A Boundary conditions sufficient to ensure uniqueness of solution?

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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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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.
 
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.
 
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.
 
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.
 
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