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Let $H(x) = x^2y\,\mathbf{i} + y^2z\,\mathbf{j} + z^2x\,\mathbf{k}\,$. Find an irrotational function $F(x)$ and a solenoidal function $G(x)$ such that $H = F + G$.

Sol. As $F(x)$ is irrotational function then $\operatorname{curl} F(x)=0$ and $G(x)$ is solenoidal then $\operatorname{div} G(x)=0\,$

How to use the above two relations to find $F$ and $G\,?$

We have $$\operatorname{div}\operatorname{curl} \,H(x)=0\,.$$

Using it, $0=\operatorname{div}\operatorname{curl}H = \operatorname{div}\operatorname{curl}F + \operatorname{div}\operatorname{curl}G$, but we get an identity ($0=0+0$ type).

Kurt G.
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user1942348
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  • Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – 8bc3 457f Jun 03 '23 at 16:03
  • Not grad, but div (or rot). What have you tried? – Ted Shifrin Jun 03 '23 at 17:53
  • @TedShifrin I am confused, how to start? I started with $div ,curl ,H(x)=0$, but it does not work. – user1942348 Jun 04 '23 at 11:32

1 Answers1

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The example is interesting because $H$ is not bounded on $\mathbb R^3$ and the usual integral formulas from the Helmholtz decomposition cannot be applied.

A solenoidal field $G$ is not one that has grad$\,G=0$ as you wrote but div$\,G=0\,.$ From $$ \operatorname{div}H=2(xy+yz+zx)\,,\quad\operatorname{curl}H=-\left(\begin{smallmatrix}y^2\\z^2\\x^2\end{smallmatrix}\right) $$ it follows that you are looking for fields $G$ and $F$ that have \begin{align} \operatorname{div}F&=2(xy+yz+zx)\,,\quad \operatorname{curl}G=-\left(\begin{smallmatrix}y^2\\z^2\\x^2\end{smallmatrix}\right)\,. \end{align} Since $\operatorname{curl}F=0$ it is of the form $F=\operatorname{grad}\Phi$ with some scalar function $\Phi\,.$ What you have to solve now is the Poisson equation \begin{align} \operatorname{div\,grad}\Phi=\Delta\Phi&=2(xy+yz+zx)\,. \end{align} Once this is done it follows that $G:=H-\operatorname{grad}\Phi$ is divergence free.

A simpler approach is the following:

Using the anti-curl formula from this post yields a field $G'$ \begin{align} G'&:=\left[-\int_0^1t\left(\begin{smallmatrix}(ty)^2\\(tz)^2\\(tx)^2 \end{smallmatrix}\right)\,dt\right]\times\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)=-\frac{1}{4}\begin{pmatrix}y^2\\z^2\\x^2\end{pmatrix}\times\begin{pmatrix}x\\y\\z\end{pmatrix}=-\frac{1}{4}\begin{pmatrix}z^3-x^2y\\x^3-y^2z\\y^3-z^2x\end{pmatrix}\\ &=-\frac{1}{4}\begin{pmatrix}z^3\\x^3\\y^3\end{pmatrix}+\frac{1}{4}H \end{align} which has the desired $\operatorname{curl}G'=-\left(\begin{smallmatrix}y^2\\z^2\\x^2\end{smallmatrix}\right)$ but is not divergence free: ($\operatorname{div}G'=(xy+yz+zx)/2$).

Still, the ansatz $$ F'=H-G'=\frac{3}{4}H+\frac{1}{4}\begin{pmatrix}z^3\\x^3\\y^3\end{pmatrix}=\frac{1}{4}\begin{pmatrix}z^3+3x^2y\\x^3+3y^2z\\y^3+3z^2x\end{pmatrix} $$ leads to a field $F'$ that has $$ \operatorname{div}F'=\frac{3}{2}(xy+yz+zx)\,. $$ Therefore, $$ \boxed{\quad F:=\frac{1}{3}\begin{pmatrix}z^3+3x^2y\\x^3+3y^2z\\y^3+3z^2x\end{pmatrix} =\frac{1}{3}\begin{pmatrix}z^3\\x^3\\y^3\end{pmatrix}+H\quad} $$ has the desired divergence $\operatorname{div}F=2(xy+yz+zx)\,.$ It can be checked that $F$ is rotation free. Finally, $G$ is $$\boxed{\quad G=H-F=\frac{1}{3}\begin{pmatrix}z^3\\x^3\\y^3\end{pmatrix}\quad} $$ which is obviously divergence-free.

The function $\Phi$ whose gradient is $F$ is

$$\Phi(x,y,z)=\frac{x^3y+y^3z+z^3x}{3}\,.$$

Kurt G.
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