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# System prepended metadata

title: Homework 6
tags: [微分方程特論]

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# Homework 6
ntu_b05202054 何信佑
https://hackmd.io/@ulynx/topdehw6
###### tags: `微分方程特論`

5/6作業
2.8(p.196) : 3, 7(a)(b)
8.1(p.567) : 2,11

## Section 2.8
### Problem 3
Given that the eigenvalue of $$y''+\lambda y=0;\ y(-\pi)=0,\ y(\pi)=0\tag{1}$$ is nonnegative. First determine whether $\lambda = 0$ is an eigenvalue; then ﬁnd the positive eigenvalues and associated eigenfunctions.
#### Solution
1. Case in which $\lambda = 0$:
    - Putting $\lambda = 0$ into the S-L problem (1), we have $y''=0$, whose general solution is $y(x)=Ax+B$. 
    - The endpoint conditions yield $y(-\pi)=-A\pi+B=0$ and $y(\pi)=A\pi+B=0$, and there follows $A=B=0$. 
    - But $y= 0\cdot x+0=0$ is a trivial solution.
    - Therefore, $\lambda =0$ is not an eigenvalue.
2. Case in which $\lambda > 0$:
    - Let $\lambda = \alpha^2 > 0$, where $\alpha\in\mathbb{R}\setminus\{0\}$, so that the S-L problem (1) becomes $y''+\alpha^2 y=0$, whose  general solution is $$y=A\cos \alpha x+B\sin \alpha x.\tag{2}$$ 
    - Applying the endpoint conditions to (2) gives $\left\{\begin{align}0&=y(-\pi)=A\cos (-\alpha\pi)+B\sin (-\alpha\pi)=A\cos \alpha x-B\sin \alpha x \\0&=y(\pi)=A\cos \alpha\pi+B\sin \alpha\pi\end{align}\right.$ 
    - In order to obtain nontrivial solution for (1), i.e. ==$A$ and $B$ are not both zero==, we require that $$\begin{align}&\qquad\begin{vmatrix}\cos\alpha \pi&-\sin \alpha \pi\\\cos\alpha \pi&\sin \alpha \pi\end{vmatrix}=0\tag{3}\\&\implies 2\cos(\alpha \pi)\sin (\alpha \pi)=\sin(2\alpha \pi)=0\\&\implies 2\alpha\pi =n\pi,\quad n=1,2,3,\ldots\\&\implies\boxed{\alpha=\dfrac{n}{2}},\quad n=1,2,3,\ldots.\end{align}$$
    - Thus, the eigenvalues of the S-L problem (1) are $$\boxed{\lambda_n=\frac{n^2}{4}\ (n=1,2,3,\ldots)}.\tag{4}$$
    - Besides, the eigenfunction associated with $\lambda_n$ is $$\boxed{y(x)=A\cos(nx/2)+B\sin(nx/2)}.\tag*{(5)$\blacksquare$}$$ 
    - <font color=red>When $n$ is odd, the endpoint conditions $\require{cancel}\cancelto{0}{A\cos(\pm n\pi/2)}+B\sin(\pm n\pi/2)=0$ implies that $B=0$; and 
when $n$ is even, the endpoint conditions $A\cos(\pm n\pi/2)+\cancelto{0}{B\sin(\pm n\pi/2)}=0$ implies that $A=0$.
So we'd better write the eigenfunction as $$\boxed{y_n(x)=\left\{\begin{array}{ll}A\cos(nx/2),& n\text{ odd};\\B\sin(nx/2),& n\text{ even}.\end{array}\right.}.\tag*{$\blacksquare$}$$</font>
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### Problem 7
Consider the eigenvalue problem $$y'' + \lambda y = 0;\ y(0) = 0,\ y(1) + y'(1) = 0;\tag{1}$$ all its eigenvalues are nonnegative. 
- **(a)** Show that $\lambda = 0$ is not an eigenvalue. 
- **(b)** Show that the eigenfunctions are the functions $\{\sin\alpha_nx\}^\infty_1$, where $\alpha_n$ is the $n$th positive root of the equation $\tan z = −z$. 
#### Solution
- **(a)** 
By letting $\lambda=0$ in the S-L problem (1), $y''=0\implies y(x)=Ax+B$
The endpoint conditions yield $$\begin{align}0&=A\cdot 0+B,\\0&=(A+B)+A.\end{align}$$ Solving this system of equations for $A$ and $B$, one obtain $A=B=0$. But this means no trivial solution. Therefore, $\lambda = 0$ is not an eigenvalue. 
- **(b)** 
    Assume $\lambda=\alpha^2>0$, where $\alpha\in\mathbb{R}\setminus\{0\}$, and the S-L problem becomes $y''+\alpha^2 y=0$ and its general solution is $$y(x)=A\cos \alpha x +B\sin\alpha x\tag{2}.$$ Applying the endpoint conditions to (2), we have $$\left\{\begin{align}0&=y(0)=A\cancelto{1}{\cos 0}+B\cancelto{0}{\sin 0} \\0&=y(1)+y'(1)=\left(A\cos \alpha +B \sin \alpha\right)+\alpha \left(-A{\sin \alpha}+B\cos\alpha\right)\end{align}\right.\tag{3}$$ The first equation in (3) indicates ==$A=0$== ($\therefore$ $B$ cannot be zero) and thus the second equation in (3) yields $$B\sin\alpha+\alpha B\cos\alpha=0\implies \sin\alpha+\alpha \cos\alpha=0.$$ 
    - If $\cos \alpha=0$, then $\sin\alpha=0$, resulting in $\alpha=0$. But we've required $\alpha\neq 0$.
    - If $\cos \alpha\neq 0$, then $\alpha$ satisfies the algebraic equation $$\tan t=-t.\tag{4}$$
    [Desmos](https://www.desmos.com/calculator/yj6cl9mnn7)
    ![](https://i.imgur.com/GhdEDvg.png)


