13.1 Two-Point Boundary Value Problems

In Section 5.3 we considered initial value problems for the linear second order equation

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 (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 (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

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

on the closed interval \([a,b]\) such that

and

The assumptions stated in this problem apply throughout this section and won’t 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 (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

on \((c,d)\) to draw conclusions about solutions of (13.1.3) on \([a,b]\).

We call \(a\) and \(b\) boundary points. The conditions (13.1.4) and (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 (13.1.2) to insure that we’re 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 (13.1.4) is an impossible condition if \(k_{1}\ne0\), or no condition at all if \(k_{1}=0\).

We abbreviate (13.1.1) as \(Ly=F\), where

and we denote

We combine (13.1.3), (13.1.4), and (13.1.5) as

This boundary value problem is homogeneous if \(F=0\) and \(k_{1}=k_{2}=0\); otherwise it’s nonhomogeneous.

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

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.

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

This implies that

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 \(B_{2}(z_{1})=B_{2}(z_{2})=0\).

Proof We’ll show that

(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

Moreover,

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 (13.1.7) and (13.1.11),

Since the determinant of a product of matrices is the product of the determinants of the matrices, (13.1.8) and (13.1.12) imply (13.1.9).

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

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

(recall (13.1.7)), which is equivalent to

From (13.1.9), this system always has a unique solution \((c_{1},c_{2})\).

(c) \(\implies\) (d): Obviously, \(y=0\) is a solution of (13.1.10). From (c) with \(F=0\) and \(k_{1}=k_{2}=0\), it’s the only solution.

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

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

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

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

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

(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 (13.1.8), which completes the proof.

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

Therefore the solution of (13.1.13) can be written as

Then

and imposing the boundary conditions yields the system

so \(c_{1}=7\) and \(c_{2}=-4\). Therefore

is the unique solution of (13.1.13)

Solution From Example 5.4.1, \(y_{p}=2e^{2x}\) is a particular solution of

Since \(\{e^{3x},e^{4x}\}\) is a fundamental set for the complementary equation, we could write the solution of (13.1.13) as

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’s 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

satisfy these requirements. Now we write the solution of (13.1.14) as

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

Therefore

and

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

Proof In Section 5.7 we saw that if

where

then \(Ly=F\). Solving for \(u_{1}'\) and \(u_{2}'\) yields

which hold if

This and (13.1.18) show that (13.1.17) is a solution of \(Ly=F\). Differentiating (13.1.17) yields

(Verify.) From (13.1.17) and (13.1.19),

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

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

We can rewrite (13.1.17) as

where

This is the Green’s function for the boundary value problem (13.1.16). The Green’s function is related to the boundary value problem (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 (13.1.20) to obtain the solution of (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 (13.1.16) has a Green’s function if and only the homogeneous boundary value problem (13.1.15) has only the trivial solution.

We leave it to you (Exercise 32) to show that the assumptions of Theorem 13.1.3 imply that the unique solution of the boundary value problem

is

Solution Here

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

and

Therefore

so Theorem 13.1.2 implies that (13.1.21) has a unique solution. Let

and

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

Since \(P_{0}=1\), (13.1.17) yields the solution

The Green’s function is

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

Proof From the proof of Theorem 13.1.3, \(y_{p}\) is a particular solution of \(Ly=F\), and

Therefore the general solution of (13.1.22) is of the form

where \(c_{1}\) and \(c_{2}\) are constants. Then

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

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 (13.1.23) holds. This completes the proof.

13.1 Exercises

1. Verify that \(B_{1}\) and \(B_{2}\) are linear operators; that is, if \(c_{1}\) and \(c_{2}\) are constants then

   \[ B_{i}(c_{1}y_{1}+c_{2}y_{2})=c_{1}B_{i}(y_{1})+c_{2}B_{i}(y_{2}),\quad i=1,2. \]

2. \(y''-y=x\),   \(y(0)=-2\),  \(y(1)=1\)

   Show answer

   \(y=-x+\dst{\frac{2}{e-1}\left(e^{x}-e^{(x-1)}\right)}\) 

3. \(y''=2-3x\),   \(y(0)=0\),  \(y(1)-y'(1)=0\)

   Show answer

   \(y=x^{2}-\dst{\frac{x^{3}}{3}}+cx\) with \(c\) arbitrary

4. \(y''-y=x\),   \(y(0)+y'(0)=3\),  \(y(1)-y'(1)=2\)

