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I'm doing a chapter on Fourier analysis and I got a rather involved problem.

Given the function $u$, \begin{align*} u = \frac1{\sqrt{\pi t}} \int_{-\infty}^\infty \exp\left(-p^2-\frac{(x-p)^2}{t}\right)dp \end{align*} I want to show that it satisfies the (heat) equation $u_t = 4 u_{xx}$ and I want to compute its boundary values... Where do I start?

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    $\begingroup$ One way: Complete the square inside the exponential and use the Gaussian integral, You will get an explicit expression of $u$. $\endgroup$ – mickep Dec 19 '15 at 8:53
  • $\begingroup$ @mickep Hmm.. I get an expression which I cannot solve: $\displaystyle\int_0^\infty \exp\left[-\frac{\left(p-\frac x{t+1}\right)^2}{\frac t{t+1}}\right] dp$... should I use a trick here ? Rewriting as $\exp(-u^2)$ doesn't give much insight either, it just makes the expression more horrible. $\endgroup$ – 1010011010 Dec 19 '15 at 9:23
  • $\begingroup$ or should I only set $u = p-\frac x{t+1}$?? Then I'm getting in trouble with the boundary values of my integral, since an integral from $-\frac x{t+1}$ doesn't make much sense for a Gaussian... At least that's my guess. $\endgroup$ – 1010011010 Dec 19 '15 at 9:26
  • $\begingroup$ $u$ is a bad name for the new variable, but indeed, you should introduce a new variable like that (notethat your integral runs from $-\infty$ to $+\infty$ and not from $0$ to $+\infty$, so it is insensible to shifting the variable). $\endgroup$ – mickep Dec 19 '15 at 10:03
  • $\begingroup$ @mickep Ah yes, so I should make the step $\int_{-\infty}^\infty \rightarrow 2 \int_0^\infty$ at a later stage in the algebra? $\endgroup$ – 1010011010 Dec 19 '15 at 10:04
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In such a chapter, it was probably derived that $$ u(t,x)=\frac{1}{\sqrt{4\pi k t}}\int_{-\infty}^{\infty}f(y)e^{-(x-y)^2/4kt}dy $$ is a solution of $$ u_{t}(t,x) = ku_{xx}(t,x),\\ \lim_{t\downarrow 0}u(t,x) = f(x), $$ where the limit in $t$ exists at any point $x$ where $f$ is continuous. The heat kernel $H(t,x)=\frac{1}{\sqrt{4\pi k t}}e^{-x^2/4kt}$ is a solution of $u_{t}(t,x)=ku_{xx}(t,x)$ that behaves as $\delta(x)$ as $t\downarrow 0$, which is why $\lim_{t\downarrow 0}u(t,x)=f(x)$ if $f$ is continuous.

You have a problem where $f(y)=e^{-y^2}$ and $k=1/4$. So your function $u$ satisfies $$ u_{t}(t,x)=\frac{1}{4}u_{xx}(t,x),\\ u(0,x) = e^{-x^2}. $$ Under fairly mild assumptions, the solution $u$ is unique. However, using the heat kernel, you can find another such solution.

If $u(t,x)$ is a solution of the heat equation, then $v(t,x)=u(t+a,x)$ is a solution of the same heat equation, but with a different initial function $v(0,x)=u(a,x)$. For example, $$ u(t,x) = \frac{1}{\sqrt{t}}e^{-x^2/t} $$ is a solution of $u_{t}=\frac{1}{4}u_{xx}$, which means $v(t,x)=u(t+1,x)$ is a solution of $v_{t}=\frac{1}{4}v_{xx}$, and $v$ satisfies $v(0,x)=e^{-x^2}$. By uniqueness of solutions, $$ \frac{1}{\sqrt{t+1}}e^{-x^2/(t+1)}=v(t,x)=\frac{1}{\sqrt{\pi t}}\int_{-\infty}^{\infty}e^{-y^2}e^{-(x-y)^2/t}dy. $$

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  • $\begingroup$ This was a question on an exam and no such derivation has ever been mentioned. Is there anything I can check for reference? $\endgroup$ – 1010011010 Dec 19 '15 at 14:21
  • $\begingroup$ @1010011010 : That would be difficult to derive if you have not seen it before. You will have to know (or derive) $\int_{-\infty}^{\infty}e^{-\alpha x^2}dx$ as a minimum, which will come up in completing the square. I figured if your text covered the heat kernel solution, then you could save doing any integral using the arguments given above. But on an exam, that will probably be too much to know/remember. So, stick to completing the square, even though you will have to know $\int_{-\infty}^{\infty}e^{-\alpha x^2}dx$, or know how to derive it. $\endgroup$ – DisintegratingByParts Dec 19 '15 at 14:27

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