After learning some distribution theory, I find that in my book, all PDEs given as examples are in free space (without any boundary conditions). I wonder if distribution theory can be used to tackle PDEs with boundary conditions.

To be more specific, let's consider this problem. Let there be a string of length $\pi$ with both ends fixed. Transverse waves can be produced on the string, satisfying the classical wave equation $$ \partial^2_t u(x,t)=c^2\partial^2_x u(x,t). $$ The boundary conditions are $u(0)=u(\pi)=0$.

Now, let impose a wired initial condition: let's pluck the string in the middle, so initially, the string is at rest, in the position $$ u(x,0)=A(\pi/2-|x-\pi/2|), A\in \mathbb R. $$ As one can see, the initial condition is not everywhere differentiable. However, $u$ can be seen as an element of $\mathcal D'(\mathbb R)$ or $\mathcal S'(\mathbb R)$, the space of (tempered) distributions. The differential equation therefore make sense in the sense of distributions.

Using Fourier transform and convolution, we can manage to get a solution, IF there are no boundary conditions. However, in this situation, I do not know how to state the boundary condition in term of distributions.

So, my question now is: can we make sense out of this problem, possibly in the sense of distributions, and solve the equation?

Edit: we can use Fourier series expansion to solve this, but then I don't feel it really a way of "understanding" how it really works - after all, the original equation ceases to make sense when it is not differentiable. I want to somehow have some formalism in making sense of the derivative of a function which is not differentiable. Possibly weak derivative?

Edit: Fourier transform over a bounded interval doesn't seem to be obvious to define; it appears that Fourier series are really easier.

  • $\begingroup$ Don't you run into overspecification if you impose additional boundary conditions in your distribution case. In the first case you have the boundary conditions $u(0,t)=u(\pi, t)=0$ valid for all $t$. In the second setting you instead have the condition $u(x,0)=..$. If you had conditions for both $u(0,t), u(\pi,t)$ and for $u(x,0)$ wouldn't your system be overdetermined and in general there would be no solution? $\endgroup$ – quarague Feb 11 '20 at 10:42
  • $\begingroup$ @quarague Sorry. Mistake. Now corrected. $\endgroup$ – Ma Joad Feb 11 '20 at 10:59
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    $\begingroup$ you could work with a series expansion $u(x,t) = \sum_{k=1}^\infty \sin(k\pi x) u_k(t)$ $\endgroup$ – daw Feb 11 '20 at 11:04
  • $\begingroup$ @daw Yes of course. But then I feel it is not really a way of "understanding" how it really works - after all, the original equation ceases to make sense when it is not differentiable. $\endgroup$ – Ma Joad Feb 11 '20 at 11:10
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    $\begingroup$ You have to use the definition of a Green function through eigenfunctions $G(x,x')=\sum_n\phi_n^*(x)\phi_n(x')/\lambda_n$, being $\lambda_n$ the corresponding eigenvalues. I think that, in a way or another, a Fourier series is needed here. $\endgroup$ – Jon Feb 11 '20 at 12:52

It is possible to use Fourier series here, by making use of the periodicity and the continuity of the initial condition. Indeed, the problem can be extended to $x$ in $\Bbb R$ by periodization (successive identical strings with fixed ends of length $\pi$). Consider the spatially $\pi$-periodic function $$u(x,t) = \sum_n c_n(t)\, \text{e}^{2\text{i}nx} \, ,$$ which is written as a spatial Fourier series. Its restriction to $x\in [0,\pi]$ may solve the problem if the boundary condition $\sum_n c_n(t) = 0$ is satisfied. Also, at time $t=0$, the Fourier coefficients $c_n(0)$ must be the Fourier coefficients of the initial condition $u(\cdot, 0)$. Similarly, their derivatives $c'_n(0)$ are the Fourier coefficients of the null function, i.e. $c'_n(0) = 0$. Injecting this function in the PDE gives $$ c_n''(t) + 4n^2c^2 c_n(t) = 0 $$ for all $n$, by uniqueness of Fourier series. Therefore, we have $c_n(t) = c_n(0)\cos(2nct)$. The same approach can be followed for other triangular signals, Gaussian signals, rectangular signals, etc.

Instead of the initial Fourier series, we could have written the Fourier transform representation $$u(x,t) = \frac1{2\pi} \int_{\Bbb R} \hat u(k,t)\, \text{e}^{\text{i}kx}\,\text d k \, ,$$ where $\hat u$ is the spatial Fourier transform of $u$.

  • $\begingroup$ Thank you! Just can't wait for your updated information on boundary conditions. The solution you give doesn't seem to ensure the boundary condition is always satisfied - however, maybe it is okay not to satisfy; if the string is distorted to infinite displacement (Dirac delta), then chances are it doesn't have to behave properly! $\endgroup$ – Ma Joad Feb 12 '20 at 3:43
  • $\begingroup$ @Jethro I couldn't make up my mind about the Dirac delta... So I finally proposed the standard Fourier series approach, which I linked to the Fourier transform in the end. $\endgroup$ – EditPiAf Feb 12 '20 at 11:31

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