Question

Find the solution u(x, y) of Laplace's equation in the rectangle 0 < x < a, 0 < y < b, that satisfies the boundary conditionsu(0, y) = 0, u(a, y) = 0, 0< y< b,u(x, 0) = 0, u(x, b) = g(x), 0 ≤ x ≤ a.

Answer 1

Answer:

$u(x,t) = \frac{g(x)}{\sin{(kb)}}\sin{(ky)} Fsinh{(kx)}$

Step-by-step explanation:

The Laplace's equation in rectangular coordinates is:

$\frac{\partial^2 u}{\partial x^2} +\frac{\partial^2 u}{\partial y^2} = 0$

For the solution, we assume that u can be write as

$u(x,y) = X(x)Y(y)$

Replace

$Y(y)\frac{d^2X}{dx^2} +X(x)\frac{d^2Y}{dy^2} =0$

Divide $X(x) Y(y)$

$\frac{1}{X(x)}\frac{d^2X}{dx^2}= -\frac{1}{Y(y)}\frac{d^2Y}{dy^2}$

The only way that this can be true, is that every term is the same constant, so we say that

$\frac{1}{X(x)}\frac{d^2X}{dx^2} =  -\frac{1}{Y(y)}\frac{d^2Y}{dy^2} = k^2$

With $k^2$ a constant.

Now, we solve every part. For $X(x)$

$\frac{d^2X}{dx^2} = k^2X(x)$

The solution is:

$X(x) = Ee^{kx} + De^{-kx}$

For $Y(y)$

$\frac{d^2Y}{dy^2} = -k^2Y(y)$

Is the differential equation for a harmonic oscillator, so the solution is

$Y(y) = A\cos{ky} + B\sin{ky}$

Now, we evaluate the boundary conditions:

$u(x,0) = 0 \rightarrow Y(0) = 0$

$Y(0) = A\cos{0} + B\sin{0} = 0; A = 0$

$Y(y) = B\sin{ky}$

The other

$u(x,b) = 0 \rightarrow Y(b) = 0$

$Y(b) = B\sin({kb}) = g(x); B = \frac{g(x)}{sin({kb})}$

$Y(y) = \frac{g(x)}{\sin({kb})} sin{(ky)}$

For $X(x)$:

$u(0,y) = 0 \rightarrow X(0) =0$

$X(0) = Ee^{0} + De^{-0} =0; E+D=0$

$E = -D$

So, $X(x) = E(e^{kx} - e^{-kx} ) = 2Esinh(kx) = Fsinh(kx))$

The other condition:

$u(a,y) = 0 \rightarrow X(a) = 0$

$X(a) = F(sinh{(ka)}) = 0$

So,

$u(x,t) = \frac{g(x)}{\sin{(kb)}}\sin{(ky)} Fsinh{(kx)}$