PDE Existence/Uniqueness Question

[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.

 

[pasmith]

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.

You have <br /> \left(\mathbf{\nabla}\sigma\right)\cdot\left(\mathbf{\nabla}V\right) + \sigma\nabla^2V<br /> = \nabla \cdot ( \sigma \nabla V) = 0<br /> 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 <br /> 0 = \int_U V \nabla \cdot (\sigma \nabla V)\,dU = <br /> \int_U \nabla \cdot( V \sigma \nabla V) - \sigma (\nabla V) \cdot (\nabla V)\,dU \\<br /> = \int_{\partial U} V \sigma \frac{\partial V}{\partial n}\,dS - \int_U \sigma \|\nabla V\|^2\,dU<br /> and as V vanishes on \partial U we have <br /> \int_U \sigma \|\nabla V\|^2\,dU = 0.<br /> 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.

 

[thegreenlaser]

Perfect, thanks!

I forgot, I do actually know that \sigma \geq 0, and I could probably restrict it to \sigma &gt; 0.