem, for a given $0<t\leq1$, is $$ X(s,t) = Y_0(s) X_0 + Y_1(s) Z(t), $$ where the matrix $Z(t)$ is determined by the condition $X_1= X(t,t)= Y_0(t) X_0 + Y_1(t) Z(t)$. The condition on unique solvability of the boundary value problems is $\det Y_1(t)\neq0$ for all $0<t\leq1$, and then $$ Z(t) = Y_1(t)^{-1}\left(X_1 - Y_0(t) X_0\right). $$ Now, $Y_1(s) = s I + O(s^2)$ as $s\to+0$. Hence, $Y_1(s)^{-1} = s^{-1} I + O(s^0)$. It follows that $Z(t) = t^{-1} (X_1 - X_0) + O(t^0)$ and $$ C = \sup_{0\leq s\leq t\leq 1}|X(s,t)| < \infty, $$ while $X(1,t) = t^{-1} Y_1(1)\left(X_1-X_0\right) + O(t^0)$ which explodes as $t\to+0$ provided that $X_0\neq X_1$.

To estimate the constant $C\geq0$ in dependence on $A$ is a different matter, as one needs to find a way to express the (nonlinear) condition $\det Y_1(t)\neq0$ for $0<t\leq 1$ in terms of $A$.

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  • $\begingroup$ Indeed, it was the question with the positive answer that I had in mind. I fixed it in my question above. Thank you very much for your help! $\endgroup$ Jan 12 '16 at 9:36

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