---
# System prepended metadata

title: Problems for Chapter 3 -- Part 2-1
tags: [應用數學三]

---

# Problems for Chapter 3 -- Part 2-1
###### tags: `應用數學三`

## Problem 1
:::warning
The Rodrigues’ formula for Hermite polynomials is $$\require{cancel}\require{color}H_n(x)=(-1)^ne^{x^2}\dfrac{\text{d}^n}{\text{d}x^n}e^{-x^2}.\tag{1}$$
- **(a)** Prove that $H_n(x)$ is a polynomial solution of Hermite's equation $$y''-2xy'+2ny=0.\tag{2}$$
- **(b)** Prove that $$|H_n(x)|\leq|H_n(ix)|.\tag{3}$$
:::
### Solution of (a)
Define the differential operator $D\equiv\frac{\text{d}}{\text{d}x}$, with $D^n\equiv\frac{\text{d}^n}{\text{d}x^n}$ for $n=1,2,\ldots$ and also $D^0\equiv\hat{1}$ (identity operator).

We will state and then verify two propositions.

Propositions
:    1. $H_n(x)$ satisfies Eqn. (2) is true for $n=0$.
:    2. Suppose that $H_n(x)$ satisfies Eqn. (2) when $n=k$, then so is the case when $n=k+1$.

Verifications
:    1. Apparently, $H_0(x)=1$  satisfies Eqn. (2).
:    2. Suppose Eqn. (2) holds when $n=k$, then  $$\begin{align}0&=\left(D^2-2xD+2\color{orangered}k\right)[H_{\color{orangered}k}(x)]\\&=\Big(-(2x-D)D+2k\Big)[H_k(x)]\\&=\Big(2-D(2x-D)+2k\Big)[H_k(x)],\tag{4}\end{align}$$ where we use the commutation relationship $$\begin{align}[D,2x-D]&=D(2x-D)-(2x-D)D\\&=2+2xD-D^2-2xD+D^2=2\\\implies -(2x-D)D &=2-D(2x-D).\end{align}$$ Acting $(2x-D)$ on Eqn. (4) yields $$\begin{align}0&=(2x-D)\Big(-D(2x-D)+2k+2\Big)[H_k(x)]\\&=-(2x-D)D[(2x-D)H_k(x)]+2(k+1)[(2x-D)H_k(x)]\\&=\Big(-(2x-D)D+2(k+1)\Big)[H_{k+1}(x)]\\&=\Big(D^2-2xD+2\left.\color{orangered}(k+1)\right.\Big)[H_{\color{orangered}k+1}(x)]\tag{5},\end{align}$$ where we use the fact that $$\begin{align}H_{n+1}(x)&=e^{x^2}(-D)^{n+1}e^{-x^2}=e^{x^2}(-D)\underbrace{\color{red}e^{-x^2}e^{x^2}}_{=\,\hat{1}}(-D)^{n}e^{-x^2}\\&=e^{x^2}(-D)\left[e^{-x^2}H_n(x)\right]=\cancel{e^{x^2}}\left[2x\cancel{e^{-x^2}}H_n(x)-\cancel{e^{-x^2}}D[H_n(x)]\right]\\&=(2x-D)[H_n(x)],\end{align}$$ which is inferred from Eqn. (1). 

With an eye on Propositions 1 and 2, we prove by induction that $H_n(x)$ satisfies Eqn. (2) for $n=0,1,2,\dots.$ $\blacksquare$


