Chapter 7 - Lecture notes 7 PDF

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Lecture notes with example practice for chapter 7...

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CHAPTER 7: POWER SERIES SOLUTION ABOUT AN ORDINARY POINT Suppose the DE a2(x)y + a1(x)y + a0y = 0  (*) has no solution that is expressible as a finite linear combination of known element function. Question 1 Does it have a power series solution of the form 

 C n (x – x0)n

n 0

Question 2 Under what conditions can we be certain that it does have such a solution? To answer these questions, we need to introduce certain basic definitions. First, let us write the DE (*) in the form y’’ + P(x)y’ + Q(x)y = 0  (**) where a1 ( x ) a 0 (x) , Q(x) = P(x) = a 2 (x) a 2 (x) Definition: 1. A function f is said to be analytic at x0 if its Taylor series about x0, (n)  f (x 0 ) (x – x0)n  n ! n 0 exist and converge to f(x) for all x in some open interval containing x0. 2.

The point x0 is called an ordinary point of (*) if both P(x) and Q(x) are analytic at x0. If either or both of these functions is not analytic at x 0, then x0 is called a singular point of (*).

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For a LDE an(x)y(n) + … + a0(x)y = g(x) with polynomial coefficients, the point x = x0 is called ordinary point if an(x0)  0. A singular point is any point x = x1 for which an(x1) = 0

e.g.1 y’’+ xy’+ (x2+2)y = 0 P(x) =x, Q(x)= x2+2 are polynomial. So they are analytic everywhere. Thus all points are ordinary points of this DE.

y =0 x

e.g.2 (x-1)y’’ +xy’ +

y’’+ (

P(x) =

1 x )y’+ y= 0 x 1 x( x  1) 1 x , Q(x) = ( x x  1) x 1

P(x) is analytic except x=1 Q(x) is analytic except at x=0 and x=1 So x=0 and x=1 are singular point. Solution About Ordinary Point We can now state a theorem concerning the existence of power series solution for (**). THEOREM If x = x0 is an ordinary point of the DE (**), then the DE(**) has two nontrivial linear independent power series solutions of the form 

 C n (x – x0)n

n 0

and these power series converge in some interval x – x0 <  (where  > 0) about x. Remarks If x0 is an ordinary point of (**), then we can write the general solution of (**) as a liner combination of these two linear independent power series. e.g.1 use the power series method to solve DE y’=y ------(1) solution: 

assume that y=



anxn

n 1



y’=  n anxn-1 n 1

substitute this into (1) 

 n 1

i.e.



anxn =

 n anxn-1 n 1

a1=a0 2a2=a1=a0 3a3=a2

a2=a0/2=a0/2! a3=a0/3x2=a0/3!

kak=ak-1 

y=

ak=a0/k!

 1 a0x =a0 (  n! n 0



n 0

xn ) n!

y=a0ex eg 2. we know that all the real numbers are ordinary points of theDE y’’+xy”+(x2+2)y=0 so the DE has two linear independent power series solutions about any point x0. that is the general solution of this DE has the form 



y=A

k 0



ak(x-x0)n + B  bk(x-x0)n k 0

where x0 is any real number and ak and bk eg 3. The DE y’’+

1 x y’+ y =0 has two singular points x=0 x ( x  1) x 1

and x=1. so this DE has two linear independent power series solutions of the form 



cn(x-x0)n

n 0

about any point x0 not equal to 0, 1 for example, it has 2 linear independent power series solutions of the form 



cn(x-2)n

n 0

eg 4. Find the power series solution of the DE y’’+xy’+(x2+2)=0-------(1) in power of x. solution: x0=0 is an ordinary point of the DE. Assume 

y=



cnxn

n 0



y’=



ncnxn-1

n 1



y’’=



n(n-1)cnxn-2

n 2

substitute these into (1)



 n 2



n(n-1)cnxn-2 +x  ncnxn-1 + x2 n 1







 n 0

n 0



ncnxn +



cnxn =0



cn-2xn +

n 2

n 1

 n 0





(n+2)(n+2)cn+2 xn +



cnxn + 2



2cnxn =0

n 0





[ (n+2)(n+2)cn+2 + (n+2)cn + cn-2] xn + (2c2 +6c3x) +c1x +(2c0+2c1x)=0

n 0

So

2c2 +2c0 =0 or c2 =-c0 6c3+3c1=0 or c3=-0.5c1 (n+2)(n+1)cn+2 +(n+2)cn +cn-2 =0 ; n 2

(n  2)Cn  Cn  2 ; n 2 (n  1)(n  2) 4C 2  C 0 3C 0 1 C4 == = C0 3X 4 12 4 5C 3  C 1 3 C5 == C1 4X5 40 1 3 6  y= C0 (1-x2 + 1 x4 +……..)+C1(x- x3 + x +…….) 4 2 40

Cn+2 = -

Eg 5. Find a power series solution of the IVP (x2 -1)y’’ +3xy’ +xy =0------(1), y(0)=4, y’(0)=6 Solution: All points except x=1,-1 are ordinary points. So we can assume a general solution of the form 



cn(x-x0)n

n 0

Since we are interested in the solution at x=0, we take x0 =0. 

