Local Analysis¶

Linear homogeneous ODE:

\[ \begin{equation} \label{eq:linear_homogeneous_ode} y^{(n)}(x) + p_{n-1}(x) y^{(n-1)}(x) + \ldots + p_1(x) y'(x) + p_0(x) y(x) = 0 \end{equation} \]

Well-posedness of the IVP for this eq. may fail at singular points of $p_k(x)$.

Classification of singular points of an ODE (i.e. singular points of coefficients)

ODE on a complex plane

\[ y = y(z), \quad z \in \mathbb{C}$$ \]

is a sol. to equation \eqref{eq:linear_homogeneous_ode}

$y'(z)$ exists in a region of $\mathbb{C}$ $\implies y$ is analytic there.

So NOT all ODE makes sense when formulated in $\mathbb{C}$.

Example

\[ y'(x) = |x| \]

is OK in $\mathbb{R}$.

\[ y(x) = \frac{1}{2} x |x| + c \]

Therefore, for $y'(z) = g(z, y(z))$ to make sense, we need $g(z, y(z))$ to be analytic in $z$.

If $y_1, y_2$ are two linearly independent solutions

\[ \begin{aligned} p_0(z) &= \frac{y_1'' y_2' - y_2'' y_1'}{y_1' y_2 - y_1 y_2'} \\ p_1(z) &= \frac{y_2'' y_1 - y_1'' y_2}{y_1' y_2 - y_1 y_2'} \end{aligned} \]

$p_0, p_1$: meromorphic functions with isolated poles

Example

\[ y'' - |z| y' + (|z| - 1) y = 0 \]

admits analytic solutions

Singular point of an ODE: non-analytic point of $p_k(z)$

Ordinary Points¶

the point where the coeff. are analytic at.

Set of ordinary points is open in $\mathbb{C}$.

e.g.

\[ y' = |z| y \]

$|z|$ is nowhere analytic $\implies$ no ordinary points.

e.g.

\[ x^5 y'' = y \]

$p_0 = -\frac{1}{x^5}$, ordinary points: $\mathbb{C} \setminus \{0\}$

Conclusion

Near an ordinary point $x_0$, all the $n$ linearly independent solutions are analytic.

Radius of convergence: $R \geq \text{dist}(x_0, \text{nearest singular point of } p_k(x))$

Example

\[ (x^2 + 1) y' + 2xy = 0 \]

$p_0 = \frac{2x}{x^2 + 1}$ is analytic near $x_0 = 0$. Singular points: $\pm \mathrm{i}$.

\[ y = C (1 + x^2)^{-1} = C \sum_{n=0}^{\infty} (-x^2)^n \]

converges for $|x| < 1$, $R = 1$.

Example

\[ \left \{ \begin{array}{l} y' = 2xy \\ y(0) = 1 \end{array} \right. \]

$p_0 = -2x$ is analytic everywhere, $x_0 = 0$ is an ordinary point.

Find a solution near $x_0 = 0$:

\[ y(x) = \sum_{n=0}^{\infty} a_n x^n \quad\text{(ansatz)} \]

Plug into the ODE:

\[ \sum_{n=0}^{\infty} n a_n x^{n-1} = \sum_{n=0}^{\infty} 2 a_n x^{n+1} \]

Compare coeff., $a_1 = 0$. For $n \geq 2$, we have the recursion relation

\[ n a_n = 2 a_{n-2} \implies a_{2k} = \frac{a_0}{k!} \]

sol

\[ y(x) = \sum_{n=0}^{\infty} \frac{a_0}{k!} x^{2k} = a_0 e^{x^2} \]

By i.c., $a_0 = 1$. So

\[y(x) = e^{x^2}\]

is the unique sol. to the IVP.

Airy Equation

\[ y'' - x y = 0 \]

Physical background: 1D time-dependent Schrödinger equation: $$ -\frac{\hbar^2}{2m} \frac{d^2}{dx2} \psi(x) + V(x) \psi(x) = i \hbar \frac{\partial}{\partial t} \psi(x, t) $$

where $\psi(x, t) = \psi(x) e^{-iEt/\hbar}, E = \hbar \omega$

\[ k^2(x) =\frac{2m (E - V(x)}{\hbar^2} \]

\[ \begin{equation} \tag{Helmholtz eq.} \implies \psi''(x) + k^2(x) \psi(x) = 0 \end{equation} \]

If $k^2(x) \sim \alpha(x - x_0)$ near $x_0$, then $x_0$ is called a turning point.

\[ \psi''(x) + \alpha(x - x_0) \psi(x) = 0 \]

$x_0$ is an ordinary point, so try the ansatz

\[ y(x) = \sum_{n=0}^{\infty} a_n x^n \]

\[ \sum_{n=2}^{\infty} a_n n(n-1) x^{n-2} = \sum_{n=0}^{\infty} a_n x^{n+1} \]

$a_0, a_1$ are arbitrary constants, determined by i.c.

