13.1: Boundary Value Problems

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$$\newcommand{\place}{\bigskip\hrule\bigskip\noindent} \newcommand{\threecol}[3]{\left[\begin{array}{r}#1\\#2\\#3\end{array}\right]} \newcommand{\threecolj}[3]{\left[\begin{array}{r}#1\$1\jot]#2\\[1\jot]#3\end{array}\right]} \newcommand{\lims}[2]{\,\bigg|_{#1}^{#2}} \newcommand{\twocol}[2]{\left[\begin{array}{l}#1\\#2\end{array}\right]} \newcommand{\ctwocol}[2]{\left[\begin{array}{c}#1\\#2\end{array}\right]} \newcommand{\cthreecol}[3]{\left[\begin{array}{c}#1\\#2\\#3\end{array}\right]} \newcommand{\eqline}[1]{\centerline{\hfill\displaystyle#1\hfill}} \newcommand{\twochar}[4]{\left|\begin{array}{cc} #1-\lambda\\#3-\lambda\end{array}\right|} \newcommand{\twobytwo}[4]{\left[\begin{array}{rr} #1\\#3\end{array}\right]} \newcommand{\threechar}[9]{\left[\begin{array}{ccc} #1-\lambda\\#4-\lambda\\#7 -\lambda\end{array}\right]} \newcommand{\threebythree}[9]{\left[\begin{array}{rrr} #1\\#4\\#7 \end{array}\right]} \newcommand{\solutionpart}[1]{\vskip10pt\noindent\underbar{\color{blue}\sc Solution({\bf #1})\ }} \newcommand{\Cex}{\fbox{\textcolor{red}{C}}\, } \newcommand{\CGex}{\fbox{\textcolor{red}{C/G}}\, } \newcommand{\Lex}{\fbox{\textcolor{red}{L}}\, } \newcommand{\matfunc}[3]{\left[\begin{array}{cccc}#1_{11}(t)_{12}(t)&\cdots _{1#3}(t)\\#1_{21}(t)_{22}(t)&\cdots_{2#3}(t)\\\vdots& \vdots&\ddots&\vdots\\#1_{#21}(t)_{#22}(t)&\cdots_{#2#3}(t) \end{array}\right]} \newcommand{\col}[2]{\left[\begin{array}{c}#1_1\\#1_2\\\vdots\\#1_#2\end{array}\right]} \newcommand{\colfunc}[2]{\left[\begin{array}{c}#1_1(t)\\#1_2(t)\\\vdots\\#1_#2(t)\end{array}\right]} \newcommand{\cthreebythree}[9]{\left[\begin{array}{ccc} #1\\#4\\#7 \end{array}\right]} 1 \ newcommand {\ dy} {\ ,\ mathrm {d}y} \ newcommand {\ dx} {\ ,\ mathrm {d}x} \ newcommand {\ dyx} {\ ,\ frac {\ mathrm {d}y}{\ mathrm {d}x}} \ newcommand {\ ds} {\ ,\ mathrm {d}s} \ newcommand {\ dt }{\ ,\ mathrm {d}t} \ newcommand {\dst} {\ ,\ frac {\ mathrm {d}s}{\ mathrm {d}t}}$$ Two-Point Boundary Values Problems We considered in Section 5.3, initial value problems for the linear second order equation \[\label{eq:13.1.1} P_{0}(x)y''+P_{1}(x)y'+P_{2}(x)y=F(x).$

Suppose $$P_{0}$$, $$P_{1}$$, $$P_{2}$$, and $$F$$ are continuous and $$P_{0}$$ has no zeros on an open interval $$(a,b)$$. From Theorem 5.3.1, if $$x_{0}$$ is in $$(a,b)$$ and $$k_{1}$$ and $$k_{2}$$ are arbitrary real numbers then Equation \ref{eq:13.1.1} has a unique solution on $$(a,b)$$ such that $$y(x_{0})=k_{1}$$ and $$y'(x_{0})=k_{2}$$. Now we consider a different problem for Equation \ref{eq:13.1.1}.

PROBLEM

Suppose $$P_{0}$$, $$P_{1}$$, $$P_{2}$$, and $$F$$ are continous and $$P_{0}$$ has no zeros on a closed interval $$[a,b]$$. Let $$\alpha$$, $$\beta$$, $$\rho$$, and $$\delta$$ be real numbers such that

$\label{eq:13.1.2} \alpha^{2}+\beta^{2}\ne 0 \; \text{and} \; \rho^{2}+\delta^{2}\ne 0,$

and let $$k_{1}$$ and $$k_{2}$$ be arbitrary real numbers. Find a solution of

$\label{eq:13.1.3} P_{0}(x)y''+P_{1}(x)y'+P_{2}(x)y=F(x)$

on the closed interval $$[a,b]$$ such that

$\label{eq:13.1.4} \alpha y(a)+\beta y'(a)=k_{1}$

and

$\label{eq:13.1.5} \rho y(b)+\delta y'(b)=k_{2}.$

The assumptions stated in this problem apply throughout this section and will not be repeated. Note that we imposed conditions on $$P_{0}$$, $$P_{1}$$, $$P_{2}$$, and $$F$$ on the closed interval $$[a,b]$$, and we are interested in solutions of Equation \ref{eq:13.1.3} on the closed interval. This is different from the situation considered in Chapter 5, where we imposed conditions on $$P_{0}$$, $$P_{1}$$, $$P_{2}$$, and $$F$$ on the open interval $$(a,b)$$ and we were interested in solutions on the open interval. There is really no problem here; we can always extend $$P_{0}$$, $$P_{1}$$, $$P_{2}$$, and $$F$$ to an open interval $$(c,d)$$ (for example, by defining them to be constant on $$(c,d]$$ and $$[b,d)$$) so that they are continuous and $$P_{0}$$ has no zeros on $$[c,d]$$. Then we can apply the theorems from Chapter 5 to the equation

$y''+\frac{P_{1}(x)}{P_{0}(x)}y'+ \frac{P_{2}(x)}{P_{0}(x)}y=\frac{F(x)}{P_{0}(x)} \nonumber$

on $$(c,d)$$ to draw conclusions about solutions of Equation \ref{eq:13.1.3} on $$[a,b]$$.

We call $$a$$ and $$b$$ boundary points. The conditions Equation \ref{eq:13.1.4} and Equation \ref{eq:13.1.5} are boundary conditions, and the problem is a two-point boundary value problem or, for simplicity, a boundary value problem. (We used similar terminology in Chapter 12 with a different meaning; both meanings are in common usage.) We require Equation \ref{eq:13.1.2} to insure that we are imposing a sensible condition at each boundary point. For example, if $$\alpha^{2}+\beta^{2}=0$$ then $$\alpha=\beta=0$$, so $$\alpha y(a)+\beta y'(a)=0$$ for all choices of $$y(a)$$ and $$y'(a)$$. Therefore Equation \ref{eq:13.1.4} is an impossible condition if $$k_{1}\ne0$$, or no condition at all if $$k_{1}=0$$.

We abbreviate Equation \ref{eq:13.1.1} as $$Ly=F$$, where

$Ly=P_{0}(x)y''+P_{1}(x)y'+P_{0}(x)y,\nonumber$

and we denote

$B_{1}(y)=\alpha y(a)+\beta y'(a) \, \text{and} \; B_{2}(y)=\rho y(b)+\delta y'(b).\nonumber$

We combine Equation \ref{eq:13.1.3}, Equation \ref{eq:13.1.4}, and Equation \ref{eq:13.1.5} as

$\label{eq:13.1.6} Ly=F,\quad \quad B_{1}(y)=k_{1},\quad \quad B_{2}(y)=k_{2}.$

This boundary value problem is homogeneous if $$F=0$$ and $$k_{1}=k_{2}=0$$; otherwise it is nonhomogeneous.

We leave it to you (Exercise 13.1.1) to verify that $$B_{1}$$ and $$B_{2}$$ are linear operators; that is, if $$c_{1}$$ and $$c_{2}$$ are constants then

$\label{eq:13.1.7} B_{i}(c_{1}y_{1}+c_{2}y_{2})=c_{1}B_{i}(y_{1})+c_{2}B_{i}(y_{2}),\quad i=1,2.$

The next three examples show that the question of existence and uniqueness for solutions of boundary value problems is more complicated than for initial value problems.

Example 13.1.1

Consider the boundary value problem

$y''+y=1,\quad y(0)=0, \quad y(\pi/2)=0. \nonumber$

The general solution of $$y''+y=1$$ is

$y=1+c_{1}\sin x+c_{2} \cos x, \nonumber$

so $$y(0)=0$$ if and only if $$c_{2}=-1$$ and $$y(\pi/2)=0$$ if and only if $$c_{1}=-1$$. Therefore

$y =1-\sin x -\cos x \nonumber$

is the unique solution of the boundary value problem.

