Boundary conditions sufficient to ensure uniqueness of solution?

  • #1
fluidistic
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I have 2 coupled PDEs:

##\nabla \cdot \vec J=0## and another one involving ##T## and partial derivatives of ##T## as well as ##\vec J##.
Where the vector field ##\vec J=-\sigma \nabla V -\sigma S \nabla T##, ##\sigma## and ##S## are tensors (2x2 matrices). ##V## and ##T## are 2D scalar fields.

The region where these PDEs hold is a square. There are Dirichlet boundary conditions for ##T## on 2 sides, and vanishing Neumann boundary conditions on the remaining 2 sides for ##T##.
Then, and here is the unusual thing, instead of directly imposing boundary conditions on ##V## (which would have effectively determined ##\vec J## uniquely), the boundary conditions are applied on ##\vec J## directly. And they are strange. On 2 sides, ##\vec J## must have vanishing normal component, meaning vanishing Neumann boundary conditions. But on the other 2 sides the requirement is that the net current entering/leaving must be equal to a particular value, ##I##.

Mathematically, ##\int _{\Gamma_1} \vec J \cdot d\vec l=I## and ##\int _{\Gamma_2} \vec J \cdot d\vec l=-I## for those two sides, which doesn't look like neither Dirichlet nor Neumann b.c.s to me, but a sort of line-integrated Neumann b.c.s.

My question is... is this enough to ensure a single, unique ##\vec J##? Or can there be two different ##\vec J## vector fields satisfying all of those conditions?

I suppose my question can be recast to whether the above conditions fully determine the scalar field ##V##.

You can assume ##T## to be uniquely determined (and possibly ignore or neglect the fact that it depends on ##\vec J##, so that the 2 coupled PDEs can be thought of as decoupled, as a first approximation).
 
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  • #2
Alright, I have thought more about this problem, and I think we can focus on a simpler one, my question still stands.

Say we are solving ##\nabla \cdot \vec J=0## where the current density is the usual ##\vec J=-\sigma \nabla V##.
This is Laplace equation ##\nabla ^2 V(x,y)=0##.

With the Neumann boundary conditions ##\frac{\partial V}{\partial y}\big |_{y=0, y=L}=0##.

And with the "strange" boundary conditions ##\int _{0}^{L} \frac{\partial V}{\partial x}\big |_{x=0} dy=I## and ##\int _{0}^{L} \frac{\partial V}{\partial x}\big |_{x=L} dy=-I##.

The question is whether these boundary conditions ensure a unique ##V(x,y)##, therefore a unique ##\vec J##.

I believe the answer is no, because I can think of 2 electrostatics potential functions (one being a non zero constant, the other being a straight line) whose boundary integrals will yield ##I## as it should, while having everything else satisfied. I am not 100% sure yet.
 
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