### Solution of (b)
$H_n(ix)=(-1)^ne^{(ix)^2}D^ne^{-(ix)^2}=(-1)^ne^{-x^2}D^ne^{x^2}$

---
## Problem 2
::: warning
The Bessel function $J_\nu(x)$ and Neumann function $N_\nu(x)$ may be written as $$J_\nu(x)=\sum^\infty_{j=0}\dfrac{(-1)^j}{j!\,(j+\nu)!}\left(\dfrac{x}{2}\right)^{\nu+2j}\tag{1}$$ and $$N_\nu(x)=\dfrac{J_\nu(x)\cos(\nu\pi)-J_{-\nu}(x)}{\sin(\nu\pi)}\tag{2}$$ where $(j+\nu)!=\Gamma(j+\nu+1)=\int^\infty_0 t^{j+\nu}e^{-t}\text{d}t$. Assume $\nu$ is real and $\nu>-1$.
- **(a)** From definitions (1) and (2) in the above, deduce the recurrence relations $$x^{-\nu}\dfrac{\text{d}}{\text{d}x}\left[x^\nu\Omega_\nu(x)\right]=\Omega_{\nu-1}(x)\tag{3}$$ and $$x^{\nu}\dfrac{\text{d}}{\text{d}x}\left[x^{-\nu}\Omega_\nu(x)\right]=-\Omega_{\nu+1}(x)\tag{4}$$ where $\Omega_\nu$ can be either $J_\nu$ or $N_\nu$.
- **(b)** Show that, between two successive positive zeros of $J_{\nu+1}(x)$ on the real axis, there is one and only one zero of $J_\nu(x)$.
- **(c\)** If $m$ is an integer, show that $$\begin{align}J_{-m}(x)&=(-1)^mJ_m(x),\tag{5a}\\N_{-m}(x)&=(-1)^mN_m(x).\tag{5b}\end{align}$$
:::
### Solution of (a)
#### Method A
1. To prove Eqn. (3) holds for $J_\nu(x)$, we multiply Eqn. (1) by $x^\nu$ and differentiate to get $$\begin{align}\dfrac{\text{d}}{\text{d}x}\left[x^\nu J_\nu(x)\right]&=\dfrac{\text{d}}{\text{d}x}\sum^\infty_{j=0}\dfrac{(-1)^j}{j!\,(j+\nu)!}\dfrac{x^{2\nu+2j}}{2^{\nu+2j}}\\
&=\sum^\infty_{j=0}\dfrac{(-1)^j(2\nu+2j)}{j!\,(j+\nu)!}\dfrac{x^{2\nu+2j-1}}{2^{\nu+2j}}\\
&=\sum^\infty_{j=0}\dfrac{(-1)^j}{j!\,(j+\nu-1)!}\dfrac{x^{2\nu+2j-1}}{2^{\nu+2j-1}}.\end{align}$$ Dividing the above by $x^\nu$ and comparing it with Eqn. (1) yields $$x^{-\nu}\dfrac{\text{d}}{\text{d}x}\left[x^\nu J_\nu(x)\right]
=\sum^\infty_{j=0}\dfrac{(-1)^j}{j!\,(j+\nu-1)!}\left(\dfrac{x}{2}\right)^{\nu+2j-1}=J_{\nu-1}(x),$$ since this is just Eqn. (1) with $\nu$ replaced by $\nu-1$. 

2. Likewise, Eqn. (4) holds for $J_\nu(x)$: $$\begin{align}\dfrac{\text{d}}{\text{d}x}\left[x^{-\nu} J_\nu(x)\right]&=\dfrac{\text{d}}{\text{d}x}\sum^\infty_{j=0}\dfrac{(-1)^j}{j!\,(j+\nu)!}\dfrac{x^{2j}}{2^{\nu+2j}}\\
&=\sum^\infty_{j=0}\dfrac{(-1)^j(2j)}{j!\,(j+\nu)!}\dfrac{x^{2j-1}}{2^{\nu+2j}}\\
&=\sum^\infty_{j=0}\dfrac{(-1)^j}{(j-1)!\,(j+\nu)!}\dfrac{x^{2j-1}}{2^{\nu+2j-1}};\\\\x^\nu\dfrac{\text{d}}{\text{d}x}\left[x^{-\nu} J_\nu(x)\right]&=\sum^\infty_{j=0}\dfrac{(-1)^j}{(j-1)!\,(j+\nu)!}\left(\dfrac{x}{2}\right)^{\nu+2j-1}\\&=\sum^\infty_{j=0}\dfrac{(-1)^{j+1}}{j!\,(j+\nu+1)!}\left(\dfrac{x}{2}\right)^{\nu+2j+1}=-J_{\nu+1}(x),\end{align}$$ where we use the fact that $\frac{1}{(j-1)!}=0$ for $j=0$ and we shift the index from  $j$ to $j+1$. 