Example 13.1.2

Consider the boundary value problem

$y''+y=1,\quad y(0)=0, \quad y(\pi)=0. \nonumber$

Again, the general solution of $$y''+y=1$$ is

$y=1+c_{1}\sin x+c_{2} \cos x, \nonumber$

so $$y(0)=0$$ if and only if $$c_{2}=-1$$, but $$y(\pi)=0$$ if and only if $$c_{2}=1$$. Therefore the boundary value problem has no solution.

Example 13.1.3

Consider the boundary value problem

$y''+y=\sin 2x,\quad y(0)=0, \quad y(\pi)=0.\nonumber$

You can use the method of undetermined coefficients (Section 5.5) to find that the general solution of $$y''+y=\sin 2x$$ is

$y=-\frac{\sin2x}{3}+c_{1}\sin x+c_{2}\cos x.\nonumber$

The boundary conditions $$y(0)=0$$ and $$y(\pi)=0$$ both require that $$c_{2}=0$$, but they don’t restrict $$c_{1}$$. Therefore the boundary value problem has infinitely many solutions

$y=-\frac{\sin2x}{3}+c_{1}\sin x,\nonumber$

where $$c_{1}$$ is arbitrary.

Theorem 13.1.1

If $$z_{1}$$ and $$z_{2}$$ are solutions of $$Ly=0$$ such that either $$B_{1}(z_{1})=B_{1}(z_{2})=0$$ or $$B_{2}(z_{1})=B_{2}(z_{2})=0,$$ then $$\{z_{1},z_{2}\}$$ is linearly dependent. Equivalently$$,$$ if  $$\{z_{1},z_{2}\}$$ is linearly independent, then

$B_{1}^{2}(z_{1})+B_{1}^{2}(z_{2})\ne 0 \; \text{and} \; B_{2}^{2}(z_{1})+B_{2}^{2}(z_{2})\ne 0.\nonumber$

Proof

Recall that $$B_{1}(z)=\alpha z(a)+\beta z'(a)$$ and $$\alpha^{2}+\beta^{2}\ne0$$. Therefore, if $$B_{1}(z_{1})=B_{1}(z_{2})=0$$ then $$(\alpha,\beta)$$ is a nontrivial solution of the system

\begin{aligned} \alpha z_{1}(a)+\beta z_{1}'(a)&= 0\\[4pt] \alpha z_{2}(a)+\beta z_{}'(a)&= 0.\end{aligned}\nonumber

This implies that

$z_{1}(a)z_{2}'(a)-z_{1}'(a)z_{2}(a)=0,\nonumber$

so $$\{z_{1},z_{2}\}$$ is linearly dependent, by Theorem 5.1.6. We leave it to you to show that $$\{z_{1},z_{2}\}$$ is linearly dependent if $$B_{2}(z_{1})=B_{2}(z_{2})=0$$.

Theorem 13.1.2

The following statements are equivalent$$;$$ that is$$,$$ they are either all true or all false$$.$$

1. There’s a fundamental set $$\{z_{1},z_{2}\}$$ of solutions of $$Ly=0$$ such that $\label{eq:13.1.8} B_{1}(z_{1})B_{2}(z_{2})-B_{1}(z_{2})B_{2}(z_{1})\ne0.$
2. If $$\{y_{1},y_{2}\}$$ is a fundamental set of solutions of $$Ly=0$$ then $\label{eq:13.1.9} B_{1}(y_{1})B_{2}(y_{2})-B_{1}(y_{2})B_{2}(y_{1})\ne0.$
3. For each continuous $$F$$ and pair of constants $$(k_{1},k_{2}),$$ the boundary value problem $Ly=F, \quad B_{1}(y)=k_{1},\quad B_{2}(y)=k_{2}\nonumber$ has a unique solution$$.$$
4. The homogeneous boundary value problem $\label{eq:13.1.10} Ly=0,\quad B_{1}(y)=0, \quad B_{2}(y)=0$ has only the trivial solution $$y=0$$.
5. The homogeneous equation $$Ly=0$$ has linearly independent solutions $$z_{1}$$ and $$z_{2}$$ such that $$B_{1}(z_{1})=0$$ and $$B_{2}(z_{2})=0.$$

We’ll show that $$\text{a}\implies \text{b}\implies \text{c}\implies \text{d}\implies \text{e}\implies \text{a}$$ of Theorem 13.1.2 .

