7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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 $u$.

$\endgroup$
  • $\begingroup$ 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

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.