Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Question:Fix $\epsilon>0$. Consider the differential equation, defined for functions $f(t,x)\in C^\infty([0,\epsilon]\times[0,\epsilon])$ defined by $$\frac{\partial}{\partial t} f(t,x)=\frac{f(t,x)^2-f(t,0)^2}{x}, x>0,$$ $$\frac{\partial}{\partial t} f(t,0)=2f(t,0) \frac{\partial f}{\partial x}(t,0).$$ I want to know if this differential equation has uniqueness for $f>0$. More precisely, suppose $f$ and $g$ are two $C^\infty$ functions satisfying the above equation with $f(0,x)=g(0,x)$, $\forall x$, and $f(t,x), g(t,x)>0$, $\forall t,x$. Must $f=g$?

Comment: The similar linear differential equation does have uniqueness. Indeed, consider, the equation $$\frac{\partial}{\partial t} f(t,x)=\frac{f(t,x)-f(t,0)}{x}, x>0,$$ $$\frac{\partial}{\partial t} f(t,0)=\frac{\partial f}{\partial x}(t,0).$$ If $f$ and $g$ are two solutions to the above equation, then $f=g$. To see this, note that since the equation is linear, we may assume $f(0,x)=0$ for all $x$, and we wish to prove that $f(t,x)=0$ for all $t$. Define $a(t)=f(t,0)$. Then, for $x>0$, $f$ satisfies the equation $$\frac{\partial}{\partial t} f(t,x)=\frac{f(t,x)-a(t)}{x}.$$ We can solve this equation explicitly. $$f(t,x)=-x^{-1}\int_0^t exp((t-s)/x) a(s) ds.$$ Since $f$ is $C^\infty$, the right hand side of the above equation is bounded in $x$. It can be shown that if such an expression is bounded as $x$ tends to $0$, $a$ must be the zero function. This means $a(t)=f(t,0)=0$ for all $t$. Standard uniqueness theorems for ODEs then show $f(t,x)=0$ for all for all $t,x$, completing the proof.

Comment: I would also be interested to know about existence, etc.

share|improve this question
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer


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

Browse other questions tagged or ask your own question.