Using Fourier transforms to solve the boundary value problem

Question

Using Fourier transforms to solve the boundary value problem $$ x =\begin{cases} u_{t}=4u_{xx}+\sin(t), & -\infty \leq x \leq \infty, t>0\\u(x,0)=e^{-x^2} \sin(x), & -\infty \leq x \leq \infty \end{cases}$$ I think I need to make the change $v=u-\cos(t)$. How to solve next?

Answer 1

You can solve by Fourier transforming in $x$. Or, you can work as Fourier did: First separate variables, form sums of solutions, and then isolate coefficients. As you say, start by subtracting $\cos(t)$. Then separate variables $$ T'(t)X(x) = 4T(t)X''(x) \\ \frac{1}{4}\frac{T'(t)}{T(t)}= \lambda = \frac{X''(x)}{X(x)}. $$ The parameter $\lambda$ must be negative in order to have bounded solutions in $X$, and that also gives decay in time rather than the opposite. So $\lambda=-s^2$ where $s$ is real. $$ X''(x)+s^2 X(x) = 0,\;\;\; T'(t)=-4s^2T(t) \\ X(x)=e^{isx},\;\; T(t) = e^{-4s^2 t}. $$ Then sum these solutions using an integral, $$ u(x,t) = \int_{-\infty}^{\infty}c(s)e^{-4s^2t}e^{isx}ds, $$ where $c(s)$ are the coefficients in the sum, analogous to discrete sums. The coefficient function is determined by the initial condition $$ e^{-x^2}\sin(x) = \int_{-\infty}^{\infty}c(s)e^{isx}ds. $$ Multiplying by $e^{-isx}$ and integrating gives $c(s)$ $$ \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-x^2}\sin(x)e^{-isx}dx=c(s). $$ The $1/2\pi$ is the standard normalization factor, essentially inherited from the discrete case. If you know the Fourier transform of $e^{-x^2}$, then you can reduce to that case \begin{align} c(s) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-x^2}\frac{e^{ix}-e^{-ix}}{2i}e^{-isx}dx \\ & = \frac{1}{4\pi i}\int_{-\infty}^{\infty}e^{-x^2}e^{-i(s-1)x}dx-\frac{1}{4\pi i}\int_{-\infty}^{\infty}e^{-x^2}e^{-i(s+1)x}dx. \end{align} You can look up the Fourier transform $\hat{f}(s)$ of $e^{-x^2}$. This gives the unknown coefficient function $$ c(s)=\frac{1}{\sqrt{2\pi}i}\{\hat{f}(s-1)-\hat{f}(s+1)\} $$ The solution is then $$ u(x,t) = \frac{1}{\sqrt{2\pi}i}\int_{-\infty}^{\infty}\{ \hat{f}(s-1)-\hat{f}(s+1)\}e^{-4s^2 t}e^{isx}ds. $$