Show that $u_x(x, y) = f′(r) \cos(\theta)$ and hence deduce that $f′(0) = 0$, which implies the Neumann boundary condition $u_r = 0$ when $r = 0$.

Question

Suppose that $u(x,y)$ is a continuously differentiable, circularly symmetric function, so that when expressed in polar coordinates, $x = r \cos(\theta)$, , $y = r \sin(\theta)$, it depends solely on the radius $r$; that is $u = f(r)$. Show that $u_x(x, y) = f′(r) \cos(\theta)$ and hence deduce that $f′(0) = 0$, which implies the Neumann boundary condition $u_r = 0$ when $r = 0$.

I asked a question related to this problem a while ago, but I never finished the problem itself.

I understood enough to know that I needed to use change of variables:

$$\begin{align} \frac{\partial u}{\partial r} &= \frac{\partial u}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial y} = \cos\theta\frac{\partial u}{\partial x} + \sin\theta \frac{\partial u}{\partial y} \\ \frac{\partial u}{\partial \theta} &= \frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta} = -r\sin\theta\frac{\partial u}{\partial x} + r\cos\theta\frac{\partial u}{\partial y} \end{align}$$

But from here I'm unsure of how to proceed.

Would anyone mind please explaining this problem?

Answer 1

As you say $u(x,y) = f(r)$ since the function $u$ is circularly symmetric and then it depends only on the radius $r$. Now we are going to compute $u_x(x,y)$:

$$\frac{\partial u}{\partial x}(x,y) = \frac{\partial f(r)}{\partial x} =\frac{d f}{d r}\frac{\partial r}{\partial x}$$

by the chain rule, since $\frac{d f}{d r}=f'(r)$ and $r^2=x^2+y^2$, a simple computation gives

$$\frac{d f}{d r}\frac{\partial r}{\partial x}=f'(r) \frac{\partial r}{\partial x} = f'(r) \frac{2x}{2\sqrt{x^2+y^2}}= f'(r) \frac{x}{r}=f'(r)\cos(\theta)$$

because $x=r\cos\theta$; hence $$\color{red}{\frac{\partial u}{\partial x}(x,y) = f'(r)\cos(\theta)}$$

Now: $$f'(0)=\frac{df}{dr}(0)=\lim_{h\to 0} \frac{f(h)-f(0)}{h}=\lim_{h\to 0}\frac{u(h_1,h_2)-u(0,0)}{h}$$ where $h=\sqrt{h_1^2+h_2^2}$.

This limit does not depend on the direction $(h_1,h_2)$, that is it does not depend on the value of $\theta$, since the function is circularity symmetric, and this implies that $f'(0)=0$. Indeed, you may notice that $$f'(0)=\lim_{h\to 0^+}\frac{u(h,0)-u(0,0)}{h}=\lim_{h\to 0^-}\frac{u(h,0)-u(0,0)}{h}=-f'(0)$$ because when $h\to 0^-$, then $(h,0)$ approaches $(0,0)$ from the negative half-line of the $x-$axis, so $\theta=\pi$ and then $\cos\theta=-1$.

Answer 2

Suppose $u(x,y) = f(r)$. Then

$$u_x(x,y) = \frac{d f(r)}{d x} = f'(r) \frac{\partial r}{\partial x} = f'(r) \frac{x}{r}= f'(r)\cos(\theta)$$

Now since $u(x,y)$ is circularly symmetric, we have $u_x(x,y)=-u_x(-x,y)$. Then $u_x(0,y)=-u_x(0,y)=0$. From the above, we then find

$$f'(0)\cos(\theta) = u_x(0,0) = 0.$$

Since the function is invariant under a rotation of the axes, we must have $f'(0)=0$.

Answer 3

Here's a different approach to the end result.

Lemma: Suppose $g$ is differentiable and even on some symmetric interval $(-a,a).$ Then $g'(0)=0.$

Proof: Set $h(x)=g(-x).$ Then $h'(x) = -g'(-x).$ Since $g$ is even, we also have $h(x) = g(x).$ It follows that $g'(x)= -g'(-x).$ Thus $g'(0)= -g'(0),$ which implies $g'(0)=0.$

Now in your problem, $u(x,0)$ is even, since $u(-x,0) = f(|x|) = u(x,0).$ Thus $u_x (0,0) = 0.$ Now for $x>0,$

$$\frac{f(x)-f(0)}{x} = \frac{u(x,0)-u(0,0)}{x} \to u_x(0,0=0) =0$$

as $x\to 0^+$ Thus $f'(0)=0$ from the right, which is the desired result.