IIT Bombay Summary of Differential Equations, Summaries of Differential Equations

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Partial differential equations

Swapneel Mahajan

www.math.iitb.ac.in/˜swapneel/

Power series

For a real number x 0 and a sequence (an) of real numbers, consider the expression ∑^ ∞

n=

an(x−x 0 )n^ = a 0 +a 1 (x−x 0 )+a 2 (x−x 0 )^2 +....

This is called a power series in the real variable x. The number an is called the n-th coefficient of the series and x 0 is called its center. For instance, ∑^ ∞

n=

n + 1 (x−1)

n (^) = 1+^1 2 (x−1)+

3 (x−1)

is a power series in x centered at 1 and with n-th coefficient equal to (^) n+1^1.

convergence. Denoting this function by f , we may write

f (x) =

∑^ ∞

n=

an(x − x 0 )n, |x − x 0 | < R.

Let us assume R > 0. It turns out that f is infinitely differentiable on the interval of convergence, and the successive derivatives of f can be computed by differentiating the power series on the right termwise. From here one can deduce that

an = f^

(n)(x 0 ) n!.

Thus, two power series both centered at x 0 take the same values in some open interval around x 0 iff all the corresponding coefficients of the two power series are equal.

Real analytic functions

Let R denote the set of real numbers. A subset U of R is said to be open if for each x 0 ∈ U , there is a r > 0 such that the open interval |x − x 0 | < r is contained in U.

Let f : U → R be a real-valued function on an open set U. We say f is real analytic at a point x 0 ∈ U if

f (x) =

∑^ ∞

n=

an(x − x 0 )n

holds in some open interval around x 0. We say f is real analytic on U if it is real analytic at all points of U. In general, we can always consider the set of all points in the domain where f is real analytic. This is called the domain of analyticity of f.

Just like continuity or differentiability, real analyticity is a local property.

Solving a linear ODE by the power

series method

Consider the initial value problem

p(x)y′′+q(x)y′+r(x)y = g(x), y(a) = y 0 , y′(a) = y 1

where p, q, r and g are real analytic functions in an interval containing the point a, and y 0 and y 1 are fixed. Theorem. Let r > 0 be less than the minimum of the radii of convergence of the functions p, q, r and g expanded in power series around a. Assume that p(x) 6 = 0 for all x ∈ (x 0 − r, x 0 + r). Then there is a unique solution to the initial value problem in the interval (x 0 − r, x 0 + r), and moreover it can be represented by a power series

y(x) =

∑^ ∞

n=

an(x − x 0 )n

whose radius of convergence is at least r.

There is an algorithm to compute the power series representation of y:

Plug a general power series into the ODE, take derivatives of the power series formally, and equate the coefficients of (x − x 0 )n^ for each n to obtain a recursive definition of the coefficients an. The an’s are uniquely determined and we obtain the desired power series solution.

In most of our examples, x 0 = 0 and the functions p, q, r and g will be polynomials of small degree.

Legendre equation

Consider the second order linear ODE:

(1 − x^2 )y′′^ − 2 xy′^ + p(p + 1)y = 0.

This is known as the Legendre equation. Here p denotes a fixed real number. The equation can also be written in the form

((1 − x^2 )y′)′^ + p(p + 1)y = 0.

This ODE is defined for all real numbers.

General solution

The coefficients (1 − x^2 ), − 2 x and p(p + 1) are polynomials (and in particular real analytic). However 1 − x^2 = 0 for x = ± 1. These are the singular points of the ODE. Our theorem guarantees a power series solution around x = 0 in the interval (− 1 , 1). It is given by

y(x)=a 0 (1− p(p 2!+1) x^2 + (p(p−2)(p 4!+1)( p+3)x^4 +... ) +a 1 (x− (p−1)( 3!p +2)x^3 + (p−1)(p−3)( 5!p +2)(p+4)x^5 +... ),

where a 0 = y(0) and a 1 = y′(0). It is called the Legendre function. The first series is an even function while the second series is an odd function.

The first few values are as follows.

n Pn(x) 0 1 1 x 2 12 (3x^2 − 1) 3 12 (5x^3 − 3 x) 4 18 (35x^4 − 30 x^2 + 3) 5 18 (63x^5 − 70 x^3 + 15x)

Their graphs in the interval (− 1 , 1) are given below.

The vector space of polynomials

The space of polynomials in one variable is a vector space with basis { 1 , x, x^2 ,... }. Further the space of polynomials carries an inner product defined by

〈f, g〉 :=

− 1 f^ (x)g(x)dx. Note that we are integrating only between − 1 and 1. This ensures that the integral is always finite. The norm of a polynomial is defined by

‖f ‖ :=

− 1 f^ (x)f^ (x)dx

Note this simple consequence of integration by parts: For differentiable functions f and g, if (f g)(b) = (f g)(a), then ∫ (^) b a

f g′dx = −

∫ (^) b a

f ′gdx.

(This process transfers the derivative from g to f .)

Orthogonality of the Legendre polynomials

Since Pm(x) is a polynomial of degree m, it follows that {P 0 (x), P 1 (x), P 2 (x),... } is a basis of the vector space of polynomials. Further, ∫ (^1) − 1

Pm(x)Pn(x)dx =

0 if m 6 = n, 2 n^2 +1 if^ m^ =^ n. Thus, the Legendre poynomials form an orthogonal basis. The norms are not 1 , hence the basis is not orthonormal.

Orthogonality can be established using the technique of derivative-transfer.

Lemma. In any inner product space, suppose {u 1 , u 2 ,... , un} and {v 1 , v 2 ,... , vn} are two orthogonal systems of vectors such that for each 1 ≤ k ≤ n, the span of u 1 ,... , uk equals the span of v 1 ,... , vk. Then ui = civi for each i for certain nonzero scalars ci.

By convention Pn(1) = 1 while qn(1) = 2nn!, so we deduce

Pn(x) = (^2) n^1 n!^ (^ dxd^ )n(x^2 − 1)n.

This is known as Rodrigues formula.

Fourier-Legendre series

A function f (x) on [− 1 , 1] is square-integrable if ∫ (^1) − 1

f (x)f (x)dx < ∞.

For instance, polynomials, continuous functions, piecewise continuous functions are square-integrable. The set of all square-integrable functions on [− 1 , 1] is a vector space. The inner product on polynomials extends to square-integrable functions. The Legendre polynomials no longer form a basis for this larger space, but they form a maximal orthogonal set. This allows us to expand any square-integrable function f (x) on [− 1 , 1] in a series of Legendre polynomials

f (x) ≈

n≥ 0

cnPn(x),

where

cn =^2 n^2 + 1

− 1

f (x)Pn(x)dx.