# IVP- Has at most one solution

Suppose $f(t, x)$ is nonincreasing with respect to $x$ for all $t \geq 0$ and $x \in \mathbb{R}$. Prove that the IVP problem $$\left\{ \begin{array}{l} x'(t)=f(t,x) \\ x(t_0)=x_0 \end{array} \right.$$ has at most one solution for $t \geq t_0$.

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## 1 Answer

Suppose that $x_1$ and $x_2$ are two different solutions. Let $h(t)=(x_1(t)-x_2(t))^2$. Then $h(t)\ge0$ for $t\ge t_0$ and $h(t_0)=0$. Take derivatives and use the equation to get $$h'(t)=2(x_1(t)-x_2(t))(f(t,x_1(t))-f(t,x_2(t))).$$ Use the fact that $f(t,x)$ is non-increasing in $x$ for each $t\ge t_0$ to deduce that $h'(t)\le0$ and that $h$ is non-increasing.

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Hey thank you so much, It make sense! You guys are so quick. –  Klara Oct 15 '12 at 15:50
you mean $(f(t,x_1(t))-f(t,x_2(t)))$ –  Neeraj Bhauryal Sep 3 '14 at 14:01
@NeerajBhauryal hankyou. I have edited the answer. –  Julián Aguirre Sep 4 '14 at 11:35