I have a question about the heat equation $\frac{\partial \varphi}{\partial t} = \frac{\partial^2 \varphi}{\partial x^2}$ with the conditions that $\varphi(x,t=0) = f_0(x)$ and $\lim_{x \rightarrow\pm \infty}\varphi(x,t) = 0$ for every $t \in \mathbb{R}$. The usual way of solving this equation is by separation of variables, i.e., $\varphi(x,t) = A(x)B(t)$. Then one gets the two ordinary differential equations

\begin{equation} \frac{1}{B}\frac{dB}{dt} = \frac{1}{A}\frac{d^2A}{dx^2} = -\gamma \mbox{,} \end{equation}

where $\gamma > 0$ in order that the condition $\lim_{x \rightarrow\pm \infty}\varphi(x,t) = 0$ is satisfied. The thing is that I can also Fourier transform the solution: $\widehat{\varphi}(k,t) = \widehat{A}(k)B(t)$, i.e., the Fourier transformed solution is also a product of a function only of $k$ and one only of $t$. However, when I Fourier transform the equation ($\partial/\partial x \leftrightarrow ik$) I get $\frac{\partial\widehat{\varphi}}{\partial t} = -k^2 \widehat{\varphi}(k,t)$, which can be readily solved as $\widehat{\varphi}(k,t) = \widehat{f_0}(k)e^{-k^2 t}$. But this solution in Fourier space cannot be represented as a product of two functions - one only of $k$ and the other one only of $t$ although it should. I oversee something. Can someone help me?

  • $\begingroup$ You may try to let $\varphi(x,t)=\int_0^\infty e^{-x^2s}K(s,t)~ds$ so that it automatically satisfies $\lim\limits_{x\to\pm\infty}\varphi(x,t)=0$ . $\endgroup$ – doraemonpaul Oct 26 '12 at 23:28
  • $\begingroup$ en.wikipedia.org/wiki/… can find at least one group (unconfirmed whether all or not) of the solution which are luckily satisfy $\lim\limits_{x\to\pm\infty}\varphi(x,t)=0$ . $\endgroup$ – doraemonpaul Oct 27 '12 at 0:29

You should not mix up the two methods (a) separation of variables and (b) Fourier transform.

The final solution of your problem will not be a product of two functions, but a superposition of such products. You use separation of variables in order to determine a sufficient supply of basis functions that can be used for the superposition.

Fourier transform with respect to the variable $x$ gives you immediately the Fourier transform $\hat\phi(k,t)$ of the final solution. Now you have to transform back. By the rules of Fourier transform the product in Fourier space will be transformed into a convolution.

  • $\begingroup$ Also note that if you're looking at a PDE in Cartesian coordinates and its equivalent in cylindrical coordinates, separation of variables may work in both, but will give you different sets of basis functions. (Then from uniqueness and the ability to match any given BCs using a superposition of the basis functions, you know that a basis is enough to express any solution on the domain.) $\endgroup$ – Evgeni Sergeev May 6 '15 at 8:22

First of all, the only solution of $A''+\gamma\,A=0$ such that $\lim_{x\to\pm\infty}A(x)=0$ is $A(x)\equiv0$. If $\gamma>0$ you get the bounded solutions of the heat equation: $$ e^{-\gamma\, t}\cos(\sqrt\gamma\, x),\quad e^{-\gamma\, t}\sin(\sqrt\gamma \,x). $$

Separation of variables does not give all solutions: only solutions that can be written as product of a function of $x$ and a function of $t$. For instance, $$ e^{- t}\cos(x)+e^{-4\,t}\sin(2\,x) $$ is a solution of the heat equation that cannot be written as $B(t)\,A(x)$.


Case $1$: $t\geq0$

Let $\varphi(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)=X''(x)T(t)$



$\begin{cases}T(t)=c_3(s)e^{-ts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore\varphi(x,t)=\int_0^\infty C_1(s)e^{-ts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-ts^2}\cos xs~ds$

Case $2$: $t\leq0$

Let $\varphi(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)=X''(x)T(t)$



$\begin{cases}T(t)=c_3(s)e^{ts^2}\\X(x)=\begin{cases}c_1(s)\sinh xs+c_2(s)\cosh xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore\varphi(x,t)=\int_0^\infty C_1(s)e^{ts^2}\sinh xs~ds+\int_0^\infty C_2(s)e^{ts^2}\cosh xs~ds$

Hence $\varphi(x,t)=\begin{cases}\int_0^\infty C_1(s)e^{-ts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-ts^2}\cos xs~ds&\text{when}~t\geq0\\\int_0^\infty C_1(s)e^{ts^2}\sinh xs~ds+\int_0^\infty C_2(s)e^{ts^2}\cosh xs~ds&\text{when}~t\leq0\end{cases}$

Note that in this question we only care about $x\in\mathbb{R}$ . Since $\sin xs=-\sin(-xs)$ , $\cos xs=\cos(-xs)$ , $\sinh xs=-\sinh(-xs)$ and $\cosh xs=\cosh(-xs)$ , $\varphi(x,t)$ has alternative form of the solution:

$\varphi(x,t)=\begin{cases}\int_0^\infty C_1(s)e^{-ts^2}\sin|x|s~ds+\int_0^\infty C_2(s)e^{-ts^2}\cos|x|s~ds&\text{when}~t\geq0\\\int_0^\infty C_1(s)e^{ts^2}\sinh|x|s~ds+\int_0^\infty C_2(s)e^{ts^2}\cosh |x|s~ds&\text{when}~t\leq0\end{cases}$

Since $\lim\limits_{x\to\pm\infty}\sin|x|s$ , $\lim\limits_{x\to\pm\infty}\cos|x|s$ , $\lim\limits_{x\to\pm\infty}\sinh|x|s$ and $\lim\limits_{x\to\pm\infty}\cosh|x|s$ do not exist, we cannot determine directly whether $\lim\limits_{x\to\pm\infty}\varphi(x,t)=0$ holds or not.

However, when we do some change of variables:

$\varphi(x,t)=\begin{cases}\int_0^\infty C_1\biggl(\dfrac{s}{|x|}\biggr)e^{-t\left(\frac{s}{|x|}\right)^2}\sin s~d\biggl(\dfrac{s}{|x|}\biggr)+\int_0^\infty C_2\biggl(\dfrac{s}{|x|}\biggr)e^{-t\left(\frac{s}{|x|}\right)^2}\cos s~d\biggl(\dfrac{s}{|x|}\biggr)&\text{when}~t\geq0\\\int_0^\infty C_1\biggl(\dfrac{s}{|x|}\biggr)e^{t\left(\frac{s}{|x|}\right)^2}\sinh s~d\biggl(\dfrac{s}{|x|}\biggr)+\int_0^\infty C_2\biggl(\dfrac{s}{|x|}\biggr)e^{t\left(\frac{s}{|x|}\right)^2}\cosh s~d\biggl(\dfrac{s}{|x|}\biggr)&\text{when}~t\leq0\end{cases}$

$\varphi(x,t)=\begin{cases}\int_0^\infty C_1\biggl(\dfrac{s}{|x|}\biggr)\dfrac{e^{-\frac{ts^2}{x^2}}\sin s}{|x|}ds+\int_0^\infty C_2\biggl(\dfrac{s}{|x|}\biggr)\dfrac{e^{-\frac{ts^2}{x^2}}\cos s}{|x|}ds&\text{when}~t\geq0\\\int_0^\infty C_1\biggl(\dfrac{s}{|x|}\biggr)\dfrac{e^{\frac{ts^2}{x^2}}\sinh s}{|x|}ds+\int_0^\infty C_2\biggl(\dfrac{s}{|x|}\biggr)\dfrac{e^{\frac{ts^2}{x^2}}\cosh s}{|x|}ds&\text{when}~t\leq0\end{cases}$

Since $\lim\limits_{x\to\pm\infty}\dfrac{e^{-\frac{ts^2}{x^2}}}{|x|}=0$ and $\lim\limits_{x\to\pm\infty}\dfrac{e^{\frac{ts^2}{x^2}}}{|x|}=0$ ,

$\therefore\varphi(x,t)=\begin{cases}\int_0^\infty C_1(s)e^{-ts^2}\sin|x|s~ds+\int_0^\infty C_2(s)e^{-ts^2}\cos|x|s~ds&\text{when}~t\geq0\\\int_0^\infty C_1(s)e^{ts^2}\sinh|x|s~ds+\int_0^\infty C_2(s)e^{ts^2}\cosh |x|s~ds&\text{when}~t\leq0\end{cases}$ automatically satisflies $\lim\limits_{x\to\pm\infty}\varphi(x,t)=0$ .

The remaining problem is how to substitute $\varphi(x,0)=f_0(x)$ to eliminate nicely on some of the $C_1(s)$ and $C_2(s)$ .


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