Assume y=



cnxn

n 0



y’=



ncnxn-1

n 1



y’’=



n(n-1)cnxn-2

n 2

substitude into (1) 





n(n-1)cnxn -

n 2

 n 2





n 1

n 0

n(n-1)cnxn-2 +3  ncnxn +  cnxn+1 =0 

-2C2 + (C0+3C1-6C3)x+  [-(n+2)(n+1)Cn+2 +n(n+2)Cn +Cn-1]xn =0 n 2

-2C2 =0

C0 +3C1-6C3=0 -(n+2)(n+1)Cn+2 +n(n+2)Cn +Cn-1 =0 ; n 2 1 1 C0 + C1 6 2 n(n  2)Cn  Cn  1 Cn+2 = ; n 2 ( n  1)( n  2) 1 8C 2  C1 = C1 C4 = 12 12 3 15C3  C 2 1 C5 = = C0 + C1 20 8 8 1 1 1 1 4 3 5 y=C0(1+ x3+ x5+…….)+ C1(x+ x3+ x + x +…….) 6 8 2 12 8 1 2 3 4 3 2 y’= C0( x + x +……)+C1(1+ x +…….) 2 8 2

C2 =0, C3=

y(0)=4  C0=4 y’(0)=6  C1=6

 y= 4(1+ 1

1 5 1 1 4 3 5 x +…….)+ 6(x+ x3+ x + x +…….) 6 8 2 12 8 11 3 1 4 =4+6x+ x + x +……. 3 2

x3+

Solution Around Singular Points Consider the DE a2(x)y + a1(x)y + a0y = 0  (*) Let x0 be a singular point of (*) Question: What type of solution can we expect? It turns out that under certain conditions we are justified in assuming a solution of the form n r y =  C n ( x  x 0 ) n 0

where r is a constant that must be determined. Thus, it is necessary to classify singular point as follow: Rewrite (6.3) as y + P(x)y + Q(x)y = 0  (**) a1 ( x ) a 0 (x) where P(x) = , Q(x) = a 2 (x) a 2 (x) Definition A singular point x0 of equation (**) is said to be a regular singular point if both (x – x0)P(x) and (x – x0)2 Q(x) are analytic at x0 that means both (x – x0)P(x) and (x – x0) Q(x) have a power series solution in (x – x0) with a positive radius of convergence.

A singular point that is not regular is said to be an irregular singular point of the equation. 2x2y’’-xy’+(x-5)=0

eg 1.

1 x 5 y’+ y=0 2x 2x 2 1 x 5 P(x)=, Q(x)= 2x 2x 2

y’’-

1. P,Qare not analytic at x=0, so x=0 is a singular point. 2. xP(x) =-

1 x 5 and x2Q(x)= are analytic at x=0. 2 2

x=0 is a regular singular point.

Eg 2. x2(x-2)2y’’ +2(x-2)y’ +(x+1)y=0 x 1 2 y’ + y=0 x 2(x  2) x 2 ( x  2 )2 2 x 1 P(x)= 2( , Q(x)= x x  2) x 2(x  2)2

y’’+

1. P,Q are not analytic at x=0, x=2, so x=0,2 are singular points. For x=0, 2

is not analytic at x=0, so x=0 is an irregular singular point. x P(x)= ( x x  2) For x=2, (x-2) P(x)=

2 x 1 are analytic at x=2. , (x-2)2 Q(x)= x2 x2

x=2 is a regular singular point.

To solve the DE (1) a2(x)y’’+a1(x)y’+a0y=0 about a regular singular point, we shall state the following theoem. THEOREM 1 (The Method of Frobenius) If x = x0 is a regular singular point of equation (1), then there exists at least one nontrivial solution of the form 

y = (x – x0)r  C n (x – x0)n n 0



=  C n (x – x0)n + r n 0

where r is a definite constant which may be determined. This solution is valid in some selected interval 0 < x – x0 <  (where  > 0) about x0. Eg 1. 2x2y’’-xy’+(x-5)=0 X=0 is a regular singular point Theorem 1 state that this DE has at least one solution of 

the form, xr  Cnxn n 0

Valid in some selected interval 0<

x...