$n = 2 \implies a_2 = 0$. The recursion relation is

\[ (n+2)(n+1) a_{n+2} = a_{n-1}, \quad n = 1, 2, \ldots \]

i.e.,

\[ a_{n+3} = \frac{a_n}{(n+2)(n+1)}, \quad n = 0, 1, 2, \ldots \]

we can get

\[ \begin{aligned} a_{3k} &= \frac{a_0}{(3k)(3k-1)(3k-3)(3k-4) \ldots 9 \cdot 8 \cdot 6 \cdot 5 \cdot 3 \cdot 2} \\ &= \frac{a_0 \Gamma(\frac{2}{3})}{9^n n! \Gamma(n + \frac{2}{3})} \\ a_{3k+1} &= \frac{a_1}{(3k+1)(3k)(3k-2)(3k-3) \ldots 10 \cdot 9 \cdot 7 \cdot 6 \cdot 4 \cdot 3} \\ &= \frac{a_1 \Gamma(\frac{4}{3})}{9^n n! \Gamma(n + \frac{4}{3})} \\ a_{3k+2} &= 0 \end{aligned} \]

So the general solution is

\[ y(x) = c_1 \sum_{k=0}^{\infty} \frac{x^{3k}}{9^k k! \Gamma(k + \frac{2}{3})} + c_2 \sum_{k=0}^{\infty} \frac{x^{3k+1}}{9^k k! \Gamma(k + \frac{4}{3})} \]

where

\[ c_1 = a_0 \Gamma(\frac{2}{3}), \quad c_2 = a_1 \Gamma(\frac{4}{3}) \]

Singular Points¶

Regular Singular Points¶

$x_0 \neq \infty$ is a regular singular point if not all $p_0(x), p_1(x), \ldots, p_{n-1}(x)$ are analytic at $x_0$, but $(x - x_0)^n p_0(x), (x - x_0)^{n-1} p_1(x), \ldots, (x - x_0) p_{n-1}(x)$ are analytic at $x_0$.

Conclusion

Near a regular singular point $x_0$, solutions are

analytic
has a pole
has an algebraic or logarithmic branch point

At least one solution has the form

\[ y(x) = (x - x_0)^{\alpha} A(x) \]

where $\alpha \notin \mathbb{Z}$ and $A(x)$ is analytic at $x_0$. $\alpha$ is called an indicial exponent.

\[ A(x) = \sum a_n (x - x_0)^n \]

then

\[ y(x) = \sum a_n (x - x_0)^{n + \alpha} \]

this is called a Frobenius series.

Example

\[ x^3 y' = (x+1) y \]

$x_0 = 0$ is an irregular singular point.

Example

\[ y(x) = C \tanh \left(\frac{x}{2}\right) = C \frac{e^x - 1}{e^x + 1} \]

poles at $x = \pm \mathrm{i} \pi$, radius of convergence $R = \pi$.

Frobenius Method for 2nd order ODEs¶

\[ y'' + \frac{p(x)}{x - x_0} y' + \frac{q(x)}{(x - x_0)^2} y = 0 \]

$p(x), q(x)$: analytic near $x_0$
$x_0$: regular singular point

Try the ansatz

\[ \begin{aligned} p(x) &= \sum_{n=0}^{\infty} p_n (x - x_0)^n \\ q(x) &= \sum_{n=0}^{\infty} q_n (x - x_0)^n \\ y(x) &= \sum_{n=0}^{\infty} a_n (x - x_0)^{n + \alpha} \end{aligned} \]

Plug into the ODE:

\[ \sum_{n=0}^{\infty} a_n (n + \alpha)(n + \alpha - 1) (x - x_0)^{n + \alpha - 2} + \frac{1}{x - x_0} \left(\sum_{n=0}^{\infty} p_n (x - x_0)^n \right) \left(\sum_{n=0}^{\infty} a_n (n + \alpha) (x - x_0)^{n + \alpha - 1} \right) + \frac{1}{(x - x_0)^2} \left(\sum_{n=0}^{\infty} q_n (x - x_0)^n \right) \left(\sum_{n=0}^{\infty} a_n (x - x_0)^{n + \alpha} \right) = 0 \]

