User:Egm6322.s09.bit.la/r4

HOMEWORK

$$\mathbf{a}\mathbf{c}-\mathbf{b}^2=\left (ac-b^2 \right )\left (\Phi _{x} \psi_{y}-\Phi _{y} \psi_{x}\right )^2$$ page 14.2

Application:



Example:

$$T^{*}(\Theta )=T_0\left (1+cos^2\Theta \right )$$ where T0 is a constant

$$T^{*}(\Theta =0)=2T_0=T^{*}\left (\Theta =2\pi \right )$$

$$T^{*}(\Theta =\frac{\pi}{2} )=T_0=T^{*}(\Theta =\frac{3\pi}{2})$$

$$T^{*}(\Theta)=\frac{3T_0}{2}+\frac{T_0}{2}cos2\Theta$$

HOMEWORK

Principles of Superposition:

$$soln=solnT^{*}= \frac{3T_0}{2}=T_{1}^{*}(\Theta )$$

$$+solnT^{*}=\frac{T_0}{2}cos2\Theta=T_{2}^{*}(\Theta )$$

HOMEWORK

$$T(r,\Theta)=T_{1}\left (r,\Theta \right )+T_{2}\left (r,\Theta \right )$$

$$T^{*}\left (\Theta \right )=T_{1}^{*}\left (\Theta \right )+T_{2}^{*}\left (\Theta \right )$$

Problem P:

PDE:

$$\operatorname{Div}( grad T )=0$$

General Solution: Equation 2 p. 20-4.

$$T\left (r=a,\Theta \right )=T^{*}(\Theta )$$

Superposition: $$P=P_{1}+ P_{2}$$

Prob P1: $$\operatorname{Div}( grad T_1 )=0$$

such that:

$$T_{1}(r=a,\Theta)=T_{1}^{*}(\Theta)$$

Prob P2: $$\operatorname{Div}( grad T_2 )=0$$

such that:

$$T_{2}(r=a,\Theta)=T_{2}^{*}(\Theta)$$

HOMEWORK

$$T_{1}(r,\Theta)=\frac{3T_0}{2}$$

For the Problem $$T_{2}:T_{2}(a,\Theta)=T_{2}^{*}(\Theta)=\frac{T_0}{2}cos2\Theta$$

HOMEWORK

A0=0, Cn=0, Dn=0 why

$$T_{2}(r,\Theta)=\sum_{n=1}^{\infty }r^{n} \left \{A_ncosn\Theta +B_nsinn\Theta \right \}$$

HW: Using b.c show:

$$A_1=0,A_2\neq 0,A_3=A_4=...A_n=0$$

$$B_n= 0,n= 1,...$$

$$A_2=\frac{T_0}{2a^2}$$

Final Solution:

$$T(r,\Theta)=T_0\left [\frac{3}{2}+\frac{1}{2}\left (\frac{r}{a^2} \right )cos2\Theta \right ]$$

In general, for arbitrary function $$T^{*}\left (\Theta \right )$$ but periodic

i.e. $$T^{*}\left (\Theta +\mathit{K}2\pi \right )=T^{*}(\Theta)$$

for all $$\Theta$$ and any K = constant



this is not periodic (not acceptable)

HOMEWORK: General solution

$$T(r,\Theta)=C_{0}+\sum_{n=1}^{\infty }r^{n} \left \{A_ncosn\Theta +B_nsinn\Theta \right\}$$

where: rn=an

$$T(a,\Theta)=C_{0}+\sum_{n=1}^{\infty }a^{n} \left \{A_{n}cosn\Theta +B_{n}sinn\Theta \right\}$$

Find the Fourier coefficients C0,An,Bn

$$C_{0}=\frac{1}{2\pi }\int_{\Theta =0}^{2\pi }T^{*}(\theta )d\theta $$

$$A_{n}=\frac{1}{2\pi }\int_{\Theta =0}^{2\pi }T^{*}(\theta )cosn\Theta d\theta $$

$$B_{n}=\frac{1}{2\pi }\int_{\Theta =0}^{2\pi }T^{*}(\theta )sinn\Theta d\theta $$

Due to orthogonality of Fourier basis function $$\left \{1,cosu\Theta,sinu\Theta \right \}$$

Homework: Verification of Insulation

$$T\left (r,\Theta \right )=T_rcos4\Theta$$

The Fourier's law or law of heat conduction:

$$\mathbf{q}=\kappa gradT $$ Where T is the temperature gradient

If there is no heat flow at $$ \Theta=0;\Theta=\frac{\pi }{2} $$ then

$$\mathbf{q}=0\Rightarrow gradT=0$$ so

$$\frac{\partial T}{\partial \Theta}=0$$

Homework: Verification of the General Solution

The general solution of the Laplace equation:

$$u(r,\theta)=A_{0}ln \ r+\sum_{n=1}^{\infty }r^{n}\left [A_{n}cos(n\theta)+B_{n}sin(n\theta) \right ]+\sum_{n=1}^{\infty }\frac{}{r^{n}}\left [C_{n}cos(n\theta)+D_{n}sin(n\theta) \right ]+C_{0}$$

Where the partial differential equation governing the problem is:

$$ \nabla ^2T=\frac{\partial^2 T}{\partial r^2}+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{1}{r^2}\frac{\partial^2 T}{\partial \Theta^2}=0$$

Applying the boundary conditions explained in the previous homework

$$T_{1}(r,\Theta)=\frac{3T_0}{2};T_{2}(r,\Theta)=\frac{T_0}{2}cos2\Theta;A_0=0;C_n=0;D_n=0$$

The resulting solution is:

$$T(r,\Theta)=C_{0}+\sum_{n=1}^{\infty }r^{n} \left \{A_ncosn\Theta +B_nsinn\Theta \right\}$$