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Derive a matrix formulation to the BVP using galerkin method: \begin{align} \frac{\partial^2f}{dx^2}+\frac{\partial^2f}{dy^2}&=0\text{ within }A\in(0,2)\times(0,2)\\ f=0\text{ on }C_1&:x=0,y=\pm2\\ \frac{\partial f}{\partial x}=K^\star\text{ on }C_2&:x=2 \end{align}

evaluate weighted integral, \begin{align} I(f(x, y))=\int \int W(x, y) \times\left(\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}\right) d y d x&=0\\ \iint\left[\frac{\partial}{\partial x}\left(W \frac{\partial f}{\partial x}\right)+\frac{\partial}{\partial y}\left(W \frac{\partial f}{\partial y}\right)-\frac{\partial W}{\partial x} \frac{\partial f}{\partial x}-\frac{\partial W}{\partial y} \frac{\partial f}{\partial y}\right] d x d y&=0\tag1 \end{align} Using green's theorem,

$$\oint\left( -W\frac{\partial f}{\partial y}dx+W\frac{\partial f}{\partial x}dy \right)-\iint \left( \frac{\partial f}{\partial x}\frac{\partial W}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial W}{\partial y} \right)\:dxdy=0$$

On the boundary $C_2$ $x=2\implies dx=0$ implies,

$$\oint W K^\star dy -\iint \left( \frac{\partial f}{\partial x}\frac{\partial W}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial W}{\partial y} \right)\:dxdy=0$$

Now letting $W_i=L_i;i=1,2,3,4$ and $\tilde f=\sum_{j=1}^4=L_j f_j$ we get,

$$\underbrace{\oint_{\Gamma {}^{[e]}} L_i K^\star dy}_{F_i} =\underbrace{\iint_{[e]} \left( \frac{\partial L_j}{\partial x}\frac{\partial L_i}{\partial x}+\frac{\partial L_j}{\partial y}\frac{\partial L_i}{\partial y} \right)\:dxdy}_{K_{ij}}$$

Imposing boundary condition, $$f=0:x=0,y=\pm 2$$

$$ \pmatrix{K_{4,4} & K_{4,6} & K_{4,8}\\ K_{6,4} & K_{6,6} & K_{6,8}\\ K_{8,4} & K_{8,6} & K_{8,8}}\pmatrix{f_4\\f_6\\f_8}=\pmatrix{F_4\\F_6\\F_8} $$

where, \begin{align} L_1 &= \frac14 (1-\eta)(1-\epsilon)\\ L_2 &= \frac14 (1-\eta)(1+\epsilon)\\ L_3 &= \frac14 (1+\eta)(1+\epsilon)\\ L_1 &= \frac14 (1+\eta)(1-\epsilon) \end{align}

for every $[e]$ I need to use $x=x_iL_i$ and $y=y_iL_i$ and get variable transformation in term of $\eta$ and $epsilon$.

For 2D case I haven't any idea how to proceed further. Like for 1D I will need to substitute the weight by shape function. But unfortunately for 2D case I am bit confused on that.

I need to calculate the $K_{ij}$ for every $[e]$, but which seems too much computation for exam.

The issue is, this type of question was on the final example, which I think very difficult to solve in that short time. Is there any tricky approach to solve these kind of problem?

N00BMaster
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  • So you have to choose sets of kernel and test functions. You could use finite elements, e.g., on a quadratic lattice with piecewise bilinear kernel functions, or on a triangular grid with piecewise linear kernel functions. Then you get to decide if you take the kernel functions also as test functions or if you take a different set. Or you could take both more global, using sine and cosine functions for both, like in a Fourier series. Or a mix of both, Fourier in one direction, with the grid in the frequency domain, and piecewise linear in the other direction. – Lutz Lehmann Oct 18 '23 at 07:02
  • I update the question by adding shape function @LutzLehmann The problem is, I need to do same calculation for every element, $1. x=x_i L_i$ and $y=y_i L_i$ $2.$ Get $K_{i,j}^{[e]}$ and $F_i$ in that $[e]$ $3.$ combine the overlap and get $K_{i,j}$ anad solve the system. But the problem is, that huge calculation seems not feasible at exam. – N00BMaster Oct 27 '23 at 07:06
  • I am also struggling with the shape function, like how to determine that for $\xi-\eta$ system here, @LutzLehmann – N00BMaster Oct 27 '23 at 17:29
  • Then it seems like you were expected to use products or "tensors" of one-dimensional basis functions, which in the most simple case are the "witches hat" functions on a regular grid. The area integrals then also factor, this could be done manually in 20-30 min. – Lutz Lehmann Oct 28 '23 at 00:03

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