Proof a

a $$\implies$$ b: Since $$\{z_{1},z_{2}\}$$ is a fundamental set of solutions for $$Ly=0$$, there are constants $$a_{1}$$, $$a_{2}$$, $$b_{1}$$, and $$b_{2}$$ such that

\begin{aligned} y_{1}=a_{1}z_{1}+a_{2}z_{2} \end{aligned}\nonumber

$\label{eq:13.1.11} y_{2} =b_{1}z_{1}+b_{2}z_{2}.$

Moreover,

$\label{eq:13.1.12} \left|\begin{array}{ccccccc} a_{1}&a_{2}\\[4pt]b_{1}&b_{2} \end{array}\right|\ne0.$

because if this determinant were zero, its rows would be linearly dependent and therefore $$\{y_{1},y_{2}\}$$ would be linearly dependent, contrary to our assumption that $$\{y_{1},y_{2}\}$$ is a fundamental set of solutions of $$Ly=0$$. From Equation \ref{eq:13.1.7} and Equation \ref{eq:13.1.11},

$\left[\begin{array}{ccccccc} B_{1}(y_{1})&B_{2}(y_{1})\\[4pt]B_{1}(y_{2})&B_{2}(y_{2}) \end{array}\right]= \left[\begin{array}{ccccccc} a_{1}&a_{2}\\[4pt]b_{1}&b_{2} \end{array}\right] \left[\begin{array}{ccccccc} B_{1}(z_{1})&B_{2}(z_{1})\\[4pt] B_{1}(z_{2})&B_{2}(z_{2}) \end{array}\right].\nonumber$

Since the determinant of a product of matrices is the product of the determinants of the matrices, Equation \ref{eq:13.1.8} and Equation \ref{eq:13.1.12} imply Equation \ref{eq:13.1.9}.

Proof b

b $$\implies$$ c: Since $$\{y_{1},y_{2}\}$$ is a fundamental set of solutions of $$Ly=0$$, the general solution of $$Ly=F$$ is

$y=y_{p}+c_{1}y_{1}+c_{2}y_{2},\nonumber$

where $$c_{1}$$ and $$c_{2}$$ are arbitrary constants and $$y_{p}$$ is a particular solution of $$Ly=F$$. To satisfy the boundary conditions, we must choose $$c_{1}$$ and $$c_{2}$$ so that

\begin{aligned} k_{1}&=B_{1}(y_{p})+c_{1}B_{1}(y_{1})+c_{2}B_{1}(y_{2})\\[4pt] k_{2}&=B_{2}(y_{p})+c_{1}B_{2}(y_{1})+c_{2}B_{2}(y_{2}),\end{aligned}\nonumber

(recall Equation \ref{eq:13.1.7}), which is equivalent to

\begin{aligned} c_{1}B_{1}(y_{1})+c_{2}B_{1}(y_{2}) &= k_{1}-B_{1}(y_{p}) \\[4pt] c_{1}B_{2}(y_{1})+c_{2}B_{2}(2_{2}) &= k_{2}-B_{2}(y_{p}). \end{aligned}\nonumber

From Equation \ref{eq:13.1.9}, this system always has a unique solution $$(c_{1},c_{2})$$.

Proof c

c$$\implies$$d: Obviously, $$y=0$$ is a solution of Equation \ref{eq:13.1.10}. From c with $$F=0$$ and $$k_{1}=k_{2}=0$$, it is the only solution.

Proof d

d $$\implies$$ e: Let $$\{y_{1},y_{2}\}$$ be a fundamental system for $$Ly=0$$ and let

$z_{1}=B_{1}(y_{2})y_{1}-B_{1}(y_{1})y_{2}\, \text{and} \; z_{2}=B_{2}(y_{2})y_{1}-B_{2}(y_{1})y_{2}.\nonumber$

Then $$B_{1}(z_{1})=0$$ and $$B_{2}(z_{2})=0$$. To see that $$z_{1}$$ and $$z_{2}$$ are linearly independent, note that

\begin{aligned} a_{1}z_{1}+a_{2}z_{2}&= a_{1}[B_{1}(y_{2})y_{1}-B_{1}(y_{1})y_{2}]+ a_{2}[B_{2}(y_{2})y_{1}-B_{2}(y_{1})y_{2}]\\[4pt] &= [B_{1}(y_{2})a_{1}+B_{2}(y_{2})a_{2}]y_{1}-[B_{1}(y_{1})a_{1}+B_{2}(y_{1})a_{2}]y_{2}.\end{aligned}\nonumber

Therefore, since $$y_{1}$$ and $$y_{2}$$ are linearly independent, $$a_{1}z_{1}+a_{2}z_{2}=0$$ if and only if

$\left[\begin{array}{ccccccc} B_{1}(y_{1})&B_{2}(y_{1})\\[4pt] B_{1}(y_{2})&B_{2}(y_{2}) \end{array}\right] \left[\begin{array}{ccccccc} a_{1}\\[4pt]a_{2} \end{array}\right]= \left[\begin{array}{ccccccc} 0\\[4pt]0 \end{array}\right].\nonumber$

