Find directions where current is maximal

Question

The current $J_i$ due to an electric field $E_i$ is given by $J_i = σ_{ij} E_j$ , where $σ_{ij} is the conductivity tensor. In a given Cartesian coordinate system,

$σ=\begin{pmatrix}2&-1&-1 \\-1&2&-1\\-1&-1&2 \end{pmatrix} $

I need to find the direction with the largest current, given the magnitude of the electric field is fixed.

I tried fixing $E_1^2+E_2^2+E_3^2=k^2$ and then calculating the magnitude of $J$: $|J|^2= 6k^2-6\Big(E_1E_2+(E_1+E_2)\sqrt{k^2-E_1^2-E_2^2}\Big)$ and then using $\nabla$ and trying to minimise the second term, but neither approach worked.

Please could I have some help? Thank you.

Answer 1

Write $J_i$ and $E_i$ as column vectors, i.e., $$ \mathbf{J}=\left( \begin{array}{c} J_1\\ J_2\\ J_3 \end{array} \right)\quad\text{and}\quad\mathbf{E}=\left( \begin{array}{c} E_1\\ E_2\\ E_3 \end{array} \right). $$ Then your question can be equivalently stated as follows:

Find $\mathbf{E}$, subject to $\left\|\mathbf{E}\right\|^2=k^2$, that maximizes $$ \left\|\mathbf{J}\right\|^2=\left\|\sigma\mathbf{E}\right\|^2. $$

Here the constant $k$ and the matrix $\sigma$ follow your notations.

This is a constrained optimization problem, where the method of Lagrange multipliers applies. Define the Lagrangian $$ \mathcal{L}=\left\|\mathbf{J}\right\|^2-\lambda\left(\left\|\mathbf{E}\right\|^2-k^2\right)=\left\|\sigma\mathbf{E}\right\|^2-\lambda\left(\left\|\mathbf{E}\right\|^2-k^2\right), $$ where $\lambda$ serves as a Lagrange multiplier here. As such, the constrained optimization becomes unconstrained: The $\mathbf{E}$ that solves the original problem must satisfy \begin{align} \frac{\partial\mathcal{L}}{\partial\mathbf{E}}&=\mathbf{0},\\ \frac{\partial\mathcal{L}}{\partial\lambda}&=0. \end{align} The matrix form of the above equations reads \begin{align} \sigma^{\top}\sigma\mathbf{E}&=\lambda\mathbf{E},\\ \mathbf{E}^{\top}\mathbf{E}&=k^2. \end{align} Obviously, the first line tells that $\lambda$ must be an eigenvalue of $\sigma^{\top}\sigma$, while $\mathbf{E}$ appears to be one of its associated eigenvectors. The second line requires that $\mathbf{E}$ must have a norm of $k$. Besides, note that these two equations yield that \begin{align} \left\|\mathbf{J}\right\|^2=\mathbf{J}^{\top}\mathbf{J}&=\mathbf{E}^{\top}\sigma^{\top}\sigma\mathbf{E}\\ &=\mathbf{E}^{\top}\left(\lambda\mathbf{E}\right)=\lambda\mathbf{E}^{\top}\mathbf{E}=\lambda k^2. \end{align} Therefore, the maximization of $\left\|\mathbf{J}\right\|$ is no more than the maximization of $\lambda$, i.e., finding the largest eigenvalue of $\sigma^{\top}\sigma$.

To sum up, the scheme for solving the constrained optimization goes as follows.

(1) Find all eigenvalues of $\sigma^{\top}\sigma$. The largest eigenvalue, denoted as $\lambda_{\text{max}}$, reveals the maximized value of $\left\|\mathbf{J}\right\|^2$ (which is $\lambda_{\text{max}}k^2$).

(2) Find an eigenvector of $\sigma^{\top}\sigma$ associated with $\lambda_{\text{max}}$ that has a norm of $k$. There could be more than one solution, and any of them drives $\left\|\mathbf{J}\right\|^2$ to its maximum $\lambda_{\text{max}}k^2$. These solutions are all possible values that $\mathbf{E}$ could take.

Hope this could be helpful for you.