# Solutions of the heat equation of the form $u(x,t)=v(x/\sqrt{t})$

Assume $n=1$ and $u(x,t)=v(\frac{x}{\sqrt{t}})$.

(a) Show that $u_t = u_{xx}$ if and only if $$v''+\frac z2 v' = 0. \tag{*}$$ Show that the general solution of $(*)$ is $$v(z)=c \int_0^z e^{-s^2/4} \, ds + d.$$

(b) Differentiate $u(x,t)=v(\frac x{\sqrt{t}})$ with respect to $x$ and select the constant $c$ properly, to obtain the fundamental solution $\Phi$ for $n=1$. Explain why this procedure produces the fundamental solution. (Hint: What is the initial condition for $u$?)

This problem is from PDE Evans, 2nd edition, Chapter 2 Exercise 13. My question is on part B: why does the procedure allow us to get the fundamental solution $\Phi$?

For context, we have (page 46 in the textbook):

DEFINITION. The function $$\Phi(x,t) := \begin{cases} \frac 1{(4\pi t)^{n/2}} e^{-\frac{|x|^2}{4t}} & (x \in \mathbb{R}^n, t > 0) \\ 0 & (x \in \mathbb{R}^n, t = 0)\end{cases}$$ is called the fundamental solution of the heat equation.

Firstly, I understand part (a). I also understand most of part (b). I followed the process exactly and obtained the fundamental solution $\Phi$ for $n=1$, with the constant $c$ chosen to be $c=\frac 1{(4\pi)^{1/2}}$. I get: $$u_x(x,t)=\frac{\partial}{\partial x} v(\frac x{\sqrt{t}}) = \frac c{t^{1/2}} e^{-\frac{|x|^2}{4t}}=\frac 1{(4\pi t)^{1/2}} e^{-\frac{|x|^2}{4t}} =: \Phi,$$ as required.

But the last part of the question asks why does this procedure work, which is asked for in the last two lines of the stated problem. I do not know why this works; I've tried to find the initial condition for $u$ per the given hint, that is, find $u(x,0)$ (substitute $t=0$), but to no avail.

• Perhaps you can notice that this should work because of the scaling in the heat equation: the limit $\Delta x,\Delta t \to 0$ is taken in such a way that $\Delta x^2/\Delta t$ is constant, hence so that $\Delta x/\sqrt{\Delta t}$ is constant. – Ian Dec 24 '14 at 3:23
• @Ian That's probably correct and very consistent with what my book says: "...the heat equation involves one derivative with respect to $t$ but two derivatives with respect to $x_i$ ($i=1,\ldots,n$). Consequently, if $u(x,t)$ solves the heat equation, so does $u(\lambda x, \lambda^2t)$ for $\lambda \in \mathbb{R}$. This scaling indicates the reatio $\frac{r^2}t$ ($r=|x|)$ is important for the heat equation and suggests that we search for a solution of it having the form $u(x,t)=v(\frac{r^2}t)=v(\frac{|x|^2}t) \quad (t > 0,x \in \mathbb{R}^n$)." – Cookie Dec 24 '14 at 3:31
• @Ian I don't exactly know why we choose the ratio $\frac{|x|^2}t$ though ... just given that we have one derivative w.r.t. $t$ and two derivatives w.r.t. $x_i$. But if you can explain that, I think I can follow your first comment from here. – Cookie Dec 24 '14 at 3:35
• @dragon Have you seen the derivation? It is very intuitive in the derivation, nothing makes any physical sense if you don't hold $\frac{\Delta x^2}{\Delta t}$ constant. Otherwise the variance of the walk either blows up or goes to zero. – Ian Dec 24 '14 at 4:03

A possible explanation: since $u(x,t)$ solves the heat equation, so does $u_x(x,t).$ And recall the initial condition for the fundamental solution is the $\delta$ distribution. And thus we have $1=u(x,0)=c\int^\infty_{-\infty} e^{\frac{s^2}{4}}\mathrm{d}s=2c\sqrt{\pi}.$ So $c=\frac{1}{\sqrt{4\pi}}$
The initial condition is $u(x,0)=\lim_{t\to 0}v(x/\sqrt t)=\lim_{z\to\infty}c\int_0^ze^{-s^2/4}ds+d=c\int_0^\infty e^{-s^2/4}ds+d=c\sqrt\pi+d$