Given PDE:

$$u_t=u_{xx}-\gamma u+g(x), -\infty \lt x \lt \infty, t \gt0, \gamma \gt 0$$ with $$u(x,0)=0, g(x)=xe^{-x^2/2}.$$

Use Fourier transform to find the solution of the given PDE.

**My Approach **

I took the Fourier transform and got the corresponding equation $$U_t(k,t)=-k^2U(k,t)-\gamma U(k,t)-ik \sqrt{2\pi}e^{-k^2/2}$$

My question is how to revert back into the original space using the inverse Fourier transformation.

  • $\begingroup$ You made a mistake for the fourier transform of $g(x)$ you should have a derivative. $\endgroup$ Jan 15, 2020 at 0:28
  • $\begingroup$ What do you mean by a derivative? $\endgroup$ Jan 15, 2020 at 0:29
  • $\begingroup$ The Fourier transform of $xh(x)$ is $i\frac {d}{dk}H(k)$ where $H(k)$ is the transform fourier of $e^{-x^2/2}$ $\endgroup$ Jan 15, 2020 at 0:32
  • $\begingroup$ Thanks for identifying my mistake. Is it okay now? $\endgroup$ Jan 15, 2020 at 0:37
  • $\begingroup$ YW Thats what I got indeed but with $\sqrt {2\pi}$ $\endgroup$ Jan 15, 2020 at 0:38

1 Answer 1


After correcting the mistake of the Fourier Transform of $g(x)$, I got this: $$U_t(k,t)=-k^2U(k,t)-\gamma U(k,t)-ik \sqrt{2\pi}e^{-k^2/2}$$ $$U_t(k,t)+(k^2+\gamma) U(k,t)=-ik \sqrt{2\pi}e^{-k^2/2}$$ Solving the DE with Integrating Factor: $$(U(k,t)e^{(k^2+\gamma)t} )'=-ik \sqrt{2\pi}e^{-k^2/2}e^{(k^2+\gamma)t} $$ Integrate: $$U(k,t)e^{(k^2+\gamma)t} =-ik \sqrt{2\pi}e^{-k^2/2}\int e^{(k^2+\gamma)t}dt +C(k)$$ $$U(k,t)e^{(k^2+\gamma)t} =-ik \frac {\sqrt{2\pi}}{(k^2+\gamma)}e^{-k^2/2} e^{(k^2+\gamma)t} +C(k)$$ $$U(k,t) =-ik \frac {\sqrt{2\pi}}{(k^2+\gamma)}e^{-k^2/2} +C(k)e^{-(k^2+\gamma)t}$$ $$U(k,0)=0 \implies C(k) =ik \frac {\sqrt{2\pi}}{(k^2+\gamma)}e^{-k^2/2}$$ $$U(k,t) =-ik \frac {\sqrt{2\pi}}{(k^2+\gamma)}e^{-k^2/2}(1 -e^{-(k^2+\gamma)t})$$


You have the condition that should simplify a little the equation: $$U(k,0)=0$$ Note that you have this result: $$\mathcal{F^{-1}}\{ikF(k)\}=f'(x)$$ And you also have: $$\mathcal{F^{-1}} \{\frac {2a}{a^2+k^2}\}=e^{-a|x|}$$

  • $\begingroup$ Yes, I got the same thing. $\endgroup$ Jan 15, 2020 at 1:02
  • $\begingroup$ Where $c(k)$ is any function of $k$ @AtulAnuragSharma $\endgroup$ Jan 15, 2020 at 1:03
  • $\begingroup$ Yeah, I got the answer up-to this part. Also, $c(k)$ can be found using the initial conditon. $\endgroup$ Jan 15, 2020 at 1:06
  • $\begingroup$ That's good @AtulAnuragSharma $\endgroup$ Jan 15, 2020 at 1:07
  • $\begingroup$ The trouble I am facing is in get the inverse Fourier. $\endgroup$ Jan 15, 2020 at 1:08

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