If this system has a nontrivial solution then so does the system

$\left[\begin{array}{ccccccc} B_{1}(y_{1})&B_{1}(y_{2})\\[4pt] B_{2}(y_{1})&B_{2}(y_{2}) \end{array}\right] \left[\begin{array}{ccccccc} c_{1} \\[4pt]c_{2} \end{array}\right]= \left[\begin{array}{ccccccc} 0\\[4pt]0 \end{array}\right].\nonumber$

This and Equation \ref{eq:13.1.7} imply that $$y=c_{1}z_{1}+c_{2}z_{2}$$ is a nontrivial solution of Equation \ref{eq:13.1.10}, which contradicts (d)

Proof e

e $$\implies$$ a. Theorem 13.1.1 implies that if $$B_{1}(z_{1})=0$$ and $$B_{2}(z_{2})=0$$ then $$B_{1}(z_{2})\ne0$$ and $$B_{2}(z_{1})\ne0$$. This implies Equation \ref{eq:13.1.8}, which completes the proof.

Example 13.1.4

Solve the boundary value problem

$\label{eq:13.1.13} x^{2}y''-2xy'+2y-2x^{3}=0,\quad y(1)=4,\quad y'(2)=3,$

given that $$\{x,x^{2}\}$$ is a fundamental set of solutions of the complementary equation

Solution

Using variation of parameters (Section 5.7), you can show that $$y_{p}=x^{3}$$ is a solution of the complementary equation

$x^{2}y''-2xy'+2y=2x^{3}=0. \nonumber$

Therefore the solution of Equation \ref{eq:13.1.13} can be written as

$y=x^{3}+c_{1}x+c_{2}x^{2}. \nonumber$

Then

$y'=3x^{2}+c_{1}+2c_{2}x, \nonumber$

and imposing the boundary conditions yields the system

\begin{aligned} c_{1}+\phantom{4}c_{2}&=\phantom{-}3\\[4pt] c_{1}+4c_{2}&=-9,\end{aligned}\nonumber

so $$c_{1}=7$$ and $$c_{2}=-4$$. Therefore

$y=x^{3}+7x-4x^{2} \nonumber$

is the unique solution of Equation \ref{eq:13.1.13}

Example 13.1.5

Solve the boundary value problem

$y''-7y'+12y=4e^{2x},\quad y(0)=3,\quad y(1)=5e^{2}.\nonumber$

Solution

From Example 5.4.1, $$y_{p}=2e^{2x}$$ is a particular solution of

$\label{eq:13.1.14} y''-7y'+12y=4e^{2x}.$

Since $$\{e^{3x},e^{4x}\}$$ is a fundamental set for the complementary equation, we could write the solution of Equation \ref{eq:13.1.13} as

$y=2e^{2x}+c_{1}e^{3x}+c_{2}e^{4x}\nonumber$

and determine $$c_{1}$$ and $$c_{2}$$ by imposing the boundary conditions. However, this would lead to some tedious algebra, and the form of the solution would be very unappealing. (Try it!) In this case it is convenient to use the fundamental system $$\{z_{1},z_{2}\}$$ mentioned in Theorem 13.1.2 e; that is, we choose $$\{z_{1},z_{2}\}$$ so that $$B_{1}(z_{1})=z_{1}(0)=0$$ and $$B_{2}(z_{2})=z_{2}(1)=0$$. It is easy to see that

$z_{1}=e^{3x}-e^{4x}\, \text{and} \;z_{2}=e^{3(x-1)}-e^{4(x-1)}\nonumber$

satisfy these requirements. Now we write the solution of Equation \ref{eq:13.1.14} as

$y=2e^{2x}+c_{1}\left(e^{3x}-e^{4x}\right)+c_{2}\left(e^{3(x-1)}-e^{4(x-1)}\right).\nonumber$

Imposing the boundary conditions $$y(0)=3$$ and $$y(1)=5e^{2}$$ yields

$3=2+c_{2}e^{-4}(e-1)\, \text{and} \; 5e^{2}=2e^{2}+c_{1}e^{3}(1-e).\nonumber$

Therefore

$c_{1}=\frac{3}{e(1-e)},\quad c_{2}=\frac{e^{4}}{e-1},\nonumber$

and

$y=2e^{2x}+\frac{3}{e(1-e)}(e^{3x}-e^{4x})+ \frac{e^{4}}{e-1}(e^{3(x-1)}-e^{4(x-1)}).\nonumber$

Sometimes it is useful to have a formula for the solution of a general boundary problem. Our next theorem addresses this question.

