Solving PDE via Fourier transform

Question

I need to find the solution to this equation using the Fourier transform: $$ \partial_t f(x,t) = t^2\partial_{xx}^2 f(x,t) -t \partial_x f(x,t)+\delta(t)\delta(x^2-1)$$ with initial condition $f(x,-1) = 0$ I was a little bit startled because the initial solution is given at $t=-1$ and not at $t=0$. I followed this procedure: for $t>0$ the equation becomes (due to the Dirac delta properties): $$ \partial_t f(x,t) = t^2\partial_{xx}^2 f(x,t) -t \partial_x f(x,t) $$ Using the Fourier transform, and omitting some easy calculations, I obtain the following ODE: $$\frac{d}{dt}\hat{f}(k,t) = -k^2t^2\hat{f}(k,t)-ikt\hat{f}(k,t) $$ The standard solution yields: $\hat{f}(k,t) = C(k)e^{-(\frac{t^3}{3}k^2+ik\frac{t^2}{2})}$ I find the coefficient $C(k)$ in the following way: $$\hat{f}(k,0) = C(k)e^0 \Rightarrow C(k) = \hat{f}(k,0)e^{-1}$$ Hence we finally get: $$ \hat{f}(k,t) = \hat{f}(k,0)e^{-1} e^{-(\frac{t^3}{3}k^2+ik\frac{t^2}{2})}$$ Now the solution tells me that $\hat{f}(k,t) = \hat{f}(k,0) e^{-(\frac{t^3}{3}k^2+ik\frac{t^2}{2})}$ without the $e^{-1}$ term: so my first question is: where am I making a mistake? The method I followed for solving that ODE is the same as ever, and it works as far as I can remember. Now another problem arose while solving this: I can't seem to find the initial solution $f(x,0)$. I need it because, without it, I wouldn't be able to use the Fourier inversion theorem: $$\hat{f}(k,0) = \frac{1}{\sqrt{2\pi}}\int f(x',0)e^{-ikx'}dx' $$ I could solve for $t>1$ but that Delta term wouldn't vanish which would make the equation much more complicated. Is there some kind of way of extending the initial solution from $t=-1$ to $t=0$? I thought that maybe the initial solution could somehow be $0$ not only at $t=-1$, but in some wider interval $[-1-\epsilon, -1+\epsilon]$. Nevertheless, this would probably work if $\epsilon \to 0$ and this doesn't seem to help me at all. Well this is as far as I could go. Hope someone can enlighten me a bit more on how to solve this. The procedure I followed seems to be the right one. The solution somehow affirms that it can be easily seen that $f(x,0) = \delta(x^2-1)$ but I can't seem to figure how it does unfortunately.

Answer 1

For $t<0$, the PDE also reduces to $$ \partial_t f(x,t) = t^2\partial_{xx}^2 f(x,t) -t \partial_x f(x,t). \tag{1} $$ Notice that $f(x,t)=0$ satisfies $(1)$ and the initial condition $f(x,-1)=0$, hence it is the solution to the PDE in the interval $-1\leq t<0$. Now, integrating both sides of $$ \partial_t f(x,t) = t^2\partial_{xx}^2 f(x,t) -t \partial_x f(x,t)+\delta(t)\delta(x^2-1) \tag{2} $$ from $t=-\epsilon$ to $t=+\epsilon$, we find $$ f(x,+\epsilon)=f(x,-\epsilon)+\int_{-\epsilon}^{+\epsilon}[t^2\partial_{xx}^2 f(x,t) -t \partial_x f(x,t)]\,dt+\delta(x^2-1).\tag{3} $$ Since $f(x,-\epsilon)=0$, and the integral on the RHS of $(3)$ vanishes in the limit $\epsilon\to 0^+$, we end up with $f(x,0^+)=\delta(x^2-1)$.