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Method of Power Series

Lots of ODE cannot be sovled using elementary integration method, so we have to give up solution of finite form and try to find solution of infinite form, like series.

To have a more global understanding, we focus on

(1)dyydx=ff(x,yy),yy(x0)=yy0

where ff(x,yy) is analytic on region RR×Rn, i.e.

ff(x,yy)=i,j1,j2,,jn=0aaij1j2jn(xx0)i(y1y10)j1(ynyn0)jn

where aaij1j2jnRn.

To simplify the notation, we denote jj=(j1,j2,,jn), and

(yyyy0)jj=(y1y10)j1(ynyn0)jn

denote

i=0,jj=0=i,j1,j2,,jn=0

Since ff(x,yy) is analytic with respect to yy, so by Picard Theorem, there exists only one solution. Now the question is, to prove that the solution is analytic, i.e. δ>0, s.t. xOδ(x0)

yy(x)=k=0cck(xx0)k

where cck=(cn1,cn2,,cnk)Rn.

Excellent Series

To prove that the solution is analytic, we have to use a method of proof, that is, using Excellent Series.

Definition of Excellent Series

Assume there are two power series

(2)i=0.jj=0aijj(xx0)i(yyyy0)jj

and

(3)i=0.jj=0Aijj(xx0)i(yyyy0)jj.

If Aijj>0 and they satisfies

|aijj|<Aijj,i,jj

Then we call series 3 is an excellent series of series 2. If series 3 converges on closed region

{(x,y):|xx0|α,|yyyy0|β}

then we call its summing function ff(x,yy) is an Excellent function of series 2.(note that in this case, series 2 also converges)

引理1: 解析函数有定义域收缩的优函数 | Lemma 1: A analytic function has an excellent function within smaller region

If f(x,yy) is analytic on region

R:{(x,y):|xx0|<α,|yyyy0|<β}

then M>0, s.t.

F(x,yy)=M(1xx0a)(1y1y10b)(1ynyn0b)

is an Excellent function of f(x,yy) on a smaller region

R0:{(x,y):|xx0|<a,|yyyy0|<b}.

Use Abel Second Theorem. Note that we can not guarrantee the convergence on boundaries, so we choose open intervals.

We can represent f(x,yy) in terms of power series

f(x,yy)=i=0.jj=0aijj(xx0)i(yyyy0)jj

in region R. Then by Abel Second Theorem, a(0,α),b(0,β) s.t.

i=0.jj=0aijjaibj1+j2++jn

is convergent, so each item of the above series can be bounded by a number M>0:

|aijj|aibj1+j2++jnM|aijj|Maibj1+j2++jn

Now, the following thing is a little tricky. Define

Aijj=Maibj1+j2++jn

Consider a power of series

(4)i=0.jj=0Aijj(xx0)i(yyyy0)jj

which is convergent because it can add up to:

F(x,yy)=M(1xx0a)(1y1y10b)(1ynyn0b)

with its range of definition R0. By definition, it is an excellent function of f(x,yy).

With the definition of Excellent function, we have to use it to formulate an excellent series of the original series. This is the following theorem.

引理2: 用上述优函数建立的微分方程有解析解 | Lemma2: ODE combined with The above Excellent Function has a solution that can be represented by Power Series

Cauchy problem

dyydx=FF(x,yy),yy(x0)=yy0

has a analytic solution yy=yy(x) on region Oρ(x0), where Fi(x,yy)=F(x,yy) that is same all over i is given from the above lemma and

ρ=a{1eb/[(n+1)aM]}.

use elementary integration method.

We let u=yi, i=1,2,,n and we only need to solve the equation

dudx=F(x,u),u(x0)=u0

where

F(x,u)=M(1xx0a)(1uu0b)n

(Here readers can see that uu0=yiyi0.)

The above ODE is a variable separation equation. So change the form and integrate on [x0,x]

bn+1(1uu0b)n+1+bn+1=aMln(1xx0a)

get u out:

u=u0+bb[aM(n+1)bln(1xx0a)+1]1n+1

That is,

yi(x)=yi0+bb[aM(n+1)bln(1xx0a)+1]1n+1,i=1,2,n

We want to use this form to get a power series. See that ln(1xx0a) can be represented by power series of (xx0) once |xx0|<a. And also (1+s)1n+1 can be represented by power series of s when |s|<1. So by combine the above two, we know yi(x) can be represented by (xx0) once |xx0|<a. To be more specific, We have to let the radius of converence to satisfy

{1ρa0aM(n+1)bln(1ρa)1{ρaρa{1eb/[(n+1)aM]}

choose

ρ=a{1eb/[(n+1)aM]}

So solution of the above ODE

yy(x)=(y1(x),y2(x),,yn(x))

can be represented by power series of (xx0) when |xx0|<ρ.

证明 | Proof

Cauchy 定理 | Cauchy Theorem

Assume ff(x,yy)=[f1(x,yy),f2(x,yy),,fn(x,yy)] is an analytic function on region R. So problem 1 has a unique analytic solution yy=yy(x) on Oρ(x), where ρ is given in Lemma 2.

  • Represent solution with power series. Show that it is unique.

  • Use an excellent series to prove the above power series convergent.

  • Represent fk(x,yy) with power series
(5)fk(x,yy)=i=0,jj=0aijjk(xx0)i(yyyy0)jj

And represent solution with power series

(6)yk(x)=yk0+i=1cik(xx0)i,k=1,2,,n

substitute 5 and 6 into ODE

dykdx=fk(x,yy),k=1,2,n

and get

i=0(i+1)ci+1k(xx0)i=i,j1,j2,,jn=0{aij1j2jnk(xx0)i×[i=1ci1(xx0)i]j1×[i=1ci2(xx0)i]j2×[i=1cin(xx0)i]jn},k=1,2,n.

Denote X=xx0, and we have

i=0(i+1)ci+1kXi=i,j1,j2,,jn=0aij1j2jnkXi(i=1ci1Xi)j1(i=1ci2Xi)j2(i=1cinXi)jn

and get cik out in terms of aijjk

c1k=a000kc2k=12!(a100k+a0100kc11+a00100kc12+a0001kc1n)=12!(a100k+a0100ka0001+a00100ka0002+a0001ka000n)

Generally, we have

cmk=Pmk(a000l,a010l,,aij1jnl)

where i+j1+j2++jnm1, 1ln. Thus, Pmk is a polynomial represented by a000l, a010l, , aij1jnl with positve operator "+". Theoretically, we can represent the solution by definite power series.

We leave the proof of this part as an additional work in Appendix at the end of the doc.

  • Prove the above series converges.

Here we formulate another ODE and use Excellent function to bound the above power series.

Since fk(x,yy) is analytic on region R, by Lemma 1, there exists an excellent function of fk(x,yy) on a smaller region R0:

Fk(x,yy)=M(1xx0a)(1y1y10b)(1ynyn0b),k=1,2n

If we represent both of them in terms of power series

fk(x,yy)=i=0.jj=0aijj(xx0)i(yyyy0)jjFk(x,yy)=i=0.jj=0Aijj(xx0)i(yyyy0)jj

then we have a relation |aijj|<Aijj, which matters in the following proof.

Now consider an ODE

dykdx=Fk(x,yy),yk(x0)=yk,k=1,2,,n,

by Lemma 2, the above ODE has an analytic solution yy=yy(x), represented by power series

yk(x)=yk0+i=1Cik(xx0)i,k=1,2,,n

Similarly, we have

Cmk=Pmk(A000l,A010l,,Aij1jnl)=Pmk(|A000l|,|A010l|,,|Aij1jnl|)Pmk(|a000l|,|a010l|,,|aij1jnl|)|cmk|

So power series i=1Cik(xx0)i is en excellent series of i=1cik(xx0)i. Since the former is convergent by Lemma 2, so the latter also converges.

附录 | Appendix: Relation between c and a

Theorem. Prove

cmk=Pmk(a000l,a010l,,aij1jnl)

where i+j1+j2++jnm1, 1ln. Thus, Pmk is a polynomial represented by a000l, a010l, , aij1jnl with positve operator "+".

Use induction.