Past exam of the differential equations 1 course of the University of Oxford / 2021 / 2 / c

Treat both sides of $a$ at once: for $\sigma \in \{-1,1\}$ set $w(t)=v(a+\sigma t)$, $0\le t\le h$. Nonnegativity givesDefine $W(t)=ct+A{\int }_0^{t}w(r)dr$. By the fundamental theorem of calculus, $W^{\prime }(t)=c+Aw(t)\le c+AW(t)$ and $W(0)=0$. The integrating factor $e^{-At}$ givesSince $w\le W$, this proves the affine Gronwall estimate

Continuous dependence of an ODE solution on parameters here means continuity of the map $\beta \mapsto y_{\beta }$ into $C([a-h,a+h])$ with its supremum norm: for every ${\beta }_0\in I$ and $\epsilon >0$, all sufficiently close $\beta \in I$ satisfySubtract the two integral equations and use the uniform Lipschitz constant $L$ on the rectangle. With $v(x)=|y_{\beta }(x)-y_{\gamma }(x)|$ we obtainPart (i), with $c=|\beta -\gamma |$ and $A=L$, therefore proves, for $L>0$,The case $\beta =\gamma$ follows from uniqueness or the same estimate with zero forcing. If $L=0$, the direct integral estimate gives $\Vert y_{\beta }-y_{\gamma }{\Vert }_{\infty }\le h|\beta -\gamma |$. In either case the map is even Lipschitz continuous, and hence continuous, throughout $I$ in the stated supremum norm.