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Let $F(x,y)\equiv (F_1(x,y), F_2(x,y))$ be a vector field on $X\times Y$, where $X, Y\subset \mathbb R$.

Let $\operatorname{div}$ denote the divergence operator, i.e., $$\operatorname{div} F = \frac{\partial F_1(x,y)}{\partial x}+ \frac{\partial F_2(x,y)}{\partial y}.$$

Suppose $F$ satisfies the following PDE

\begin{cases} \operatorname{div} F(x,y) = g(x,y), & (x,y)\in X^{\circ}\times Y^{\circ},\\ F(x,y) = m(x,y), & (x,y)\in \partial (X\times Y), \end{cases}

where $g(x,y)$ and $m(x,y)$ are known functions, and $A^{\circ}$ denotes interior of a set $A$ and $\partial A$ denotes the boundary.

What are the different ways to (numerically approximate) solve $F(x,y)$? (When) is the solution unique?

user3428708
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  • "best" is pretty objective when it comes to numerical schemes. It completely depends on the geometry of your domain as to which might be the most "efficient method", whatever that means. – DaveNine May 11 '18 at 00:32
  • thank you DaveNine. I have slightly modified the question. I am new to this field so I am struggling. thank you! – user3428708 May 11 '18 at 00:56
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    Are you sure your second condition is on $\partial X \times \partial Y$? If $X, Y$ are intevals, this set has only four points. –  May 11 '18 at 01:30
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    @JohnMa: I think $\partial (X \times Y) = \partial X \times Y \cup X \times \partial Y$. user3428708 got it wrong, I'll warrant. – Robert Lewis May 11 '18 at 01:35
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    You don't have enough equations to solve for $F$. – Matthew Cassell May 11 '18 at 01:36

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