$n = 0$ i.e. $(x - x_0)^{\alpha - 2}$ term:

\[ (\underset{P(\alpha)}{\underbrace{\alpha(\alpha - 1) + p_0 \alpha + q_0}}) a_0 = 0 \]

$a_0 \neq 0$, arbitrary constant
$P(\alpha) = 0$

$(x - x_0)^{n + \alpha - 2}$ terms:

$$

If $P(\alpha) = 0$ & $P(\alpha + n) \neq 0$, we can get $a_n$ from solving the recursion relation. Then we get Frobenius series sol.

From $P(\alpha) = 0$, we get two roots $\alpha_1, \alpha_2$.

If $\Re \alpha_1 > \Re \alpha_2$, then $P(\alpha_1 + n) \neq 0 \implies y(x) = \sum a_n (x - x_0)^{n + \alpha_1}$.

Modified Bessel Equation

\[ y'' + \frac{1}{x} y' - \left(1 + \frac{\nu^2}{x^2}\right) y = 0 \]

where $\nu \in \mathbb{C}$ is a constant.

It is obvious that $x_0 = 0$ is a regular singular point.

\[ p(x) = 1, \quad q(x) = -(x^2 + \nu^2). \]

Plug in $y(x) = \sum_{n=0}^{\infty} a_n x^{n + \alpha}$,

\[ a_0 [\alpha(\alpha - 1) + \alpha - \nu^2] = 0 \]

i.e.,

\[ P(\alpha) = \alpha^2 - \nu^2 = 0 \implies \alpha = \pm \nu \]

Without loss of generality, assume $\Re \nu \geq 0$.

\[ P(\alpha_1 + 1) = (\alpha_1 + 1)^2 - \nu^2 \neq 0 \]

where $\alpha_1 = \nu$. $P(\alpha_1 + n) \neq 0$.

\[ \begin{aligned} \implies & a_1 = 0, \, a_3 = a_5 = \ldots = a_{2k+1} = 0, \\ &a_{2n} = \frac{a_{2n-2}}{(\nu + 2n)^2 - \nu^2} \\ \implies & a_{2n} = \frac{a_0 \Gamma(\nu + 1)}{2^{2n} n! \Gamma(n + \nu + 1)} \end{aligned} \]

then

\[ y(x) = a_0 \Gamma(\nu + 1) \sum_{n=1}^{\infty} \frac{(x/2)^{2n + \nu}}{n! \Gamma(n + \nu + 1)} \]

Set

\[ a_0 = \left(\frac{1}{2} \right)^{\nu} \frac{1}{\Gamma(\nu + 1)} \]

we get the modified Bessel function of the first kind

\[ y(x) = \sum_{n=0}^{\infty} \frac{(x/2)^{2n + \nu}}{n! \Gamma(n + \nu + 1)} := I_{\nu}(x) \]

Radius of convergence: $R = \infty$.

For $\nu =

Irregular Singular Points¶

Some Notations for Asymptotic Analysis

when $x \to x_0$,

$f(x) \ll g(x) \iff f(x) = \omicron(g(x))$
$\displaystyle f(x) \sim g(x) \iff \lim_{x \to x_0} \frac{f(x)}{g(x)} = 1$

Example

\[ x^2 y'' + (1 + 3x) y' + y = 0 \]

We can derive that one sol. $y(x) \sim C_2 x^{-1} e^{1/x}$ as $x \to 0^+$.

\[ x^3 y'' = y \]

\[ \begin{aligned} y_1(x) &\sim C_1 x^{\frac{3}{4}} e^{2 x^{-\frac{1}{2}}} \\ y_2(x) &\sim C_2 x^{\frac{3}{4}} e^{-2 x^{-\frac{1}{2}}} \end{aligned} \]

$e^{x^{-1}}, e^{2 x^{-1/2}}, e^{-2 x^{-1/2}}$ describes the essential singularity of the sol. at $x_0 = 0$, called controlling factors.

Inspired by the exponential controlling factors, we look for sol. of the form

\[ y(x) = e^{S(x)} \]

Plug into the ODE,

$$

Method of Dominant Balance:

\[ x^3 y'' = y \]

$y(x) = e^{S(x)} \implies (S'' + S'^2) x^3 = 1$