Hi fellow math enthusiasts,
I am currently working on some research to do with the electric field induced within the
brain via magnetic stimulation.
I am trying to solve the partial differential equation in 2D cartesian co-ordinates
\begin{equation}
\nabla \cdot \underline{\sigma} \vec{\nabla}\phi = – \nabla \cdot \left(\underline{\sigma} \frac{\partial \vec{A}}{\partial t}\right)
\end{equation}
For a magnetic vector potential of $\vec{A}=-\frac y2 \hat{i}+\frac x2\hat{j}$ and the magnetic vector potential can be separated so the function of time does not need to be known.
From above this can be written as
$$
\sigma_x\frac{\partial^2\phi}{\partial x^2}+\sigma_y\frac{\partial^2\phi}{\partial y^2}=0
$$
and after a change of variables $\gamma=\left(\frac{\sigma_x}{\sigma_y}\right)^{\frac12}y$ revels Laplace's equation
$$
\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial \gamma^2}=0
$$
The surface to which Laplace's equation is to be solved in is the rectangle bounded by the lines $x=a,\ x=-a$ and $y=b,\ y=-b$ subject to Neumann boundary conditions
\begin{align}
\left.\frac{\partial\phi}{\partial x}\right|_{x=a}&=\frac y2 \quad &\left.\frac{\partial\phi}{\partial x}\right|_{x=-a}&=\frac y2\\
\left.\frac{\partial\phi}{\partial y}\right|_{y=b}&=-\frac x2 \quad &\left.\frac{\partial\phi}{\partial y}\right|_{y=-b}&=-\frac x2 \\
\end{align}
Which cannot be solved although my math prof. said that since it is a linear PDE, each boundary condition can be solved independently with the other three Neumann BC set to zero and the final answer is a linear combination of the four solutions. Which I have tried to no avail. Any thoughts?
Thank you
Best Answer
You don't need to solve four different boundary value problems: two are enough.
Then add. I'll work through part 1.
The reason to have homogeneous BCs on a pair of opposite sides is that we can build them into the solution by using a basis of eigenfunctions of the second derivative operator. In problem 1, we work with eigenfunctions of $\partial^2/\partial x^2$ on $(-a,a)$. The eigenfunctions for Neumann (or Dirichlet) conditions are trigonometric: sines, cosines, and their linear combinations. In this case, we are to expand $-x/2$, which is an odd function. So we will not need cosines.
Sine function $\sin \beta x$ satisfies the boundary condition $(\sin \beta x)'\bigg|_{x=a}=0$ when $\beta = \frac{2n+1}{2a}\pi$ for some integer $n\ge 0$. I write $\beta_n = \frac{2n+1}{2a}\pi$ below.
Having taken care of the homogeneous boundary conditions, we turn to the PDE itself. The building blocks (separated solutions) for the Laplace equation are of the form "(hyperbolic trig) times (trig)". Namely, any function of the form $$ u(x,y)=\sum_{n=0}^\infty (A_n\cosh \beta_n y+B_n\sinh\beta_n y) \sin\beta_n x \tag{1} $$ solves the PDE.
Having taken care of the PDE, we turn to the remaining, inhomogeneous boundary conditions. The goal is to choose $A_n,B_n$ so that $u$ satisfies them. Observing that $\partial u/\partial y$ is subjected to the same condition on top and bottom sides, we conclude that it should be an even function of $y$, hence $u$ must be an odd function of $y$. This simplifies (1) to $$ u(x,y)=\sum_{n=0}^\infty B_n\sinh\beta_n y \, \sin\beta_n x $$ with the derivative $$ u_y(x,y)=\sum_{n=0}^\infty \beta_n B_n\cosh\beta_n y \, \sin\beta_n x \tag{2}$$
When $y=b$, the sum (2) must match $-x/2$.
The coefficients of $-x/2$ in the basis $\{\sin \beta_n x\}$ are $$ \frac{1}{a}\int_{-a}^a -\frac{x}{2}\sin \beta_n x\,dx = \frac{(-1)^{n+1}}{a \beta_n^2} $$ unless I messed up the integration. Hence, $\beta_n B_n = (-1)^{n+1}/(a \beta_n^2)$, which yields $B_n= (-1)^{n+1}/(a\beta_n^3)$. Solution: $$ u(x,y)= \sum_{n=0}^\infty (-1)^{n+1}\frac{8a^2}{(2n+1)^3\pi^3} \sinh \frac{2n+1}{2a}\pi y \, \sin \frac{2n+1}{2a}\pi x \tag{3}$$ This completes part 1.
Luckily for you, Part 2 is so similar that relabeling $x\leftrightarrow y$, $a\leftrightarrow b$ is all you need. The solution is $$ v(x,y)= \sum_{n=0}^\infty (-1)^{n+1}\frac{8b^2}{(2n+1)^3\pi^3} \sinh \frac{2n+1}{2b}\pi x \, \sin \frac{2n+1}{2b}\pi y \tag{4}$$ The sum $u+v$ is $\phi$ that you wanted.
Some remarks. The advantage of homogeneous BCs restrict the space of solutions to a linear subspace. When we look for separated solutions $X_n(x)Y_n(y)$, we can use $X_n$ (or $Y_n$, in part 2) which satisfy the homogeneous BCs; this way, and sum $\sum X_nY_n$ will satisfy them as well. Simply put, $0$ is the only number which we can add to itself, as many times as we want, without changing it.
Summary of the process: