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 Skip Navigation LinksMath Help > Calculus > Differential Equations > Legendre's Differential Eq

Legendre's Differential Equation: (1−x2)y''−2xy'+n(n+1)y=0

The solutions of this DE are called Legendre Functions.  If n is a non-negative integer, then the Legendre Functions are called Legendre Polynomials, Pn(x).

Legendre's DE is a second order ordinary differential equation, so two sets of functions are needed to form the general solution.  Legendre Polynomials of the second kind are called Qn(x).  Then, the general solution to this DE is

y(x) = AnPn(x)+BnQn(x)

However, Qn(x) has singularities at ±1, so Bn is forced to be zero if the solution is meant to have a meaningful physical interpretation.

Rodriques' Formula

Pn(x)= 1
——
2nn!
· dn
——
dxn
(x2−1)n

Recurrence Relations

(n+1)Pn+1(x)=(2n+1)xPn(x)−nPn−1(x)

(2n+1)Pn(x)=Pn+1'(x)−Pn−1'(x)

(x2−1)Pn'(x)=nxPn(x)−nPn−1(x)

Pn−1'(x)=xPn'(x)−nPn(x)

Pn+1'(x)=xPn'(x)+(n+1)Pn(x)

Legendre Polynomials for small n

  Pn(x) Qn(x)
0 P0(x)=1 Q0(x)=(1/2)ln((x+1)/(x−1))
1 P1(x)=x Q1(x)=−1+(1/2) x ln((x+1)/(x−1))
2 P2(x)=(1/2)(3x2−1) Q2(x)=(−3/2)x+(1/4)(−1+3x2) ln((x+1)/(x−1))
3 P3(x)=(1/2)(5x3−3x) Q3(x)=(2/3)−(5/2)x2−(3/4)x(1−(5/3)x2) (ln((x+1)/(x−1))

To find the recurrence relation for coefficients of Pn, begin with the first recurrence relation, above, reproduced here:

(n+1)Pn+1(x)=(2n+1)xPn(x)-nPn-1(x)

Now, by replacing n+1 with n, and then divide by n to get

Pn(x)=((2n-1)/n) xPn-1(x) + ((1-n)/n) Pn-2(x)

This gives the triangle of coefficients (using Microsoft Excel), emphasizing the formula in cell D5 (which is replicated to all the cells below the n=0 row) as follows:

 ABCDEFGHIJKLMNO
1n Coefficients, constant first, increasing degree
20 1            
31 01           
42 -0.501.5          
53 0-1.5
=(2*$A5-1)/$A5*C4+
(1-$A5)/$A5*D3
02.5         
64 0.3750-3.7504.375        
75 01.8750-8.7507.875       
86 -0.31306.5630-19.688014.438      
97 0-2.188019.6880-43.313026.813     
108 0.2730-9.844054.1410-93.844050.273    
119 02.4610-36.0940140.7660-201.094094.961   
1210 -0.246013.5350-117.3050351.9140-427.3240180.426  
1311 0-2.707058.6520-351.9140854.6480-902.1290344.449 
1412 0.2260-17.5960219.9460-997.0902029.790-1894.4710660.194

Now, A060818 gives the denominators of the coefficients, so we can make a new table, multiplying each coefficient by the appropriate denominator to give a table of numerators.  Here, the formulas in cells B5 and D5 are emphasized to aid your understanding:

 ABCDEFGHIJKLMNO
1n A060818
(denom.)
A100258:
Numerators of Coefficients, constant first, increasing degree
2011            
31101           
422-103          
532
=2^(FLOOR($A5/2,1)+
FLOOR($A5/4,1)+
FLOOR($A5/8,1))
0-3
=Sheet1!D5*
$B5
05         
64830-30035        
7580150-70063       
8616-501050-3150231      
97160-3503150-6930429     
108128350-1260069300-1201206435    
11912803150-46200180180-25740012155   
1210256-63034650-300300900900-109395046189  
13112560-6930150150-9009002187900-230945088179 
141210242310-1801802252250-1021020020785050-19399380676039

Other facts about Legendre Polynomials

|Pn(x)|≤1

Pn(x) is an odd function if n is odd, and so in particular Pn(0)=0 if n is odd.

Pn(x) is an even function if n is even.

If n is even, Pn(0)=(−1)n/2 (1·3·5·...·(n−1)) / (2·4·6·...·n).  

The proof by induction begins with the recurrence relation for the Pn in terms of Pn-1 and Pn-2.

Pn(x)=((2n-1)/n) xPn-1(x) + ((1-n)/n) Pn-2(x)

Now, this recurrence relation gets a lot simpler when x=0:

Pn(0)= ((1-n)/n) Pn-2(0)

and using recursion starting with P0(0)=1, P1(0)=0, the result follows.  

Pn(1)=1

Pn(−1)=(−1)n 

Pn(x) = (1/π)

  π

 0

(x+sqrt(x2−1)cosφ)n

  1

 −1

f(x)Pn(x)dx = (−1/2)n/n!

  1

 −1

(x2−1)nf(n)(x) dx,

where f(x) is differentiable n times on −1≤x≤1.  f(n)(x) denotes the nth derivative of f(x).

Internet references

Engineering Fundamentals: Legendre Polynomial  

Wikipedia: Legendre polynomials 

OEIS: Search for Legendre Polynomial, A100258 

Related pages in this website

Chebyshev's Differential Equations: First kind (1−x²)y''−xy'+n²y=0, Second kind (1−x²)y''−3xy'+n(n+2)y=0; Chebyshev's Polynomials

 

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