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 $A$?



ODE estimate for boundary value problem
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 $A$.