3. As for $N_{\nu}$, we find $$\begin{align}N_{\nu-1}(x)&=\dfrac{J_{\nu-1}(x)\cos[(\nu-1)\pi]-J_{-\nu+1}(x)}{\sin[(\nu-1)\pi]}\\&=\dfrac{-J_{\nu-1}(x)\cos(\nu\pi)-J_{-\nu+1}(x)}{-\sin(\nu\pi)}\\
&=\dfrac{x^{-\nu}\frac{\text{d}}{\text{d}x}\left[x^\nu J_\nu(x)\right]\cos(\nu\pi)-x^{-\nu}\frac{\text{d}}{\text{d}x}\left[x^{\nu} J_{-\nu(x)}\right]}{\sin(\nu\pi)}\\
&=x^{-\nu}\dfrac{\text{d}}{\text{d}x}\dfrac{x^\nu\left[J_\nu(x)\cos(\nu\pi)-J_{-\nu}(x)\right]}{\sin(\nu\pi)}\\
&= x^{-\nu}\dfrac{\text{d}}{\text{d}x}\left[x^\nu N_\nu(x)\right]\end{align}$$ and $$\begin{align}N_{\nu+1}(x)&=\dfrac{J_{\nu+1}(x)\cos[(\nu+1)\pi]-J_{-\nu-1}(x)}{\sin[(\nu+1)\pi]}\\&=\dfrac{-J_{\nu+1}(x)\cos(\nu\pi)-J_{-\nu-1}(x)}{-\sin(\nu\pi)}\\
&=\dfrac{-x^{\nu}\frac{\text{d}}{\text{d}x}\left[x^{-\nu} J_\nu(x)\right]\cos(\nu\pi)+x^{\nu}\frac{\text{d}}{\text{d}x}\left[x^{-\nu} J_{-\nu(x)}\right]}{\sin(\nu\pi)}\\
&=-x^{\nu}\dfrac{\text{d}}{\text{d}x}\dfrac{x^{-\nu}\left[J_\nu(x)\cos(\nu\pi)-J_{-\nu}(x)\right]}{\sin(\nu\pi)}\\
&=-x^{\nu}\dfrac{\text{d}}{\text{d}x}\left[x^{-\nu} N_\nu(x)\right],\end{align}$$ which respectively coincide with Eqs. (3) and (4) for $N_\nu(x)$. $\blacksquare$

#### Method B
The general contour integration definition for Bessel functions is $$\Omega_\nu(x)=\dfrac{1}{2\pi i}\oint_C\dfrac{e^{\frac{x}{2}\left(z-\frac{1}{z}\right)}}{z^{\nu+1}}\text{d}z$$

### Solution of (b)
#### Proof of Existence
Multiplying Eqn. (3) by $x^\nu$and shifting the order from $\nu$ to $\nu+1$, we have $$\dfrac{\text{d}}{\text{d}x}\left[x^{\nu+1}J_{\nu+1}(x)\right]=x^{\nu+1}J_{\nu}(x).\tag{6}$$

According to Rolle’s theorem, between two consecutive zeros of a continuous function lies a zero of it’s derivative. Taking the continuous function in Eq. (6) to be $x^{\nu+1}J_{\nu+1}(x)$, between two of its consecutive zeros lies at least one zero of its derivative, i.e. of $x^{\nu+1}J_{\nu}(x)$. 

Since $\nu>-1$ and our interval in interest is $x\in(0,\infty)$, a zero of $x^{\nu+1}J_{\nu+1}(x)$ must be a zero of $J_{\nu+1}(x)$, and a zero of $x^{\nu+1}J_{\nu}(x)$ must be a zero of $J_{\nu}(x)$.


Therefore, between two consecutive positive zeros of $J_{\nu+1}(x)$ lies at least one zero of $J_{\nu}(x)$.

