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I asked the following question at Math Stackexchange a while ago here but did not get a correct answer.

Let $f(x,t)$ and $G(x,t)$ be smooth functions from $\mathbb R^2\to\mathbb R$.

The PDE $$\dfrac{\partial}{\partial t}f(x,t)=2f(x,t) \dfrac{\partial}{\partial x}G(x,t)+G(x,t)\dfrac{\partial}{\partial x}f(x,t)$$ applies on all of $\mathbb R^2$. Furhermore, let us impose the condition $$f(x,0)=0, \forall x\in \mathbb R$$

Is it necessarily true that $f(x,t)=0$ for all $(x,t)\in\mathbb R^2$?

I will comment that it is easy to show this is true if $f$ is assumed to be analytic, but it seems rather difficult if I don't have this assumption. Through some basic PDE tricks (method of characteristics, etc) it is possible to show a local version of this result: that for a fixed $x$, there is a small $\epsilon(x)$ such that $f(x,t)=0$ for all $0<t<\epsilon(x)$, but this is not good enough for my purposes.

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The answer is "No, it is not necessarily true that $f(x,t)=0$ for all $(x,t)\in\mathbb{R}^2$".

Here is a counterexample: Let $G(x,t) = x^2$ and let $h:\mathbb{R}\to\mathbb{R}$ be any smooth function that vanishes to infinite order at $0\in\mathbb{R}$. Now define $f:\mathbb{R}^2\to\mathbb{R}$ by requiring that $f(x,t) = 0$ when $xt\le1$ (which includes the 'initial line' $t=0$ as well as the line $x=0$), while $$ f(x,t) = \frac{1}{x^4}\,h\left(\frac{xt-1}{x}\right) $$ when $xt\ge1$. Then $f$ is smooth and satisfies the equation and initial conditions, but $f$ need not vanish in the regions where $xt > 1$.

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    $\begingroup$ Wow, this is a clever counterexample. If you don't mind, could I get some insight on how you found it? Was it just a matter of trial and error, or did you use some known PDE tools? $\endgroup$ – Darren Ong Jul 19 '17 at 2:14
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    $\begingroup$ @DarrenOng: It was mainly the method of characteristics. The main point is to choose a function $G$ such that not all of the integral curves of the vector field $\partial_t - G\,\partial_x$ pass through the $x$-axis (which is where your initial condition is specified) and such that it has a nontrivial first integral. $G$ has to be nonlinear for this, but, fortunately, $G = x^2$ works, with a first integral $x/(1-xt)$, whose level sets are hyperbolae, with one branch passing through the $x$-axis. Then one finds the general solution using the method of characteristics and an integrating factor. $\endgroup$ – Robert Bryant Jul 19 '17 at 9:08

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