# Expressing $\frac{\partial u}{\partial t}=k\frac{\partial^2 u}{\partial x^2}+S(x,t)$ as an integral involving $S(x,t)$

I am trying to express the solution of$$\frac{\partial u}{\partial t}=k\frac{\partial^2 u}{\partial x^2}+S(x,t) \tag{1}$$ for $$-\infty< x<\infty$$ with initial condition $$u(x,0)=0$$ as an integral involving the source term $$S(x,t)$$.

Taking the Fourier transform of $$(1)$$ w.r.t $$x$$ reduced the PDE to the ODE $$\frac{d}{dt}\hat{u}(w,t)+kw^2\hat{u}(w,t)=\hat{S}(w,t), \tag{2}$$ where I have defined $$\hat{u}(w,t)=\mathcal{F}_x(u(x,t))$$ (where $$\mathcal{F}_x$$ denotes the Fourier transform w.r.t $$x$$). Using the integrating factor $$e^{kw^2t}$$, the solution to $$(2)$$ is, $$\hat{u}(w,t)=\int_{0}^{t} e^{kw^2(t'-t)}\hat{S}(w,t') \ dt'.$$ But I am unsure of how to find $$u(x,t)$$ . I was thinking of simplifying the following $$u(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\int_{0}^{t} e^{kw^2(t'-t)}\hat{S}(w,t') \ dt'\right) e^{iwx} \ dw.$$

Note the answer is $$u(x,t)=\frac{1}{\sqrt{2\pi}}\int_{0}^{t}\left(\int_{-\infty}^{\infty}\frac{\exp(-(x-x')^2/(4k(t-t')))}{\sqrt{2k(t-t')}}S(x',t') \ dx'\right) \ dt$$

• You have a sign error in $(2)$ – Dylan Apr 20 at 9:17
• @Dylan Thanks for spotting this. However, this makes a marginal difference in solving my problem. – Stuart-James Burney Apr 21 at 8:31