@Chen Siyi

November 23, 2020

Note8 Bessel Equations

Note8 Bessel EquationsBessel Equations of Order $v$ Find the Indical and Recurrence EquationsFind the First Independent SolutionFind the First Independent Solution with the Larger $r_1$ The Bessel Function of the First KindFind the Second Independent Solution ( $v \notin \mathbb{N}$ )Find the Second Independent Solution ( $v \in \mathbb{N}$ )Reduction of OrderThe Other MethodThe Bessel Function of the Second KindReduce Differential Equations to Bessel Equation $x^{2} y^{\prime \prime}+x y^{\prime}-\left(x^{2}+v^{2}\right) y=0$ $x^{2} y^{\prime \prime}+x y^{\prime}+\left(a^{2} x^{2}-v^{2}\right) y=0$ $x^{2} y^{\prime \prime}+a x y^{\prime}+\left(x^{2}-v^{2}\right) y=0$ * For thoughts: $y^{\prime \prime}-x y=0$

Let's apply the Method of Frobenius to solve Bessel equations.
And analyze the solutions (Bessel functions).

$v$

x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-v^{2}\right) y=0

Having a regular singular point at 0.
The Method of Frobenius can be applied.

Find the Indical and Recurrence Equations

Choose the Frobenius ansatz

x(t)=t^{r} \sum_{k=0}^{\infty} a_{k} t^{k} \quad \quad \quad a_0 \neq 0

Besides,

xp(x) = 1, \quad \quad p_0 = 1

x^2q(x)=x^2 - v^2, \quad \quad q_0 = -v^2,\quad q_2 = 1

Setting

F(x):=x(x-1)+p_{0} x+q_{0} = x^2 - v^2

We get the indicial equation and recurrence equations

\begin{aligned} F(r) &= r^2 - v^2 = 0 \\ a_{m} F(r+m) &=-\sum_{k=0}^{m-1}\left(q_{m-k}+(r+k) p_{m-k}\right) a_{k}, \quad m \geq 1 \end{aligned}

Which gives us

\begin{aligned} &r^2 - v^2 = 0 \\ &a_{1}((r+1)^2 - v^2) = 0\\ &a_{m} = -\frac{a_{m-2}}{(m+r+v)(m+r-v)}, \quad m \geq 2 \end{aligned}

$r_1 = v$ $r_2 = -v$ .

$r_1 - r_2 = 2v \notin \mathbb{N}$ , two independent solutions would be found easily.
$r_1 - r_2 = 2v \in \mathbb{N}$ , we may use the special technique.
However, we will see actually for Bessel Equations, the condition is slightly less strict:
$v \notin \mathbb{N}$ $r_1$ $r_2$ give two independent solutions.

Find the First Independent Solution

$r_1$

LARGER $r_1 = v$ , we have

\begin{aligned} &a_{1}((v+1)^2 - v^2) = 0\\ &a_{m} = -\frac{a_{m-2}}{(m+2v)m}, \quad m \geq 2 \end{aligned}

$a_1 = a_3 = a_5 = \dots = 0$ and

a_{2 k}=\frac{(-1)^{k} a_{0}}{2^{2 k} k !(1+v)(2+v) \cdots(k+v)}

Question:
$v$ may not be an integer. Don't write as factories.
Then how do you simpliy this solution?

The Bessel Function of the First Kind

Recall Euler Gamma function's property:

\Gamma(s+1)=s \Gamma(s)

So it gives

(1+v)(2+v) \cdots(k+v)=\frac{\Gamma(k+1+v)}{\Gamma(1+v)}

$a_0 = \frac{2^{-v}}{\Gamma(1+v)}$ , we will have the first independent solution be $v$

J_{v}(x)=\left(\frac{x}{2}\right)^{v} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k ! \Gamma(k+1+v)}\left(\frac{x}{2}\right)^{2 k}

Question:
$x$ $J_v(x)$ defined?

