University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg37

Mtg 37: Thurs, 17Nov09

Page 37-1

P.36-4 continued

$P_n(x)$

$P_1(x)=x$

$Q_n(x)$

From P.18-1 :

$\begin{array}{c}{Q}_1(x)=\frac{1}{2}xlog\left(\frac{1+x}{1-x}\right)-1=x{\tanh }^{-1}x-1\end{array}$

HW show $Q_1(x)=\frac{1}{2}xlog\left(\frac{1+x}{1-x}\right)-1=x{\tanh }^{-1}x-1$

ref K p33 for $Q_2,Q_3...$

$\begin{array}{c}{Q}_n(x)=P_n(x){\tanh }^{-1}x-2\sum _{j=1,3,5}^J\frac{2n-2j+1}{(2n-j+1)j}{P}_{n-j}(x)\end{array}$

$Q_0(x)={\tanh }^{-1}(x)$ is odd

HW: Use Eq(2) to show when $Q_n$ is even or odd, depending on "n"

HW: Plot $\{P_0,P_1,...,P_4\}$ and $\{Q_0,Q_1,...,Q_4\}$

Legendre function $L_n(x)=P_n(x)$ or $Q_n(x)$ solution of Legendre solution

Page 37-2

$\begin{array}{c}⟨L_n,L_m⟩=0\end{array}$

for $n\ne m$

for $L_n=P_n,⟨L_n,L_n⟩=⟨P_n,P_n⟩=\frac{2}{2n+1}$

for $⟨L_n,L_n⟩=⟨P_n,Q_n⟩=$ HW

$⟨L_n,L_m⟩=⟨P_n,P_m⟩=$ Eq.(3) P.33-1

$⟨L_n,L_m⟩=⟨P_n,Q_m⟩=0$ for $n\ne m$

Proof: Legendre equation, Eq.(1) P.14-2 : 2) $[(1-x^2)y^{\prime }]+n(n+1)y=0$

Where

$\begin{array}{c}[(1-x^2)y^{\prime }]=(1-x^2)y^{″}-2x{y}^{\prime }\end{array}$

$\Rightarrow [(1-x^2){L_n}^{\prime }]+n(n+1)L_n=0$

Multiply by $L_m$ and integrate from -1 to +1:

$\underset{-1}{\overset{1}{\int }}{L}_m[(1-x^2){L_n}^{\prime }]dx+n(n+1)\underset{-1}{\overset{1}{\int }}{L}_m{L}_{n}dx=0$

Where $L_m[(1-x^2){L_n}^{\prime }]=\alpha$

Page 37-3

Integrate $\alpha$ by parts:

$\begin{array}{c}-\underset{-1}{\overset{1}{\int }}(1-x^2){L_n}^{\prime }{L_m}^{\prime }dx+(n)(n+1)⟨L_n,L_m⟩=0\end{array}$

Interchange n and m:

$\begin{array}{c}-\underset{-1}{\overset{1}{\int }}(1-x^2){L_m}^{\prime }{L_n}^{\prime }dx+(m)(m+1)⟨L_m,L_n⟩=0\end{array}$

Eq(1)-Eq(2):

$[n(n+1)-m(m+1)]⟨L_n,L_m⟩=0$

Where $[n(n+1)-m(m+1)]\ne 0$ since $n\ne m$

$\Rightarrow ⟨L_n,L_m⟩=0$ when $n\ne m$

cf.K.p41

References

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