User:Egm6322.s12.TEAM1.KIM/HW2

Problem 3: Legendre homogeneous solution
 Solved without assistance 

Given
$$\displaystyle {{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)={{\tanh }^{-1}}x $$ (3.1) $$\displaystyle {{Q}_{1}}(x)=\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1=x{{\tanh }^{-1}}x-1 $$ (3.2)

Find
Show Eq (3.1) and Eq (3.2)

Solution
Order 0 Legendre equation is that $$\displaystyle(1-{{x}^{2}}){y}''-2x{y}'=0$$ Let z = y' $$\displaystyle {z}'+\frac{2x}{{{x}^{2}}-1}z=0$$ To know the solution, I use integrating factor method. $$\displaystyle h(x) =\exp \left[ \int_ – ^{x}{\frac{2x}{{{x}^{2}}-1}dx} \right] =\exp \left[ \int_ – ^{x}{\frac{1}{1-{{x}^{2}}}d(1-{{x}^{2}})} \right] =\exp [\log (1-{{x}^{2}})] =1-{{x}^{2}}$$  From the previous lecture (PEA1), $$\displaystyle z ={{h}^{-1}}(x)\int_ – ^{x}{h(x)*0dx} =\frac{1-{{x}^{2}}}$$  $$ y =\int{zdx}+{{k}_{1}} ={{k}_{1}}+\int_ – ^{}{\frac{1-{{x}^{2}}}dx} ={{k}_{1}}+\frac{2}\int_ – ^{}{\left( \frac{1}{1+x}+\frac{1}{1-x} \right)dx} ={{k}_{1}}+{{k}_{2}}\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) $$ $$\displaystyle Q_0$$ is one of homogeneous solutions. $$\displaystyle {{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$  Now, We have to find what tanhx is. $$\displaystyle \tanh x = \frac{\sinh x}{\cosh x} = \frac {e^x - e^{-x}} {e^x + e^{-x}} = \frac{e^{2x} - 1} {e^{2x} + 1}$$ $$ x=\frac{1}{2}\log \left( \frac{1+\tanh x}{1-\tanh x} \right)$$ $$ {{\tanh }^{-1}}x=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) $$ These show that $$\displaystyle {{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)={{\tanh }^{-1}}x$$

Order 1 Legendre equation is that $$\displaystyle (1-{{x}^{2}}){y}''-2x{y}'+2y=0$$ $$\displaystyle {y}''+\frac{2x}{({{x}^{2}}-1)}{y}'+\frac{2}{(1-{{x}^{2}})}y=0$$  using Integrating Factor Method, $$\displaystyle h(x)={{u}_{1}}^{2}(x)\exp \left[ \int_ – ^ – {{{a}_{1}}(x)dx} \right] ={{[{{P}_{1}}(x)]}^{2}}\exp \left[ \int_ – ^ – {\frac{2x}{{{x}^{2}}-1}dx} \right] ={{x}^{2}}\exp \left[ \int_ – ^ – {\frac{1}{1-{{x}^{2}}}d(1-{{x}^{2}})} \right] ={{x}^{2}}(1-{{x}^{2}})$$ $$\displaystyle {{Q}_{1}}(x) ={{u}_{2}}(x) ={{u}_{1}}(x)\int_ – ^ – {\frac{1}{h(x)}dx} =x\int{\frac{1}{{{x}^{2}}(1-{{x}^{2}})}dx} =x\int{\frac{1}+\frac{1}{(1-{{x}^{2}})}dx} =x\int{\frac{1}{(1-{{x}^{2}})}dx+}x\int{\frac{1}dx} =\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 $$ These show that $$\displaystyle {{Q}_{1}}(x)=\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1=x{{\tanh }^{-1}}x-1$$

Given
$$Q_n(x) = P_n(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {2n-2j+1} {(2n-j+1)j} P_{n-j}(x)$$ $$\displaystyle J = 1+2[\frac {n-1}{2}]$$ $$\displaystyle P_0(x) = 1$$ $$\displaystyle P_1(x) = x$$ $$\displaystyle P_2(x) = \frac {1} {2} (3x^2-1)$$ $$\displaystyle P_3(x) = \frac {1} {2} (5x^3-3x)$$ $$\displaystyle P_4(x) = \frac {1} {8} (35x^4 - 30x^2 + 3)$$

Find
Verify $$\displaystyle Q_0, \,\, Q_1, \,\, Q_2 $$ using above equation.

Solution
When n = 0, $$\displaystyle J = 1+2[\frac {0-1}{2}] = 0$$ $$\displaystyle Q_0(x) = P_0(x) tanh^{-1}(x) = tanh^{-1}(x) $$ When n = 1, $$\displaystyle J = 1+2[\frac {1-1}{2}] = 1$$ $$\displaystyle Q_1(x) = P_1(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^1 \frac {2-2+1} {(2-1+1)1} P_{1-1}(x)$$ $$\displaystyle Q_1(x) = P_1(x) tanh^{-1}(x) - 1 = xtanh^{-1}(x) - 1$$ When n = 2, $$\displaystyle J = 1+2[\frac {2-1}{2}] = 2$$ $$\displaystyle Q_2(x) = P_2(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^2 \frac {2n-2j+1} {(2n-j+1)j} P_{n-j}(x)$$ $$\displaystyle Q_2(x) = (\frac 1 2 (3x^2-1)) tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^2 \frac {2(2)-2(j)+1} {(2(2)-(j)+1)(j)} P_{2-j}(x)$$ $$\displaystyle Q_2(x) = (\frac 1 2 (3x^2-1)) tanh^{-1}(x) - (\frac 3 2) (x) $$

Given
$$Q_n(x) = P_n(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {2n-2j+1} {(2n-j+1)j} P_{n-j}(x)$$ $$\displaystyle J = 1+2[\frac {n-1}{2}]$$

Find
Show that Qn is odd or even depending on the index n.

Solution
$$\displaystyle P_2k(x) $$ is even $$\displaystyle P_{2k+1}(x)$$ is odd

Hence, Let 2k = n = even number k = 1,2,3......  $$Q_2k(x) = P_{2k}(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {4k-2j+1} {(4k-j+1)j} P_{2k-j}(x)$$ $$\displaystyle tanh^{-1}(x)$$ is always odd. $$\displaystyle P_{2k-j}(x) $$ is odd.

$$\displaystyle Q_2k(x) = (even)(odd) - (odd) $$ When n = even, Q(x) is odd.

Let 2k+1 = n = odd number k = 1,2,3......  $$Q_{2k+1}(x) = P_{2k+1}(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {2(2k+1)-2j+1} {(2(2k+1)-j+1)j} P_{2(2k+1)-j}(x)$$ $$\displaystyle tanh^{-1}(x)$$ is always odd. $$\displaystyle P_{2(2k+1)-j}(x) $$ is even. $$\displaystyle Q_{2k+1}(x) = (odd)(odd) - (even) $$ When n = odd, Q(x) is even.