Given a vector field $\mathbf{H}$, find a vector field $\mathbf{F}$ and a scalar field g, such that $\mathbf{H}$ = curl(F) + ∇(g).

Question

Let$\;\mathbf{H}(x,y,z) = x^2y\mathbf{i}+y^2z\mathbf{j}+z^2x\mathbf{k}$. Find a vector field $\mathbf{F}$ and a scalar field g, such that $\mathbf{H}$ = curl(F) + ∇(g).

I took divergence on both sides which gave me $2xy+2yz+2xz=∇^2g$. I took curl on both sides which gave me three horrible equations one of which I have written down below:

$$\frac{\partial^2f_1}{\partial y^2}-\frac{\partial^2f_2}{\partial x\partial y}-\frac{\partial^2f_3}{\partial x\partial z}+\frac{\partial^2f_1}{\partial z\partial y}=-y^2$$ where I have taken $$\mathbf{F}=f_1\mathbf{i}+f_2\mathbf{j}+f_3\mathbf{k}$$

I tried to assign some values hoping to work out the rest but this gave me nothing. Please help.

Answer 1

I don't know if there is a 'canonical' way to approach this problem. Here's my thinking.

---

The critical observation is that if there exists $F$ and $g$ such that $H = \nabla \times F + \nabla g$, then $$\nabla\cdot H = \nabla\cdot\nabla F + \nabla^2 g = \nabla^2g$$ as div of a curl of a vector field $F$ is zero.

Now $\nabla \cdot H = 2(xy + yz + zx)$. As $2(xy + yz + xz) = \nabla^2 g = g_{xx} + g_{yy} + g_{zz}$ it's not to hard to find one such $g$: $$g(x,y,z) = x^2yz + y^2xz + z^2xy$$ In which case $\nabla g = (2xyz + y^2z + z^2y, 2xyz + x^2z + z^2x, 2xyz + x^2y + y^2z)$.

If $H = \nabla \times F + \nabla g$ for some $F$, then

$$\begin{align*} \nabla \times F & = H - \nabla g \\ & = (x^2y - y^2z - z^2y - 2xyz, \\ & \ \ \ \ \ \ \ y^2z - x^2z - z^2x - 2xyz, \\ & \ \ \ \ \ \ \ z^2x - x^2y - y^2x - 2xyz)\end{align*}$$

...which if you stare at long enough and play around, you can convince yourself there is no field $F$ which gives this result.

So better, we should find an $g$ where terms from $\nabla g$ cancels out terms from $H$ instead of adding to them. For example,

$$g(x,y,z) = \frac{1}{3} \left( x^3y + y^3z + z^3x \right)$$

We still have $\nabla^2 g = 2(xy + yz + xz)$. But also

$$\nabla g = \frac{1}{3}(3x^2y + z^3, 3y^2z + z^3,3z^2x + y^3 )$$

Thus as desired, $\nabla \times F$ is the relatively simple $$\nabla \times F = H - \nabla g = -\frac{1}{3}(z^3, x^3, y^3) $$

Such an $F$ is given by $F(x,y,z) = -\frac{1}{3}(x^3z, y^3x, z^3y)$.

Whew.

Answer 2

Below a slightly different approach which first gives $\mathbf{F}$ and then $G$.

The problem at hand is quite similar to that of finding the vector and scalar potentials in terms of the electric and magnetic fields in electrodynamics. In that case there are standard solutions. But here, since $% \mathbf{H}$ is not decaying for large $\mathbf{x}$, they do not apply. But it seems that a suitable $\mathbf{F}$ can easily be found. Thus we have$$ \mathbf{H}=\left( \begin{array}{c} x_{1}^{2}x_{2} \\ x_{2}^{2}x_{3} \\ x_{3}^{2}x_{1} \end{array} \right) =\partial _{\mathbf{x}}\times \mathbf{F}+\partial _{\mathbf{x}}G.. $$ We can assume $ \partial _{\mathbf{x}}\cdot \mathbf{F}=0. $ Now

\begin{eqnarray*} \partial _{\mathbf{x}}\times \mathbf{H} &=&\left( \begin{array}{c} -x_{2}^{2} \\ -x_{3}^{2} \\ -x_{1}^{2} \end{array} \right) =\partial _{\mathbf{x}}\times (\partial _{\mathbf{x}}\times \mathbf{F% })=\mathbf{-\partial }_{\mathbf{x}}^{2}\mathbf{F}, \\ \partial _{\mathbf{x}}\cdot \mathbf{H} &\mathbf{=}% &2(x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1})=\mathbf{-\partial }_{\mathbf{x}}^{2}G. \end{eqnarray*} Thus, $\mathbf{-\partial }_{\mathbf{x}}^{2}F_{1}=-x_{2}^{2}$ and assuming $ F_{1}=F_{1}(x_{2})$ we find $F_{1}(x_{2})=a+bx_{2}+cx_{2}^{2}+dx_{2}^{3}+ \frac{1}{12}x_{2}^{4}$. Setting the coefficients $a,b,c,d$ equal to $0$, we obtain $$ \mathbf{F}(\mathbf{x})=\frac{1}{12}\left( \begin{array}{c} x_{2}^{4} \\ x_{3}^{4} \\ x_{1}^{4} \end{array} \right) , $$ which is divergence-free. Now $$ \partial _{\mathbf{x}}\times \mathbf{F}=\left( \begin{array}{c} -\partial _{3}F_{2} \\ -\partial _{1}F_{3} \\ -\partial _{2}F_{1} \end{array} \right) =-\frac{1}{3}\left( \begin{array}{c} x_{3}^{3} \\ x_{1}^{3} \\ x_{2}^{3} \end{array} \right) , $$ and $$ \mathbf{H}-\partial _{\mathbf{x}}\times \mathbf{F}=\left( \begin{array}{c} x_{1}^{2}x_{2} \\ x_{2}^{2}x_{3} \\ x_{3}^{2}x_{1} \end{array} \right) +\frac{1}{3}\left( \begin{array}{c} x_{3}^{3} \\ x_{1}^{3} \\ x_{2}^{3} \end{array} \right) =\left( \begin{array}{c} x_{1}^{2}x_{2}+\frac{1}{3}x_{3}^{3} \\ x_{2}^{2}x_{3}+\frac{1}{3}x_{1}^{3} \\ x_{3}^{2}x_{1}+\frac{1}{3}x_{2}^{3} \end{array} \right) =\partial _{\mathbf{x}}G. $$ \begin{eqnarray*} \partial _{1}G &=&x_{1}^{2}x_{2}+\frac{1}{3}x_{3}^{3}\Longrightarrow G=\frac{ 1}{3}x_{1}^{3}x_{2}+\frac{1}{3}x_{3}^{3}x_{1}+M \\ \partial _{2}G &=&x_{2}^{2}x_{3}+\frac{1}{3}x_{1}^{3}=\frac{1}{3} x_{1}^{3}+\partial _{2}M \\ \partial _{2}M &=&x_{2}^{2}x_{3}\Rightarrow M=\frac{1}{3}x_{2}^{3}x_{3}+N \end{eqnarray*} Setting $N=0$, $$ G=\frac{1}{3}(x_{1}^{3}x_{2}+x_{3}^{3}x_{1}+x_{2}^{3}x_{3})$$