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My question goes along a similar line as this other one that was left unanswered because the OP did not have enough equations to produce a solution.

I want to solve for the vector field $\mathbf f(\mathbf r)$, defined over $V\subset\mathbb R^3$, that satisfies

$$\boldsymbol\nabla\cdot \mathbf f-\mathbf g(\mathbf r)\cdot\mathbf f=h(\mathbf r), \text{ and}$$ $$\boldsymbol\nabla\times \mathbf f=\mathbf 0.$$

Also, we assume that $\mathbf f|_{\partial V}$ is known.

How would one go about solving it? I saw this webpage that showed how to solve a similar kind of equations but with scalar field, and I am at a loss on how one would generalize that method to a vector field.

I've thought that one could possibly exploit the fact $\mathbf f$ is conservative to write the equations instead as

$$\nabla^2\Phi-\mathbf g\cdot\boldsymbol\nabla\Phi=h$$

but I don't see how that would help.

Chaotic
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  • If the solving method is too involved, but you do know of a reference where they show how it's solved, then I'd be very grateful if you could tell me where to find it. – Chaotic May 02 '22 at 04:48
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    Solve $$\nabla \cdot \vec{f} - \vec{g} \cdot \vec{f} = h$$ using the method of characteristics. Then noting that $\nabla \times \vec{f} = 0 \implies \vec{f} = \nabla \phi$ if your domain is simply connected, you can simply integrate componentwise to determine $\phi$. – Matthew Cassell May 02 '22 at 05:25
  • Thank you for commenting, but how would you apply the method of characteristic when there are 3 unknowns? The equations in the question can be rewritten in similar style as the page you link: $$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=c(x, y, z, u, v, w)$$

    and $$\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}, \frac{\partial v}{\partial z}=\frac{\partial w}{\partial y}, \frac{\partial u}{\partial z}=\frac{\partial w}{\partial x}$$

    The page only shows how to solve when there is 1 unknown $u$, but here we have 3 coupled ones.

     – Chaotic May 02 '22 at 18:39
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    Sorry, I made a mistake, I forgot $f$ was vector valued in the first place but then proceeded to use the fact it was vector valued. This is why you shouldn't do maths late at night. The best approach would to use what you did; set $f = \nabla \phi$ and solve $$\nabla^{2} \phi - \vec{g} \cdot \nabla \phi = h$$ which is a time-independent convection-diffusion equation, a type of Poisson equation. – Matthew Cassell May 03 '22 at 01:22
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    Ohh I see, that makes sense. Thank you for taking the time to update the answer. I also did not know it had a proper name, that will also help me greatly in finding solution methods and understanding it in general. – Chaotic May 03 '22 at 01:43

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