Taking the Fourier and Laplace transform of the Heat Equation

I am trying to solve the PDE $$u_t(x,t)-ku_{xx}(x,t)=S_0\delta(x)\delta(t)$$ subject to the initial condition $$u(x,0)=\delta(x)$$.

Performing a Fourier transform on the PDE with respect to $$x$$ $$(x\rightarrow w)$$ which produces $$\frac{\partial}{\partial t}\hat{u}(w,t)+kw^2\hat{u}(w,t)=\frac{\delta(t)S_0}{\sqrt{2\pi}}.$$

We can then take a Laplace transform of this ODE with respect to $$t$$ $$(t\rightarrow s)$$ which produces $$s\hat{\bar{u}}(w,s)+kw^2\hat{\bar{u}}(w,s)=\frac{S_0}{\sqrt{2\pi}}+1, \tag{1}$$ as $$\mathcal{F}_x(u(x,0))=\mathcal{F}_x(\delta(x))\implies \hat{u}(w,0)=\frac{1}{\sqrt{2\pi}}$$ (using the definition with a factor of $$\frac{1}{\sqrt{2\pi}}$$).

My question is, is it even possible to invert $$(1)$$? This is a necessary step to solving the PDE.

Note that this post is similar to Problem with Heat Equation and Laplace Transform

Edit:

The $$+1$$ on the RHS appears as $$\mathcal{L}_t\left(\frac{\partial}{\partial t}\hat{u}(w,t)\right)=s\mathcal{L}_t(\hat{u}(w,s))-\hat{u}(w,0)$$

• I literally just answered this in a comment in the linked post. Also, after taking the Laplace transform, I'm not sure where the $+1$ comes from on the RHS. Apr 11, 2019 at 10:59
• Does inverting in s not leave you a guassian in $\omega$, which I think has a gaussian inverse fourier transform?
– Paul
Apr 11, 2019 at 11:00
• @Mattos I've uploaded an edit. This should follow on from the link post, in which case the initial condition was $u(x, 0) = U_0\delta(x)$. Am I wrong?
– user557493
Apr 11, 2019 at 11:10
• @Bell No everything you did looks fine, I'm just tired and didn't realise what was going on. Yes, in the other post there is just a constant factor of $U_{0}$ in the initial condition that you don't have. Now just follow what I wrote on the other post in the comments about how to invert. Apr 11, 2019 at 11:33
• @Mattos Thanks mate, I'm expecting to get a Guassian. I just need to solve $$u(x,t)=(S_0+1)\mathcal{F}^{-1}_x\left(\frac{1}{\sqrt{2\pi}}\exp(-kw^2t)\right).$$
– user557493
Apr 11, 2019 at 11:36