The Wave Equation and Linear Combinations

Question

Say you are trying to solve the wave equation,

$$\frac{\partial^2y}{\partial t^2}=c^2\frac{\partial^2y}{\partial x^2}$$

If the boundary is finite, you find that you need an infinite sum of solutions to fit some initial condition. $y = \sum y_n$.

Now if the medium is infinite, you don't have any boundary conditions, but require that the solution be bounded. In this case, you solve the wave equation using separation of variables as before, but with no boundary condition (since it is infinite). You notice that the eigenvalues fill out the entire space; they aren't discrete as like the $y_n$'s but continuous (my book talks about solving Sturm-Liouville, i.e. differential equation problems, in the midst of solving the wave equation. That's where the eigenvalue comes in).

Anyways, in this continuous eigenvalue case, we don't take a sum, but an integral. That is, using previous knowledge of the finite boundary case, we attempt solutions of the form,

$$\int_0 ^\infty y_w dw$$

to fit the initial condition. My question is, how does $\sum y_n \rightarrow \int y_w dw$? Specifically, why is there a $dw$? I know you need that differential to tell you what you are integrating over, but in the finite medium case, it was $\sum y_n$ and not $\sum y_n n$. If the latter was true, then I'd see how $n \rightarrow dw$, but there is no $n$.

I know that if the eigenvalues are continuous, like $\left\{ 0.00001,0.00002,... \right\}$, it would be impossible to add up all the solutions corresponding to each eigenvalue. So we take a little chuck of eigenvalues, a $dw$. Then the integral allows us to move across the real line and evaluate $y_w$ in each of those chunks. But then we aren't adding up each $y_w$, but adding up $y_w dw$. Does finding an area not change the problem? The wave equation is linear, so it admits a linear combination of solutions. $y_w$ is a solution, not $y_w dw$?

Answer 1

I think about this way:

For discrete case we have

$$y = \sum_{n=1}^\infty y_n = \lim\limits_{N\to \infty}\sum_{n=1}^N y_n = \lim\limits_{N\to \infty}\sum_{n=1}^N \Big( y_n \big((n+1) - n \big) \Big), $$ so in fact in the discrete case your $y = \sum \big( y_n \Delta n\big)$, where $\Delta n = (n+1) - n = 1$.

Let us decrease $\Delta n$ by dividing it by some number greater than zero, e.g. introduce $\Delta_k n : = \frac{(n+1)-n}{k} = \frac{1}{k}$. Note that $$ \lim\limits_{N\to \infty}\sum_{n=1}^N \Big( y_n \big((n+1) - n \big) \Big) = \lim\limits_{N\to \infty}\sum_{n=1}^N \Big( y_n \Delta n \Big) = \lim\limits_{N\to \infty}\sum_{n=1}^{kN} \Big( y_n \Delta_k n \Big) . $$ Thus, taking limit with $k \to \infty$ we get $\lim\limits_{k \to\infty} \Delta_k n = \lim\limits_{k \to\infty} \frac{(n+1)-n}{k} \approx dw$, where $dw$ is the infinitesimal increment of $w$ - the variable covering entire real positive semi axis. Thus we have $$ y = \lim\limits_{\substack{N\to \infty\\k\to \infty}}\sum_{n=1}^{kN} \Big( y_n \Delta_k n \Big) \approx \int_0^\infty y_w \,dw. $$

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PS To be more rigorous, one would need to define analogues of lower and upper Riemman sums w.r.t. spectrum and who that refinement of the interval does not change limit of these sums. Basically, follow the definition of Riemman integral. I believe it is possible to show that refinement does not change the limit of the sum using orthogonality property of spectral basis. However, it might take some time to work out all the technicalities.

PPS I apologize that my answer is a bit hand wavy and not rigorous enough, but I hope it will manage to convey my understanding of the process lying behind this transition.