User:Egm6322.s09.bit.sahin/r4

Another Application:

A domain which is a quadrant of annulus is subjected a boundary conditions that

temperature at $$r=b$$ is $$T\left (r=b,\theta \right )=T_{b}cos4\theta $$

temperature at $$r=a$$ is $$T\left (r=a,\theta \right )=T_{a}cos4\theta$$



where $$T_{a}$$ and $$T_{b}$$ are given constants. Also, the boundaries at $$\theta=0$$ and $$\theta =\pi/2$$ are insulated which means no heat flow at these boundaries. According to the Fourier's law:

$$\underline{q}=\underline{\kappa } \cdot gradT$$

here $$\underline{q}$$ denotes heat flux tensor. Relevant to the insulated conditions,

$$\underline{q}=0\Rightarrow gradT=0\Leftrightarrow \frac{\partial T}{\partial \theta }=0$$ on $$\theta=0,\theta =\pi/2 $$

The general solution of the Laplace Eq. is

$$T\left (r,\theta \right )=A_{0}lnr+\sum_{n=1}^{\infty }r^{n}\left (A_{n}cosn\theta +B_{n}sinn\theta  \right )+  \sum_{n=1}^{\infty }\frac{1}{r^{n}}\left (C_{n}cosn\theta +D_{n}sinn\theta  \right )+C_{0}$$

We can eliminate the following terms that do not satisfy the boundary conditions:

1) $$A_{0}lnr+C_{0}$$, indipendent of $$\theta$$

2) $$B_{n}sinn\theta$$, $$D_{n}sinn\theta$$, cannot satisfy the $$\frac{\partial T}{\partial \theta }\left (r,\theta =0 \right )=0$$

To show the second one, let's differentiate the general solution with respect to $$\theta$$

$$\frac{\partial T}{\partial \theta } =\sum_{n=1}^{\infty }r^{n}n\left (-A_{n}sinn\theta +B_{n}cosn\theta \right )+\sum_{n=1}^{\infty }\frac{1}{r^{n}}n\left (-C_{n}sinn\theta +D_{n}cosn\theta  \right )$$

Since $$sinn\theta=0$$ and $$cosn\theta=1$$ at $$\theta=0$$, $$B_{n}$$ and $$C_{n}$$ must be zero to satify the condition that $$\frac{\partial T}{\partial \theta }=0$$.

Thus the solution has the following form

$$T\left (r,\theta \right )=\sum_{n=1}^{\infty }\left (A_{n}r^{n}+\frac{C_{n}}{r_{n}} \right )cosn\theta$$

Using the boundary conditions, we have

$$T_{a}cos4\theta =\sum_{n=1}^{\infty }\left (A_{n}a^{n}+\frac{C_{n}}{a_{n}} \right )cosn\theta$$

$$T_{b}cos4\theta =\sum_{n=1}^{\infty }\left (A_{n}b^{n}+\frac{C_{n}}{b_{n}} \right )cosn\theta$$

Since the only term in the boundary condition is the term with $$cos4\theta$$, boundary conditions can only be satisfied for n=4, all other $$A_{n} $$ and $$C_{n} $$ must be zero. Then we have,

$$\begin{bmatrix} a^{8} & 1 \\ b^{8} & 1 \end{bmatrix}\begin{Bmatrix} A_{4}\\C_{4} \end{Bmatrix}=\begin{Bmatrix} a^{4}T_{a}\\b^{4}T_{b}

\end{Bmatrix} $$

Eventually the solution for the temperature distribution is

$$T\left (r,\theta \right )=\left \{\frac{a^{4}b^{4}T_{b}}{\left (b^{8}-a^{8} \right )}\left [\frac{r^{4}}{a^{4}}-\frac{a^{4}}{r^{4}} \right ]-\frac{a^{4}b^{4}T_{a}}{b^{8}-a^{8}}\left [\frac{r^{4}}{b^{4}}-\frac{b^{4}}{r^{4}} \right ] \right \}cos4\theta$$