I don't know how to start this one

• Oct 7th 2010, 09:16 PM
Jimmy_W
I don't know how to start this one
Find the solutions to $\bigtriangledown^2 u = 0$in two and three dimensions subject to the following boundary conditions

(a) $u(x) = u_0$ for $||x|| = a$ and $u = u_1$ for $||x|| = b$

(b) $u(x) = u_0$ for $||x|| = a$ and $\bigtriangledown u \cdot n = k$ for $||x|| = b$

Note that the boundary conditions do not depend on $\theta$or $\phi$ (in 3D)
• Oct 8th 2010, 05:28 AM
Jester
Since the boundary conditions are independent of $\theta$ or $\theta$ and $\phi$, I might suggest switching the PDEs into polar coordinates and looking for solution in terms of the radius only.
• Oct 18th 2010, 05:15 AM
Jimmy_W
Quote:

Originally Posted by Jimmy_W
Find the solutions to $\bigtriangledown^2 u = 0$in two and three dimensions subject to the following boundary conditions

(a) $u(x) = u_0$ for $||x|| = a$ and $u = u_1$ for $||x|| = b$

(b) $u(x) = u_0$ for $||x|| = a$ and $\bigtriangledown u \cdot n = k$ for $||x|| = b$

Note that the boundary conditions do not depend on $\theta$or $\phi$ (in 3D)

I'm having trouble understanding exactly what the initial conditions are saying here.

If I had to solve this PDE with conditions (another question I have):

$u(0,y) = 0$

$u(1,y) = 0$

$u(x,0) = 1$

$u(x,1) = 1$

I would do it as follows:

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

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

$X''(x)Y(y) + X(x) Y''(y) = 0$

$- \frac{X''(x)}{X(x)} = \frac{Y''(y)}{Y(y)} = k^2$

So

$X''(x) = -k^2 X(x)$ and $Y''(y) = k^2 Y(y)$

Solving these,

$X(x) = C_1 \cos(kx) + C_2 \sin(kx)$ and $Y(y) = C_3 e^{ky} + C_4 e^{-ky}$

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

$u(x,y) = C_1 \cos(kx) + C_2 \sin(kx)) (C_3 e^{ky} + C_4 e^{-ky})$ .........(Eqn 1)

Then applying all conditions and skipping a few steps the solution is

$\displaystyle u(x,y) = \sum_{n=1}^{\infty} 2 \frac{1 - (-1)^n}{n \pi (e^{n \pi} - e^{-n \pi})} \sin (n \pi x) (e^{n \pi y} - e^{-n \pi y})$

But I'm getting confused with the initial conditions here, and how to go from Eqn (1) onwards with the intitial conditions onwards i.e.

(a) $u(x) = u_0$ for $||x|| = a$ and $u = u_1$ for $||x|| = b$

(b) $u(x) = u_0$ for $||x|| = a$ and $\bigtriangledown u \cdot n = k$ for $||x|| = b$

For the 3d case, I have

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

Assume $u(x,y,z) = X(x)Y(y)Z(z)$

Then

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

Dividing through

$\displaystyle \frac{1}{X(x)} \frac{d^2 X(x)}{dx^2} + \frac{1}{Y(y)} \frac{d^2 Y(y)}{dy^2} + \frac{1}{Z(z)} + \frac{d^2 Z(z)}{dz^2}$

Which results in 3 ODE's

$\frac{d^2 X(x)}{dx^2} - aX(x) = 0$

$\frac{d^2 Y(x)}{dy^2} - bY(y) = 0$

$\frac{d^2 Z(z)}{dz^2} + (a+b)Z(z) = 0$

which have solutions

$X(x) = C_1 e^{\pm \sqrt{a}x}$

$Y(x) = C_2 e^{\pm \sqrt{a}y}$

$Z(z) = C_3 e^{\pm i \sqrt{a+b}z}$

So, due to superposition,

$u(x,y,z) = C e^{\pm \sqrt{a}x} e^{\pm \sqrt{a}y} e^{\pm i \sqrt{a+b}z}$

which is all I have...I don't understand the form of the initial conditions.
• Oct 18th 2010, 05:16 AM
Jimmy_W
Quote:

Originally Posted by Danny
Since the boundary conditions are independent of $\theta$ or $\theta$ and $\phi$, I might suggest switching the PDEs into polar coordinates and looking for solution in terms of the radius only.

Here is my attempt for general solution in polar co-ordinates in 2d..

$\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0$

Assume $u(r, \theta) = R(r)\Theta (\theta)$

Skipping some steps, I get

$R(r) = C_1 r^k + C_2 r^{-k}$

and

$\Theta(\theta) = C_3 \cos(k \theta) + C_4 \sin (k \theta)$

so

$u(r, \theta) = (C_1 r^k + C_2 r^{-k})(C_3 \cos(k \theta) + C_4 \sin (k \theta))$

I may be able to continue, but again, I'm having trouble interpreting the initial conditions (a) and (b).