Legendre Differential Equation
Legendre Differential Equation
The Legendre equation appears in Unit-1 (variable coefficient equations) especially in engineering physics and boundary value problems.
Standard form:
\[ (1 - x^2) y'' - 2x y' + n(n+1) y = 0 \quad \text{where } n = 0,1,2,\dots \]It is a second-order linear equation with variable coefficients. Solutions are Legendre Polynomials \( P_n(x) \) when n is a non-negative integer (polynomial solutions).
Rodrigue's Formula:
\[ P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2 - 1)^n \]Recurrence Relations:
\[ (n+1) P_{n+1}(x) = (2n+1) x P_n(x) - n P_{n-1}(x) \] \[ P_n(-x) = (-1)^n P_n(x) \quad \text{(even/odd property)} \]
Example 1: n=0
\[ (1-x^2)y'' - 2x y' + 0 \cdot y = 0 \implies (1-x^2)y'' - 2x y' = 0 \]
Let \( v = y' \), then \( (1-x^2) v' - 2x v = 0 \)
\( \frac{dv}{dx} = \frac{2x v}{1-x^2} \implies \frac{dv}{v} = \frac{2x dx}{1-x^2} \)
\( \ln v = -\ln|1-x^2| + c \implies v = \frac{c_1}{1-x^2} \)
\( y' = c_1 (1-x^2)^{-1} \implies y = c_1 \tanh^{-1} x + c_2 \)
Polynomial solution: \( P_0(x) = 1 \)
\[ (1-x^2)y'' - 2x y' + 0 \cdot y = 0 \implies (1-x^2)y'' - 2x y' = 0 \]
Let \( v = y' \), then \( (1-x^2) v' - 2x v = 0 \)
\( \frac{dv}{dx} = \frac{2x v}{1-x^2} \implies \frac{dv}{v} = \frac{2x dx}{1-x^2} \)
\( \ln v = -\ln|1-x^2| + c \implies v = \frac{c_1}{1-x^2} \)
\( y' = c_1 (1-x^2)^{-1} \implies y = c_1 \tanh^{-1} x + c_2 \)
Polynomial solution: \( P_0(x) = 1 \)
Example 2: n=1
\[ (1-x^2)y'' - 2x y' + 2y = 0 \]
Using Rodrigue: \( P_1(x) = \frac{1}{2^1 \cdot 1!} \frac{d}{dx}(x^2-1) = x \)
Verify: \( y = x \)
\( y' = 1 \), \( y'' = 0 \)
Left side: \( (1-x^2)(0) - 2x(1) + 2x = -2x + 2x = 0 \)
\( P_1(x) = x \)
\[ (1-x^2)y'' - 2x y' + 2y = 0 \]
Using Rodrigue: \( P_1(x) = \frac{1}{2^1 \cdot 1!} \frac{d}{dx}(x^2-1) = x \)
Verify: \( y = x \)
\( y' = 1 \), \( y'' = 0 \)
Left side: \( (1-x^2)(0) - 2x(1) + 2x = -2x + 2x = 0 \)
\( P_1(x) = x \)
Example 3: n=2 using recurrence
We know \( P_0 = 1 \), \( P_1 = x \)
For n=1: \( 2 P_2 = (2\cdot1 +1) x P_1 - 1 \cdot P_0 \)
\( 2 P_2 = 3x^2 - 1 \implies P_2(x) = \frac{3x^2 - 1}{2} \)
\( P_2(x) = \frac{1}{2}(3x^2 - 1) \)
We know \( P_0 = 1 \), \( P_1 = x \)
For n=1: \( 2 P_2 = (2\cdot1 +1) x P_1 - 1 \cdot P_0 \)
\( 2 P_2 = 3x^2 - 1 \implies P_2(x) = \frac{3x^2 - 1}{2} \)
\( P_2(x) = \frac{1}{2}(3x^2 - 1) \)
Example 4: n=3 using Rodrigue's formula
\[ P_3(x) = \frac{1}{2^3 \cdot 3!} \frac{d^3}{dx^3} (x^2-1)^3 \]
\( (x^2-1)^3 = x^6 - 3x^4 + 3x^2 - 1 \)
1st deriv: \( 6x^5 - 12x^3 + 6x \)
2nd deriv: \( 30x^4 - 36x^2 + 6 \)
3rd deriv: \( 120x^3 - 72x \)
\( P_3(x) = \frac{1}{8 \cdot 6} (120x^3 - 72x) = \frac{120x^3 - 72x}{48} = \frac{5x^3 - 3x}{2} \)
\( P_3(x) = \frac{1}{2}(5x^3 - 3x) \)
\[ P_3(x) = \frac{1}{2^3 \cdot 3!} \frac{d^3}{dx^3} (x^2-1)^3 \]
\( (x^2-1)^3 = x^6 - 3x^4 + 3x^2 - 1 \)
1st deriv: \( 6x^5 - 12x^3 + 6x \)
2nd deriv: \( 30x^4 - 36x^2 + 6 \)
3rd deriv: \( 120x^3 - 72x \)
\( P_3(x) = \frac{1}{8 \cdot 6} (120x^3 - 72x) = \frac{120x^3 - 72x}{48} = \frac{5x^3 - 3x}{2} \)
\( P_3(x) = \frac{1}{2}(5x^3 - 3x) \)
Example 5: n=4 and n=5 (quick using recurrence)
From recurrence:
n=3: \( 4 P_4 = (6+1)x P_3 - 3 P_2 = 7x \cdot \frac{5x^3-3x}{2} - 3 \cdot \frac{3x^2-1}{2} \)
After calculation:
\( P_4(x) = \frac{1}{8}(35x^4 - 30x^2 + 3) \)
\( P_5(x) = \frac{1}{8}(63x^5 - 70x^3 + 15x) \)
From recurrence:
n=3: \( 4 P_4 = (6+1)x P_3 - 3 P_2 = 7x \cdot \frac{5x^3-3x}{2} - 3 \cdot \frac{3x^2-1}{2} \)
After calculation:
\( P_4(x) = \frac{1}{8}(35x^4 - 30x^2 + 3) \)
\( P_5(x) = \frac{1}{8}(63x^5 - 70x^3 + 15x) \)
Example 6: Non-homogeneous Legendre Equation
Solve \( (1-x^2)y'' - 2x y' + 6y = 8x^2 \) (n=2 case)
Homogeneous solution: \( y_c = c_1 P_2(x) + c_2 Q_2(x) \)
But for polynomial RHS, use Variation of Parameters with \( y_1 = P_2(x) = \frac{3x^2-1}{2} \), and second solution \( Q_2(x) = P_2(x) \ln\left|\frac{1+x}{1-x}\right| \) or use series.
Easier: Assume particular solution of form \( y_p = Ax^2 + B \) (since RHS is degree 2 and P_2 has degree 2)
Substitute:
\( y_p = Ax^2 + B \)
\( y_p' = 2Ax \), \( y_p'' = 2A \
Solve \( (1-x^2)y'' - 2x y' + 6y = 8x^2 \) (n=2 case)
Homogeneous solution: \( y_c = c_1 P_2(x) + c_2 Q_2(x) \)
But for polynomial RHS, use Variation of Parameters with \( y_1 = P_2(x) = \frac{3x^2-1}{2} \), and second solution \( Q_2(x) = P_2(x) \ln\left|\frac{1+x}{1-x}\right| \) or use series.
Easier: Assume particular solution of form \( y_p = Ax^2 + B \) (since RHS is degree 2 and P_2 has degree 2)
Substitute:
\( y_p = Ax^2 + B \)
\( y_p' = 2Ax \), \( y_p'' = 2A \