User:Egm6321.f09.team3.diggs/HW5

= Problem 1: Solving for $$ \Delta \Psi $$ =

Given
The differential equation:

$$ \Delta \Psi = \frac{1}{h_1 \cdot h_2 \cdot h_3} \sum \frac {\partial}{\partial \xi} [ \frac{h_1 \cdot h_2 \cdot h_3}{(h_i)^2} \cdot \frac {\partial \psi}{\partial \xi}] $$

where

$$ (h_1,h_2,h_3) = (1, r \cdot cos \theta, r) $$

$$ (\xi_1, \xi_2, \xi_3) = (r, \theta, \phi) $$

Find
$$ \Delta \Psi_i $$ for i = (1, 2, 3)

Solution
for i = 1:

$$ \Delta \Psi_1 = \frac{1}{r^2 \cdot cos \theta} \cdot \frac {\partial}{\partial r} [ \frac{r^2 \cdot cos \theta}{(1)^2} \cdot \frac {\partial \psi}{\partial r}] $$

for i = 2:

$$ \Delta \Psi_2 = \frac{1}{r^2 \cdot cos \theta} \cdot \frac {\partial}{\partial \theta} [ \frac{r^2 \cdot cos \theta}{(r \cdot cos \theta)^2} \cdot \frac {\partial \psi}{\partial \theta}] $$

for i = 3:

$$ \Delta \Psi_3 = \frac{1}{r^2 \cdot cos \theta} \cdot \frac {\partial}{\partial \phi} [ \frac{r^2 \cdot cos \theta}{(r)^2} \cdot \frac {\partial \psi}{\partial \phi}] $$

Combined:

$$ \Delta \Psi = \frac{1}{r^2 \cdot cos \theta} [\frac {\partial}{\partial r} [ r^2 \cdot cos \theta \cdot \frac {\partial \psi}{\partial r}] + \frac {\partial}{\partial \theta} [ \frac{1}{cos \theta} \cdot \frac {\partial \psi}{\partial \theta}]  +  \frac {\partial}{\partial \phi} [ cos \theta \cdot \frac {\partial \psi}{\partial \phi}] $$

Simplified:

$$ \Delta \Psi = \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \frac{\partial \psi}{\partial r}) + \frac{1}{r^2 \cdot cos \theta} \frac{\partial }{\partial \theta}(\frac {1}{cos \theta} \frac {\partial \psi}{\partial \theta}) + \frac{1}{r^2} \frac{\partial ^2 \psi}{\partial \phi^2} $$

= Problem 2: Prove $$ \lambda (\lambda +1) = k $$ =

Given
The differential equation:

$$ r^2R'' + 2rR' - kR = 0 $$

where

$$ R(r) = r^{\lambda} $$

Find
Prove $$ \lambda (\lambda +1) = k $$

Solution
First, take the derivatives of R:

$$R' = \lambda r^{\lambda - 1}$$

$$R'' = \lambda (\lambda - 1) r^{\lambda - 2}$$

Substitute in:

$$ r^2 \cdot \lambda (\lambda - 1) r^{\lambda - 2} + 2r \cdot \lambda r^{\lambda - 1} - k \cdot r^{\lambda} = 0 $$

Simplify:

$$ \lambda (\lambda - 1) r^{\lambda} + 2 \lambda r^{\lambda} - k r^{\lambda} = 0 $$

Divide by $$ r^{\lambda} $$:

$$ \lambda^2 - \lambda + 2\lambda = k $$

Simplify:

$$ \lambda(\lambda + 1) = k $$, as expected!

= Contributing Team Members =

Angela Diggs

= Play area =

$$ 0 = -\dfrac{dp}{dx} + \mu \dfrac{\partial^2 w}{\partial x^2} + \mu \dfrac{\partial^2 w}{\partial y^2} $$

$$x* = \frac{x}{L}$$

$$y* = \frac{y}{L}$$

$$w* = \frac{w}{-\frac{L^2}{\mu} \dfrac{dp}{dx}}$$

$$\nabla^2{w*} = \dfrac{\partial^2 w*}{\partial x*^2} + \dfrac{\partial^2 w*}{\partial y*^2} = -1$$

$$\nabla^2 w* = -1$$

$$\nabla^2 W = \nabla^2 w* - \nabla^2 f = 0$$

$$\nabla^2 f = -1$$

$$f = \frac{1}{2} (a^2 - y*^2)$$

$$f = \frac{1}{2} (a^2 - x*^2)$$

$$\nabla^2 w* = \nabla^2 W = \nabla^2 f$$

$$w* = W + f$$

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

$$w*(x,y=a) = 0$$

$$\dfrac{\partial w*(x,0)}{\partial y*} = 0$$

$$w*(x=1,y) = 0$$

$$\dfrac{\partial w*(0,y)}{\partial x*} = 0$$

$$0 = \nabla^2 W = \nabla^2[X(x)Y(y)] = XY + YX$$

$$\frac{Y}{Y} = -\frac{X}{X} = -\alpha$$

$$\frac{Y}{Y} = \frac{X}{X} = -\alpha$$

$$Y(y) = Acos(\alpha y) + B sin (\alpha y)$$

$$X(x) = C \exp(\alpha x) + D \exp(-\alpha x)$$

$$W(x,y) = (Acos(\alpha y) + B sin (\alpha y)) (C \exp(\alpha x) + D \exp(-\alpha x))$$

$$\dfrac{\partial Y(0)}{\partial y*} = 0 = B$$

$$\dfrac{\partial X(0)}{\partial x*} = 0 => C = D $$

$$W(x,y) = Acos(\alpha y) [\exp(\alpha x) + \exp(-\alpha x)]$$

$$W(x,a) = 0 => A cos(\alpha a) = 0$$

$$\alpha_n = \frac{(2n-1) \pi}{2a}$$

$$W(x,y) = \sum A_n cos(\alpha_n y) (2 cosh(\alpha_n x))$$

$$W(1,y) = \sum A_n cos(\alpha_n y) (2 cosh(\alpha_n)) = 0$$

$$A_n = \frac{2}{\alpha_n^3 a} \frac{a\alpha_n cos(\alpha_n a) - sin(\alpha_n a)}{cosh(\alpha_n)}$$

$$W(x,y) = \sum \frac{4}{\alpha_n^3 a} \frac{a\alpha_n cos(\alpha_n a) - sin(\alpha_n a)}{cosh(\alpha_n)}cos(\alpha_n y) cosh(\alpha_n x)$$

$$w*(x,y) = \frac{1}{2} (a^2 - y^2) + \sum \frac{4}{\alpha_n^3 a} \frac{a\alpha_n cos(\alpha_n a) - sin(\alpha_n a)}{cosh(\alpha_n)}cos(\alpha_n y) cosh(\alpha_n x)$$

$$W(x,y) = \sum A_n cos(\alpha_n y) cos(\alpha_n x)$$

= in error =

$$w*(x,y) = \frac{1}{2}(a^2 - y*^2) + \sum A_n cos(\alpha_n y) (\exp(\alpha_n x) + \exp(\alpha_n x))$$

$$w*(1,y) = \frac{1}{2}(a^2 - y*^2) + \sum A_n cos(\alpha_n y) (1 + \exp(-2\alpha_n)) = 0$$

$$\frac{1}{2}(y*^2 - a^2) = \sum A_n cos(\alpha_n y) (1 + \exp(-2\alpha_n))$$