$v = 1$ as example, we have

J_{1}(x)=\frac{x}{2} \sum_{k=0}^{\infty} \frac{(-1)^{k} x^{2 k}}{2^{2 k}(k+1) ! k !}

$v \notin \mathbb{N}$ )

$2v$ is not an integerSMAllER $r_2 = -v$ , we have

\begin{aligned} &a_{1}((v-1)^2 - v^2) = 0, \quad a_{1}(2v - 1) = 0\\ &a_{m} = -\frac{a_{m-2}}{(m-2v)m}, \quad m \geq 2 \end{aligned}

$a_1 = a_3 = a_5 = \dots = 0$ and

a_{2k} = \frac{(-1)^{k} a_0}{2^{2 k} k !(1-v)(2-v) \cdots(n-v)}

Similarly,

(1-v)(2-v) \cdots(k-v) = \frac{\Gamma(k+1-v)}{\Gamma(1-v)}

$a_0 = \frac{2^{-v}}{\Gamma(1+v)}$ , the second independent solution will be $-v$

J_{-v}(x)=\left(\frac{x}{2}\right)^{-v} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k ! \Gamma(k+1-v)}\left(\frac{x}{2}\right)^{2 k}

Then the general solution is

y(x)=C_{1} J_{v}(x)+C_{2} J_{-v}(x)

$2v$ is an odd integer, which means $v$ is not an integer, the above results also holds.

$v$ is not an integer, the above results will hold.

$v \in \mathbb{N}$ )

Reduction of Order

$y_2(x) = c(x) · J_ν(x)$ , then

\begin{aligned} & x^{2} y_{2}^{\prime \prime}+x y_{2}^{\prime}+\left(x^{2}-\nu^{2}\right) y_{2}=0 \\ \Rightarrow & x^{2}\left(c^{\prime \prime}(x) J_{\nu}(x)+2 c^{\prime}(x) J_{\nu}^{\prime}(x)+c(x) J_{\nu}^{\prime \prime}(x)\right) \\ &+x\left(c^{\prime}(x) J_{\nu}(x)+c(x) J_{\nu}(x)\right)+\left(x^{2}-\nu^{2}\right) c(x) \cdot J_{\nu}(x)=0 \\ \Rightarrow & x^{2} J_{\nu}(x) c^{\prime \prime}(x)+\left(2 x^{2} J_{\nu}^{\prime}(x)+x J_{\nu}(x)\right) c^{\prime}(x)=0 \\ \Rightarrow & \ln \left|c^{\prime}(x)\right|=\left(-2 \ln \left|J_{\nu}(x)\right|-\ln |x|\right) \\ \Rightarrow & c^{\prime}(x)=\frac{1}{x \cdot J_{\nu}^{2}(x)} \\ \Rightarrow & c(x)=\int \frac{d x}{x \cdot J_{\nu}^{2}(x)} \end{aligned}

So a second independent solution is given as

y_{2}(x)=J_{\nu}(x) \int \frac{d x}{x \cdot J_{\nu}^{2}(x)}

The Other Method

x_{2}(t)=\left.\frac{\partial}{\partial r}\left(t^{r} \sum_{k=0}^{\infty} a_{k}(r) t^{k}\right)\right|_{r=r_{2}}=c \cdot x_{1}(t) \ln t+t^{r_{2}} \sum_{k=0}^{\infty} a_{k}^{\prime}\left(r_{2}\right) t^{k}

\frac{a_{2 k}^{\prime}(r)}{a_{2 k}(r)}=\frac{d}{d r} \ln \left|a_{2 k}(r)\right|

Practice:
Using 5 minites to try solving out the second solution by yourself.
Do you find any problems?

$a^{\prime}_{2k}(r_2)$ , let's find these new constants in another way. Assume

y_{2}(x)=a J_{v}(x) \ln x+x^{-v}\left[\sum_{k=0}^{\infty} c_{k} x^{k}\right], \quad x>0

$y_2\prime$ $y_2\prime \prime(x)$ $J_v(x)$ $a,c_0,c_1,\dots$

Let's try with the Bessel Equation of order 1.

y_{2}(x)=a J_{1}(x) \ln x+x^{-1}\left[\sum_{k=0}^{\infty} c_{k} x^{k}\right], \quad x>0

$J_1(x)$ is a solution, we can simplify the equation to be

2 a x J_{1}^{\prime}(x)-c_1+\sum_{k=2}^{\infty}(k^2-2k)c_k x^{k-1}+\sum_{k=0}^{\infty} c_{k} x^{k+1}=0

$J_1 (x)$ then

a\left[\sum_{k=0}^{\infty} \frac{(-1)^{k}(2 k+1) x^{2 k+1}}{2^{2 k}(k+1) ! k !}\right]-c_{1} +\sum_{k=0}^{\infty}\left[\left(k^{2}+2k\right) c_{k+2}+c_{k}\right] x^{k+1} = 0

$c_1 = 0$ .

