Is solutions of ODE depends continuously on the parameter

Question

Im not understanding the claim that the author here say that the solution of the ODE (37) is a continuous function of the parameter $t$, is it a standard result in ODE in general or is it just for this type of ODE. Furthermore, how to prove that the solution $h(x,t)$ is continuous function of $t$? Thanks

Answer 1

Set $g=[tP+(1-t)]h'$, then your equation is equivalent to the first order system \begin{align} g'&=-tQh\\ h'&=[tP+(1-t)]^{-1}g \end{align} and the standard fixed-point integral equation $y=F_t(y)$, $y=\pmatrix{g\\h}$. Now if $P>0$ and $P,Q$ are continuous, then the fixed-point operators have a uniform global Lipschitz constant $L=\sup |Q|+[\min(1,\inf P)]^{-1}$, and the operators to parameters $s,t\in[0,1]$ have a bounded difference $\|F_t-F_s\|$ proportional to $|t-s|$. In a suitably modified weighted supremum norm $\|y\|_L=\sup e^{-2L|x|}\|y(x)\|$ the fixed-point operators have uniformly a contraction constant $q_t\le q=\frac12$.

For the perturbation of fixed-point operators we know that if the contraction constant is uniformly $q<1$ and the unique fixed points are $y_t=F_t(y_t)$, then by the a-priory bound on the fixed point iteration for $y_t$ $$ \|y_t-y_s\|\le\frac1{1-q}\|F_t(y_s)-y_s\|=\frac1{1-q}\|F_t(y_s)-F_s(y_s)\| \le\frac{M\,|t-s|}{1-q}\|y_s\|. $$ This proves Lipschitz continuity of the map of the parameter to the solution at parameter $s$, and thus in the interval $[0,1]$.