Past exam of the differential equations 1 course of the University of Oxford / 2023 / 1

Fix any $t_0\ge 0$ and compare $y=R$ with $y=0$. The quotient required in a global Lipschitz condition in the dependent variable isConsequently,It is nevertheless locally Lipschitz in $y$: for $|y|,|v|\le R$,Thus the Picard-Lindelöf theorem gives local existence and uniqueness for the initial value problem; the remaining work establishes existence on the entire half-line for the specified small data.

Write $A(t)={\int }_0^{t}a(s)ds$. Multiplying the ordinary differential equation by the integrating factor $e^{A(t)}$ and using the product rule givesIntegration from $0$ to $t$ therefore yields the Volterra integral equationConversely, a continuous function satisfying this integral equation becomes continuously differentiable after multiplication by $e^{A(t)}$, by the fundamental theorem of calculus, and satisfies the original initial value problem.

For $s>t$, the coefficient bound gives $A(s)-A(t)\ge C_1(s-t)$. Thus for $0\le s\le t$ the positive kernel satisfiesThe starting Picard iterate is continuous and has supremum norm $M_0=|b|$. If the $n$th Picard iterate is bounded and continuous, its integral defines another continuous function, andTaking the supremum norm and applying mathematical induction proves boundedness of every iterate and

Put $B=|b|>0$ and $\gamma =4B/C_1<1$. Initially $M_0=B<2B$. If $M_n<2B$, the preceding supremum norm estimate givesHence mathematical induction gives

The difference of two successive Picard iterates factors asThe preceding bound and the kernel estimate therefore giveMoreover $D_1:=\Vert y_1-y_0{\Vert }_{\infty }\le B^2/C_1$. Summing the resulting geometric series, for $m>n$,Thus there is a uniform limit $y$ on $[0,\infty )$, continuous by the uniform limit theorem, with $\Vert y{\Vert }_{\infty }\le 2B$. The integral passes to this limit becauseTherefore $y$ satisfies the Volterra integral equation, and hence the initial value problem. This proves global existence by Global Picard iteration with a decaying kernel. The same estimate gives the useful error boundIf $b=0$, every iterate is zero and the convergence assertion is immediate.

Let $y$ be any continuous function solving the integral equation on $[0,\infty )$, and retain $B=|b|>0$. A first-exit argument avoids assuming in advance that its supremum norm is finite. If $|y|$ ever reached $2B$, continuity would give a first such time $T>0$. On $[0,T]$ we would have $|y|\le 2B$, so the kernel estimate would implya contradiction. Therefore $|y(t)|<2B$ for every finite $t$.

That pointwise strict inequality alone would not give a strict bound on the supremum norm. Apply the integral equation once more, now using $|y|\le 2B$ globally:

Existence was proved by Picard iteration. If $y$ and $v$ are any two continuous solutions with the same $b\ne 0$, the previous part makes both bounded by $2B$. Subtracting their Volterra integral equations givesSince $\gamma <1$, this forces $y=v$.

For $b=0$, zero is a solution. On any finite interval $[0,T]$, another continuous solution has a finite bound $K_T={\max }_{[0,T]}|y|$. Since $K(t,s)\le 1$, its integral equation impliesThe Gronwall inequality gives $y=0$ on every such interval. Hence