1. y
V =0 a
V =V0 x b b
V =0 z
𝛻 2 𝑉=0

3. 𝑌𝑑 2 (𝑋) 𝑋𝑌𝑑 𝑥 2 + 𝑋𝑑 2 (𝑌) 𝑋𝑌𝑑 𝑦 2 = 0
𝛻 2 𝑉=0
𝜕 2 𝑉 𝜕 𝑥 𝜕 2 𝑉 𝜕 𝑦 2 = 0
𝑉 𝑥,𝑦 =𝑋 𝑥 𝑌(𝑦)
𝜕 2 (𝑋𝑌) 𝜕 𝑥 𝜕 2 (𝑋𝑌) 𝜕 𝑦 2 = 0
𝑌𝜕 2 (𝑋) 𝜕 𝑥 𝑋𝜕 2 (𝑌) 𝜕 𝑦 2 = 0
𝑌𝑑 2 (𝑋) 𝑑 𝑥 𝑋𝑑 2 (𝑌) 𝑑 𝑦 2 = 0
𝑌𝑑 2 (𝑋) 𝑋𝑌𝑑 𝑥 𝑋𝑑 2 (𝑌) 𝑋𝑌𝑑 𝑦 2 = 0
𝑑 2 (𝑋) 𝑋𝑑 𝑥 𝑑 2 (𝑌) 𝑌𝑑 𝑦 2 = 0

4. 𝑑 2 (𝑋) 𝑋𝑑 𝑥 2 = 𝐶 1 and 𝑑 2 (𝑌) 𝑌𝑑 𝑦 2 = 𝐶 2
𝑑 2 (𝑋) 𝑋𝑑 𝑥 𝑑 2 (𝑌) 𝑌𝑑 𝑦 2 = 0
𝑋 𝑥 =𝐴 𝑒 𝑘𝑥 +𝐵 𝑒 −𝑘𝑥
Y 𝑦 =𝐶𝑠𝑖𝑛(𝑘𝑦)+𝐷𝑐𝑜𝑠(𝑘𝑦)
𝑑 2 (𝑋) 𝑋𝑑 𝑥 2 = 𝐶 1 and 𝑑 2 (𝑌) 𝑌𝑑 𝑦 2 = 𝐶 2
𝐶 1 = 𝑘 2 and 𝐶 2 = −𝑘 2
Completely general up to this point
𝑑 2 (𝑋) 𝑋𝑑 𝑥 2 = 𝑘 2 and 𝑑 2 (𝑌) 𝑌𝑑 𝑦 2 = −𝑘 2

5. Y 𝑦 =𝐶𝑠𝑖𝑛(𝑘𝑦)+𝐷𝑐𝑜𝑠(𝑘𝑦)
𝑋 𝑥 =𝐴 𝑒 𝑘𝑥 +𝐵 𝑒 −𝑘𝑥
Y 𝑦 =𝐶𝑠𝑖𝑛(𝑘𝑦)+𝐷𝑐𝑜𝑠(𝑘𝑦)
Boundary conditions
Symmetric in x:
⟹𝐴=𝐵
𝑋 𝑥 =𝐴 (𝑒 𝑘𝑥 + 𝑒 −𝑘𝑥 )
𝑋 𝑥 =2𝐴 (𝑒 𝑘𝑥 + 𝑒 −𝑘𝑥 ) 2 =2𝐴 cosh (𝑘𝑥)
⟹𝑉 𝑥,𝑦 =2𝐴 cosh (𝑘𝑥)(𝐶𝑠𝑖𝑛 𝑘𝑦 +𝐷𝑐𝑜𝑠 𝑘𝑦 )
⟹𝑉 𝑥,𝑦 = cosh (𝑘𝑥)(𝐶𝑠𝑖𝑛 𝑘𝑦 +𝐷𝑐𝑜𝑠 𝑘𝑦 )

