Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $t \geq t_0$.

share|improve this question

1 Answer 1

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 $$ h'(t)=2(x_1(t)-x_2(t))(f(t,x_1(t))-f(t,x_1(t))). $$ 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.

share|improve this answer
Hey thank you so much, It make sense! You guys are so quick. –  Klara Oct 15 '12 at 15:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.