@Chen Siyi

November 16, 2020

Note7 Power Series Solutions

For homogeneous linear ODEs with variable coefficients, sometimes finding an explicit solution is difficult, then we use the method of power series ansatz to solve/approximate solutions.

Recall: homogeneous, linear, ordinary, variable coefficients.

Summary of Power Series Ansatz

1. Analyze the equation, decide whether we can use power series ansatz around some point
2. Choose which form of ansatz to use
3. Plug into the ansatz, get recurrence relationship of the coefficients
4. Set initial value of coeffiencients. solve for coefficients to get one or more independent solutions
5. If not enough independent solutions are found, using reduction of order to find more solutions
6. Obtain the general solution

ODE with Analytic Coefficients

Where $P(t)$ and $Q(t)$ are analytic in a neiborhood of $t_0$.

“ a neighborhood of $t_0$” contains $t_0$

Then we can choose the ansatz

Accordingly,

$x\prime(t) = \sum_0^\infin ka_k (t-t_0)^{k-1}$

$x\prime \prime(t) = \sum_0^\infin k(k-1)a_k (t-t_0)^{k-2}$

Plug the three equations back, we can obtain the relationship of the coefficients {$a_0$, $a_1$, $a_2$, ...}.

Depending on the situation, after setting values for first $n$ terms (always 2), we can solve 1 to $n$(expected) independent solutions.

If not enough indepedent solutions are found, sometimes we can use reduction of order to find more.

Comments:

• The solutions found should be valid within its radius of convergence

Radius of Convergence of a Power Series:

• $\frac{1}{R} = \lim_{n \rightarrow \infin} \frac{|c_{n+1}|}{|c_n|}$
• $\frac{1}{R} = \lim_{n \rightarrow \infin} |c_n|^{1/n}$

ODE with Coefficents having Singular Points

The general form of a homogeneous linear second-order ODE with variable coefficients:

It is said to have a singular point at $t_0$ if $P(t_0) = 0$.

Generally around singular points, it's hard to decide or find continuous solutions. But there're some specific cases we can deal with.

Regular Singular Points

is said to have a regular singular point at $t_0$ if the functions $(t − t_0)p(t)$ and $(t − t_0)^2q(t)$ are analytic in a neighborhood of $t_0$. A singular point which is not regular is said to be irregular.

The general claim is: if an equation has a regular sigular point at $t_0$, then we can assume $\begin{array}{l} p(t)=\frac{p_{-1}}{t-t_{0}}+\sum_{j=0}^{\infty} p_{j}\left(t-t_{0}\right)^{j} \\ q(t)=\frac{q_{-2}}{\left(t-t_{0}\right)^{2}}+\frac{q_{-1}}{t-t_{0}}+\sum_{j=0}^{\infty} q_{j}\left(t-t_{0}\right)^{j} \end{array}$ and use the ansatz $x(t)=(t-t_0)^{r} \sum_{k=0}^{\infty} a_{k} (t-t_0)^{k}$ to find solutions.

Euler's Equation

Analysis:

This is exactly the case where the equation $x^{\prime \prime}+\alpha \frac{1}{t} x^{\prime}+\beta \frac{1}{t^2} x=0, \quad \alpha, \beta \in \mathbb{R}$ is having a regular singular point at $t = 0$.

Then we can choose the ansatz

Inserting back and solve for $r$ we get

• $(\alpha-1)^{2}-4 \beta > 0$

  $$x\left(t ; c_{1}, c_{2}\right)=c_{1} t^{r_{1}}+c_{2} t^{r_{2}}, \quad c_{1}, c_{2} \in \mathbb{R}$$

• $(\alpha-1)^{2}-4 \beta = 0$ , $r_{1}=r_{2}=\frac{1-\alpha}{2}$, need to use reduction of order

  $$x\left(t ; c_{1}, c_{2}\right)=c_{1} t^{r_{1}}+c_{2} t^{r_{1}} \ln t, \quad c_{1}, c_{2} \in \mathbb{R}$$

Reduction of order:

For equation $y\prime \prime + p(t)y \prime + q(t)y = 0$, and a known solution $y_1(x)$, let $y_2(x) = v(x)y_1(x)$, then you can solve for $v(x)$ using

$$y_1(t)v\prime \prime + (2y_1\prime (t) + p(t)y_1(t))v\prime = 0$$

• $(\alpha-1)^{2}-4 \beta < 0$

  After getting $x_{1}(t)=t^{r_{1}}=t^{\lambda}(\cos (\mu \ln t)+i \sin (\mu \ln t))$. $x_{2}(t)=t^{r_{1}}=t^{\lambda}(\cos (\mu \ln t)-i \sin (\mu \ln t))$, further have

  $$x\left(t ; c_{1}, c_{2}\right)=c_{1} t^{\lambda} \cos (\mu \ln t)+c_{2} t^{\lambda} \sin (\mu \ln t), \quad c_{1}, c_{2} \in \mathbb{R}$$

The Method of Frobenius

Basic Method

If it has a regular singular point at $t = 0$, then we can write out

$p_j$ and $q_j$ are known constants for us

We choose the Frobenius ansatz

Accordingly,

$x^{\prime}(t)=\sum_{k=0}^{\infty}(r+k) a_{k} t^{r+k-1}$

$x^{\prime \prime}(t)=\sum_{k=0}^{\infty}(r+k)(r+k-1) a_{k} t^{r+k-2}$

Plug back into the equations we then get

Setting

We get the indicial equation and recurrence equations to solve for $a_k$

With the recurrence equations, you can usually generate out a easier recurrence equation.

For good and different $r_i$ solved by the indical equation, llus some assumed initial values for $a_0$, $a_1$, ..., we are possible to solve for all $a_k$.

If everything goes fine, with $r_1 \neq r_2$ are two GOOD solutions, you get two INDEPENDENT solutions.

Question

Find the series solution to the below equation in the vicinity of $x_0=0$

$$2 x^{2} y^{\prime \prime}+7 x(x+1) y^{\prime}-3 y=0$$

Answer

Find a Second Independent Solution

But things can go wrong if

• $r_1 = r_2$: then need further work to obtain another solution (how? Later.)

• $r_1 = r_2 + N$: then though $r_1$ gives a solution, for $r_2$, due to $F(r_2 + N) = F(r_1) = 0$,

  • if the right-side of the recurrence equation vanishes for $F(r_2 + m) = F(r_2 + N)$,

    then $a_N$ is arbitrary, by setting $a_N$ as zero when dealing with $r_1$, and as an arbitrary non-zero number when dealing with $r_2$, we can further find a second independent solution. Though we can also use another general method (how? Later.)

  • if the right-side of the recurrence equation doesn't vanish, need further work to obtain another solution (how? Later.)

Noticing the above 3 cases have one thing in common: $r_1 = r_2 + N$, $N \in \mathbb{N}$ including 0. There's a general method for the above cases.

The recurrence equations can give a relationship $a_k(r)$. Then we have

where the constant $c∈R$ may vanish. If $r_1 =r_2$, then $c=1$.

And a tricky way to find $a_{2 k}^{\prime}(r_2)$ is to use

But this method may still fail sometimes... why? Then are there other methods?

Concrete examples are in Note8.