6. ⟹𝑉 𝑥,𝑦 = cosh (𝑘𝑥)(𝐶𝑠𝑖𝑛 𝑘𝑦 +𝐷𝑐𝑜𝑠 𝑘𝑦 )
⟹𝑉 𝑥,0 = 0= 𝐷𝑐𝑜𝑠 0 )
⟹𝑉 𝑥,𝑦 =𝐶cosh (𝑘𝑥)𝑠𝑖𝑛 𝑘𝑦
⟹𝑉 𝑥,𝑎 =0=𝐶cosh (𝑘𝑥)𝑠𝑖𝑛 𝑘𝑎
⟹𝑘𝑎=𝑛𝜋
⟹𝑉 𝑥,𝑦 =𝐶cosh (𝑛𝜋𝑥/𝑎)𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎

7. ⟹𝑉 𝑥,𝑦 =𝐶cosh (𝑛𝜋𝑥/𝑎)𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎
𝑛=0,1,2,3…
One boundary condition to go

8. ⟹𝑉 𝑥,𝑦 =𝐶cosh (𝑛𝜋𝑥/𝑎)𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎
𝑛=0,1,2,3…
⟹𝑉 𝑏,𝑦 = 𝑉 0 =𝐶cosh (𝑛𝜋𝑏/𝑎)𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎
Never going to happen
⟹𝑉 𝑥,𝑦 = 𝐶 𝑛 cosh (𝑛𝜋𝑥/𝑎)𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎
⟹𝑉 𝑥,𝑦 = 𝑛=1 ∞ 𝐶 𝑛 cosh (𝑛𝜋𝑥/𝑎)𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎
⟹𝑉 𝑏,𝑦 = 𝑉 0 = 𝑛=1 ∞ 𝐶 𝑛 cosh (𝑛𝜋𝑏/𝑎)𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎

9. ⟹ 𝑉 0 = 𝑛=1 ∞ 𝐶 𝑛 cosh (𝑛𝜋𝑏/𝑎)𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎

10. ⟹ 𝑉 0 = 𝑛=1 ∞ 𝐶 𝑛 cosh (𝑛𝜋𝑏/𝑎)𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎

11. 0 𝑎 𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎 𝑠𝑖𝑛 𝑛 ′ 𝜋𝑦/𝑎 𝑑𝑦 =0 𝑓𝑜𝑟 𝑛≠ 𝑛 ′
⟹ 0 𝑎 𝑉 0 𝑠𝑖𝑛 𝑛 ′ 𝜋𝑦/𝑎 𝑑𝑦= 𝑛=1 ∞ 𝐶 𝑛 cosh (𝑛𝜋𝑏/𝑎) 0 𝑎 𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎 𝑠𝑖𝑛 𝑛 ′ 𝜋𝑦/𝑎 𝑑𝑦
0 𝑎 𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎 𝑠𝑖𝑛 𝑛 ′ 𝜋𝑦/𝑎 𝑑𝑦 =0 𝑓𝑜𝑟 𝑛≠ 𝑛 ′
= 𝑎 2 𝑓𝑜𝑟 𝑛= 𝑛 ′
Orthogonal, complete set of functions

12. 0 𝑎 𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎 𝑠𝑖𝑛 𝑛 ′ 𝜋𝑦/𝑎 𝑑𝑦 =0 𝑓𝑜𝑟 𝑛≠ 𝑛 ′
⟹ 0 𝑎 𝑉 0 𝑠𝑖𝑛 𝑛 ′ 𝜋𝑦/𝑎 𝑑𝑦= 𝑛=1 ∞ 𝐶 𝑛 cosh (𝑛𝜋𝑏/𝑎) 0 𝑎 𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎 𝑠𝑖𝑛 𝑛 ′ 𝜋𝑦/𝑎 𝑑𝑦
0 𝑎 𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎 𝑠𝑖𝑛 𝑛 ′ 𝜋𝑦/𝑎 𝑑𝑦 =0 𝑓𝑜𝑟 𝑛≠ 𝑛 ′
= 𝑎 2 𝑓𝑜𝑟 𝑛= 𝑛 ′
⟹ 0 𝑎 𝑉 0 𝑠𝑖𝑛 𝑛 ′ 𝜋𝑦/𝑎 𝑑𝑦= 𝐶 𝑛 ′ cosh ( 𝑛 ′ 𝜋𝑏/𝑎) 𝑎 2
⟹ 0 𝑎 𝑉 0 𝑠𝑖𝑛 𝑛 𝜋𝑦/𝑎 𝑑𝑦= 𝐶 𝑛 cosh ( 𝑛 𝜋𝑏/𝑎) 𝑎 2
0 𝑎 𝑉 0 𝑠𝑖𝑛 𝑛 𝜋𝑦/𝑎 𝑑𝑦= 𝑉 0 0 𝑎 𝑠𝑖𝑛 𝑛 𝜋𝑦/𝑎 𝑑𝑦=0 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
= 2𝑎 𝑉 0 𝑛𝜋 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑

