Monday Math 135

Suppose we have a three-dimensional scalar field a(r) and three-dimensional solenoidal (divergence-free) vector field V(r), both going to zero (sufficiently fast) as |r|→∞. Can we then construct a vector field F(r) such that \mathbf{\nabla}\cdot\mathbf{F}=a and \mathbf{\nabla}\times\mathbf{F}=\mathbf{V}; that is to say, can we construct a field with a specified divergence and curl?

As noted in final remarks here, the answer is yes, we can, and given our boundary conditions, it is unique. More specifically, Helmholtz’s theorem, also known as the “fundamental theorem of vector calculus,” detailed here, tells us our field F may be decomposed into irrotational (zero curl) and solenoidal (zero divergence) components. We further noted that the irrotational component may be expressed as the gradient of a scalar field, and the solenoidal term as the curl of a vector field; we thus expressed the decomposition as
\mathbf{F}=-\mathbf{\nabla}\phi+\mathbf{\nabla}\times\mathbf{A}
Since \mathbf{\nabla}\times\mathbf{A} is the solenoidal term, the divergence of F is just that of the irrotational term, and so our divergence condition is
\begin{array}{rcl}\mathbf{\nabla}\cdot\mathbf{F}&=&a\\\mathbf{\nabla}\cdot\left(-\mathbf{\nabla}\phi\right)&=&a\\\nabla^2\phi&=&-a\end{array},
Poisson’s equation.
Next, we noted here, that in three-dimensions, the Green’s function for the Laplacian with our boundary conditions is G(\mathbf{r},\mathbf{r}')=-\frac{1}{4\pi\left|\mathbf{r}-\mathbf{r}'\right|}. So, we see that
\phi(\mathbf{r})=\iiint{G(\mathbf{r},\mathbf{r}')a(\mathbf{r}')}\,d^3r'=\frac{1}{4\pi}\iiint\frac{a(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r',
since
\begin{array}{rcl}\nabla_r^2\iiint{G(\mathbf{r},\mathbf{r}')a(\mathbf{r}')}d^3r'&=&\iiint\nabla_r^2\left(G(\mathbf{r},\mathbf{r}')\right)a(\mathbf{r}')d^3r'\\&=&\iiint\delta(\mathbf{r}-\mathbf{r}')a(\mathbf{r}')d^3r'\\&=&a(\mathbf{r})\end{array}.

Now, let us consider the curl of our field. As the first term is irrotational, we then get curl equation
\begin{array}{rcl}\mathbf{\nabla}\times\mathbf{F}&=&\mathbf{V}\\\mathbf{\nabla}\times\left(\mathbf{\nabla}\times\mathbf{A}\right)&=&\mathbf{V}\\\mathbf{\nabla}\left(\mathbf{\nabla}\cdot\mathbf{A}\right)-\nabla^2\mathbf{A}&=&\mathbf{V}\end{array}.
Next, let us consider if our solution A is solenoidal; then the above would reduce to
\nabla^2\mathbf{A}=-\mathbf{V},
the vector Poisson’s equation, whose x, y, and z components are each (scalar) Poisson’s equations analogous to the one for φ. So, we then try solutions like in our previous part:
A_x(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{V_x(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'
A_y(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{V_y(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'
A_z(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{V_z(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r',
or combining into a vector equation,
\mathbf{A}(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'.
Now, we are just left to confirm that this is solenoidal.
First, we note that \mathbf{\nabla}\left(\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}\right)=-\mathbf{\nabla}'\left(\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}\right), where \mathbf{\nabla}' is the del operator with respect to the primed coordinates.
Then, we see
\begin{array}{rcl}\mathbf{\nabla}\cdot\mathbf{A}&=&\mathbf{\nabla}\cdot\left(\frac{1}{4\pi}\iiint\frac{\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'\right)\\&=&\frac{1}{4\pi}\iiint\mathbf{V}(\mathbf{r}')\mathbf{\nabla}\cdot\left(\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}\right)\,d^3r'\\&=&-\frac{1}{4\pi}\iiint\mathbf{V}(\mathbf{r}')\mathbf{\nabla}'\cdot\left(\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}\right)\,d^3r'\end{array}.

Now, recall vector integration by parts; specifically, for scalar function φ(r) and vector function v(r), then one has on volume V with bounding surface S:
\iiint_{V}\mathbf{\nabla}\phi\cdot\mathbf{v}\,dV=\iint_{S}\phi\mathbf{v}\cdot\mathbf{\hat{n}}dS-\iiint_{V}\phi\mathbf{\nabla}\cdot\mathbf{v}\,dV.
Here, our vector field is V(r‘), and our scalar field is \frac1{\left|\mathbf{r}-\mathbf{r}'\right|}, also as a function of r‘. Then, we see that

\begin{array}{rcl}\iiint_{V'}\mathbf{V}(\mathbf{r}')\mathbf{\nabla}'\cdot\left(\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}\right)\,d^3r'&=&\iint_{S'}\frac{\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\cdot\mathbf{\hat{n}}'dS'-\iiint_{V'}\frac{\mathbf{\nabla}'\cdot\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'\\&=&\iint_{S'}\frac{\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\cdot\mathbf{\hat{n}}'dS'\end{array},
as V is solenoidal, so the numerator in the integrand of the right-hand-side volume integral is zero. Now, if V(r) goes to zero faster than |r|-1 as |r|→∞, then the surface integral goes to zero as the volume integral is expanded to all space, and thus, our solution is solenoidal, as we assumed, and so satisfies \mathbf{\nabla}\times\left(\mathbf{\nabla}\times\mathbf{A}\right)=\mathbf{V}.
Thus, we have (unique) solution
\mathbf{F}=-\mathbf{\nabla}\phi+\mathbf{\nabla}\times\mathbf{A},
where
\phi(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{a(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r',
and
\mathbf{A}(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'.

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