# Solving the wave equation, with boundary conditions, in the sense of distributions (Generalized functions)

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.

• 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? – quarague Feb 11 '20 at 10:42
• @quarague Sorry. Mistake. Now corrected. – Ma Joad Feb 11 '20 at 10:59
• you could work with a series expansion $u(x,t) = \sum_{k=1}^\infty \sin(k\pi x) u_k(t)$ – daw Feb 11 '20 at 11:04
• @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. – Ma Joad Feb 11 '20 at 11:10
• 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. – 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 $$u$$.