#### Proof of Uniqueness
Assume there are *at least two* (distinct) zeros of $J_{\nu}(x)$, say, $\alpha_1<\cdots<\alpha_n$ ($n\geq2$), all lying between two consecutive positive zeros of $J_{\nu+1}(x)$, say, $a$ and $b$. In other words, $0<a<\alpha_1<\cdots<\alpha_n<b$.

From Eqn. (4), we have $$\dfrac{\text{d}}{\text{d}x}\left[x^{-\nu}J_{\nu}(x)\right]=-x^{-\nu}J_{\nu+1}(x).\tag{7}$$ By Rolle’s theorem again, we conclude that there exists at least one zero of $J_{\nu+1}(x)$ between $\alpha_i$ and $\alpha_{i+1}$ ($i=1,2,\dots,n-1$), and thus there at least $n-1$ zeros between $a$ and $b$, which contradicts to our assumption that $a$ and $b$ are *consecutive* zeros.

Therefore, our assumption must be must be false. Its negation is true: there are *less than two*. The number of zeros is *exactly one*. $\blacksquare$

#### Review: Rolle's Theorem
Provided that 
1. $f(x)$ is continuous on $[a,b]$, 
2. $f(x)$ is differentiable on $(a,b)$,
3. $f(a)=f(b)$;  
then there exists $\xi\in(a,b)$ such that $f'(\xi)=0$.
#### Reference 
- [UDEL: Zeros of Bessel Functions](http://www.math.udel.edu/~angell/zeros_bessel.pdf) p.3
- [北京大学数学物理方程：柱函数](https://astrojacobli.github.io/Docs/Mathematical_Method/chpp17.pdf) p.6
### Solution of (c\)
1. Replacing $\nu$ in Eqn. (1) with $\pm m\in\mathbb{Z}$ gives $$J_{\pm  m}(x)=\sum^\infty_{j=0}\dfrac{(-1)^j}{j!\,(j\pm m)!}\left(\dfrac{x}{2}\right)^{\pm m+2j}.$$ Especially, we obtain Eqn. (5a): $$\begin{align}J_{-m}(x)&=\sum^\infty_{j=\color{orangered}0}\dfrac{(-1)^j}{j!\,(j-m)!}\left(\dfrac{x}{2}\right)^{-m+2j}\\
&=\sum^\infty_{j=\color{orangered}m}\dfrac{(-1)^j}{j!\,(j-m)!}\left(\dfrac{x}{2}\right)^{-m+2j}\\
&=\sum^\infty_{j=\color{green}0}\dfrac{(-1)^{\color{green}j+m}}{\left.\color{green}(j+m)\right.!\,\left.\color{green}j\right.!~~}\left(\dfrac{x}{2}\right)^{-m+2\color{green}(j+m)}\\
&=(-1)^m\sum^\infty_{j=0}\dfrac{(-1)^j}{(j+m)!\,j!}\left(\dfrac{x}{2}\right)^{m+2j}\\&=(-1)^mJ_m(x).\end{align}$$ Notice that we use the fact that $\frac{1}{(j-m)!}=0$ for $j<m$; also notice that it is valid to shift the index of summation from $j$ to $j+m$  only if $m$ is an integer.

