The electrostatic potential U(x,y) in a rectangular region (0≤x≤20m, 0≤ y ≤40m) is governed by the 2 dimensional Laplace partial differential equation: d^2 U(x,y)/dx^2 + d^2 U(x,y)/dy^2 =0.
Determine Potential U(x,y) in the rectangular region 0 ≤ x ≤ 20 and 0 ≤ y ≤ 40 for the following boundary conditions:
U(0, y) = U(20, y) = U(x,0) = 0 Volts and U(x,40) = 110V
Answers
Potential U inside the
rectangular area (0<= x<=a, 0<= y <= b) is found as:
U(x,y)
= U₁(x,y) + U₂(x,y) + U₃(x,y) + U₄ (x,y) --- (1)
U₂(x,y) is found by setting Potential on boundaries x=0, x=a, y=0 to
zero and finding the solution to the Laplace 2-dimensional Equation.
Others in (1) are found by setting appropriate boundary conditions to 0.
We are given conditions:
U₁(0,y)=0, U₂(x,b) = 110 V, U₃(a,y)=0 , U₄(x,0) = 0
a = 20 units, b
= 40 units
Laplace PDE 2-dim of 2nd degree:
δ²U(x,y)/δx² + δ²U(x,y)/δy² = 0 --- (2)
Simple solutions exist like U₂(x,y) = a (x² - y²) or, U(x,y) = P(x)+Q(y),
P(x), Q(y) are 2nd degree polynomials
in x, y respectively.
But the boundary conditions cannot be satisfied by
these solutions.
So we use separation of variables method.
U₂(x,y) = X(x) * Y(y) ---- (3)
So differentiating and substituting in (2),
we get:
X''(x)
Y(y) + Y''(y) X(x) = 0 -- (4)
=> X''(x) /X(x) = - Y''(y)/ Y(x)
= λ² (say) constant ≠ 0.
We choose the (+ or -) sign of λ² in such a
way that in the direction with both boundaries are zero, we get Sine as
solution. In the direction of non-zero boundary value, we get Sinh as
solution.
X''(x) + λ² X(x) = 0 --- (5) Y''(y) - λ² Y(y) = 0 --- (6)
Solutions are: X(x) = A Cos(λx) + B Sin(λx)
---(7)
Y(y) = C
Cosh(λx) + D Sinh(λx) ---- (8)
Apply boundary conditions now.
U₂(0,y) = 0 => X(0) = 0
=> A=0.
U₂(x,0)= 0 => Y(0)= 0
=> C = 0 => U₂(x,y)=
BD Sin(λx) Sinh(λy) -- (9)
U₂(a,y) = 0 => X(a) = 0
Sin(λa) = 0 => λ = nπ/a, n = 1, 2,3, 4....
U₂(x,b) = 110V => Sin(nπ x/a) Sinh(2πb/a) = 110 ---
(10)
This cannot be solved with one value
of n. Fourier sources is to be used.
Let U₂ⁿ (x,y) =
Sin(nπ x/a) Sinh (2π y/a)
[tex]U(x,y)=\Sigma_{n=1}^\infty\ c_n\
Sin(\frac{2\pi}{a}x)\ Sinh(\frac{2\pi}{a}y) ---- (11)\\\\c_n=\frac{2}{a\
Sinh(\frac{2\pi}{a}b)} \int_0^a {U_2(x,b)*Sin(\frac{n\pi}{a}x)} \,
dx\\\\=\frac{2*110}{a\ Sinh(\frac{2\pi}{a}b)} \int_0^a {Sin(\frac{n\pi}{a}x)} \,
dx\\\\=\frac{220}{a\ Sinh(\frac{2\pi}{a}b)}\frac{a}{n\pi}
[-Cos(\frac{n\pi}{a}x)}]_0^a\\\\=\frac{220*2}{n\pi\ Sinh(\frac{2\pi}{a}b)} ---
(12)\\\\[/tex]
Finally, the solution
:
[tex]U(x,y)=\Sigma_{n=1}^\infty\
\frac{440}{n\pi\ Sinh(\frac{2\pi}{a}b)}\ Sin(\frac{2\pi}{a}x)\
Sinh(\frac{2\pi}{a}y) ---- (13)\\\\U(x,y)=\Sigma_{n=1}^\infty\ \frac{440}{n
\pi\ Sinh(4\pi)}\ Sin(\frac{\pi}{10}x)\ Sinh(\frac{\pi}{10}y) ----(14)\\ [/tex]
