# Non-linear partial differential equation

I would like to find out if there is any specific method -apart from numerics- for finding solutions of a non-linear PDE of the form

$$\nabla \times \mathbf{A} = \pm\lambda\mathbf{A} \tag{1}$$

under the constraint

$$\nabla\lambda \times \mathbf{A} = 0 \tag{2}$$

where λ is either an unknown real constant or an unknown scalar function ($\lambda(\mathbf{r})$) of a vector variable in $\mathbb{R}^3$ and $\mathbf{A}$ is a vector field in same space, which may be restricted in a curved manifold. This problem occurs in eigenrotation flows (Euler fluids), and their application in Parallel Electric and Magnetic components of the associated Maxwell fields. There is a large literature on this problem which starts with Eugenio Beltrami at the end of 19th century and Viktor Trkal at the beginning of the 20th century. Recent applications in plasma and E/M started with Chandrsekhar, Kendal, Lakhtakia and others for the case of solenoidal fields ($\nabla\cdot\mathbf{A} = 0$). In the non-solenoidal case, taking the divergence of (1) results in the lhs to

$$\nabla\cdot\mathbf{A} = -\nabla(\log\lambda)\cdot\mathbf{A} \tag{3}$$

Condition (2) has the strange feature that the nth application of the Curl operator results in the simplest group structure (although the problem is nonlinear!)

$$\nabla\times\nabla\times ...\nabla \mathbf{A} = {(\pm)}^n{\lambda}^n\mathbf{A} \tag{4}$$

where n is the number of application of the Curl operator. The associated nonlinear wave equation will then will have an additional "source" term of the form $\nabla(\nabla\cdot\mathbf{A})$. At first hand, it appears that the only solution in a Euclidean space must have constant λ but in the case of curved manifolds with a connection the problem is not trivial.

Concerning the generality of (4) in relation to eigenrotation fields that are not Beltrami it should be mentioned that a more natural choice appears to have the form

$$\nabla\lambda \times \mathbf{A} \propto \mathbf{A}_{\perp} = \kappa(\mathbf{r})\bar{\mathbf{R}} \mathbf{A} \tag{5}$$

where the rhs denotes an appropriate projection on the plane of $\nabla\lambda$ and $\mathbf{A}$ with $\kappa$ a proportionality scalar and $\bar{\mathbf{R}}$ the relative rotation matrix.

In such a case one obtains

$$\nabla\times\nabla\times\mathbf{A} = (\kappa\bar{\mathbf{R}} \pm \lambda)\mathbf{A} \tag{6}$$

Subsequent applications of the Curl operator won't work the same as in (4) unless one can prove that there is a Ladder of functions {$\kappa_{i}$} all satisfying condition (2) or (5) or at least $\kappa = c\lambda$.

Addendum: connection w. Wave Equation (08/05/2011)

It is notable that furthe analysis of the additional source term $\nabla(\nabla\cdot\mathbf{A})$ using known vector identities as found in tables (see http://en.wikipedia.org/wiki/Vector_calculus_identities#Properties) results in two "convective" like derivatives. Using the double curl we also obtain from (1) or (6)

$$\nabla^2\mathbf{A} + \epsilon(\mathbf{r})\mathbf{A} = \mathbf{J} \tag{7}$$

where the source has the explicit form

$$\mathbf{J} = (\nabla\phi\cdot\nabla)\mathbf{A} + (\mathbf{A}\cdot\nabla)\nabla\phi \tag{8}$$

with $\lambda = e^{-\phi}$ and $\epsilon(\mathbf{r})$ is either $\pm\lambda(\mathbf{r})$ if condition (2) is satisfied or $(1+\lambda)\kappa\bar{\mathbf{R}} \pm \lambda$ if condition (5) holds. The situation now resembles a vector potential in a material with a complicated variable refractive index and possibly a very unusual optical rotation (the last may not be realizable in an actual situation though due to physical restrictions).

• Also: in what sense do you want "solutions"? – Willie Wong May 7 '11 at 12:20
• Is $\lambda$ prescribed? Or is it an unknown? If the former, you don't have a non-linear PDE: the eqution would be linear. – Willie Wong May 7 '11 at 12:59
• Sorry for the brakets, they were mistyped as parnthesis instead of '{}'. These been corrected I consider the second part of your question. Apparently the Divergence of a Curl should be zero so that the lhs takes the form $\nabla(\lambda\mathbf{A})=0$ which is then expanded as $\nabla\lambda\cdot\mathbf{A}+\lambda\nabla\cdot\mathbf{A}=0$ hence we get (3). Regarding solutions of the above, it would be enough to know if there are any special closed manifolds on which the above gets simplified or becomes solvable in the most general sense - if such a solution exist at all that is. – rtheo May 7 '11 at 13:05
• Indeed, $\lambda$ is an unknown to be defined from (1) and (2). Thank you for the clarification. It has now been better specified in the main text – rtheo May 7 '11 at 15:29
• The book Nonlinear Waves in Integrable and Nonintegrable Systems by Jianke Yang may help. – dustin Sep 7 '13 at 16:30

The eigenvectors of the curl operator $$\nabla \times {\bf A}$$ in Cylindrical coordinates are the Bessel functions $J_0( a r)$, you will have to plug back this in to see how $a$ depends on $\lambda$. But all this assumes $\lambda$ is a constant. In your problem, you don't have $\lambda$ constant, but special cases are sometimes useful.

Unless I'm missing something, the constraint (2)

$\nabla \lambda \times \mathbf{A} = 0$

means that the vector field must be of the form

$\mathbf{A} = f \nabla \lambda$

where f is a scalar function. Inserting into (1) yields

$\nabla f \times \nabla \lambda = f \lambda \nabla \lambda$.

The only possible solutions are $\nabla\lambda = 0$, so $\lambda$ has to be a constant. Together with the other answer about the eigenvectors of the curl operator, maybe this solves your problem.

Hint:

Given

$$\nabla \times \textbf{A} = \pm \lambda \textbf{A}. \tag{1}$$

Then we obtain

$$\nabla \times \Big( \nabla \times \textbf{A} \Big) = \pm \lambda \nabla \times \textbf{A}, \tag{2}$$

whence

$$\nabla \Big( \nabla \cdot \textbf{A} \Big) - \nabla^2 \textbf{A} = \lambda^2 \textbf{A}, \tag{3}$$

therefore

$$\nabla \times \left[ \nabla \Big( \nabla \cdot \textbf{A} \Big) - \nabla^2 \textbf{A} \right] = \nabla \times \left[ \lambda^2 \textbf{A} \right], \tag{4}$$

so

$$\underbrace{\nabla \times \left[ \nabla \Big( \nabla \cdot \textbf{A} \Big) \right]}_0} = \nabla \times \left[ \lambda^2 \textbf{A} - \nabla^2 \textbf{A} \right], \tag{5}$$

yielding

$$\nabla \times \left[ \lambda^2 \textbf{A} - \nabla^2 \textbf{A} \right] = 0. \tag{6}$$

But

$$\nabla \times \textbf{V} = 0 \Rightarrow \textbf{V} = \nabla \phi, \tag{6}$$

thus

$$\nabla^2 \textbf{A} - \lambda^2 \textbf{A} = \nabla \Phi.$$

Question is how to solve this...