Theorem 13.1.3

Suppose the homogeneous boundary value problem

$\label{eq:13.1.15} Ly=0,\quad B_{1}(y)=0,\quad B_{2}(y)=0$

has only the trivial solution$$.$$ Let $$y_{1}$$ and $$y_{2}$$ be linearly independent solutions of $$Ly=0$$ such that $$B_{1}(y_{1})=0$$ and $$B_{2}(y_{2})=0,$$ and let

$W=y_{1}y_{2}'-y_{1}'y_{2}.\nonumber$

Then the unique solution of

$\label{eq:13.1.16} Ly=F,\quad B_{1}(y)=0,\quad B_{2}(y)=0$

is

$\label{eq:13.1.17} y(x)= y_{1}(x)\int_{x}^{b}\frac{F(t)y_{2}(t)}{P_{0}(t)W(t)}\,dt+ y_{2}(x)\int_{a}^{x}\frac{F(t)y_{1}(t)}{P_{0}(t)W(t)}\,dt.$

Proof

In Section 5.7 we saw that if

$\label{eq:13.1.18} y=u_{1}y_{1}+u_{2}y_{2}$

where

\begin{aligned} u_{1}'y_{1}+u_{2}'y_{2}&=0\\[4pt] u_{1}'y_{1}'+u_{2}'y_{2}'&=F,\end{aligned}\nonumber

then $$Ly=F$$. Solving for $$u_{1}'$$ and $$u_{2}'$$ yields

$u_{1}'=-\frac{Fy_{2}}{P_{0}W}\, \text{and} \; u_{2}'=\frac{Fy_{1}}{P_{0}W},\nonumber$

which hold if

$u_{1}(x)=\int_{x}^{b}\frac{F(t)y_{2}(t)}{P_{0}(t)W(t)}\,dt \, \text{and} \; u_{2}(x)=\int_{a}^{x}\frac{F(t)y_{1}(t)}{P_{0}(t)W(t)}\,dt.\nonumber$

This and Equation \ref{eq:13.1.18} show that Equation \ref{eq:13.1.17} is a solution of $$Ly=F$$. Differentiating Equation \ref{eq:13.1.17} yields

$\label{eq:13.1.19} y'(x)=y_{1}'(x)\int_{x}^{b}\frac{F(t)y_{2}(t)}{P_{0}(t)W(t)}\,dt+ y_{2}'(x)\int_{a}^{x}\frac{F(t)y_{1}(t)}{P_{0}(t)W(t)}\,dt.$

(Verify.) From Equation \ref{eq:13.1.17} and Equation \ref{eq:13.1.19},

$B_{1}(y)=B_{1}(y_{1})\int_{a}^{b}\frac{F(t)y_{2}(t)}{P_{0}(t)W(t)}\,dt=0\nonumber$

because $$B_{1}(y_{1})=0$$, and

$B_{2}(y)=B_{2}(y_{2})\int_{a}^{b}\frac{F(t)y_{1}(t)}{P_{0}(t)W(t)}\,dt=0\nonumber$

because $$B_{2}(y_{2})=0$$. Hence, $$y$$ satisfies Equation \ref{eq:13.1.16}. This completes the proof.

We can rewrite Equation \ref{eq:13.1.17} as

$\label{eq:13.1.20} y=\int_{a}^{b} G(x,t)F(t)\,dt,$

where

\begin{aligned} G(x, t)=\left\{\begin{array}{ll}{\frac{y_{1}(t) y_{2}(x)}{P_{0}(t) W(t)}} & {, \quad a \leq t \leq x} \\[4pt] {\frac{y_{1}(x) y_{2}(t)}{P_{0}(t) W(t)}} & {, \quad x \leq t \leq b}\end{array}\right. \end{aligned}\nonumber

This is the Green’s function for the boundary value problem Equation \ref{eq:13.1.16}. The Green’s function is related to the boundary value problem Equation \ref{eq:13.1.16} in much the same way that the inverse of a square matrix $${\mathbf A}$$ is related to the linear algebraic system $${\mathbf y}={\mathbf A}{\mathbf x}$$; just as we substitute the given vector $${\mathbf y}$$ into the formula $${\mathbf x}={\mathbf A}^{-1}{\mathbf y}$$ to solve $${\mathbf y} ={\mathbf Ax}$$, we substitute the given function $$F$$ into the formula Equation \ref{eq:13.1.20} to obtain the solution of Equation \ref{eq:13.1.16}. The analogy goes further: just as $${\mathbf A}^{-1}$$ exists if and only if $${\mathbf A}{\mathbf x} ={\mathbf 0}$$ has only the trivial solution, the boundary value problem Equation \ref{eq:13.1.16} has a Green’s function if and only the homogeneous boundary value problem Equation \ref{eq:13.1.15} has only the trivial solution.

