Consider Laplace's equation in a rectangle as shown in the following figure. The boundary conditions are shown in the figure. The problem is solved in the case of a1 =a2=1.
Is there a way to simplify this problem to a simplified version such that can be solved analytically? (i.e. a rectangle with one constant $a$ using conformal mapping or any other method?).
$$\nabla.a\nabla V = 0$$
A numerical solution of the Laplace equation (in a rectangular geometry) can be obtained with an equivalent resistor network. Key internet references are found here:
In our case, the domain of interest in subdivided in rectangles:
program hesam;Here is a link to the complete (Delphi Pascal) unit that does the FEM Calculations: Below is the $\,40\times 30\,$ grid that has been used for sample calculations and a contour map of some results with $V_0=0$ , $V_1=1$ and random $a_1,a_2$.
Uses Laplace;
procedure test; var k : integer; begin Starten; {Initialize } for k := 0 to 9 do begin { FEM Calculations } Rekenen(Random,Random,0,1); { Store in 'results' file } Opschrijven(k); end; end;
begin test; end.
Sample output ('result' file) - can you see where it is? - :
x y V[x,y] 20 0 5.00000000000000E-0001 20 1 5.00000000000000E-0001 20 2 5.00000000000000E-0001 20 3 4.99999999999999E-0001 20 4 4.99999999999999E-0001 20 5 4.99999999999999E-0001 20 6 4.99999999999999E-0001 20 7 4.99999999999999E-0001 20 8 4.99999999999999E-0001 20 9 5.00000000000000E-0001 20 10 5.00000000000001E-0001 20 11 5.00000000000001E-0001 20 12 5.00000000000001E-0001 20 13 5.00000000000002E-0001 20 14 5.00000000000002E-0001 20 15 5.00000000000002E-0001 20 16 5.00000000000003E-0001 20 17 5.00000000000003E-0001 20 18 5.00000000000003E-0001 20 19 5.00000000000002E-0001 20 20 5.00000000000002E-0001 20 21 5.00000000000003E-0001 20 22 5.00000000000002E-0001 20 23 5.00000000000002E-0001 20 24 5.00000000000001E-0001 20 25 5.00000000000001E-0001 20 26 5.00000000000001E-0001 20 27 5.00000000000001E-0001 20 28 5.00000000000000E-0001 20 29 5.00000000000000E-0001 20 30 5.00000000000000E-0001
This is solvable with a Finite Fourier Transform in the $x$ coordinate. Without loss of generality, let $V_0=-1$ and $V_1=1$ (achievable with a simple shift and scale of the $V$ function). If we let $$ v_n(y) = \frac1b\int_{-b}^b V(x,y) \sin \left(\frac{n\pi x}{b}\right)dx $$ where we are neglecting the $\cos$ terms through the (anti-)symmetry of the system, then we have $$ V(x,y) = \sum_{n=1}^\infty v_n(y) \sin\left(\frac{n\pi x}b\right) $$ Now, we can see, through integration by parts, that $$ \frac1b\int_{-b}^b \frac{\partial^2 V}{\partial x^2}\sin\left(\frac{n\pi x}b\right)dx = -\frac{2n\pi(-1)^n}{b^2}-\frac{n^2\pi^2}{b^2}v_n(y) $$ Now, we're going to get rid of the $a$ variation by introducing boundaries. Assuming that the boundaries occur at $y_1$ and $y_2$, where $-c<y_1<y_2<c$, we will have the following boundary conditions at each of these boundaries: $$ V_1 = V_2 $$ and $$ a_1\mathbf{j}\cdot \nabla V_1 = a_2\mathbf{j}\cdot \nabla V_2 \qquad \text{or} \qquad a_1\frac{\partial V_1}{\partial y} = a_2\frac{\partial V_2}{\partial y} $$ which effectively says that the flux across the boundary from one side is equation to the flux across the boundary into the other side (if we interpret the equation as the steady state diffusion equation).
Now, as we can transform both of these equations in $x$ without changing their form, we can continue to apply these equations to the $v_n(y)$ functions. Applying our transform within each of the domains gives, $$ \frac{\partial^2 v_n}{\partial y^2}-\frac{n^2\pi^2}{b^2}v_n(y)=\frac{2n\pi(-1)^n}{b^2} $$ which has the general solution form $$ v_n(y) = \frac{2(-1)^{n+1}}{n\pi} + A_n\cosh\left(\frac{n\pi y}{b}\right)+B_n\sinh\left(\frac{n\pi y}{b}\right)\tag{1} $$ I'm not going to complete the process, but it's straightforward from here - transform the upper and lower boundaries so that they are boundaries in $v_n(y)$, then determine the $A_n$ and $B_n$ for each of the three domains by matching boundaries according to the above boundary condition setup. Note that the constant in (1) effectively produces $V_0(x,y) = \frac{x}b$, and is thus essentially accounting for the left and right boundaries.
