
# 13.2E: Sturm-Liouville Problems (Exercises)

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In Exercises $$\ref{exer:13.2.1}$$–$$\ref{exer:13.2.7}$$ rewrite the equation in Sturm–Liouville form $$($$with $$\lambda=0)$$. Assume that $$b,$$ $$c,$$ $$\alpha,$$ and $$\nu$$ are contants.

[exer:13.2.1] $$y''+by'+cy=0$$

[exer:13.2.2] $$x^{2}y''+xy'+(x^{2}-\nu^{2})y=0$$ (Bessel’s equation)

[exer:13.2.3] $$(1-x^{2})y''-xy'+\alpha^{2}y=0$$ (Chebyshev’s equation)

[exer:13.2.4] $$x^{2}y''+bxy'+cy=0$$ (Euler’s equation)

[exer:13.2.5] $$y''-2xy'+2\alpha y=0$$ (Hermite’s equation)

[exer:13.2.6] $$xy''+(1-x)y'+\alpha y=0$$ (Laguerre’s equation)

[exer:13.2.7] $$(1-x^{2})y''-2xy'+\alpha(\alpha+1)y=0$$ (Legendre’s equation)

[exer:13.2.8] In Example [example:13.2.4} we found that the eigenvalue problem $x^{2}y''+xy'+\lambda y=0,\quad y(1)=0,\quad y(2)=0 \tag{A}\nonumber$is equivalent to the Sturm-Liouville problem $\tag{B} (xy')'+\frac{\lambda}{x}y=0,\quad y(1)=0,\quad y(2)=0.\nonumber$ Multiply the differential equation in (B) by $$y$$ and integrate to show that $\lambda\int_{1}^{2}\frac{y^{2}(x)}{x}\,dx=\int_{1}^{2}x(y'(x))^{2}\,dx.\nonumber$ Conclude from this that the eigenvalues of (A) are all positive.

[exer:13.2.9] Solve the eigenvalue problem $y''+2y'+y+\lambda y=0,\quad y(0)=0, \quad y(1)=0.\nonumber$

[exer:13.2.10] Solve the eigenvalue problem $y''+2y'+y+\lambda y=0,\quad y'(0)=0, \quad y'(1)=0.\nonumber$

[exer:13.2.11] $$y''+\lambda y=0$$, $$y(0)+2y'(0)=0$$,$$y(2)=0$$

[exer:13.2.12] $$y''+\lambda y=0$$, $$y'(0)=0$$,$$y(1)-2y'(1)=0$$

[exer:13.2.13] $$y''+\lambda y=0$$, $$y(0)-y'(0)=0$$,$$y'(\pi)=0$$

[exer:13.2.14] $$y''+\lambda y=0$$, $$y(0)+2y'(0)=0$$,$$y(\pi)=0$$

[exer:13.2.15] $$y''+\lambda y=0$$, $$y'(0)=0$$,$$y(2)-y'(2)=0$$

[exer:13.2.16] $$y''+\lambda y=0$$, $$y(0)+y'(0)=0$$,$$y(2)+2y'(2)=0$$

[exer:13.2.17] $$y''+\lambda y=0$$, $$y(0)+2y'(0)=0$$,$$y(3)-2y'(3)=0$$

[exer:13.2.18] $$y''+\lambda y=0$$, $$3y(0)+y'(0)=0$$,$$3y(2)-2y'(2)=0$$

[exer:13.2.19] $$y''+\lambda y=0$$, $$y(0)+2y'(0)=0$$,$$y(3)-y'(3)=0$$

[exer:13.2.20] $$y''+\lambda y=0$$, $$5y(0)+2y'(0)=0$$,$$5y(1)-2y'(1)=0$$

[exer:13.2.21] Find the first five eigenvalues of the boundary value problem

$y''+2y'+y+\lambda y=0,\quad y(0)=0,\quad y'(1)=0\nonumber$

with errors not greater than $$5\times 10^{-8}$$. State the form of the associated eigenfunctions.

