[Homework problem:] In my assignment, I was given a problem where I had to solve a PDE using Fourier transform. It goes like this:

Solve the PDE $$u_t=u_{xx}$$ subject to the initial conditions: $u(0, t)=0$ and $u(x, 0)=f(x)$.

My attempt at a solution:

Using Fourier transform on both sides of the original equation, I get:

$$\frac{\partial{\tilde{u}}}{\partial{t}}=-k^2\tilde{u}(k,t)$$ [Here I have used the property of Fourier transform of derivatives which states that $\cal{F}(f^\prime(x))=-ik\cal{F}(f(x))$.]

Let $\tilde{u}(k, t)=U(t)$ since here we are differentiating with respect to $t$ only, so $\tilde{u}$ is effectively a function of $t$ only.

Then we get an ODE in $t$ that can be easily solved: $$\frac {dU}{dt}=-k^2U$$ whose solution is of the form $U=Ce^{-k^2t}$, where $C$ is an integration constant.

Now, we see that $C=\tilde{u}(k, 0)$ actually, and we are given that $u(x, 0)=f(x)$, so $C$ must be $\tilde{f}(k)$, or $$\tilde{u}(k, t)=\tilde{f}(k)e^{-k^2t}.$$

Then: $$u(x, t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\tilde{u}(k, t)e^{ikx}dk$$ $$=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\tilde{f}(k)e^{ikx}e^{-k^2t}dk$$ Now, $$\tilde{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$$ $$\Rightarrow u(x, t)= \frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)e^{-ikx}e^{ikx} e^{-k^2t}dkdx$$ $$=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)dx\int_{-\infty}^{\infty}e^{-k^2t} dk$$

But I am stuck here since this only gives me the integral form, I can't evaluate the exact solution from here. Can I use the condition $u(0, t)=0$ somehow to get the solution? Or do I need the exact functional form of $f(x)$ to find the exact solution?

  • $\begingroup$ There is no exact solution. It depends on $f(x)$ $\endgroup$ – BioPhysicist Oct 10 '19 at 11:53
  • 1
    $\begingroup$ Be careful with the variables! You have two $x$:s that you have treated as one. One is a variable of integration (when calculating $\tilde{f}$) and one is a free variable (the one in $u(x,t)$). $\endgroup$ – md2perpe Oct 10 '19 at 17:16

Without knowing $f$ the closest we can come is writing the solution as an integral or a convolution: $$ u(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\tilde{f}(k)e^{ikx}e^{-k^2t}dk = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x')e^{-ikx'}dx'\right)e^{ikx}e^{-k^2t}dk \\ = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x') \left(\int_{-\infty}^{\infty} e^{-ik(x-x')} e^{-k^2t} dk \right) dx' = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x') \, \sqrt{\frac{\pi}{t}} e^{-\frac{(x-x')^2}{4t}} dx' \\ = \frac{1}{2\sqrt{\pi t}} \int_{-\infty}^{\infty} f(x') \, e^{-\frac{(x-x')^2}{4t}} dx' = \left(f*\frac{1}{2\sqrt{\pi t}} e^{-\frac{\bullet^2}{4t}}\right)(x) $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.