multivariable-calculus
SO as stated, I am trying to find a vector $\vec F$such that
$$\nabla \times \vec F=0$$
$$\nabla \cdot \vec F=0$$
The way I go about it is:
Becasue curl is 0, we know that $\vec F=\nabla f$ so the divergence equation then becomes
$$\nabla ^2f=0$$
Which I then say $f=A(x)B(y)C(z)$, which results in me getting the following function f:
$$f=\cos(x)\cosh(y)\cos(z)$$
so $\vec F=\nabla f$,
Which has 0 curl, but nonzero div. Sad face
The claim is true. Any $3\times3$ matrix can be expressed as $$ A= \lambda I+ a b^T + B $$ where $\lambda$ is real, $a$ and $b$ are 3-vectors and $B$ is skew (so that $Bx=c\times x$ for some vector $c$).
To prove this, choose an orthogonal matrix $Q$ to diagonalize the symmetric part of $A$. Then $Q^TAQ=D+K$ where $D$ is diagonal and $K$ is skew. If the diagonal entries of $D$ are not all distinct then it is easy to write $D=\lambda I+\hat a \hat b^T$ and we finish as below. If the entries are all distinct, we can suppose that $Q$ was chosen so that the largest eigenvalue of $D$ is first, the smallest second and the middle last. Then for some positive $\mu$ and $\nu$, the matrix $D$ can be written $$ D = \lambda I + \mu \begin{pmatrix} 1 & 0 & 0\cr 0 &-\nu^2&0\cr 0&0&0 \end{pmatrix} =\lambda I + \hat a\hat b^T+\hat K, $$ with $$ \hat a= \mu\begin{pmatrix}1\cr \nu\cr0\end{pmatrix}, \quad \hat b= \begin{pmatrix} 1 \cr -\nu\cr 0 \end{pmatrix}, \quad \hat K=\mu\begin{pmatrix} 0&\nu&0\cr -\nu & 0&0\cr 0&0&0 \end{pmatrix} . $$ Let $a=Q\hat a$, $b=Q\hat b$, and $B=Q(\hat K+K)Q^T$, and you're done.
The complete answer is given on pages 22-27 of my 2011 vector calculus notes. I think many good calculus text include these heuristic arguments, I found them in Thomas' calculus a few editions back. Long story short, what you should really do to understand is to prove Greene's and Stokes' Theorems, this will give you deeper insight into the nature of your question. Let me summarize the method here:
All of this said, I really would rather give the less helpful answer that $d$ is the natural exterior derivative operation on the exterior algebra of $\mathbb{R}^3$ and it is just the case that: 1. $df = \omega_{\nabla f}$, 2. $d\omega_{\vec{F}} = \Phi_{\nabla \times \vec{F}}$ and 3. $d \Phi_{\vec{G}} = \nabla \cdot G dx \wedge dy \wedge dz$
where $\omega_{\langle a,b,c \rangle} = adx+bdy+cdz$ and $\Phi_{\langle a,b,c \rangle} = ady \wedge dz+bdz \wedge dz+cdx \wedge dy$. Therefore, gradient, curl and divergence are just different levels of the cohomological operator which ultimately reveals the deeper shape of space.
Best Answer
There are many ways to generate harmonic functions.
One way is as the poster suspected by looking at functions of the form:
$$f(x) = \cos(p x) \cosh(q y) \cos( r z)$$
By direct substitution, it is easy to see $$\nabla^2 f(x) = (q^2 - p^2 - r^2) f(x)$$
If one has choose $p, q, r$ such that $q^2 - p^2 - r^2 = 0$, e.g.
$$p = r = 1 \quad\text{ and }\quad q = \sqrt{2},$$ then we get a $f$ that is harmonic, i.e. $\displaystyle\nabla^2 f = 0$.
We can also generate polynomials that is harmonic. The simplest way is to observe:
$$\begin{cases} \frac{\partial^2}{\partial x^2} (x \pm iy)^n = \;\;n(n-1)(x \pm iy)^{n-2}\\ \frac{\partial^2}{\partial y^2} (x\pm iy)^n = -n(n-1)(x\pm iy)^{n-2} \end{cases}$$
This implies $\displaystyle\quad\nabla^2 ( x \pm i y)^n = 0\quad$ and hence $$\phi(x) = (x + iy)^n + (x - iy)^n$$
is a polynomial that is harmonic. By a orthogonal transformation of $(x,y,z)$ and through linear combinations, you can generate other polynomials of degree $n$ that is harmonic and homogenous in $(x,y,z)$.