We leave it to you (Exercise 13.1.32) to show that the assumptions of Theorem 13.1.3 imply that the unique solution of the boundary value problem

$Ly=F,\quad B_{1}(y)=k_{1},\quad B_{2}(y)=k_{2}\nonumber$

is

$y(x)=\int_{a}^{b}G(x,t)F(t)\,dt +\frac{k_{2}}{B_{2}(y_{1})}y_{1}+ \frac{k_{1}}{B_{1}(y_{2})}y_{2}.\nonumber$

Example 13.1.6

Solve the boundary value problem

$\label{eq:13.1.21} y''+y=F(x).\quad y(0)+y'(0)=0,\quad y(\pi)-y'(\pi)=0,$

and find the Green’s function for this problem.

Solution

Here

$B_{1}(y)=y(0)+y'(0) \, \text{and} \; B_{2}(y)=y(\pi)-y'(\pi).\nonumber$

Let $$\{z_{1},z_{2}\}=\{\cos x,\sin x\}$$, which is a fundamental set of solutions of $$y''+y=0$$. Then

\begin{aligned} B_{1}(z_{1})&=(\cos x-\sin x)\big|_{x=0}=\phantom{-}1\\[4pt] B_{2}(z_{1})&=(\cos x+\sin x)\big|_{x=\pi}=-1\end{aligned}\nonumber

and

\begin{aligned} B_{1}(z_{2})&=(\sin x+\cos x)\big|_{x=0}=1\\[4pt] B_{2}(z_{2})&=(\sin x-\cos x)\big|_{x=\pi}=1.\end{aligned}\nonumber

Therefore

$B_{1}(z_{1})B_{2}(z_{2})-B_{1}(z_{2})B_{2}(z_{1})=2,\nonumber$

so Theorem 13.1.2 implies that Equation \ref{eq:13.1.21} has a unique solution. Let

$y_{1}=B_{1}(z_{2})z_{1}-B_{1}(z_{1})z_{2}= \cos x-\sin x\nonumber$

and

$y_{2}=B_{2}(z_{2})z_{1}-B_{2}(z_{1})z_{2}= \cos x+\sin x. \nonumber$

Then $$B_{1}(y_{1})=0$$, $$B_{2}(y_{2})=0$$, and the Wronskian of $$\{y_{1},y_{2}\}$$ is

$W(x)= \left|\begin{array}{rrccccc} \cos x-\sin x& \cos x+ \sin x\\[4pt] -\sin x-\cos x&-\sin x+\cos x \end{array}\right|=2.\nonumber$

Since $$P_{0}=1$$, Equation \ref{eq:13.1.17} yields the solution

\begin{aligned} y(x) &=\frac{\cos x-\sin x}{2} \int_{x}^{\pi} F(t)(\cos t+\sin t)dt \\[4pt] &+\frac{\cos x+\sin x}{2}\int_{0}^{x}F(t)(\cos t-\sin t)dt. \end{aligned}\nonumber

The Green’s function is

\begin{aligned} G(x, t)=\left\{\begin{array}{ll}{\frac{(\cos t-\sin t)(\cos x+\sin x)}{2}} & {, \quad 0 \leq t \leq x} \\[4pt] {\frac{(\cos x-\sin x)(\cos t+\sin t)}{2}} & {, \quad x \leq t \leq \pi}\end{array}\right. \end{aligned}\nonumber

We’ll now consider the situation not covered by Theorem 13.1.3 .

Theorem 13.1.4

Suppose the homogeneous boundary value problem

$\label{eq:13.1.22} Ly=0,\quad B_{1}(y)=0,\quad B_{2}(y)=0$

has a nontrivial solution $$y_{1},$$ and let $$y_{2}$$ be any solution of $$Ly=0$$ that isn’t a constant multiple of $$y_{1}.$$ Let $$W=y_{1}y_{2}'-y_{1}'y_{2}.$$ If

$\label{eq:13.1.23} \int_{a}^{b}\frac{F(t)y_{1}(t)}{P_{0}(t)W(t)}\,dt=0,$

then the homogeneous boundary value problem

$\label{eq:13.1.24} Ly=F, \quad B_{1}(y)=0,\quad B_{2}(y)=0$

has infinitely many solutions$$,$$ all of the form $$y=y_{p}+c_{1}y_{1},$$ where

$y_{p}=y_{1}(x) \int_{x}^{b}\frac{F(t)y_{2}(t)}{P_{0}(t)W(t)}\,dt+ y_{2}(x) \int_{a}^{x}\frac{F(t)y_{1}(t)}{P_{0}(t)W(t)}\,dt \nonumber$

and $$c_{1}$$ is a constant$$.$$ If

$\int_{a}^{b}\frac{F(t)y_{1}(t)}{P_{0}(t)W(t)}\,dt \ne0, \nonumber$

then Equation \ref{eq:13.1.24} has no solution.

