Solve Laplace\'s equation with homogeneous vertical boundary conditions: (^2 u)/

Question

Solve Laplace's equation with homogeneous vertical boundary conditions:

(^2u)/(x^2) + (^2u)/(y^2) = 0 ,0<x<a, 0<y<b

u(0,y) = 0,                 u(a,y) = 0, 0<y<b

u(x,0) = f(x),             u(x,b) = g(x), 0<x<a

We try for a separable solution u(x,y) = X(x)Y(y), plugging XY into the PDE for u we get: X’’Y + X’’Y = 0

The PDE can be separated into an ODE in X and an ODE in Y (placing constants in the DE for Y): X’’/X = -Y’’/Y =

The boundary conditions in the PDE translate into initial conditions for the differential equation in X:

Since these differential equations are independent of each other, they can be separated

DE in X: X’’ + X = 0

IC's in X:    X(0) = 0      X(a) = 0

We have solved this S-L problem before, without the unknown constant, the solution is X(x) = ______________________

Therefore, = _____________ and the differential equation in Y becomes:

DE in Y: ____________________ = 0

Using A and B as the constants, the solution for Y is Y(y) = ______________

So the PDE admits a series solution of the form:

u(x,y)= Ancosh((n/a)y) + Bnsinh((n/a)y) sin((n/a)x)

Now we can apply the lateral boundary conditions:

u(x,0)= Ansin((n/a)x)= f(x)

u(x,b)= Ancosh((n/a)b) + Bnsinh((n/a)b) sin((n/a)x)=g(x)

The first series we recognize as a Fourier series on [0,a], so that

An = _______________________________

The second series is also a Fourier series with coefficient

Ancosh((n/a)b) + Bnsinh((n/a)b) = ________________________

Which we solve for

Bn = __________________________________________

Explanation / Answer

Here we consider the following boundary value problem: Let K and L be positive real numbers. Let f1 and f2 be real functions defined on [0,K] and let g1 and g2 be real functions defined on [0,L]. Find the real function u defined on a rectangle {(x,y):0xK,0yL} which satisfies the Laplace PDE

2ux2+2uy2=0(1)

and the boundary conditions

u(x,0)u(0,y)=f1(x),=g1(y),u(x,L)u(K,y)=f2(x),0xK,=g2(y),0yL.(2)(3)

Here we consider the following boundary value problem: Let K and L be positive real numbers. Let f1 and f2 be real functions defined on [0,K] and let g1 and g2 be real functions defined on [0,L]. Find the real function u defined on a rectangle {(x,y):0xK,0yL} which satisfies the Laplace PDE

2ux2+2uy2=0(1)

and the boundary conditions

u(x,0)u(0,y)=f1(x),=g1(y),u(x,L)u(K,y)=f2(x),0xK,=g2(y),0yL.(2)(3)

2ux2+2uy2=0,u(x,0)=0,u(x,L)=0,0xK.(8)

A(x)B(y)+A(x)B(y)=0.

A(x)A(x)=B(y)B(y)=

A(x)=A(x),B(y)=B(y).

A(x)B(0)=0,A(x)B(L)=0,0xK,

B(y)=B(y),B(0)=0,B(L)=0.(9)

A(x)=nA(x).(10)

n=(nL)2,sin(nLy),nN.(11)

cosh(nLx),sinh(nLx),nN.

sinh(nL(Kx))sinh(nLK),sinh(nLx)sinh(nLK),nN.(12)

sin(nLy)sinh(nL(Kx))sinh(nLK),sin(nLy)sinh(nLx)sinh(nLK),nN.(13)

u1(x,y)=n=1ansin(nLy)sinh(nL(Kx))sinh(nLK)+n=1bnsin(nLy)sinh(nLx)sinh(nLK)(14)

n=1ansin(nLy)=g1(y)

n=1bnsin(nLy)=g2(y)

an=2LL0g1(y)sin(nLy)dy,nN,

bn=2LL0g2(y)sin(nLy)dy,nN.he solution of the given boundary value problem (1), (2), (3) is the sum u1(x,y)+u2(x,y), that is

u(x,y)=n=1ansin(nLy)sinh(nL(Kx))sinh(nLK)+n=1bnsin(nLy)sinh(nLx)sinh(nLK)+n=1cnsin(nKx)sinh(nK(Ly))sinh(nKL)+n=1dnsin(nKx)sinh(nKy)sinh(nKL)

anbncndn=2LL0g1(y)sin(nLy)dy,nN,=2LL0g2(y)sin(nLy)dy,nN,=2KK0f1(x)sin(nKx)dx,nN,=2KK0f2(x)sin(nKx)dx,nN.

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