# PDE Existence/Uniqueness Question

1. May 1, 2014

### thegreenlaser

My equation is:

$$\left(\mathbf{\nabla}\sigma\right)\cdot\left(\mathbf{\nabla}V\right) + \sigma\nabla^2V = 0$$

If I'm given V(r) on the boundary of some volume, and I know σ(r) inside the volume, is there a unique solution V(r) inside that volume for any arbitrary (well-behaved) function σ(r)?

I suspect the answer is yes, but I've never taken a formal PDE class, so I wanted to double-check.

Edit: just so it's clear, I don't need to know how to solve for V, I just need to know that it's possible to find V in principle.

2. May 2, 2014

### pasmith

You have $$\left(\mathbf{\nabla}\sigma\right)\cdot\left(\mathbf{\nabla}V\right) + \sigma\nabla^2V = \nabla \cdot ( \sigma \nabla V) = 0$$ which if $\sigma$ is given is a linear PDE for $V$. Hence if $V_1$ and $V_2$ satisfy this PDE in a volume $U$ with $V_1 = V_2$ on $\partial U$, then $V = V_1 - V_2$ satisfies the PDE with $V = 0$ on $\partial U$.

Hence $$0 = \int_U V \nabla \cdot (\sigma \nabla V)\,dU = \int_U \nabla \cdot( V \sigma \nabla V) - \sigma (\nabla V) \cdot (\nabla V)\,dU \\ = \int_{\partial U} V \sigma \frac{\partial V}{\partial n}\,dS - \int_U \sigma \|\nabla V\|^2\,dU$$ and as $V$ vanishes on $\partial U$ we have $$\int_U \sigma \|\nabla V\|^2\,dU = 0.$$ Hence if $\sigma$ is everywhere strictly positive (or strictly negative) we can conclude that $\|\nabla V\|^2 = 0$ everywhere in $U$, from which it follows that $V$ is constant on $U$, and in view of the boundary conditions $V = 0$. Hence $V_1 = V_2$.

It follows that if $\sigma$ is everywhere strictly positive or everywhere strictly negative and a solution exists, then that solution is unique.

3. May 2, 2014

### thegreenlaser

Perfect, thanks!

I forgot, I do actually know that $\sigma \geq 0$, and I could probably restrict it to $\sigma > 0$.