# PDE involving curl

I need to solve the equation: $$\operatorname{curl} \left(\operatorname{curl}(\mathbf{u}(x,y,z)) \right)=\mathbf{v}(x,y,z)$$ Because $$\operatorname{curl}(\operatorname{curl}(\mathbf{f}))=-\nabla^2 \mathbf{f}+\nabla(\nabla \cdot \mathbf{f})$$ the previous equation is equivalent to: $$-\nabla^2 \mathbf{u}(x,y,z)+\nabla(\nabla \cdot \mathbf{u}(x,y,z))=\mathbf{v}(x,y,z)$$ Given $\mathbf{v}(x,y,z)$ how can I solve this PDE in $\mathbf{u}(x,y,z)$? Thanks in advance.

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## 2 Answers

In general you can't. Assuming your vector fields are to be defined on all of ${\mathbb R}^3$, the criterion for $\nabla \times {\bf w}) = {\bf v}$ to be solvable is $\nabla \cdot {\bf v} = 0$. If that is satisfied, $\bf w$ is called a vector potential for $\bf v$. Now if there is a vector potential for $\bf v$ there are lots of them, because you can add the gradient of any function. In this case, you want $\bf w$ to have a vector potential $\bf u$, so you want to use the "Coulomb gauge condition" $\nabla \cdot {\bf w} = 0$. Then let $\bf u$ be any vector potential for $\bf w$.

One way to find a vector potential $\bf w$ for $\bf v$ satisfying the Coulomb gauge is to start with a vector potential satisfying an easier gauge condition, e.g. $w_1 = 0$. This can be found by integrating: $v_2 = - \frac{\partial}{\partial x} w_3$ so $w_3 = - \int v_2 \ dx + g(y,z)$, $v_3 = \frac{\partial}{\partial x} w_2$ so $w_2 = \int v_3\ dx + h(y,z)$, and choose $g$ and $h$ so $v_1 = \frac{\partial w_3}{\partial y} - \frac{\partial w_2}{\partial z} = 0$. If the divergence of this $\bf w$ is $\psi(x,y,z)$, you now want to replace $\bf w$ by $\bf w + \nabla \phi$ where $\nabla^2 \phi = -\psi$: thus you solve a Poisson equation for $\phi$.

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Do you mean "If that is satisfied, $\mathbf w$ is called a vector potential for $\mathbf v$"? –  Rahul Mar 16 '12 at 18:11
Yes, thanks for the correction. –  Robert Israel Mar 16 '12 at 21:12

Split your second order equation into the system $${\rm curl\,}\mathbf{u}=\mathbf{w}, \qquad {\rm curl\,}\mathbf{w}=\mathbf{v}.$$ Since your $\mathbf{v}$ must have divergence 0, you can use Biot-Savart to find a value of $$\mathbf{w}(\mathbf{r}) =\int_{R^3}\mathbf{v}(\mathbf{r'})\times\frac{\mathbf{r-r'}}{4\pi|\mathbf{r-r'}|^3}d\mathbf{r'}$$ which also has divergence 0, assuming the integral converges. This solves the second equation. You might repeat that for $\mathbf{u}$ if the integrals converge, but since you don't need $\mathbf{u}$ to have divergence 0, it suffices to use a version of the Poincare lemma which says that $$\mathbf{w}(\mathbf{r})={\rm curl\,}\left( \int_0^1\mathbf{w}(t\mathbf{r})\times t\mathbf{r}\,dt \right) +\int_0^1 t^2\mathbf{r}\,{\rm div\,}\mathbf{w}(t\mathbf{r})\,dt$$ where I'm writing $t\mathbf{r}=(tx,ty,tz)$. Since our $\mathbf{w}$ has divergence 0, you can set $\mathbf{u}(\mathbf{r}) =\int_0^1\mathbf{w}(t\mathbf{r})\times t\mathbf{r}\,dt$ which gives the first equation $\mathbf{w}={\rm curl\,}\mathbf{u}.$ I think that does it if I didn't mess up.

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You have to use \mathbf{...} or {\bf ...} instead of \bf{...} unless you want everything after that point to also become bold. I fixed it in your post. –  Rahul Mar 16 '12 at 18:17