    From the graph above we see that each point of intersection represents a real solution of Eq. (4), to which we assign an integer $n$, so that the solution set of Eq. (4) possesses the following strict partial order: $$\cdots<\alpha_{-n}<\cdots<\alpha_{-1}<\alpha_0<\alpha_1<\cdots<\alpha_n<\cdots.$$ Because $\tan t$ and $-t$ are odd functions, we have both $\alpha_0=0$ (which should be excluded) and $\alpha_{-n}=-\alpha_n$ for each $n=1,2,3,\ldots.$ Here we designate $$\alpha_1<\alpha_2<\cdots<\alpha_n<\cdots$$ as a sequence $\{\alpha_n\}^\infty_{n=0}$.

    Now that we've let $\lambda=\alpha^2$, we can let $\lambda_n=(\alpha_n)^2$ to construct a new sequence $\{\lambda_n\}^\infty_{n=1}$ such that $$\lambda_n\in\left\{\alpha^2\middle|\tan \alpha=-\alpha~\text{and}~\ \alpha\in\mathbb{R}\setminus\{0\}\right\}$$ and $$\lambda_1<\lambda_2<\cdots<\lambda_n<\cdots$$ where $\lambda_n$ is the $n$th eigenvalue of the S-L problem (1). Moreover, its associated eigenfunction is $y_n(x)=B\sin\alpha_n x$. (To within a constant factor $B$, all of these eigenfunctions form a function sequence $\{\sin\alpha_nx\}^\infty_{n=1}$.) $\blacksquare$
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## Section 9.1
### Problem 2
Verify that the eigenvalues and eigenfunctions for the indicated Sturm–Liouville problem are those listed: $$y'' + \lambda y = 0\ (0<x<L),\quad y(0) = y'(L) = 0\tag{1};$$ $$ \lambda_n= \dfrac{(2n − 1)^2\pi^2}{4L^2},\quad
 y_n(x) =\sin\frac{(2n − 1)\pi x}{2L},\quad n \geq 1\tag{2}$$

#### Solution
1. Case in which $\lambda=0$:
    Eq. (1) becomes $y''=0\implies y(x)=Ax+B$.
    The endpoint conditions: 
    - $y'(L)=A=0$;
    - $y(0)=A\cdot 0+B=0\implies B=0$;
    $\therefore~y(x)=0$ (trivial).
2. Case in which $\lambda>0$
    Let $\lambda=\alpha^2$, where $\alpha\in\mathbb{R}\setminus\{0\}$.
    Eq. (1) becomes $y''+\alpha^2 y=0\implies y(x)=A\cos\alpha x+B\sin\alpha x$.
    The endpoint conditions: 
    - $y(0)=A\cancelto{1}{\cos 0}+B\cancelto{0}{\sin 0}=0\implies A=0$; 
    - $y'(L)=-\cancelto{0}{A}\alpha \sin(\alpha L)+B\alpha \cos(\alpha L)=0$
    
