Regularity of solution of ODE $\frac{d}{dx}\left(\frac{1}{n^2}v'\right)+k^2 v = 0$

Question

Given ODE: $$ \frac{d}{dx}\left(\frac{1}{n^2}v'\right)+k^2 v = 0, x \in [a,b] $$ where $n$ is a discontinuous function in $[a,b]$ such that $n^2$ is discontinuous and non-zero (take a step function for example), and boundary conditions $v(a) = \alpha, v(b) = \beta$ are given. What would be the regularity of the solution of the ODE ? I expect $v$ to be just differentiable, however I am unable to find any online notes or textbooks on this ODE (does this ODE have a special name?). Also, how does one make sense of the term " $\frac{d}{dx}\left(\frac{1}{n^2}v'\right)$ "? They say the derivative is defined in the "weak sense", but that doesn't help me solve this ODE.

Answer 1

I would say your equation falls within the framework of Sturm-Livouille theory, though usually $p = \frac1{n^2}$ is not discontinuous in common examples from there.

For simplicity I will assume that $n$ is bounded from above and away from zero, that is $$ 0 < \lambda \leq n(x) \leq \Lambda < \infty$$ for almost every $x$. Without this assumption the analysis becomes more delicate, as the equation may degenerate, and the regularity would then depend more on the precise structure of $n$. Additionally I will assume that $k$ is also bounded.

I claim that a unique solution $v$ always exists, but as alluded to in the comments, in general you can only expect $\frac{v'}{n^2}$ to be continuous. This however poses the problem that $v$ may not be twice differentiable, so one needs to understand the equation through a weak formulation. To do this one needs to do the following:

1. Derive a weak formulation, and show any $C^2$ solution, if it exists, will satisfy it. 2.Consider the existence of solutions for this weak solution.
2. Show the claimed regularity for weak solution.

These steps are fairly standard, but are usually treated in the generality of $n$-dimensional elliptic PDEs. I don't know of a canonical reference for the ODE case, but it is treated for instance in Chapter 8 of

Brezis, Haim, Functional analysis, Sobolev spaces and partial differential equations, Universitext. New York, NY: Springer (ISBN 978-0-387-70913-0/pbk; 978-0-387-70914-7/ebook). xiii, 599 p. (2011). ZBL1220.46002.

I will refer you to the above text rather than writing out the lengthy details, and only highlight the important steps in in setting.

Step 1: In this case the weak formulation one obtains is $$ \int_a^b \frac1{n^2} v' \varphi' - k v \varphi \,\mathrm{d}x, $$ which is required to hold for all $\varphi \in C^1([a,b])$ such that $v(a) = v(b) = 0$ (and also for all $\varphi \in H^1_0((a,b))$ in the language of Sobolev spaces).

Here we've multiplied the equation by $\varphi$, integrated over $(a,b)$, and then integrated by parts to move the outer derivative on $\varphi$. Now this formulation makes sense even if $v$ is not twice differentiable.

Step 2: Now we can show this integral problem admits solutions in the Sobolev space $H^1((a,b))$. Here things are a little complicated, as $k^2 \geq 0$ means that the problem may admit non-unique solutions.

To see this, set $n \equiv 1$, $k \equiv 1$, $(a,b) = (0,\pi)$ and impose boundary conditions $v(0)=v(1) = 0$. Then $v = \lambda \sin x$ solves $$ v'' + v = 0. $$ Instead, one needs to apply the Fredholm alternative which gives conditions for when unique solvability holds, and when it fails. The point of this step however, is that it justifies this approach of weak formulations as it gives a better understanding on when we have existence and uniqueness.

Step 3: Finally we turn to regularity. This involves a bootstrapping argument, where we notice that since our solution $v$ lies a-priori in $H^1((a,b))$, by Sobolev embedding $v$ is continuous. Then $k^2v$ is bounded, so we infer from the equation that $$ \frac{\mathrm{d}}{\mathrm{d}x} \frac{v'}{n^2} \text{ is bounded } \implies \frac{v'}{n^2} \text{ is continuous}. $$ The above is more of a heuristic argument since we needs to establish this carefully using the weak formulation, but this is possible to do. Hence we arrive at the claimed regularity of $v$.

If $n$ is more regular one can further bootstrap and show that $v$ is $C^2$. If not however, we're stuck as if $v'/n^2$ is continuous but $n^2$ isn't, then $v'$ must also be discontinuous. This is a local property however, and for step functions one can show that $v$ will be $C^2$ away from the discontinuity points of $n$.