Divergence structure equation

Question

Consider Laplace's equation with potential function $c$: $$-\Delta u + cu = 0, \tag{$*$}$$ and the divergence structure equation $$-\operatorname{div}(aDv)=0, \tag{$**$}$$ where the function $a$ is positive.

(a) Show that if $u$ solves $(*)$ and $w > 0$ also solves $(*)$, then $v:=u/w$ solves $(**)$ for $a:=w^2$.

(b) Conversely, show that if $v$ solves $(**)$, then $u:=va^{1/2}$ solves $(*)$ for some potential $c$.

I only know so far that as $u$ and $w$ solve $(*)$, we have $$-\Delta u + cu=0 \text{ and } -\Delta w + cw = 0.$$ How may I prove the divergence structure equation? I am confused with differentiating with the $\text{div}$ operator.

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My work so far for part (b):

If $v$ solves $(**)$, then $-\operatorname{div}(aDv)=0$. So

\begin{align*} D \cdot(a Dv)=a^{1/2}\Delta v - v \Delta (a^{1/2})=0. \end{align*}

Also, given $u := va^{1/2}$, I find $$\Delta u = \frac 12a^{-1/2} DaDv + a^{1/2} \Delta v + \frac 12 D(va^{-1/2})+\frac 12va^{-1/2}\Delta a.$$

How can I apply $-\operatorname{div}(aDv)=0$ to the expression $-\Delta u+cu$ and prove that it is equal to $0$?

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More work so far in part (b):

\begin{align} \require{cancel} \Delta u &= \nabla \cdot \nabla(a^{1/2}v) \\ &= \nabla \cdot \left[\frac{v\nabla a+2a\nabla v}{2a^{1/2}} \right] \\ &= \frac{(\nabla \cdot (v\nabla a+2a\nabla v))(2a^{1/2})- (v\nabla a+2a\nabla v)(\nabla\cdot(2a^{1/2}))}{4a} \\ &=\frac{(\nabla \cdot (v\nabla a)+2\cancelto{0}{\nabla \cdot (a\nabla v)})(2a^{1/2})- (v\nabla a+2a\nabla v)(\nabla\cdot(2a^{1/2}))}{4a} \\ &= \frac{(\nabla v \cdot \nabla a+v \cdot \Delta a)(2a^{1/2})- (v\nabla a+2a\nabla v)(2(\nabla\cdot a^{1/2}))}{4a} \end{align}

Answer 1

For part b) we take $u=\sqrt{a}v$ and calculate

$$\Delta u = \nabla\cdot(\nabla(\sqrt{a}v)) = \nabla \cdot \left(\frac{v\nabla a + 2 a\nabla v}{2\sqrt{a}}\right) = \\ \frac{1}{2\sqrt{a}}\nabla \cdot \left(v\nabla a \right) + \frac{1}{2\sqrt{a}}\nabla \cdot \left(2a\nabla v \right) + (v\nabla a + 2 a\nabla v)\cdot \nabla\left(\frac{1}{2\sqrt{a}}\right)$$

The first term on the bottom line above is

$$\frac{1}{2\sqrt{a}}\nabla \cdot \left(v\nabla a \right) = \frac{1}{2\sqrt{a}}(\nabla v \cdot \nabla a + v\Delta a)$$

The second term is zero since $v$ satisfy $\nabla \cdot \left(v\nabla a \right) = 0$ and finally the third term is

$$(v\nabla a + 2 a\nabla v)\cdot \nabla\frac{1}{2\sqrt{a}} = -\frac{1}{4a\sqrt{a}}(v\nabla a + 2 a\nabla v)\cdot \nabla a = -\frac{1}{4a\sqrt{a}}(v(\nabla a)^2 + 2 a\nabla v\cdot \nabla a)$$

Adding the three terms above we get

$$\Delta u = v\left(\frac{1}{2\sqrt{a}}(\Delta a) -\frac{1}{4a\sqrt{a}}((\nabla a)^2)\right) = u\left(\frac{2a\Delta a-(\nabla a)^2}{4a^2}\right)$$

which shows that $u = \sqrt{a}v$ satisfy the equation $\Delta u = cu$ where

$$c =\frac{2a\Delta a-(\nabla a)^2}{4a^2} $$

Answer 2

We have \begin{align} w^2 Dv &= w^2 D(u/w) \\ &= w^2 \left(\frac{w Du - u Dw}{w^2} \right) \\ &=wDu-uDw. \end{align} So \begin{align} D \cdot (w^2 Dv) &= D \cdot (w Du-uDw) \\ &= w \Delta u-u \Delta w \\ &= w(cu)-u(cw) \\ &=0. \end{align} Thus, $-\operatorname{div}(w^2 Dv)=0$.