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math121b:02-26

2020-02-26, Wednesday

Today we begin Chapter 12, the series solution to ODE.

Consider a linear differential equation P(x)y(x)=0 P(x) y(x) = 0 where P(x)P(x) is some differential operator in xx.

The philosophy is that, assume the solution exists and is analytic around x=0x=0, namely, we can do Taylor series of y(x)y(x) around x=0x=0, then y=n=0anxn y = \sum_{n=0}^\infty a_n x^n then, we can try to figure out what is the relations between ana_n.

Examples y=2xy y' = 2x y.

A nasty equation: Legendre equation

(1x2)y2xy+l(l+1)y=0 (1-x^2) y'' - 2 xy' + l(l+1) y = 0 where ll is a constant. If we plug in y=nanxny = \sum_n a_n x^n, we get an+2=(ln)(l+n+1)(n+2)(n+1)an a_{n+2} = -\frac{(l-n)(l+n+1)}{(n+2)(n+1)} a_n Hence, if we know a0a_0 and a1a_1, we know all the subsequence aia_i.

The general solution is y(x)=a0(1l(l+1)2!x2+)+a1(x(l1)(l+2)3!x3+) y(x) = a_0 \left( 1 - \frac{l(l+1)}{2!}x^2 + \cdots \right) + a_1 \left( x - \frac{(l-1)(l+2)}{3!} x^3 + \cdots \right)

We can get that, the series converges for x<1|x|<1.

If ll is an integer, then one of the series converges. If l1+l2=1l_1 + l_2 = -1, then l1l_1 and l2l_2 gives the same solution. That is why we use l(l+1)l(l+1) to label the different solutions.

Eigenvalue problem

We note that, for special values of ll, we can have a convergent solution for Legendre equation near x2=1x^2=1. In this example, if ll takes some special value (integers), then the solution space has some nice solutions (polynomial solution). Roughly speaking, such special values are called eigenvalues, and such solutions are called eigenfunctions (or eigenvectors).

Rodrigue Formula

Pl(x)=12ll!(x)l(x21)l P_l(x) = \frac{1}{2^l l!} (\d_x)^l (x^2 -1)^l Let's show that it satisfies the Legendre equation.

math121b/02-26.txt · Last modified: 2020/02/28 13:28 by pzhou