$\left(k^{2}+2\right) c_{k+2}+c_{k}$ $k$ $c_1 = c_3 = \dots = 0$ .

And from setting the coefficients of odd powers as 0, we have

\left[(2 k+1)^{2}-1\right] c_{2 k+2}+c_{2 k}=a\frac{(-1)^{k+1}(2 k+1)}{2^{2 k}(k+1) ! k !}, \quad k=0,1,2,3, \ldots

$k = 0$ , we have

0\cdot c_2 + c_0 = -a

$c_0 = -a$ $c_0 = 1$ $a = -1$ . Then

\left[(2 k+1)^{2}-1\right] c_{2 k+2}+c_{2 k}=\frac{(-1)^{k}(2 k+1)}{2^{2 k}(k+1) ! k !}, \quad k=1,2,3, \ldots

$k = 1$ , we get

\left(3^{2}-1\right) c_{4}+c_{2}=(-1) 3 /\left(2^{2} \cdot 2 !\right)

$c_2$ can be selected in arbitrary, and then we fix the second independent solution.

$c_2 = \frac{1}{2^2}$ , and then we would be possible to simplify:

c_{2 m}=\frac{(-1)^{m+1}\left(H_{m}+H_{m-1}\right)}{2^{2 m} m !(m-1) !}

$H_{m}(x):=\sum_{i = 1}^m \frac{1}{i}$ $H_0 = 0$ , is the Harmonic Numbers. So in conclusion we obtain:

y_{2}(x)=-J_{1}(x) \ln x+\frac{1}{x}\left[1-\sum_{m=1}^{\infty} \frac{(-1)^{m}\left(H_{m}+H_{m-1}\right)}{2^{2 m} m !(m-1) !} x^{2 m}\right], \quad x>0

The Bessel Function of the Second Kind

$v$ $J_v(x)$ $y_2(x)$ . In our specific case here of order 1, we set the Bessel function of the second kind of order 1 as

Y_{1}(x)=\frac{2}{\pi}\left[-y_{2}(x)+(\gamma-\ln 2) J_{1}(x)\right]

$v$ $J_v(x)$ $J_{-v}(x)$ :

Y_{v}(x)=\frac{J_{v}(x) \cos \pi v-J_{-v}(x)}{\sin \pi v}

And then the general solution can be written as

y(x)=C_{1} J_{v}(x)+C_{2} Y_{v}(x)

Reduce Differential Equations to Bessel Equation

$x^{2} y^{\prime \prime}+x y^{\prime}-\left(x^{2}+v^{2}\right) y=0$

Exercise:
Show that the general solution of this equation can be expressed as
$y(x)=C_{1} J_{v}(-i x)+C_{2} Y_{v}(-i x)$

$x^{2} y^{\prime \prime}+x y^{\prime}+\left(a^{2} x^{2}-v^{2}\right) y=0$

Exercise:
Show that the general solution of this equation can be expressed as
$y(x)=C_{1} J_{v}(ax)+C_{2} Y_{v}(ax)$

$x^{2} y^{\prime \prime}+a x y^{\prime}+\left(x^{2}-v^{2}\right) y=0$

Exercise:
Show that the general solution of this equation can be expressed as
$y(x)=x^{\frac{1-a}{2}}\left[C_{1} J_{n}(x)+C_{2} Y_{n}(x)\right]$
Hint:
$y(x)=x^{\frac{1-a}{2}} z(x)$

$y^{\prime \prime}-x y=0$

*Exercise:
Show that the general solution of this equation can be expressed as
$y(x)=C_{1} \sqrt{x} J_{\frac{1}{3}}\left(\frac{2}{3} i x^{\frac{3}{2}}\right)+C_{2} \sqrt{x} J_{-\frac{1}{3}}\left(\frac{2}{3} i x^{\frac{3}{2}}\right)$
Hint:
$\frac{d^2y}{dx^2}$

Note8 Bessel Equations

Bessel Equations of Order v

Find the Indical and Recurrence Equations

Find the First Independent Solution

Find the First Independent Solution with the Larger r_1

The Bessel Function of the First Kind

Find the Second Independent Solution (v \notin \mathbb{N})

Find the Second Independent Solution (v \in \mathbb{N})