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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

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To be precise, you are not trying to find a VECTOR $F$ such that ... You are trying to find a vector field $F$ such that ... – Old John Sep 8 '13 at 0:00
Think from a physics point of view No curl meaning no rotational spin in the 3rd axis and no divergence what kinda of vector or process will behave like that then you can get a function easily – MRK Sep 8 '13 at 0:00
You are pretty close, try $\cos(x)\cosh(\sqrt{2}y)\cos(z)$ instead! – achille hui Sep 8 '13 at 0:23
@achillehui that did it. I wasnt accounting for the fact that each cos should be cos(kx) and that the sum of the k^2 has to be 0! If you make that an answer Iwill select it as best – yankeefan11 Sep 8 '13 at 0:36 You can try the functions in the table, if your domain of interest does not have the origin. – Shuhao Cao Sep 8 '13 at 5:42
up vote 2 down vote accepted

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 $(x,y,z)$.

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i remember thinking some where that all harmonic functions have zero curl

more precisely all conservative vector fields are harmonic also follow Clairaut's Theorem

which is second partials in reverse order are equal

and for divergence to e zero the vector field must be constant meaning no rate of change in all directions

an example in 3D


F= yz i + xz j + xy K

Curl = 0 ( check this by computing but it is 0)

Div = 0

so F sub xy = F sub yx so they are conservative so no curl and clearly no divergence

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Can you expand on this, and perhaps include a proof? As it stands, this would likely be appropriate for a comment, not an answer. – user61527 Sep 8 '13 at 0:07
i have included some details – MRK Sep 8 '13 at 0:08
Please correct your first sentence. Gradient of a harmonic function! – Ted Shifrin Sep 8 '13 at 0:23
I made a mistake on the previous one i did not show properly check i made an edit – MRK Sep 8 '13 at 0:24

yankeefan11, you were pretty close! Using the bold face font for vectors, we have as a given that

$\nabla \times \mathbf{F} = 0, \tag{1}$

which does in fact imply the existence of $\phi$ such that

$\mathbf F = \nabla \phi, \tag{2}$

and now using the divergence condition we find we must have

$\nabla^2 \phi = \nabla \cdot \mathbf F = 0. \tag{3}$

Running things in reverse, we see that for any harmonic $\phi$, setting

$\mathbf F = \nabla \phi \tag{4}$

produces an $\mathbf F$ with

$\nabla \cdot \mathbf F = 0 \tag{5}$


$\nabla \times \mathbf F = 0 \tag{6}$

as well. There are lots of solutions; the field gets considerably thinner when boundary conditions are added.

Fiat Lux!

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