    For a nontrivial solution of (1), we require $B\neq 0$, so $$\begin{align}&\qquad B\alpha \cos(\alpha L)=0\implies \alpha \cos(\alpha L)=0\\&\implies \xcancel{\alpha=0}~\text{or}~\cos(\alpha L)=0\implies \alpha L=\left(n-\frac{1}{2}\right)\pi,\quad n=1,2,3,\ldots\\&\implies \alpha_n=\frac{(2n-1)\pi}{2L},\quad n=1,2,3,\ldots\\&\implies\boxed{ \lambda_n=\alpha_n^2=\frac{(2n-1)^2\pi^2}{4L^2},\quad n=1,2,3,\ldots}\end{align}$$ are the eigenvalues of the S-L Problem. And the associated eigenfunctions (dropping the constant $B$) are $$\boxed{y_n(x)=\sin(\alpha x)=\sin\frac{(2n-1)\pi x}{2L},\quad n=1,2,3,\ldots}$$

3. The second part of Theorem 3 in Sec. 9.2 states that 
    > Moreover if $q(x)\geq0$ on $[a,b]$ and the coefficients $\alpha_1$, $\alpha_2$, $\beta_1$, and $\beta_2$ in (9) are all nonnegative, the eigenvalues are all nonegative);
    
    so we don't need to consider the case in which $\lambda<0$.

4. We still need to confirm $\lambda_n$ and $y_n$ are *indeed* eigenvalues and eigenfunctions of the S-L problem (1): for each $n=1,2,3,\ldots$, $$\begin{align}&\phantom{=1}y_n''(x)+\lambda_n y_n(x)\\&=\dfrac{d^2}{dx^2}\left[\sin\dfrac{(2n − 1)\pi x}{2L}\right]+\dfrac{(2n − 1)^2\pi^2}{4L^2}\sin\dfrac{(2n − 1)\pi x}{2L}\\&=-\left[\dfrac{(2n − 1)\pi}{2L}\right]^2\sin\dfrac{(2n − 1)\pi x}{2L}+\dfrac{(2n − 1)^2\pi^2}{4L^2}\sin\dfrac{(2n − 1)\pi x}{2L}=0\end{align}$$ and $$y_n(0)=\sin\frac{(2n-1)\pi \cdot 0}{2L}=0,$$  $$y_n'(L)=\dfrac{(2n-1)\pi}{2L}\underbrace{\cos\frac{(2n-1)\pi \cancel{L}}{2\cancel{L}}}_{=0}=0.\tag*{$\blacksquare$}$$
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### Problem 11
Show that $\lambda_0= 0$ is an eigenvalue of the regular S-L problem $$\begin{array}{ll}y'' + \lambda y = 0& (0<x<L);\\y(0) = 0,& hy(L)-y'(L)=0\ (h>0),\end{array}\tag{1}$$ if and only if $hL = 1$, in which case the associated eigenfunction is $y_0(x) = x$.
#### Solution
$y''=0\implies y(x)=Ax+B$
The endpoint conditions:
    $y(0)=A\cdot 0 +B=0\implies B=0\tag{2a}$
    $hy(L)-y'(L)=h(AL+B)-A=0\tag{2b}$
1. **claim**: 
    > If $hL=1$, then $\lambda_0= 0$ is an eigenvalue of (1) and $y_0(x)=x$ is the associated eigenfunction.

    **proof**: 
    If $hL=1$, then (2b) becomes $h(AL+B)-A=\cancel{hLA}+hB-\cancel{A}=0$, which is a redundant constraint regarding (2a); $A$ can be arbitrary nonzero real number. 

    $y(x)=Ax$ is a nontrivial solution of (1) when $\lambda=0$, i.e. $\lambda=0$ is an eigenvalue of (1) with the associated eigenfunction being $y(x)=x$ (to within a factor $A$). $\blacktriangle$
    
2. **claim**:
    > If $\lambda_0= 0$ is an eigenvalue of (1) and $y_0(x)=x$ is the associated eigenfunction, then $hL=1$.

    **proof**:
    If $\lambda_0= 0$ is an eigenvalue of (1) and $y_0(x)=x$ is the associated eigenfunction, then $y(x)=x$ must satisfiy each equation in (1). Substituting $y(x)=x$ into the first two equations in (1), we have $$\begin{align}y_0''+\lambda_0y_0&=x''+0=0,\\ y_0(0)&=0,\end{align}$$ as expected. Because $y(x)=x$ must also satisfy $hy(L)-y'(L)=0$, it is necessary that $hy(L)-y'(L)=hL-1=0$, which implies $hL=1$. $\blacktriangle$ 