I'm solving this equation: $\frac{\partial u}{\partial t}= k \frac{\partial^2 u}{\partial x^2}$, with the following conditions:
- $ t\geq0$
- $-\infty<x<\infty$
- $ u(x,0) = f(x)$
- $ \mathscr{F} \{u(x,t)\} =U(s,t)$
Firts I need to find $U(s,t)$, then I have to find an general expression of $u(x,t)$ in terms of $f(x)$ using the convolution theorem, and then make $f(x)=1$ to find $u(x,t)$
The way I am solving the problem is:
$\frac{\partial u}{\partial t}= k\frac{\partial^2 u}{\partial x^2}$, with $u(x,t)=\mathscr{T}(t)\mathscr{X}(x)$
$\frac{\partial \mathscr{T}(t)\mathscr{X}(x)}{\partial t}= k\frac{\partial^2 \mathscr{T}(t)\mathscr{X}(x)}{\partial x^2}$
$\mathscr{X}\frac{\partial \mathscr{T}}{\partial t}= k\mathscr{T}\frac{\operatorname{d}^2 \mathscr{X}}{\operatorname{d} x^2}$ , now divided by $\frac{1}{u(x,t)}$
$\frac{1}{\mathscr{T}}\frac{\partial \mathscr{T}(t)}{\partial t} - \frac{k}{\mathscr{X}}\frac{\operatorname{d}^2 \mathscr{X}}{\operatorname{d} x^2}=0$ , Since the only way for these equations to be zero is that they are both equivalent to a constant.
$\frac{1}{\mathscr{T}}\frac{\partial \mathscr{T}(t)}{\partial t} =\frac{k}{\mathscr{X}}\frac{\operatorname{d}^2 \mathscr{X}}{\operatorname{d} x^2}= -m^2$, therefore we obtain two equations:
$\frac{\partial \mathscr{T}}{\partial t}+\mathscr{T}m^2=0$ and $\frac{\operatorname{d}^2 \mathscr{X}}{\operatorname{d} t^2}+\frac{\mathscr{X}m^2}{k}=0$, therefore the solutions are:
$\mathscr{T}(t)=Ae^{-mt}$ and $\mathscr{X}(x)=B\cos(\frac{m}{\sqrt{k}})+C\sin(\frac{m}{\sqrt{k}})$
${u(x,t)}= Ae^{-mt}[B\cos(\frac{m}{\sqrt{k}})+C\sin(\frac{m}{\sqrt{k}})]$, Now to find out $u(s,t)$, do the following:
$U(s,t)= \mathscr{F} \{u(x,t)\} $
$U(s,t)= \mathscr{F} \{Ae^{-mt}[B\cos(\frac{m}{\sqrt{k}})+C\sin(\frac{m}{\sqrt{k}})]\}$
$U(s,t)= ABe^{-mt}\mathscr{F}\{\cos(\frac{m}{\sqrt{k}})\} + ACe^{-mt}\mathscr{F}\{\sin(\frac{m}{\sqrt{k}})\} $
$U(s,t)=ABe^{-mt}[\pi\delta(s+\frac{m}{\sqrt{k}})+\pi\delta(s-\frac{m}{\sqrt{k}})]+ACe^{-mt}[i\pi\delta(s-\frac{m}{\sqrt{k}})-\pi\delta(s+\frac{m}{\sqrt{k}})]$
Now I don't have idea how to find $u(x,t)$ in terms of $f(x)$ using the convolution theorem.