Uniqueness of Solutions to Differential Equations

The convergence of the Picard iteration shows the existence part of our existence and uniqueness theorem. Now we prove the uniqueness part.

Let’s say that $u(t)$ and $v(t)$ are both solutions of the differential equation — $u'(t)=F(u(t))$ and $v'(t)=F(v(t))$ — and that they both satisfy the initial condition — $u(0)=v(0)=a$ — on the same interval $J=[-c,c]$ from the existence proof above. We will show that $u(t)=v(t)$ for all $t\in J$ by measuring the $L^\infty$ norm of their difference:

$\displaystyle Q=\lVert u-v\rVert_\infty=\max\limits_{t\in J}\lvert u(t)-v(t)\rvert$

Since $J$ is a closed interval, this maximum must be attained at a point $t_1\in J$ . We can calculate

$\displaystyle\begin{aligned}Q&=\lvert u(t_1)-v(t_1)\rvert\\&=\left\lvert\int\limits_0^{t_1}u'(s)-v'(s)\,ds\right\rvert\\&\leq\int\limits_0^{t_1}\lvert F(u(s))-F(v(s))\rvert\,ds\\&\leq\int\limits_0^{t_1}K\lvert u(s)-v(s)\rvert\,ds\\&\leq cKQ\end{aligned}$

but by assumption we know that $cK<1$ , which makes this inequality impossible unless $Q=0$ . Thus the distance between $u$ and $v$ is $0$ , and the two functions must be equal on this interval, proving uniqueness.

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May 9, 2011 - Posted by John Armstrong | Analysis, Differential Equations

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