A Boundary conditions sufficient to ensure uniqueness of solution?

[fluidistic]

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).

 

[fluidistic]

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.

 

[pasmith]

The problem is linear, so the difference between two solutions must satisfy homogenous boundary conditions (ie. with I = 0). Can we conclude that \nabla V must vanish identically subject to these?

By the divergence theorem (applied with 0 \leq z \leq 1 and \frac{\partial}{\partial z} \equiv 0), \begin{split}<br /> \int_A \|\nabla V\|^2\,dA &amp;= \int_A \nabla\cdot(V\nabla V) - \nabla^2 V\,dA \\<br /> &amp;= \oint_{\partial A} V\frac{\partial V}{\partial n}\,ds \\<br /> &amp;= \int_0^{L_y} \left[V \frac{\partial V}{\partial x}\right]_{x=0}^{x=L_x} \,dy <br /> + \underbrace{\int_0^{L_x} \left[ V \frac{\partial V}{\partial y}\right]_{y=0}^{y=L_y}\,dx}_{=\,0}<br /> \end{split} and I don't see how the given conditions at x = 0 and x = L<br /> _x force this to be zero.

 

[pasmith]

Looking for separable solutions with X&#039;&#039;/X = - Y&#039;&#039;/Y = C, I find that C = 0 gives only V = \mbox{constant} as a solution, and for C \neq 0 we must have V(x,y) = f(x)\cos(n\pi y/L_y) for integer n \geq 1 so that f&#039;&#039; = \frac{n^2 \pi^2}{L_y^2} f. It follows that <br /> \int_0^{L_y} \frac{\partial V}{\partial x}\,dy = f&#039;(x)\left[ \frac{L_y}{n\pi} \sin \left(\frac{n\pi y}{L_y}\right)\right]_0^{L_y} = 0 for any choice of f.

 

[fluidistic]

Thank you very much pasmith, this confirm my fears. The solution is therefore not only not unique, but there are infinitely many (as many as there are fs, apparently).
Meanwhile I could make progress on my physical problem, and indeed, the boundary conditions aren't as "open" as the ones I thought would hold. They are different, leading to a unique solution. Very nice. I had never seen this question asked before.