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.


1 Answer 1


The answer is yes: f=g. I wrote up a paper with a more general result here. The idea is the following. If $f$ were assumed to be of Laplace transform type $$ f(t,x) = \frac{1}{x} \int_0^\infty e^{-w/x} A(t,w)\: dw, $$ then, by taking the inverse Laplace transform, one finds that $A(t,w)$ satisfies a PDE which is much easier to study.

But not every solution is of Laplace transform type! What is proved, though, is that every solution is of Laplace transform type, modulo an error which can be controlled. In particular, the above equation does not have existence: only initial conditions $f(0,x)$ which are of Laplace transform type modulo an appropriate error give rise to solutions.

Incidentally, there is a slight error in the above question. Where I said "we can solve this equation explicitly," I didn't include the initial condition. One should treat $f(\epsilon,x)$ as the initial condition, and solve backwards in time.

  • 1
    $\begingroup$ Also, I meant to mention that this is closely related to, but not the same as, Barry Simon's influential work: A new approach to inverse spectral theory. I. Fundamental formalism, Ann. of Math. (2) 150 (1999), no. 3, 1029–1057. See the paper I linked to for further details on the connections. $\endgroup$ Sep 23, 2015 at 12:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .