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

\[ \begin{equation} \frac{d \pmb{y}}{dx}=\pmb{f}(x,\pmb{y}), \quad \pmb{y}(x_0)=\pmb{y}_0 \label{eq-cauchy} \end{equation} \]

where \(\pmb{f}(x,\pmb{y})\) is analytic on region \(R\subset \mathbb{R}\times \mathbb{R}^n\), i.e.

\[ \pmb{f}(x,\pmb{y})=\sum_{i,j_1,j_2,\cdots,j_n=0}^\infty \pmb{a}_{i{j_1}{j_2}\cdots{j_n}}(x-x_0)^i(y_1-y_{10})^{j_1}\cdots(y_n-y_{n0})^{j_n} \]

where \(\pmb{a}_{i{j_1}{j_2}\cdots{j_n}}\in \mathbb{R}^n\).

To simplify the notation, we denote \(\pmb{j}=(j_1,j_2,\cdots,j_n)\), and

\[ (\pmb{y}-\pmb{y}_0)^{\pmb{j}}=(y_1-y_{10})^{j_1}\cdots(y_n-y_{n0})^{j_n} \]

denote

\[ \sum_{i=0,\pmb{j}=0}^\infty=\sum_{i,j_1,j_2,\cdots,j_n=0}^\infty \]

Since \(\pmb{f}(x,\pmb{y})\) is analytic with respect to \(\pmb{y}\), so by Picard Theorem, there exists only one solution. Now the question is, to prove that the solution is analytic, i.e. \(\exists \delta>0\), s.t. \(x\in O_\delta(x_0)\)

\[ \pmb{y}(x)=\sum_{k=0}^{\infty}\pmb{c}_k(x-x_0)^k \]

where \(\pmb{c}_k=(c_{n1},c_{n2},\cdots,c_{nk}) \in \mathbb{R}^n\).

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

\[ \begin{equation} \sum_{i=0.\pmb{j}=0}^\infty a_{i\pmb{j}}(x-x_0)^i(\pmb{y}-\pmb{y}_0)^{\pmb{j}} \label{series1} \end{equation} \]

and

\[ \begin{equation} \sum_{i=0.\pmb{j}=0}^\infty A_{i\pmb{j}}(x-x_0)^i(\pmb{y}-\pmb{y}_0)^{\pmb{j}}. \label{series2} \end{equation} \]

If \(A_{i\pmb{j}}>0\) and they satisfies

\[ |a_{i\pmb{j}}|<A_{i\pmb{j}}, \quad \forall i,\pmb{j} \]

Then we call series \(\ref{series2}\) is an excellent series of series \(\ref{series1}\). If series \(\ref{series2}\) converges on closed region

\[ \{(x,y): |x-x_0|\leq \alpha,|\pmb{y}-\pmb{y}_0|\leq\beta\} \]

then we call its summing function \(\pmb{f}(x,\pmb{y})\) is an Excellent function of series \(\ref{series1}\).(note that in this case, series \(\ref{series1}\) 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 \(f(x,\pmb{y})\) is analytic on region

\[ R: \{(x,y): |x-x_0|<\alpha, |\pmb{y}-\pmb{y}_0|<\beta\} \]

then \(\exists M>0\), s.t.

\[ F(x,\pmb{y})=\frac{M}{\left(1-\frac{x-x_0}{a}\right)\left(1-\frac{y_1-y_{10}}{b}\right)\cdots\left(1-\frac{y_n-y_{n0}}{b}\right)} \]

is an Excellent function of \(f(x,\pmb{y})\) on a smaller region

\[ R_0:\{(x,y): |x-x_0|<a,|\pmb{y}-\pmb{y}_0|<b\}. \]

HintsProof

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

We can represent \(f(x,\pmb{y})\) in terms of power series

\[ f(x,\pmb{y})=\sum_{i=0.\pmb{j}=0}^\infty a_{i\pmb{j}}(x-x_0)^i(\pmb{y}-\pmb{y}_0)^{\pmb{j}} \]

in region \(R\). Then by Abel Second Theorem, \(\exists a\in(0,\alpha), b\in(0,\beta)\) s.t.

\[ \sum_{i=0.\pmb{j}=0}^\infty a_{i\pmb{j}}a^i b^{j_1+j_2+\cdots+j_n} \]

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

\[ |a_{i\pmb{j}}|a^i b^{j_1+j_2+\cdots+j_n}\leq M \Rightarrow |a_{i\pmb{j}}|\leq \frac{M}{a^i b^{j_1+j_2+\cdots+j_n}} \]

Now, the following thing is a little tricky. Define

\[ A_{i\pmb{j}}=\frac{M}{a^i b^{j_1+j_2+\cdots+j_n}} \]

Consider a power of series

\[ \begin{equation} \sum_{i=0.\pmb{j}=0}^\infty A_{i\pmb{j}}(x-x_0)^i(\pmb{y}-\pmb{y}_0)^{\pmb{j}} \label{series-exce} \end{equation} \]

which is convergent because it can add up to:

\[ F(x,\pmb{y})=\frac{M}{\left(1-\frac{x-x_0}{a}\right)\left(1-\frac{y_1-y_{10}}{b}\right)\cdots\left(1-\frac{y_n-y_{n0}}{b}\right)} \]

with its range of definition \(R_0\). By definition, it is an excellent function of \(f(x,\pmb{y})\).

