Fourier transform in PDEs

Question

I was trying to solve this PDE: $\partial_t f(x,t) = -e^{-t}f(x,t)-t\partial_x f(x,t), -\infty < x < \infty, f(x,0) = \frac{e^{-x^2}}{1+x^2}$ I used the Fourier transform $$f(x,t) = \frac{1}{\sqrt{2\pi}}\int \hat{f}(k,t)e^{ikx} dk$$ where $\hat{f}(k,t) = \frac{1}{\sqrt{2\pi}} \int f(x,t)e^{-ikx}dx$ and, after some easy calculations, I obtained the following: $$ \hat{f}(k,t) = \hat{f}(k,0)e^{-(1-e^{-t})}e^{-ikt^2/2} $$ Now I just needed to compute $$\hat{f}(k,0) = \frac{1}{\sqrt{2\pi}} \int f(x,0)e^{-ikx}dx = \frac{1}{\sqrt{2\pi}}\int \frac{1}{1+x^2}e^{-x^2-ikx}dx $$ But I've got absolutely no idea how to solve it. No complex techniques, it's not a Gaussian integral of the form $\int e^{-ay^2}e^{-iky}dy$ Is there some way of easily solving this? Or is there some kind of trick not to solve this and get straight to the expression $f(x,t)$? This is an exercise taken from an exam, so one needn't make excessively long calculations, due to the limited amount of time. If someone could help me, I'd be much grateful

Answer 1

Yes, you can get to the solution another way. The PDE is first order linear, $$ f_t+tf_x=e^{-t}f, $$ that can be solved by characteristics. On each curve where $dx/dt = t$, that is, on each curve $x(t) = a+\tfrac{1}{2}t^2$ you have $$ \frac{d}{dt}f(a+\tfrac{1}{2}t^2,t) = f_t+tf_x = -e^{-t}f. $$ Integrate with respect to $t$ to get $$ f(a+\tfrac{1}{2}t^2,t) = f(a,0)e^{e^{-t}-1}. $$ Then solve $x = a+\tfrac{1}{2}t^2$ for $a$ in terms of $x$ and $t$ to get $$ f(x,t) = f(x-\tfrac{1}{2}t^2,0)e^{e^{-t}-1}. $$ Now you can use your initial values.

Edit: the logic is this: we assumed a solution $f$ exists, and found a formula for it. That is a uniqueness argument. You should plug into the PDE to show that the found $f$ really is a solution.

Answer 2

You were almost there: plugging your second and third equations into the first one, we obtain \begin{align} f(x,t) &= \frac{1}{\sqrt{2\pi}}\int \hat{f}(k,0)e^{-(1-e^{-t})-\frac{i}{2}kt^2+ikx} \,dk \\ &=\frac{1}{2\pi}e^{e^{-t}-1}\int e^{-\frac{i}{2}kt^2+ikx}\int f(y,0)e^{-iky}\,dy\,dk. \tag{1} \end{align} Changing the order of integration, we find \begin{align} f(x,t) &=\frac{1}{2\pi}e^{e^{-t}-1}\int f(y,0)\int e^{-ik(y-x+\frac{1}{2}t^2)}\,dk\,dy \\ &=\frac{1}{2\pi}e^{e^{-t}-1}\int f(y,0)\,2\pi\delta(y-x+\frac{1}{2}t^2)\,dy \\ &=e^{e^{-t}-1}f(x-\frac{1}{2}t^2,0), \tag{2} \end{align} which is the same result found by @Bob Terrell using the method of characteristics.