[exer:13.2.22] Solve the eigenvalue problem for

$x^{2}y''-2xy'+2y+\lambda x^{2}y=0, \quad y(1)=0,\quad y(2)=0.\nonumber$

[exer:13.2.23] Find the first five eigenvalues of

$x^{2}y''-2xy'+2y+\lambda x^{2}y=0, \quad y'(1)=0,\quad y(2)=0\nonumber$

with errors no greater than $$5\times 10^{-8}$$. State the form of the associated eienfunctions.

[exer:13.2.24] Find the first five eigenvalues of

$x^{2}y''-2xy'+2y+\lambda x^{2}y=0, \quad y(1)=0,\quad y'(2)=0\nonumber$

with errors no greater than $$5\times 10^{-8}$$. State the form of the associated eienfunctions.

[exer:13.2.25] Consider the Sturm-Liouville problem

$\tag{A} y''+\lambda y=0,\quad y(0)=0,\quad y(L)+\delta y'(L)=0.\nonumber$

# a

Show that (A) can’t have more than one negative eigenvalue, and find the values of $$\delta$$ for which it has one.

# b

Find all values of $$\delta$$ such that $$\lambda=0$$ is an eigenvalue of (A).

# c

Show that $$\lambda=k^{2}$$ with $$k>0$$ is an eigenvalue of (A) if and only if

$\tag{B} \tan kL=-\delta k.\nonumber$

# d

For $$n=1$$, $$2$$, …, let $$y_{n}$$ be an eigenfunction associated with $$\lambda_{n}=k_{n}^{2}$$. From Theorem [thmtype:13.2.4}, $$y_{m}$$ and $$y_{n}$$ are orthogonal over $$[0,L]$$ if $$m\ne n$$. Verify this directly. (B).

[exer:13.2.26] Solve the Sturm-Liouville problem

$y''+\lambda y=0,\quad y(0)+\alpha y'(0)=0,\quad y(\pi)+\alpha y'(\pi)=0,\nonumber$

where $$\alpha\ne0$$.

[exer:13.2.27] Consider the Sturm-Liouville problem

$\tag{A} y''+\lambda y=0,\quad y(0)+\alpha y'(0)=0,\quad y(1)+(\alpha-1) y'(1)=0,\nonumber$

where $$0<\alpha<1$$.

# a

Show that $$\lambda=0$$ is an eigenvalue of (A), and find an associated eigenfunction.

# b

Show that (A) has a negative eigenvalue, and find the form of an associated eigenfunction.

# c

Give a graphical argument to show that (A) has infinitely many positive eigenvalues $$\lambda_{1} < \lambda_{2}<\cdots <\lambda_{n}<\cdots$$, and state the form of the associated eigenfunctions.

[exer:13.2.28] Show that $$\lambda=0$$ is an eigenvalue of (SL) if and only if $\alpha(\rho L+\delta)-\beta\rho =0.\nonumber$

[exer:13.2.29] The point of this exercise is that (SL) can’t have more than two negative eigenvalues.

# a

Show that $$\lambda$$ is a negative eigenvalue of (SL) if and only if $$\lambda=-k^{2}$$, where $$k$$ is a positive solution of

$(\alpha\rho-\beta\delta k^{2})\sinh kL+k(\alpha\delta-\beta\rho)\cosh kL.\nonumber$

# b

Suppose $$\alpha\delta-\beta\rho=0$$. Show that (SL) has a negative eigenvalue if and only if $$\alpha\rho$$ and $$\beta\delta$$ are both nonzero. Find the negative eigenvalue and an associated eigenfunction.

# c

Suppose $$\beta\rho-\alpha\delta\ne0$$. We know from Section 11.1 that (SL) has no negative eigenvalues if $$\alpha\rho=0$$ and $$\beta\delta=0$$. Assume that either $$\alpha\rho\ne0$$ or $$\beta\delta\ne0$$. Then we can rewrite (A) as

$\tanh kL= \frac{k(\beta\rho-\alpha\delta)}{\alpha\rho-\beta\delta k^{2}}.\nonumber$

By graphing both sides of this equation on the same axes (there are several possibilities for the right side), show that it has at most two positive solutions, so (SL) has at most two negative eigenvalues.

