$\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + 1 = 0$
with the following boundary conditions:
$\phi(\pm 1,y)=0 \ and \ \phi(x,\pm 1) = 0$
I was able to solve this problem using Galerkin method with Weak Form. I was able to use integration by parts to transfer an order of differentiation to the test function v which I used to be the variation of $\phi$. Using this method I used basic class $C^0$ Lagrange Equations ($N(1) = (1-\xi)/2 ,N(2) = (1+\xi)/2$) four element with an overall nine nodes. The four elements would all connect such that the only unknown nodal value would be node 5 (the center node). Here is a layout of the mesh. I was able to obtain the solution of the center node to be $\phi_5 = 0.375$.The problem arose when I tried to extend the process to higher order continuity equations (i.e. $C^2$) to establish a minimum conforming space as denoted in this textbook. The Finite Element Method for Boundary Value Problems Mathematics and Computations by Karan S. Surana and J.N. Reddy. The higher order hierarchical approximation functions require there to be more degrees of freedom on each of the nodes, but I wasn't able to get the correct solution at node 5 for the function value as I previously did. I believe it might be a mesh problem, but I was hoping to have a theoretical solution to compared to my numerical results. I was able to successfully solve this problem for a convection diffusion equation in in one dimension, but I cannot seem to solve a two dimensional problem the way I originally did it. I can provide more work for those truly interested in the problem via other means than on here.
Try $$ \phi(x,y)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{n,m}\sin(n\pi x)\sin(m\pi y). $$
The endpoint conditions are satisfied. The coefficients $A_{n,m}$ must satisfy $$ \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{n,m}(n^2+m^2)\sin(n\pi x)\sin(m\pi y)=1, $$ which leads to the orthogonality relations $$ A_{n,m}(n^2+m^2)\int_{0}^{1}\sin^2(n\pi x)dx\int_{0}^{1}\sin^2(m\pi y)dy\\ =\int_{0}^{1}\sin(n\pi x)dx\int_{0}^{1}\sin(m\pi y)dy. $$ So, $$ \phi(x,y) =\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{\int_{0}^{1}\sin(n\pi x)dx\int_{0}^{1}\sin(m\pi y)dy}{(n^2+m^2)\int_{0}^{1}\sin^2(n\pi x)dy\int_{0}^{1}\sin^2(m\pi y)dy}\sin(n\pi x)\sin(m\pi y) $$