Proof of Existence and Uniqueness of IVP (Exam Question)

Question

I am trying to practise some past exam questions but there are no solutions online. Here attached is the following question. This is an undergraduate course. Could anyone help me how to solve this? I have spent many many hours on this and have absolutely no idea.

Answer 1

This really depends on what level the answer is expected. The easy answer is that the right side is infinitely continuously differentiable, thus Lipschitz on any bounded subset of the domain, and thus a local solution exists for some $\delta>0$.

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If those 15 marks are a sizeable amount of the total, if they represent 5-15 minutes of the answer time, probably something more is expected. The $x$-derivative of the right side is $t\cos(tx)$, so that over some interval $[-a,a]$ you get a global $x$-Lipschitz constant $L=a$ independent of vertical constraints. Independent of any constraints one gets the bounds of the right side itself. As $1+\sin(tx)\in[0,2]$, any solution of the IVP has to live inside the double cone $|x(t)-x_0|\le 2|t-0|$. As there is no restriction in $x$ direction required for these bounds, the vertical dimension of the box $(t,x)\in[-a,a]\times[x_0-b,x_0+b]$ can be chosen arbitrarily, such as $b=2a$.

The contraction constant for the Picard iteration in the unmodified supremum norm is $q=La=a^2$. The only constraint so far is that $q<1$ is required, so set for instance $a=\frac23$, then there exists a solution for $δ=a$.

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Using the actual variable Lipschitz constant $L(t)=|t|$ gives for the Picard iteration operator $P[x](t)=x_0+\int_0^t f(s,x(s))\,ds$ the perturbation inequality \begin{align} |P[y](t)-P[x](t)|&\le \int_0^t L(s)|y(s)-x(s)|\,ds \\\implies \sup_{t\in[-a,a]}e^{-t^2/2}|P[y](t)-P[x](t)|&\le (e^{a^2/2}-1)\sup_{t\in[-a,a]}e^{-t^2/2}|y(t)-x(t)| \end{align} using the integrating factor $\exp(\int_0^tL(s)\,ds=\exp(t^2/2)$. In the correspondingly modified weighted supremum norm the iteration is now contracting for $a<\sqrt{2\ln2}=1.177410...$, so $δ=1$ is admissible.