[exer:13.2.30] The point of this exercise is that (SL) has infinitely many positive eigenvalues $$\lambda_{1}<\lambda_{2}<\cdots<\lambda_{n}<\cdots$$, and that $$\lim_{n\to\infty}\lambda_{n}=\infty$$.

# a

Show that $$\lambda$$ is a positive eigenvalue of (SL) if and only if $$\lambda=k^{2}$$, where $$k$$ is a positive solution of

$\tag{A} (\alpha\rho+\beta\delta k^{2})\sin kL+k(\alpha\delta-\beta\rho)\cos kL=0.\nonumber$

# b

Suppose $$\alpha\delta-\beta\rho=0$$. Show that the positive eigenvalues of (SL) are $$\lambda_{n}=(n\pi/L)^{2}$$, $$n=1$$, $$2$$, $$3$$, ….

Now suppose $$\alpha\delta-\beta\rho\ne0$$. From Section 11.1, if $$\alpha\rho=0$$ and $$\beta\delta=0$$, then (SL) has the eigenvalues

$\lambda_{n}=[(2n-1)\pi/2L]^{2},\quad n=1,2,3, \dots\nonumber$

(why?), so let’s suppose in addition that at least one of the products $$\alpha\rho$$ and $$\beta\delta$$ is nonzero. Then we can rewrite (A) as

$\tag{B} \tan kL= \frac{k(\beta\rho-\alpha\delta)} {\alpha\rho-\beta\delta k^{2}}.\nonumber$

By graphing both sides of this equation on the same axes (there are several possibilities for the right side), convince yourself of the following:

# c

If $$\beta\delta=0$$, there’s a positive integer $$N$$ such that (B) has one solution $$k_{n}$$ in each of the intervals

$\tag{C} \left((2n-1)\pi/L, (2n+1)\pi/L)\right),\quad n=N,N+1,N+2,\dots,\nonumber$

and either

$\lim_{n\to\infty}\left(k_{n}-\frac{(2n-1)\pi}{2L}\right) =0 \text{\quad or\quad} \lim_{n\to\infty}\left(k_{n}-\frac{(2n+1)\pi}{2L}\right)=0.\nonumber$

# d

If $$\beta\delta\ne0$$, there’s a positive integer $$N$$ such that (B) has one solution $$k_{n}$$ in each of the intervals (C) and

$\lim_{n\to\infty}\left(k_{n}-\frac{n\pi}{N}\right)=0.\nonumber$

[exer:13.2.31] The following Sturm–Liouville problems are genera1izations of Problems 1–4 of Section 11.1.

Problem 1: $$(p(x)y')'+\lambda r(x)y=0$$, $$y(a)=0$$,$$y(b)=0$$

Problem 2: $$(p(x)y')'+\lambda r(x)y=0$$, $$y'(a)=0$$,$$y'(b)=0$$

Problem 3: $$(p(x)y')'+\lambda r(x)y=0$$, $$y(a)=0$$,$$y'(b)=0$$

Problem 4: $$(p(x)y')'+\lambda r(x)y=0$$, $$y'(a)=0$$,$$y(b)=0$$

Prove: Problems 1–4 have no negative eigenvalues. Moreover, $$\lambda=0$$ is an eigenvalue of Problem 2 with associated eigenfunction $$y_{0}=1$$, but $$\lambda=0$$ isn’t an eigenvalue of Problems 1, 3, and 4.

[exer:13.2.32] Show that the eigenvalues of the Sturm–Liouville problem

$(p(x)y')'+\lambda r(x)y=0,\quad \alpha y(a)+ \beta y'(a)=0,\quad \rho y(b)+\delta y'(b)\nonumber$

are all positive if $$\alpha\beta\le 0$$,   $$\rho\delta\ge 0$$, and $$(\alpha\beta)^{2}+(\rho\delta)^{2}>0$$.