Proof

From the proof of Theorem 13.1.3 , $$y_{p}$$ is a particular solution of $$Ly=F$$, and

$y_{p}'(x)=y_{1}'(x)\int_{x}^{b}\frac{F(t)y_{2}(t)}{P_{0}(t)W(t)}\,dt +y_{2}'(x)\int_{a}^{x}\frac{F(t)y_{1}(t)}{P_{0}(t)W(t)}\,dt. \nonumber$

Therefore the general solution of Equation \ref{eq:13.1.22} is of the form

$y=y_{p}+c_{1}y_{1}+c_{2}y_{2}, \nonumber$

where $$c_{1}$$ and $$c_{2}$$ are constants. Then

\begin{aligned} B_{1}(y)&= B_{1}(y_{p}+c_{1}y_{1}+c_{2}y_{2})= B_{1}(y_{p})+c_{1}B_{1}(y_{1})+c_{2}B_{1}y_{2}\\[4pt] &=B_{1}(y_{1})\int_{a}^{b}\frac{F(t)y_{2}(t)}{P_{0}(t)W(t)}\,dt +c_{1}B_{1}(y_{1})+c_{2}B_{1}(y_{2})\\[4pt] &=c_{2}B_{1}(y_{2})\end{aligned}\nonumber

Since $$B_{1}(y_{1})=0$$, Theorem 13.1.1 implies that $$B_{1}(y_{2})\ne0$$; hence, $$B_{1}(y)=0$$ if and only if $$c_{2}=0$$. Therefore $$y=y_{p}+c_{1}y_{1}$$ and

\begin{aligned} B_{2}(y)&=B_{2}(y_{p}+c_{1}y_{1}) =B_{2}(y_{2})\int_{a}^{b}\frac{F(t)y_{1}(t)}{P_{0}(t)W(t)}\,dt+c_{1}B_{2}(y_{1})\\[4pt] &=B_{2}(y_{2})\int_{a}^{b}\frac{F(t)y_{1}(t)}{P_{0}(t)W(t)}\,dt,\end{aligned}\nonumber

since $$B_{2}(y_{1})=0$$. From Theorem 13.1.1 , $$B_{2}(y_{2})\ne0$$ (since $$B_{2}(y_{1}=0$$). Therefore $$Ly=0$$ if and only if Equation \ref{eq:13.1.23} holds. This completes the proof.

Example 13.1.7

Applying Theorem 13.1.4 to the boundary value problem

$\label{eq:13.1.25} y''+y=F(x),\quad y(0)=0,\quad y(\pi)=0$

explains the Examples 13.1.2 and 13.1.3 . The complementary equation $$y''+y=0$$ has the linear independent solutions $$y_{1}=\sin x$$ and $$y_{2}=\cos x$$, and $$y_{1}$$ satisfies both boundary conditions. Since $$P_{0}=1$$ and

$W= \left|\begin{array}{crccccc} \sin x&\cos x \\[4pt]\cos x&-\sin x \end{array}\right|=-1, \nonumber$

Equation \ref{eq:13.1.23} reduces to

$\int_{0}^{\pi} F(x)\sin x\,dx=0.\nonumber$

From Example 13.1.2 , $$F(x)=1$$ and

$\int_{0}^{\pi} F(x)\sin x\,dx=\int_{0}^{\pi} \sin x\,dx=2,\nonumber$

so Theorem 13.1.3 implies that Equation \ref{eq:13.1.25} has no solution. In Example 13.1.3 ,

$F(x)=\sin 2x=2\sin x \cos x\nonumber$

and

$\int_{0}^{\pi}F(x)\sin x\,dx=2\int_{0}^{\pi}\sin^{2}x\cos x\,dx =\frac{2}{3}\sin^{3}x\bigg|_{0}^{\pi}=0,\nonumber$

so Theorem 13.1.3 implies that Equation \ref{eq:13.1.25} has infinitely many solutions, differing by constant multiples of $$y_{1}(x)=\sin x$$.

This page titled 13.1: Boundary Value Problems is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform.