Mtg 37: Wed, 10 Nov 10

Heat Problem continued P.36-3

Eq.(1)P.36-2 : ${\displaystyle R_{n}(r)\ }$

${\displaystyle r\rightarrow \ 0,r^{-(n+1)}\rightarrow \ +\infty \ }$

${\displaystyle n\geq 0\Rightarrow n+1>0\ }$

${\displaystyle \Rightarrow \ }$ set:

${\displaystyle \displaystyle B_{n}=0}$ for physically meaningful solution

(1)

Eq.(1)P.36-3 : ${\displaystyle \Theta _{n}(\theta \ )\ }$

${\displaystyle \theta \rightarrow \pm {\frac {\pi \ }{2}}\Rightarrow \mu =\sin \theta \rightarrow \pm 1\ }$

${\displaystyle \left|Q_{n}(\mu )\right|\rightarrow +\infty \ }$

${\displaystyle \Rightarrow \ \ }$ set:

${\displaystyle \displaystyle D_{n}=0}$ for physically meaningful solution

(2)

HW7.1

Plot ${\displaystyle \left\{P_{0},P_{1},P_{2},P_{3}\right\}\ }$ in Figure 1

and ${\displaystyle \left\{Q_{0},Q_{1},Q_{2},Q_{3}\right\}\ }$ in Figure 2.

Observe ${\displaystyle P_{n}(\mu )\ }$ and ${\displaystyle Q_{n}(\mu )\ }$ as ${\displaystyle \mu \rightarrow \pm 1\ }$.

END HW7.1

Combine Eq.(1), Eq.(2) and Eq.(2)P.36-3 :

${\displaystyle \Psi (r,\theta )=\sum _{n=0}^{\infty }(A_{n}r^{n})[C_{n}P_{n}(\sin \theta )]=\sum _{n=0}^{\infty }{\bar {A}}_{n}r^{n}P_{n}(\sin \theta )\ }$ ,

where ${\displaystyle {\bar {A}}_{n}=A_{n}C_{n}\ }$ and ${\displaystyle \sin \theta =\mu \ }$.

${\displaystyle \displaystyle \Psi (r,\theta )=\sum _{n=0}^{\infty }A_{n}r^{n}P_{n}(\sin \theta )}$

(1)

which is similar to the Fourier-Legendre series (see below), since the basis functions involve the Legendre polynomials ${\displaystyle \displaystyle P_{n}(\mu )}$, with ${\displaystyle \displaystyle \mu =\sin \theta }$.

NOTE: Fourier series

${\displaystyle f(\theta )=1\cdot a_{0}+\sum _{n=1}^{\infty }a_{n}\cos n\omega \theta +\sum _{n=1}^{\infty }b_{n}\sin n\omega \theta \ }$

Basis functions ${\displaystyle =\left\{1,\cos n\omega \theta ,\sin n\omega \theta \right\}\ }$ , where ${\displaystyle n=1,2,3,...\ }$ linearly independent.
END NOTE

NOTE: Vector space ${\displaystyle \displaystyle \mathbb {R} ^{n}}$

${\displaystyle \left\{\mathbf {b} _{1},\mathbf {b} _{2},...,\mathbf {b} _{n}\right\}\in \mathbb {R} ^{n}\ }$ basis vectors.

${\displaystyle \left\{\mathbf {b} _{1},\mathbf {b} _{2},...,\mathbf {b} _{n}\right\}\ }$ are linearly independent.

Consider: ${\displaystyle \mathbf {v} \in \mathbb {R} ^{n}\Rightarrow \ \mathbf {v} =\sum _{i=1}^{n}a_{i}\mathbf {b} _{i}=a_{i}b_{i}\ }$

Find ${\displaystyle a_{i}\ }$ for ${\displaystyle i=1,2,...,n\ }$

${\displaystyle \mathbf {v} \cdot \mathbf {b} _{j}=\sum _{i}a_{i}\mathbf {b} _{i}\cdot \mathbf {b} _{j}\ }$ , for ${\displaystyle j=1,2,...,n\ }$

${\displaystyle \Rightarrow \left[(\mathbf {b} _{i}\cdot \mathbf {b} _{j})\right]\left\{a_{i}\right\}=\left\{(\mathbf {v} \cdot \mathbf {b} _{j})\right\}\ }$

where ${\displaystyle [(\mathbf {b} _{i}\cdot \mathbf {b} _{j})]\in \mathbb {R} ^{n\times n}\ }$ is the Gram Matrix ${\displaystyle \mathbf {\Gamma } \ }$ and is n rows by n columns, ${\displaystyle \left\{a_{i}\right\}\ }$ is a n row by 1 column matrix and ${\displaystyle \left\{(\mathbf {v} \cdot \mathbf {b} _{j})\right\}\ }$ is a n row by 1 column matrix.

Theorem: ${\displaystyle \left\{(\mathbf {b} _{1},\cdots ,\mathbf {b} _{n})\right\}\ }$ is linearly independent iff the determinant of ${\displaystyle \mathbf {\Gamma \ } \neq 0\ }$ , where iff is defined as "if and only if" ${\displaystyle (\iff )\ }$
END NOTE

Generalize to functions: Inner (scalar) product of two functions:

${\displaystyle g:\left[a,b\right]\rightarrow \ \mathbb {R} \ }$

${\displaystyle h:\left[a,b\right]\rightarrow \ \mathbb {R} \ }$

where ${\displaystyle \left[a,b\right]\ }$ is the domain and ${\displaystyle \mathbb {R} \ }$ is the range.

${\displaystyle \left\langle g,h\right\rangle =\int _{a}^{b}g(x)h(x)\,dx\ }$

Consider: ${\displaystyle f(\theta )=\sum _{n=0}^{\infty }A_{n}P_{n}(\sin \theta )\ }$

Question: Find ${\displaystyle A_{n},n=0,1,...,\infty \ }$

${\displaystyle \displaystyle {\begin{aligned}\left\langle f,P_{m}\right\rangle =\sum _{n=0}^{\infty }A_{n}\left\langle P_{n},P_{m}\right\rangle \end{aligned}}}$ for ${\displaystyle m=0,1,...,\infty \ }$

(1)

${\displaystyle \displaystyle \left[\left\langle P_{n},P_{m}\right\rangle \right]\left\{A_{n}\right\}=\left\{\left\langle f,P_{m}\right\rangle \right\}}$

(2)

where ${\displaystyle \left[\left\langle P_{n},P_{m}\right\rangle \right]\ }$ is the Gram Matrix ${\displaystyle \mathbf {\Gamma \ } (P_{0},P_{1},P_{2},\cdots )\ }$ that has infinite rows and columns.

${\displaystyle \left\{A_{n}\right\}\ }$ is an infinite row by 1 column matrix and ${\displaystyle \left\{\left\langle f,P_{m}\ \right\rangle \right\}\ }$ is an infinite row by 1 column matrix .

${\displaystyle \displaystyle {\mathcal {F}}:=\left\{P_{0},P_{1},P_{2},...\right\}}$

(2a)

${\displaystyle \displaystyle \left\langle f,P_{m}\right\rangle =\int _{\theta =-{\frac {\pi }{2}}}^{\frac {\pi }{2}}f(\theta )P_{m}(\sin \theta )\,d\mu }$

(3)

where ${\displaystyle \sin \theta \ =\mu \ \ }$

Similarly for ${\displaystyle \left\langle P_{n},P_{m}\right\rangle \ }$

${\displaystyle \displaystyle \left\langle P_{n},P_{m}\right\rangle =\int _{\mu =-1}^{\mu =1}P_{n}(\mu )P_{m}(\mu )\,d\mu \ ={\frac {2}{2n+1}}\delta _{mn}}$

(4)

where ${\displaystyle \delta _{mn}\ }$ is the Kronecker delta Ref Eq.(2)P.33-2

${\displaystyle \Rightarrow \ \mathbf {\Gamma \ } (P_{0},P_{1},...)\ }$ is diagonal and ${\displaystyle \Gamma \ ^{-1}\ }$ can be found easily

Use ${\displaystyle \mu \ =\sin \theta \ \ }$ as integrating variable instead fo ${\displaystyle \theta \ \ }$ due to Eq.(4)P.34-4 .

Rewrite Eq.(3)P.34-4 as:

${\displaystyle \displaystyle \left\langle f,P_{m}\right\rangle =\int _{\mu =-1}^{\mu =1}f(\mu )P_{m}(\mu )\,d\mu }$

(1)

${\displaystyle \displaystyle \Rightarrow \ A_{n}={\frac {1}{\left\langle P_{n},P_{n}\right\rangle }}\left\langle f,P_{n}\right\rangle }$

(2)

${\displaystyle \displaystyle {\mathcal {F}}}$ is complete, i.e., any continuous function ${\displaystyle \displaystyle f}$ can be expressed as an infinite series of functions in ${\displaystyle \displaystyle {\mathcal {F}}}$:

${\displaystyle \displaystyle f(\mu )=\sum _{n=0}^{\infty }A_{n}P_{n}(\mu )}$

(3)

Similar to Fourier series, Eq.(3) is an equality, not an approximation, due to competeness of ${\displaystyle \displaystyle {\mathcal {F}}}$.

Example:

${\displaystyle \displaystyle f(\theta )=T_{0}\cos ^{4}\theta =T_{0}(1-\mu ^{2})^{2}}$

(4)

${\displaystyle \displaystyle A_{n}={\frac {2n+1}{2}}\int _{\mu =-1}^{\mu =1}T_{0}(1-\mu ^{2})^{2}P_{n}(\mu )\,d\mu }$ for ${\displaystyle n=0,1,2,...\ }$

(5)