13. ⟹𝑉 𝑥,𝑦 = 𝑛=1,3,5… ∞ 𝐶 𝑛 cosh (𝑛𝜋𝑥/𝑎)𝑠𝑖𝑛 𝑛𝜋𝑦/𝑎
⟹𝑉 𝑥,𝑦 = 𝑛=1,3,5… ∞ 4 𝑉 0 𝑛𝜋 cosh (𝑛𝜋𝑥/𝑎) cosh (𝑛𝜋𝑏/𝑎) sin (𝑛𝜋𝑦/𝑎)

14. ⟹𝑉 𝑥,𝑦 = 𝑛=1,3,5… ∞ 4 𝑉 0 𝑛𝜋 cosh (𝑛𝜋𝑥/𝑎) cosh (𝑛𝜋𝑏/𝑎) sin (𝑛𝜋𝑦/𝑎)
4 𝑉 0 𝜋 cosh (𝜋𝑥/𝑎) cosh (𝜋𝑏/𝑎) sin (𝜋𝑦/𝑎) + 4 𝑉 0 3𝜋 cosh (3𝜋𝑥/𝑎) cosh (3𝜋𝑏/𝑎) sin (3𝜋𝑦/𝑎)

15. Matlab code to visualize V(x,y)
[x,y]=meshgrid(-50:1:50,0:1:100);
v=0;
for i=1:100
v=v+(4./pi)*(cosh(((2*i-1)*pi*x)/100).*sin((2*i-1)*pi*y/100))/((2*i-1)*cosh((2*i-1)*pi/2));
surf(v);
view(-27,38);
end
surf(v)
axis([ ])
view(-27,38)

16. Numerical technique to solve the same problem (relaxation method)
Uses mean-value theorem
Create a 100  100 grid for V(x,y) or V(i,j)
Impose the boundary conditions first by setting initial values for certain V(i,j)
Generate V(i,j) for the entire grid by using mean value theorem:
V(i,j)= (V(i-1,j-1)+V(i,j-1)+V(i+1,j-1)+V(i+1,j)+V(i+1,j+1)+V(i,j+1)+V(i-1,j+1)+V(i-1,j))/8;
5. Iterate many times till you reach equilibrium (hence relaxation).
V(i-1,j+1)
V(i,j+1)
V(i+1,j+1)
V(i,j)
V(i-1,j)
V(i+1,j)
V(i-1,j-1)
V(i,j-1)
V(i+1,j-1)

17. Numerical technique to solve the same problem (relaxation method): MATLAB code
%create 100 by 100 grid for V(i,j)
for i=1:100
for j=1:100
V(i,j)=0;
end
%Impose the boundary conditions first by setting initial values for certain V(i,j)
V(i,1)=1;
V(i,100)=1;
surf(V)
view(-27,38);
pause %pause for effect
for k = 1:5000 %iterate 5000 times
for i = 2:99
for j = 2:99
V(i,j)= (V(i-1,j-1)+V(i,j-1)+V(i+1,j-1)+V(i+1,j)+V(i+1,j+1)+V(i,j+1)+V(i-1,j+1)+V(i-1,j))/8; %mean value theorem
view(-27,38)
pause(0.005)