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I'm trying to solve the following problem:

Assume that we are given $f(t,x) \in C^{\infty}(\mathbb R \times \mathbb R^n)$ such that $f(t, \cdot) \in \mathcal S(\mathbb R^n)$ for all $t>0$ (where $\mathcal S(\mathbb R^n)$ is Schwartz space). Assume furthermore that $f$ is a solution to the heat equation on $\mathbb R^n$ with $$ \begin{align} \partial_tf(t,x) &= \Delta f(t,x) \qquad t>0 \\ \lim_{t\to 0} f(t,x) &= g(x) \end{align} $$ for some $g\in \mathcal S(\mathbb R^n)$.

I want to show that under these conditions we must have $f(t,x) = (K_t\ast g)(x)$, where $\ast$ denotes convolution and $$ K_t(x) = \frac{1}{(4\pi t)^{n/2}}e^{-\langle x, x\rangle/4t}$$ is the heat kernel.

What I'm having trouble with is the following: The idea is to consider the Fourier transform $\hat f$ of $f$ and to derive the ordinary differential equation

$$\partial_t \hat f(t,k) = - k^2 \hat f(t,k)$$

But how do I even know that $\hat f(t,k)$ is differentiable with respect to $t$? What I would like to do is differentiate under the integral sign here:

$$\partial_t \hat f(t,k) =\partial_t \int_{\mathbb R^n} f(t,x)e^{-ikx}\, dx$$

But I don't know how to justify it. Could anyone help me out?

Thanks a lot! =)

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1 Answer 1

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You cannot, in general, interchange the differentiation and the integration here. I take the following example from this MO answer. Let $\phi(x)$, $x\in\mathbb{R}$, be an arbitrary bump function ($\phi \in C^\infty_c(\mathbb{R})$). Consider the function $f(x,t)$ defined to be identically 0 if $t \not\in (-\pi/2,\pi/2)$ and $f(x,t) = \phi(x - \tan t)$ otherwise. One checks that this function is indeed $C^\infty(\mathbb{R}^2)$: for each $|t| \geq \pi/2$, and for any $x$, one can find a small neighborhood of $(x,t)$ such that $f$ vanishes identically there. And $f(x,t)$ is a composition of smooth functions on $(-\pi/2,\pi/2)\times \mathbb{R}$.

Notice also that for any $k$, the function $\partial_t^kf(t,x)$, restricted to a fixed time slice $t = t_0$, is Schwartz (and in fact has compact support).

Now, taking the spatial Fourier transform you get that $\hat{f}(t,\xi) = e^{i \tan(t) \xi} \hat{\phi}(\xi)$ for $t\in (-\pi/2, \pi/2)$. Computing the time derivative you get $$ |\frac{d}{dt} \hat{f}(t,\xi)| = |\xi| \sec^2(t) |\hat{\phi}(\xi)| $$ which is not continuous at $t = \pm \pi/2$.


As a side remark, in Stein and Shakarchi's Fourier Analysis, the analogous statement to the one you want to prove is given with the additional assumption that the restrictions $f(t,x)$ to $t$ are uniformly Schwartz, which then imply that one can interchange the differentiation with integral.

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@Sam: you are right. Please see edit for counterexample. –  Willie Wong Jan 19 '12 at 15:32
    
In fact, using the assumption that $\int\phi \neq 0$, you have that $\hat{f}(t,0) = c \neq 0$ if $t\in (-\pi/2,\pi/2)$ and $=0$ otherwise. –  Willie Wong Jan 19 '12 at 15:35
    
Thank you very much for this counterexample. Great. I have also looked through a couple of books on Fourier analysis and they all seem to assume that $f$ is uniformly bounded in $t$ in some way. I also found a counterexample due to Tychonoff (in Körner's book) for the case $f\in C^\infty$. But none so far, for the case considered here. So I thought maybe satisfying the heat equation by itself is a strong enough condition to force the function $f$ to be sufficiently well-behaved on small time-intervals. But if it is, it certainly doesn't seem straight-forward to prove. Strange exercise... –  Sam Jan 19 '12 at 16:22
    
It just doesn't feel like you have enough regularity: if you slightly modify the counterexample, you have that the space you are considering does not embed into $L^1_tL^p_x$ for any $p$, even on compact time intervals. So even energy methods are ruled out. Where did you find this given as an exercise? –  Willie Wong Jan 19 '12 at 18:30
    
This is an exercise from my course on mathematical methods of physics (undergrad course). More precisely, it's from a collection of 17 exercises meant as preparation for our upcoming exam; the exam will consist of six exercises - four of which will be taken from the collection. I have already found one problem that asks us to prove an incorrect statement, so it may well be that there is an error in this one as well... Certainly all evidence seems to point this way. –  Sam Jan 19 '12 at 19:06

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