Show that $\int^1_0\left(r\left(\frac{dg}{dr}\right)^2+\frac{g^2}r\right) \operatorname{dr}\ge1$ for any differentiable function $g(r)$.

Question

This is a continuation from a question asked earlier that sadly I couldn't go on to do the second bit, so I must ask once more: Asked earlier that I now have satisfactory answers to is:

Let $S$ be a bounded region of $\mathbb R^2$ and $\partial S$ be its boundary. Let $u$ be the unique solution to Laplace’s equation in $S$, subject to the boundary condition $u=f$ on $\partial S$, where $f$ is a specified function. Let $w$ be any smooth function with $w=f$ on $\partial S$. By writing $w=u+\delta$ show that $$\int_S|\nabla w|^2 \operatorname{dA}\ge\int_S|\nabla u|^2 \operatorname{dA} \space\space(\star)$$

I would now like to ask the second bit (to reiterate, I am not asking for an answer the above bit but for the below bit):

Let $S$ be the unit disc in $\mathbb R^2$. By considering functions of the form $g(r)\cos\theta$ on both sides of $(\star)$, where $r$ and $\theta$ are polar coordinates, deduce that $$\int^1_0\left(r\left(\frac{dg}{dr}\right)^2+\frac{g^2}r\right) \operatorname{dr}\ge1$$ for any differentiable function $g(r)$ satisfying $g(1)=1$ and for which the integral converges at $r=0$.

I assume that when it says "consider functions of the form" it is talking about for $u$ and $w$. The first part $$\int_S|\nabla w|^2 \operatorname{dA}\ge\int_S|\nabla u|^2 \operatorname{dA}$$ I think gives $$\int^1_0|\nabla w|^2r\operatorname{dr}\ge\int^1_0|\nabla u|^2r\operatorname{dr}$$

Now using $\nabla^2u=0$ on $S$ with $u=g(r)\cos\theta$ I think I have to solve $$\frac1r\cos\theta\left(\frac{dg}{dr}+r\frac{d^2g}{dr^2}\right)-\frac1{r^2}\cos\theta=0$$ and I have that this gives $g(r)=\frac12(\log r)^2+k\log r+c$ and $g(1)=1$ gives $c=1$.

I don't know if any of this is right or useful.

Assuming that any of this is correct I don't see how to find $c$ or what I would substitute in for $w$.

Thanks for any help and sorry if this was painfully incorrect.

Answer 1

First of all, you made a mistake computing $\nabla^2u$. It should be

$$\frac{1}{r}\left(\frac{d}{dr}r\frac{dg}{dr}\right)-\frac{g}{r^2} = 0$$

which has a general solution $ar + \frac{b}{r}$. The nice solution satisfying boundary condition is $g = r$, and $u$ becomes $r \cos \theta$.

Compute $\nabla u = (\frac{\partial u}{\partial r}, \frac{1}{r}\frac{\partial u}{\partial \theta}) = (\cos \theta, -\sin\theta)$. Obviously, $|\nabla u|^2 = 1$, and $\int_S|\nabla u|^2\operatorname{dA} = \pi$.

Now take an arbitrary function $w = g(r)\cos\theta$.

Compute $\nabla w = (\frac{d g}{dr}\cos\theta,-\frac{g}{r}\sin\theta)$, and $|\nabla w|^2 = (\frac{d g}{dr})^2 \cos^2\theta + (\frac{g}{r})^2\sin^2\theta$. Observe that $\int_S |\nabla w|^2 \operatorname{dA} = \int_S (\frac{d g}{dr})^2 \cos^2\theta + (\frac{g}{r})^2\sin^2\theta\ r \operatorname{dr} \operatorname{d\theta} = \pi\int_0^1 (r(\frac{d g}{dr})^2 + \frac{g^2}{r}) \operatorname{dr}$.

Conclude.