If $u(r, \theta)$ is a solution of Laplace’s equation show that $u(\frac{1}{r}, \theta)$ is also a solution.

Question

Suppose that $u(r, θ)$ is a solution of Laplace’s equation. Show that $u(\frac{1}{r}, θ)$ is also a solution.

So far, I know that if $u$ satisfies Laplace's equation, then $$\Delta u = u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta \theta}$$ since $u$ is given in polar coordinates. I let $w = u(\frac{1}{r}, \theta)$. I want $w$ to also satisfy Laplace's Equation. Thus $\Delta w = w_{rr} + \frac{1}{r}w_r + \frac{1}{r^2}w_{\theta \theta}$

I computed the partial derivatives of $w$ and plugged them into the above equation but can't seem to get the right result.

Thanks for the help!

Answer 1

Consider a function $v(r, \theta) = u\left(q, \theta\right),$ with $q = \frac{1}{r}.$

Then, using the chain rule:

1. $v_{r} = q_r u_q = -\displaystyle\frac{1}{r^2} u_q$
2. $v_{rr} = \displaystyle\frac{2}{r^3} u_q + \frac{1}{r^4} u_{qq}$
3. $v_{\theta\theta} = u_{\theta\theta}$

Then:

$$v_{rr} + \frac{1}{r}v_r + \frac{1}{r^2}v_{\theta \theta} = \frac{2}{r^3} u_q + \frac{1}{r^4} u_{qq} -\displaystyle\frac{1}{r^3} u_q + \frac{1}{r^2}u_{\theta\theta} = \frac{1}{r^4}\left(u_{qq} + ru_q + r^2 u_{\theta\theta}\right).$$

Therefore, since $r = \frac{1}{q}$, then:

$$v_{rr} + \frac{1}{r}v_r + \frac{1}{r^2}v_{\theta \theta} = q^4\left(u_{qq} + \frac{1}{q}u_q + \frac{1}{q^2} u_{\theta\theta}\right) .$$

Answer 2

Hint

Prove that the general solution of the above equation is as follows$$u(r,\theta)=c_0+c_1\ln r+\sum_{n=1}^{\infty}(a_nr^n+b_nr^{-n})(d_n\cos n\theta+e_n\sin n\theta)$$by using the Separation of Variables method.

Answer 3

Let's try distinguishing between $r$ as the parameter to $u$ and the variable. So let's define $w(r', \theta) = u(\frac{1}{r'}, \theta)$. Now observe:

$$w_{r'} = u_r \frac{dr}{dr'} = u_r \frac{-1}{r'^2}$$ $$w_{r'r'} = \frac{d}{dr'} \left (u_r \right ) \frac{-1}{r'^2} + u_r \frac{d}{dr'} \left (\frac{-1}{r'^2} \right ) = u_{rr} \frac{1}{r'^4} + u_r \frac{2}{r'^3}$$ $$w_{\theta \theta} = u_{\theta \theta}$$

So now we see:

$$w_{r'r'} + \frac{1}{r'} w_{r'} + \frac{1}{{r'}^2} w_{\theta \theta} = u_{rr} \frac{1}{r'^4} + u_r \frac{2}{r'^3} + \frac{1}{r'} u_r \frac{-1}{r'^2} + \frac{1}{r'^2}u_{\theta \theta}$$ and using the fact that $r = \frac{1}{r'}$ $$= r^4 u_{rr} + r^3 u_r + r^2 u_{\theta \theta} = r^4 (u_{rr} + \frac{1}{r} u_r + \frac{1}{r^2} u_{\theta \theta}) = r^4 ( 0) = 0$$

That is, $w$ satisfies the relation:
$$w_{r'r'} + \frac{1}{r'} w_{r'} + \frac{1}{{r'}^2} w_{\theta \theta} = 0$$

I'll admit it became difficult to maintain the change of variables in the problem, so don't beat yourself up.