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[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?

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  • $\begingroup$ There is no exact solution. It depends on $f(x)$ $\endgroup$ – BioPhysicist Oct 10 '19 at 11:53
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    $\begingroup$ 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
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Without knowing $f$ the closest we can come is writing the solution as an integral or a convolution: $$ u(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\tilde{f}(k)e^{ikx}e^{-k^2t}dk = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x')e^{-ikx'}dx'\right)e^{ikx}e^{-k^2t}dk \\ = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x') \left(\int_{-\infty}^{\infty} e^{-ik(x-x')} e^{-k^2t} dk \right) dx' = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x') \, \sqrt{\frac{\pi}{t}} e^{-\frac{(x-x')^2}{4t}} dx' \\ = \frac{1}{2\sqrt{\pi t}} \int_{-\infty}^{\infty} f(x') \, e^{-\frac{(x-x')^2}{4t}} dx' = \left(f*\frac{1}{2\sqrt{\pi t}} e^{-\frac{\bullet^2}{4t}}\right)(x) $$

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