2. From Eqn. (2), with $\nu$ replaced by $-m$, we obtain Eqn. (5b): $$\begin{align}N_{-m}(x)&=\dfrac{J_{-m}(x)\cos(-m\pi)-J_{m}(x)}{\sin(-m\pi)}\\&=\dfrac{(-1)^mJ_m(x)\cos(m\pi)-(-1)^{-m}J_{-m}(x)}{-\sin(m\pi)}\\&=(-1)^m\dfrac{J_{m}(x)\cos(m\pi)-J_{-m}(x)}{\sin(m\pi)}\\&=(-1)^m,N_m(x)\end{align}$$ where we use Eqn. (5a) in the second equality. $\blacksquare$
---
## Problem 3
::: warning
Show that all the roots of $J_\nu(z)$ with $\nu>-1$ are on the real axis.
:::
### Solution
#### Method A
1. From the complex Bessel equation (complex version from Eqn.(3.43) in the handout) $$\dfrac{1}{z}\dfrac{\text{d}}{\text{d}z}\left[z\dfrac{\text{d}J_\nu(z)}{\text{d}z}\right]+\left(1-\dfrac{\nu^2}{z^2}\right)J_{\nu}(z)=0,\tag{1}$$ we use the transformation $z=kt\left(t\in\mathbb{R}\setminus\{0\},~z\in\mathbb{C}\setminus\{0\}\right)$ and multiply Eqn. (1) by constant $k^2$ to obtain $$\dfrac{1}{t}\dfrac{\text{d}}{\text{d}t}\left[t\dfrac{\text{d}J_\nu(kt)}{\text{d}t}\right]+\left(k^2-\dfrac{\nu^2}{t^2}\right)J_{\nu}(kt)=0.\tag{2}$$ Put complex numbers $k=a$ and $k=b$ in Eqn. (2): $$\begin{align}&\dfrac{1}{t}\dfrac{\text{d}}{\text{d}t}\left[t\dfrac{\text{d}J_\nu(at)}{\text{d}t}\right]+\left(a^2-\dfrac{\nu^2}{t^2}\right)J_{\nu}(at)=0,\tag{2a}\\&\dfrac{1}{t}\dfrac{\text{d}}{\text{d}t}\left[t\dfrac{\text{d}J_\nu(bt)}{\text{d}t}\right]+\left(b^2-\dfrac{\nu^2}{t^2}\right)J_{\nu}(bt)=0.\tag{2b}\end{align}$$ Multiply Eqs. (2a) and (2b) by $tJ_{\nu}(bt)$ and $tJ_{\nu}(at)$, respectively, and integrate their difference from $0$ to $x$ over $t$:  $$\begin{align}\left(a^2-b^2\right)t^2J_{\nu}(at)J_{\nu}(bt)\\=-tJ_\nu(bt)&\dfrac{\text{d}}{\text{d}t}\left[t\dfrac{\text{d}J_\nu(at)}{\text{d}t}\right]+tJ_\nu(at)\dfrac{\text{d}}{\text{d}t}\left[t\dfrac{\text{d}J_\nu(bt)}{\text{d}t}\right]\\\\
\implies\left(a^2-b^2\right)tJ_{\nu}(at)J_{\nu}(bt)\\=-J_\nu(bt)&\dfrac{\text{d}}{\text{d}t}\left[t\dfrac{\text{d}J_\nu(at)}{\text{d}t}\right]+J_\nu(at)\dfrac{\text{d}}{\text{d}t}\left[t\dfrac{\text{d}J_\nu(bt)}{\text{d}t}\right]\\\\
\implies\left(a^2-b^2\right)\int^x_0tJ_\nu(at)J_\nu(bt)\text{d}t&=t\left[ J_\nu(at)\dfrac{\text{d}J_\nu(bt)}{\text{d}t}-J_\nu(bt)\dfrac{\text{d}J_\nu(at)}{\text{d}t}\right]^{t=x}_{t=0}\\
&=x\left[ J_\nu(ax)\dfrac{\text{d}J_\nu(bx)}{\text{d}x}-J_\nu(bx)\dfrac{\text{d}J_\nu(ax)}{\text{d}x}\right],\tag{3}\end{align}$$ where we use the fact that $J_\nu(0)=0$ for $\nu>-1$, which is seen in the series representation of Bessen function.

2. Because the coefficients of Eqn. (4) are real, $$[J_{\nu}(z)]^*=J_{\nu}(z^*)$$ Let $\alpha$ be a complex zero of $J_\nu(z)$. i.e. $J_\nu(\alpha x)=0$. Given that the point $z=\alpha$ is a zero of $J_\nu(z)$, then $$[J_\nu(\alpha x)]^*=J_\nu\left(\alpha^*x\right)=0.$$

3. Plug $a=\alpha$, $b=\alpha^*$ and $x=1$ in Eqn. (3), we have $$\begin{align}\left(\alpha^2-{\alpha^*}^2\right)\int^1_0tJ_\nu(\alpha t)J_\nu(\alpha^*t)\text{d}t=J_\nu(\alpha x)J_\nu\dfrac{\text{d}J_\nu(\alpha^*x)}{\text{d}x}-J_\nu(\alpha^*x)\dfrac{\text{d}J_\nu(\alpha x)}{\text{d}x}=0,\end{align}$$ while $$\int^1_0tJ_\nu(\alpha t)J_\nu(\alpha^*t)\text{d}t=\int^1_0t|J_\nu(\alpha t)|^2\text{d}t>0,$$ so $$\alpha^2-{\alpha^*}^2=0\implies \alpha=\pm \alpha^*,$$ which means $\alpha$ is either real or pure imaginary.

