User:Egm6321.f12.team4.harris/Homework2 R2.1

Statement
Verify that

where $$ y_H^1 $$ and $$ y_H^2 $$ are homogeneous solutions of the Legendre Differential Operator $$ L_2(y):= (1-x^2)y''-2xy'+n(n+1)=0 $$ for n=1.

Given
{| style="width:100%" border="0" The Legendre Differential Operator for n=1 is given as
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Solution
To solve this problem, one must substitute the linearly independent homogeneous solutions and their first and second derivatives back into the Legendre Differential Operator for n=1 and verify that they are equal to zero.

Looking at the first homogeneous solution (Equation 2.1.2), its first and second derivatives are $$ y'=\frac{d}{dx}(x)=1 $$ (2.1.4) $$ y''=\frac{d^2}{dx^2}(x)=\frac{d}{dx}(1)=0 $$ (2.1.5)
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Substituting Equations 2.1.4 and 2.1.5 back into Equation 2.1.1, one can see that $$ 0=(1-x^2)(0)-2x(1)+2x $$

Thus $$ L_2(y_H^1)=0 $$

Looking at the second homogeneous solution (Equation 2.1.3), its first derivative is

$$\displaystyle y'=\frac{d}{dx}[\underbrace{\frac{x}{2}\underbrace{ln(\frac{1+x}{1-x})}_{\frac{d}{dx}(ln(u(x)))=\frac{1}{u(x)}u'(x)}}_{chain rule}-1] $$

$$ \displaystyle \frac{1}{2}ln(\frac{1+x}{1-x})+\frac{x}{2}(\frac{1-x}{1+x})\underbrace{(\frac{d}{dx}(\frac{1+x}{1-x}))}_{ \frac{d}{dx}(\frac{u(x)}{v(x)})=\frac{v(x)u'(x)-u(x)v'(x)}{v(x)^2}} $$

$$ \displaystyle \frac{1}{2}ln(\frac{1+x}{1-x})+\frac{x}{2}(\frac{1-x}{1+x})[\frac{(1-x)(1)-(1+x)(-1))}{(1-x)^2}] $$

$$ \displaystyle \frac{1}{2}ln(\frac{1+x}{1-x})+\frac{x}{2}(\frac{1-x}{1+x})(\frac{2}{(1-x)^2}) $$

$$ \displaystyle y'=\frac{1}{2}ln(\frac{1+x}{1-x})+\frac{x}{1-x^2} $$
 * (2.1.6)

The second derivative is

$$ \displaystyle \frac{d^2}{dx^2}[\frac{x}{2}ln(\frac{1+x}{1-x})-1]=\frac{d}{dx}[\frac{1}{2}ln(\frac{1+x}{1-x})+\frac{x}{1-x^2}] $$

Using the same derivation rules listed above

$$ \displaystyle \frac{1}{2}(\frac{1-x}{1+x})\frac{d}{dx}(\frac{1+x}{1-x})+\frac{(1-x^2)(1)-x(-2x)}{(1-x^2)^2} $$

$$ \displaystyle \frac{1}{2}(\frac{1-x}{1+x})(\frac{2}{(1-x)^2})+\frac{1+x^2}{(1-x^2)^2} $$

$$ \displaystyle \frac{2}{(1-x^2)^2} $$
 * (2.1.7)

Substituting Equations 2.1.6 and 2.1.7 back into Equation 2.1.1, one can see that $$ \displaystyle 0=(1-x^2)(\frac{2}{(1-x^2)^2})-2x(\frac{1}{2}ln(\frac{1+x}{1-x})+\frac{x}{1-x^2})+2(\frac{x}{2}ln(\frac{1+x}{1-x})-1) $$

$$ \displaystyle \frac{2}{1-x^2}-(xln(\frac{1+x}{1-x})+\frac{2x^2}{1-x^2})+(xln(\frac{1+x}{1-x})-2) $$ $$ \displaystyle \frac{2}{1-x^2}-\frac{2x^2}{1-x^2}-2(\frac{1-x^2}{1-x^2})$$ $$ \displaystyle \frac{2-2x^2-2+2x^2}{1-x^2}=0 $$

Thus $$ L_2(y_H^2)=0 $$

Therefore it is proven that $$ L_2(y_H^1)=L_2(y_H^2)=0. $$

Author and References

 * Solved and Typed by -- Kaitlin Harris