Value at $(0,0)$ of the solution to a 2D linear elliptic PDE

Question

The setup is the following.

Let $D=(-\frac{1}{2} , \frac{1}{2})^2 \subset \mathbb{R}^2$, $\partial D$ be its boundary. Let $a, b\in \mathbb{R}$ and $f$ be continuous on $\partial D$. Also we assume that $f$ is periodic in the sense that $f(\frac{1}{2},x_2) = f(-\frac{1}{2},x_2)$ and $ f(x_1, \frac{1}{2})= f(x_1, -\frac{1}{2})$. Consider the equations $$\Delta u+ax_1 \frac{\partial u}{\partial x_1}+bx_2 \frac{\partial u}{\partial x_2} = 0 \text{ , on } D \text{;}$$ $$u|_{\partial D} = f \text{ .}$$

I would like to know if there is an explicit or approximate solution to this PDE.

Actually, I am only interested in the value $u(0,0)$ or its approximation.

Thank you!

Answer 1

This equation appears like it can be solved by using separation of variables. Assume there exists a separable solution $$u(x_1,x_2) = X_1(x_1)X_2(x_2)$$ Plugging this into the original PDE allows the two variables to be separated with eigenvalue of $\lambda_n$ $$X_1''+ax_1X'_1-\lambda_n X_1=0\quad\&\quad X_2''+bx_2X_2'+\lambda_n X_2=0$$ These equations are not easily solvable. They can be solved assuming power series solution, but Mathematica is much more clever in recognizing the power series to be previously studied functions $$X_1(x_1;\lambda_n)=e^{-\frac{ax_1^2}{2}}\left[g_1(x_2)\;H_{\frac{\lambda_n}{a}-1}\left(\frac{\sqrt{a}\;x_1}{\sqrt{2}}\right)+g_2(x_2)\;_1F_1\left(\frac{a-\lambda_n}{2a};\frac{1}{2};\frac{ax_1^2}{2}\right)\right]$$ $$X_2(x_2;\lambda_n) = e^{-\frac{bx_2^2}{2}}\left[h_1(x_1)\;H_{-\frac{\lambda_n}{b}-1}\left(\frac{\sqrt{b}\;x_2}{\sqrt{2}}\right)+h_2(x_1)\;_1F_1\left(\frac{b+\lambda_n}{2b};\frac{1}{2};\frac{bx_2^2}{2}\right)\right]$$ where $H_i$ are the Hermite polynomials and $_1F_1$ is the Kummer confluent hypergeometric function of the first kind (I've never studied this one) and $g$ and $h$ are arbitrary functions of $x_2$ and $x_1$ respectively. $g$ and $h$ are dependent on $f$, so they can be found when $f$ is known.

The most general solution would be to sum over all the eigenfunctions $$u = \sum_nX_1(x_1;\lambda_n)X_2(x_2;\lambda_n)$$ however, since you are interested in $u(0,0)$ we can understand a little more without knowing $f$. Using $$_1F_1(\cdot;\cdot;0) = 1\quad\&\quad H_i(0) = \frac{\sqrt{\pi}\;2^i}{\Gamma\left(\frac{1-i}{2}\right)}$$ we can see that the solution at the origin would look like $$u(0,0) = \sum_n\left[g_1(0)\frac{\sqrt{\pi}\;2^{-\frac{\lambda_n}{a}-1}}{\Gamma\left(\frac{\lambda_n}{2a}\right)}+g_2(0)\right]\left[h_1(0)\frac{\sqrt{\pi}\;2^{\frac{\lambda_n}{b}-1}}{\Gamma\left(1-\frac{\lambda_n}{2b}\right)}+h_2(0)\right]$$ Again, $g$ and $h$ are dependent on $f$. Not sure if this approach is what you were looking for. I think there may be a more elegant approach.