4. If $\alpha$ is pure imaginary, then let $\alpha=i\beta$, with $\beta$ being real and non-zero, but when $\nu>-1$, then the summation$$\require{cancel}\begin{align}J_\nu(i\beta)&=\sum^\infty_{j=0}\dfrac{(-1)^j}{j!\,(j+\nu)!}\left(\dfrac{i\beta}{2}\right)^{\nu+2j}\\&=\left(\dfrac{i\beta}{2}\right)^\nu\sum^\infty_{j=0}\dfrac{\cancel{(-1)^j}\cdot \cancel{i^{2j}}}{j!\,(j+\nu)!}\left(\dfrac{\beta}{2}\right)^{2j},\end{align}$$ is always larger than $0$. In other words, $\alpha$ cannot be pure imaginary. Therefore, $\alpha$ is real. $\blacksquare$

#### Reference 
- [北京大学数学物理方程：柱函数](https://astrojacobli.github.io/Docs/Mathematical_Method/chpp17.pdf) p.4-5
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## Problem 4
:::warning
If $Y_\nu(x)$ and $Z_\nu(x)$ are any two solutions of Bessel's equation $$\dfrac{1}{x}\dfrac{\text{d}}{\text{d}x}\left(x\dfrac{\text{d}R(x)}{\text{d}x}\right)+\left(1-\dfrac{\nu^2}{x^2}\right)R(x)=0,\tag{1}$$ show that $$Y_\nu(x)Z'_\nu(x)-Y'_\nu(x)Z_\nu(x)=\dfrac{A_\nu}{x},\tag{2}$$ in which $A_\nu$ may depend on $\nu$ but is independent of $x$.
:::
### Solution
Eqn. (1) in its standard form is $$\dfrac{\text{d}^2R(x)}{\text{d}x^2}+\dfrac{1}{x}\dfrac{\text{d}R(x)}{\text{d}x}+\left(1-\dfrac{\nu^2}{x^2}\right)R(x)=0.\tag{3}$$ Since $\frac{1}{x}$ and $1-\frac{\nu^2}{x^2}$ are continuous on $x\in(0,\infty)$, the Abel's identity states that the ***Wronskian*** of any two solutions $Y_\nu(x)$ and $Z_\nu(x)$ of Eqn. (3), that is the function defined by the determinant $$W(Y_\nu,Z_\nu)(x)\equiv\begin{vmatrix}Y_\nu(x)&Z_\nu(x)\\Y'_\nu(x)&Z'_\nu(x)\end{vmatrix}=Y_\nu(x)Z'_\nu(x)-Y'_\nu(x)Z_\nu(x),\qquad x\in(0,\infty),$$ satisfies the relation $$W(Y_\nu,Z_\nu)(x)=C_\nu\exp\left(-\int^x_{x_0}\dfrac{1}{t}\text{d}t\right)=C_\nu e^{-\ln(x/x_0)}=\dfrac{C_\nu x_0}{x},\qquad x\in(0,\infty)$$ for every point  $x_0\in(0,\infty)$. If we let $A_\nu\equiv C_\nu x_0$, then Eqn. (2) is proved. $\blacksquare$
#### Reference
- [Wikipedia: Proof of Abel's identity](https://en.wikipedia.org/wiki/Abel%27s_identity#Proof)
- [Stackexchange: How can you find the Wronskian of Bessel Functions?](https://reurl.cc/E7L370)
- [北京大学数学物理方程：柱函数](https://astrojacobli.github.io/Docs/Mathematical_Method/chpp17.pdf) p.4
