I'm sorry if this question is too easy. Since I'm a beginner in this field, I want a clear example for it. Here's the problem:

Use Fourier transforms to solve $$\frac{\partial u}{\partial t}=2\frac{\partial^2 u}{\partial x^2}, \qquad x>0,t>0$$ if $u(0,t) = 0$, $u(x,0)=e^{-x}$ and $u(x,t)$ is bounded.

I can solve the PDE of degree 1, and $\frac{\partial^2 u}{\partial x^2}$ seems too strange for me. And one more hard thing for me is that I don't know how to utilize the boundary conditions. So help me with this problem. Thanks


2 Answers 2


The Fourier series and the Fourier transform techniques are essentially the same, except that you use a discrete sum to sum all separated solutions in one case, and you use an integral sum to sum all separated solutions in the other case. In either case, the process starts with finding the separated solutions $$ U(x,t)=X(x)T(t) \\ U_{t} = 2 U_{xx} \\ X(x)T'(t) = 2X''(x)T(t) \\ \frac{T'(t)}{2T(t)} = \frac{X''(x)}{X(x)} \\ \frac{T'(t)}{2T(t)} = \lambda, \;\; \lambda=\frac{X''(x)}{X(x)} $$ The parameter $\lambda$ must be non-positive in order for the solutions to not explode in time. So $\lambda=-s^2$ where $s$ is assumed to be real. The general solutions are $$ T(t) = e^{-2s^2 t},\;\;\; X(x) = \sin(sx)\;\;\; -\infty < s < \infty. $$ The $\sin(sx)$ solutions are chosen so that $X(0)=0$. All solutions for $s > 0$ are duplicated when $s < 0$. So only $s > 0$ is needed. The general solution involves a coefficient function $c(s)$ that depends on $s$: $$ u(x,t) = \int_{0}^{\infty}c(s)e^{-2s^2 t}\sin(sx)ds. $$ The coefficient function $c(s)$ is determined by the condition $u(x,0)=e^{-x}$: $$ e^{-x} = u(x,0) = \int_{0}^{\infty}c(s)\sin(sx)ds. $$ The inverse Fourier sin transform gives $c(s)$: $$ \frac{2}{\pi}\int_{0}^{\infty}e^{-x}\sin(sx)dx = c(s). $$ Therefore, \begin{align} c(s)& = \frac{2}{\pi}\int_{0}^{\infty}\frac{e^{x(is-1)}-e^{x(-is-1)}}{2i}dx \\ & = \left.\frac{1}{\pi i}\left( \frac{e^{x(is-1)}}{is-1}+\frac{e^{x(-is-1)}}{is+1}\right)\right|_{x=0}^{\infty} \\ & = -\frac{1}{\pi i}\left(\frac{1}{is-1}+\frac{1}{is+1}\right) \\ & = -\frac{2}{\pi}\frac{s}{-s^2-1} \\ & = \frac{2}{\pi}\frac{s}{1+s^2}. \end{align} The solution $u$ is $$ u(x,t) = \frac{2}{\pi}\int_{0}^{\infty}\frac{s}{1+s^2}e^{-2s^2t}\sin(sx)ds. $$


I will show you the way using the Laplace transform. Setting $s=j\omega$ will give you the Fourier transform.

  1. Step multiply everything with $e^{-st}$ and integrate from $t=0$ to $t=\infty$.

$$\int_0^{\infty}\frac{\partial u}{\partial t}e^{-st}dt=2\int_{0}^{\infty}\frac{\partial^2 u}{\partial x^2}e^{-st}dt$$

  1. We will now use integration by parts on the left integral and we will take out the $\dfrac{\partial^2}{\partial x^2}$ of the right integral because the integration is with respect to $t$.

$$ue^{-st}\biggl|_0^{\infty} +s\int_0^{\infty}ue^{-st}dt=2\frac{\partial^2 }{\partial x^2}\int_{0}^{\infty}ue^{-st}dt$$

  1. Now, we see the same integral on both sides. Let us call it $L_s(x)$, the Laplace transform with respect to $t$. Note, that it is only a function of $x$ and $s$ is just a parameter. Additionally we assume $u(x,t)e^{-st}|_\infty =0$. $$ue^{-st}\biggl|_0^{\infty} +sL_s(x)=2\frac{\partial^2 }{\partial x^2}L_s(x)$$ $$-u(x,t=0) +sL_s(x)=2L_s''(x).$$

This is an ODE if you plug in $u(x,t=0)=e^{-x}$.


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