User:Egm6321.f12.team5.yi/report2

R2.1 Legendre Differential Equation

Given

Consider the following function
 * {| style="width:100%" border="0"

$$L_2(y):=(1-x^2)y''-2xy'+n(n+1)y=0$$ $$n=1 \Rightarrow(1-x^2)y''-2xy'+2y=0$$ And two linearly-independent homogeneous solutions:
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$y^1_H(x)=x:=P_1(x)$$ $$y^2_H(x)=\frac{x}{2} \log \left(\frac{1+x}{1-x} \right)-1:=Q_1(x)$$
 * style="width:95%" |
 * 
 * }

Find

Verify that
 * {| style="width:100%" border="0"

$$L_2(y^1_H)=L_2(y^2_H)=0$$
 * style="width:95%" |
 * 
 * }

Solution

To make the problem easier, we can separate the proof into two part:
 * {| style="width:100%" border="0"

$$L_2(y^1_H)=0$$ (2.1.1)
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$L_2(y^2_H)=0$$ (2.1.1) Since $$y^1_H(x)=x$$, we can find that:
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$${y^1_H}'=1$$ (2.1.3)
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$${y^1_H}''=0$$ (2.1.4) Thus:
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$L_2(y^1_H)|_{n=2}=(1-x^2){y^1_H}''-2x{y^1_H}'+2y^1_H=(1-x^2)\times 0-2x\times 1+2x=0$$ (2.1.5)
 * style="width:95%" |
 * 
 * }

Since
 * {| style="width:100%" border="0"

$$y^2_H(x)=\frac{x}{2} \log \left(\frac{1+x}{1-x} \right)-1=\frac{x}{2} \log (1+x)-\frac{x}{2} \log (1-x)-1$$ (2.1.6) we can find that:
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$${y^2_H}'=\frac{1}{2}log(1+x)-\frac{1}{2}log(1-x)+\frac{x}{2(x+1)}-\frac{x}{2(x-1)}$$ (2.1.7)
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$${y^2_H}''=\frac{1}{x+1}-\frac{1}{x-1}+\frac{x}{2(x-1)^2}-\frac{x}{2(x+1)^2}$$ (2.1.7) Thus:
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$L_2(y^2_H)|_{n=2}=(1-x^2){y^2_H}''-2x{y^2_H}'+2y^1_H$$ $$=(1-x^2)(\frac{1}{x+1}-\frac{1}{x-1}+\frac{x}{2(x-1)^2}-\frac{x}{2(x+1)^2})-2x(\frac{1}{2}log(1+x)-\frac{1}{2}log(1-x)+\frac{x}{2(x+1)}-\frac{x}{2(x-1)})+2(\frac{x}{2} \log (1+x)-\frac{x}{2} \log (1-x)-1)$$ $$=(2+\frac{x^2+x}{2-2x}+\frac{x^3-x}{x(1+x)^2})-(xlog(1+x)-xlog(1-x)+\frac{x^2}{2+1}-\frac{x^2}{x-1})+(xlog(1+x)-xlog(1-x)-2)$$ $$=0$$ (2.1.8)
 * style="width:95%" |
 * 
 * }

So we get:
 * {| style="width:100%" border="0"

$$L_2(y^1_H)=L_2(y^2_H)=0$$
 * style="width:95%" |
 * 
 * }