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

\[ \frac{d \pmb{y}}{dx}=\pmb{F}(x,\pmb{y}), \quad \pmb{y}(x_0)=\pmb{y}_0 \label{eq-cauchy-prime} \]

has a analytic solution \(\pmb{y}=\pmb{y}(x)\) on region \(O_\rho(x_0)\), where \(F_i(x,\pmb{y}) = F(x,\pmb{y})\) that is same all over \(i\) is given from the above lemma and

\[ \rho=a\{1-e^{b/[(n+1)aM]}\}. \]

HintsProof

use elementary integration method.

We let \(u=y_i\), \(i=1,2,\cdots,n\) and we only need to solve the equation

\[ \frac{d u}{dx}=F(x,u), \quad u(x_0)=u_0 \]

where

\[ F(x,u) = \frac{M}{\left(1-\frac{x-x_0}{a}\right)\left(1-\frac{u-u_{0}}{b}\right)^n} \]

(Here readers can see that \(u-u_0=y_i-y_{i0}\).)

The above ODE is a variable separation equation. So change the form and integrate on \([x_0,x]\)

\[ \frac{-b}{n+1}\left(1-\frac{u-u_0}{b}\right)^{n+1} +\frac{b}{n+1}= -aM\ln\left(1-\frac{x-x_0}{a}\right) \]

get \(u\) out:

\[ u = u_0 + b- b\left[\frac{aM(n+1)}{b} \ln\left(1-\frac{x-x_0}{a}\right) + 1\right]^{\frac{1}{n+1}} \]

That is,

\[ y_i(x) = y_{i0} + b- b\left[\frac{aM(n+1)}{b} \ln\left(1-\frac{x-x_0}{a}\right) + 1\right]^{\frac{1}{n+1}}, \quad \forall i=1,2\cdots,n \]

We want to use this form to get a power series. See that \(\ln{\left(1-\frac{x-x_0}{a}\right)}\) can be represented by power series of \((x-x_0)\) once \(|x-x_0|<a\). And also \((1+s)^{\frac{1}{n+1}}\) can be represented by power series of \(s\) when \(|s|<1\). So by combine the above two, we know \(y_i(x)\) can be represented by \((x-x_0)\) once \(|x-x_0|<a\). To be more specific, We have to let the radius of converence to satisfy

\[ \begin{cases} \displaystyle 1-\frac{\rho}{a} \geq 0 \\ \displaystyle \frac{aM(n+1)}{b}\ln\left(1-\frac{\rho}{a}\right)\leq 1 \end{cases} \Rightarrow \begin{cases} \rho\leq a \\ \rho\leq a\{1-e^{b/[(n+1)aM]}\} \end{cases} \]

choose

\[ \rho=a\{1-e^{b/[(n+1)aM]}\} \]

So solution of the above ODE

\[ \pmb{y}(x)=(y_1(x),y_2(x),\cdots, y_n(x)) \]

can be represented by power series of \((x-x_0)\) when \(|x-x_0|<\rho\).

证明 | Proof

Cauchy 定理 | Cauchy Theorem

Assume \(\pmb{f}(x,\pmb{y})=[f_1(x,\pmb{y}), f_2(x,\pmb{y}),\cdots, f_n(x,\pmb{y})]\) is an analytic function on region \(R\). So problem \(\ref{eq-cauchy}\) has a unique analytic solution \(\pmb{y} = \pmb{y}(x)\) on \(O_\rho(x)\), where \(\rho\) is given in Lemma 2.

HintsProof

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

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

• Represent \(f_k(x,\pmb{y})\) with power series

\[ \begin{equation} f_k(x,\pmb{y})=\sum_{i=0,\pmb{j}=0}^\infty a_{i\pmb{j}}^k (x-x_0)^i(\pmb{y}-\pmb{y}_0)^{\pmb{j}} \label{eq-f} \end{equation} \]

And represent solution with power series

\[ \begin{equation} y_k(x) = y_{k0} + \sum_{i=1}^\infty c_i^k(x-x_0)^i,\quad k=1,2,\cdots,n \label{eq-y} \end{equation} \]

substitute \(\ref{eq-f}\) and \(\ref{eq-y}\) into ODE

\[ \frac{d y_k}{dx} = f_k(x,\pmb{y}),\quad k=1,2\cdots,n \]

and get

\[ \begin{align*} \sum_{i=0}^\infty (i +1) c_{i+1}^k(x-x_0)^i = \sum_{i,j_1,j_2,\cdots,j_n=0}^\infty \Bigg\{ & a^k_{i{j_1}{j_2}\cdots{j_n}}(x-x_0)^i \times \\ &\left[\sum_{i=1}^\infty c_i^1(x-x_0)^i\right]^{j_1} \times \\ &\left[\sum_{i=1}^\infty c_i^2(x-x_0)^i \right]^{j_2} \times \\ &\cdots \\ & \left[\sum_{i=1}^\infty c_i^n(x-x_0)^i \right]^{j_n}\Bigg\},\quad k=1,2\cdots,n. \end{align*} \]

Denote \(X = x-x_0\), and we have

\[ \sum_{i=0}^\infty (i +1) c_{i+1}^kX^i = \sum_{i,j_1,j_2,\cdots,j_n=0}^\infty a^k_{i{j_1}{j_2}\cdots{j_n}}X^i \left(\sum_{i=1}^\infty c_i^1X^i\right)^{j_1} \left(\sum_{i=1}^\infty c_i^2X^i\right)^{j_2} \cdots \left(\sum_{i=1}^\infty c_i^nX^i\right)^{j_n} \]

and get \(c_i^k\) out in terms of \(a^k_{i\pmb{j}}\)

\[ \begin{align*} c_1^k &= a^k_{00\cdots 0}\\ c_2^k &= \frac{1}{2!} (a^k_{10\cdots 0}+a^k_{010\cdots 0}c^1_1 + a^k_{0010\cdots 0}c^2_1+\cdots a^k_{00\cdots01}c^n_1) \\ &= \frac{1}{2!} (a^k_{10\cdots 0} + a^k_{010\cdots 0}a^1_{00\cdots 0} + a^k_{0010\cdots 0}a^2_{00\cdots 0}+\cdots a^k_{00\cdots01}a^n_{00\cdots 0}) \end{align*} \]

Generally, we have

\[ c^k_m=P_m^k(a^l_{00\cdots 0}, a^l_{01\cdots 0},\cdots,a^l_{i{j_1}\cdots{j_n}}) \]

where \(i+j_1+j_2+\cdots+j_n\leq m-1\), \(1\leq l\leq n\). Thus, \(P_m^k\) is a polynomial represented by \(a^l_{00\cdots 0}\), \(a^l_{01\cdots 0}\), \(\cdots\), \(a^l_{i{j_1}\cdots{j_n}}\) 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 \(f_k(x,\pmb{y})\) is analytic on region \(R\), by Lemma 1, there exists an excellent function of \(f_k(x,\pmb{y})\) on a smaller region \(R_0\):

\[ F_k(x,\pmb{y}) = \frac{M}{\left(1-\frac{x-x_0}{a}\right)\left(1-\frac{y_1-y_{10}}{b}\right)\cdots\left(1-\frac{y_n-y_{n0}}{b}\right)}, \quad k=1,2\cdots n \]

If we represent both of them in terms of power series

\[ \begin{align*} f_k(x,\pmb{y})&=\sum_{i=0.\pmb{j}=0}^\infty a_{i\pmb{j}}(x-x_0)^i(\pmb{y}-\pmb{y}_0)^{\pmb{j}} \\ F_k(x,\pmb{y})&=\sum_{i=0.\pmb{j}=0}^\infty A_{i\pmb{j}}(x-x_0)^i(\pmb{y}-\pmb{y}_0)^{\pmb{j}} \end{align*} \]

then we have a relation \(|a_{i\pmb{j}}|<A_{i\pmb{j}}\), which matters in the following proof.

Now consider an ODE

\[ \frac{d y_k}{dx} = F_k(x,\pmb{y}),\quad y_k(x_0)=y_k,\quad k=1,2,\cdots,n,\quad \]

by Lemma 2, the above ODE has an analytic solution \(\pmb{y}=\pmb{y}(x)\), represented by power series

\[ y_k(x) = y_{k0} + \sum_{i=1}^\infty C_i^k(x-x_0)^i,\quad k=1,2,\cdots,n \]

Similarly, we have

\[ \begin{align*} C^k_m&=P_m^k(A^l_{00\cdots 0}, A^l_{01\cdots 0},\cdots,A^l_{i{j_1}\cdots{j_n}})\\ &=P_m^k(|A^l_{00\cdots 0}|, |A^l_{01\cdots 0}|,\cdots,|A^l_{i{j_1}\cdots{j_n}}|)\\ &\geq P_m^k(|a^l_{00\cdots 0}|, |a^l_{01\cdots 0}|,\cdots,|a^l_{i{j_1}\cdots{j_n}}|)\\ &\geq |c_m^k| \end{align*} \]

So power series \(\sum\limits_{i=1}^\infty C_i^k(x-x_0)^i\) is en excellent series of \(\sum\limits_{i=1}^\infty c_i^k(x-x_0)^i\). Since the former is convergent by Lemma 2, so the latter also converges.

附录 | Appendix: Relation between \(c\) and \(a\)

Theorem. Prove

\[ c^k_m=P_m^k(a^l_{00\cdots 0}, a^l_{01\cdots 0},\cdots,a^l_{i{j_1}\cdots{j_n}}) \]

where \(i+j_1+j_2+\cdots+j_n\leq m-1\), \(1\leq l\leq n\). Thus, \(P_m^k\) is a polynomial represented by \(a^l_{00\cdots 0}\), \(a^l_{01\cdots 0}\), \(\cdots\), \(a^l_{i{j_1}\cdots{j_n}}\) with positve operator "+".

HintsProof

Use induction.