# What is the physical significance of scale invariance (under appropriate boundary conditions) in the 1D Helmholtz equation?

So we were given this problem in mathematical physics (Context is that we're learning about Sturm-Liouville):

Consider the 1D Helmholtz equation with $$k^2>0$$ $$\frac{\partial ^2y}{\partial x^2}+k^2 y=0$$ on the interval $$0\leq x \leq L$$ subject to the boundary conditions $$ay(0)+by'(0)=0$$ $$ay(L)+by'(L)=0$$ What are the forms of the eigenfunctions and eigenvalues?

It was easy enough to get that the eigenfunctions are of the form

$$A \left(\sin (k x)-\frac{b k }{a}\cos (k x)\right)$$

with eigenvalues

$$k^2=\left(\frac{\pi n}{L}\right)^2$$

There's this $$A$$ left out in front. I get how in quantum mechanics maybe this is useful as a normalizer for probabilities. But is there a physical significance beyond the fact that it makes the math nicer? What does this result mean when applied to a classical system, like a string? Or does it just not apply?

• I'm somewhat confused by your question. Are you asking what the amplitude of the solution represents? Nov 13, 2020 at 19:51

One physical interpretation of the Helmholtz equation is, as you mention, a string which is fixed at $$x=0$$ and $$x = L$$ (if $$b=0$$ in your equations) with an oscillation amplitude $$y(x)$$. If you knew your material of the string, you would know the propagation speed $$c$$ and you could calculate the eigenfrequency $$f$$ from $$k = \frac{2\pi f}{c}$$
The solution $$y(x)$$ that you state has an amplitude $$A$$, which can in principle be anything, as long nobody told you how the string was excited initially. As an example, if I hold the string in a position $$y(x) = 2\sin(\pi \frac{x}{L})$$ and release, the string will oscillate with an amplitude $$A = 2$$ and with a frequency $$f = \frac{ck}{2\pi} = \frac{ c}{2 L}$$. That is, a half standing wave between $$x=0$$ and $$x=L$$.