   Show answer

   \(y=-x+2e^{x}+e^{-(x-1)}\) 

5. \(y''+4y=1\),   \(y(0)=3\),  \(y(\pi/2)+y'(\pi/2)=-7\)

   Show answer

   \(y= \dst{\frac{1}{4}+\frac{11}{4}\cos2x+\frac{9}{4}\sin2x}\)

6. \(y''-2y'+y=2e^{x}\),   \(y(0)-2y'(0)=3\),  \(y(1)+y'(1)=6e\)

   Show answer

   \(y=(x^2+13-8x)e^{x}\) 

7. \(y''-7y'+12y=4e^{2x}\),   \(y(0)+y'(0)=8\),  \(y(1)=-7e^{2}\) (see Example 13.1.5)

   Show answer

   \(y=2e^{2x}+\dst{\frac{3(5e^{3x}-4e^{4x})}{e(15-16e)} +\frac{2e^{4}(4e^{3(x-1)}-3e^{4(x-1)})}{16e-15}}\)

8. State a condition on \(F\) such that the boundary value problem

   \[ y''=F(x), \quad y(0)=0, \quad y(1)-y'(1)=0 \]

   has a solution, and find all solutions.

   Show answer

   \(\dst{\int_{a}^{b}tF(t)\,dt=0}\)  \(y=\dst{-x\int_{x}^{1}F(t)\,dt-\int_{0}^{x}tF(t)\,dt+c_{1}x}\) with \(c_{1}\) arbitrary

9. 1. State a condition on \(a\) and \(b\) such that the boundary value problem

      \begin{equation} y''+y=F(x),\quad y(a)=0,\quad y(b)=0 \tag{A} \end{equation}

      has a unique solution for every continuous \(F\), and find the solution by the method used to prove Theorem 13.1.3

   2. In the case where \(a\) and \(b\) don’t satisfy the condition you gave for (a), state necessary and sufficient on \(F\) such that (A) has a solution, and find all solutions by the method used to prove Theorem 13.1.4.

   Show answer

   (a)  \(b-a\ne k\pi\)  (\(k=\) integer)

         \(y=\dst{\frac{\sin(x-a)}{\sin(b-a)}\int_{x}^{b}F(t)\sin(t-b)\,dt +\frac{\sin(x-b)}{\sin(b-a)}\int_{a}^{x}F(t)\sin(t-a)\,dt}\)

        (b)  \(\dst{\int_{a}^{b}F(t)\sin(t-a)\,dt=0}\)

           \(y=-\dst{\sin(x-a)\int_{x}^{b}F(t)\cos(t-a)\,dt -\cos(x-a)\int_{a}^{x}F(t)\sin(t-a)\,dt}\)

           \(+c_{1}\sin(x-a)\) with \(c_{1}\) arbitrary

10. Follow the instructions in Exercise 9 for the boundary value problem

    \[ y''+y=F(x),\quad y(a)=0,\quad y'(b)=0. \]
    Show answer

    (a)  \(b-a\ne(k+1/2)\pi\)  (\(k=\) integer)

            \(y=-\dst{ \frac{\sin(x-a)}{\cos(b-a)}\int_{x}^{b}F(t)\cos(t-b)\,dt- \frac{\cos(x-b)}{\cos(b-a)}\int_{x}^{b}F(t)\sin(t-a)\,dt}\)

         (b)  \(\dst{\int_{a}^{b}F(t)\sin(t-a)\,dt=0}\)

            \(y=-\dst{\sin(x-a)\int_{x}^{b}F(t)\cos(t-a)\,dt -\cos(x-a)\int_{a}^{x}F(t)\sin(t-a)\,dt}\)

            \(+c_{1}\sin(x-a)\) with \(c_{1}\) arbitrary

11. Follow the instructions in Exercise 9 for the boundary value problem

    \[ y''+y=F(x),\quad y'(a)=0,\quad y'(b)=0. \]
    Show answer

    (a)  \(b-a\ne k\pi\) (\(k=\) integer)

            \(y=\dst{\frac{\cos(x-a)}{\sin(b-a)}\int_{x}^{b}F(t)\cos(t-b)\,dt+ \frac{\cos(x-b)}{\sin(b-a)}\int_{a}^{x}F(t)\cos(t-a)\,dt}\)

            (b) \(\dst{\int_{a}^{b}F(t)\cos(t-a)\,dt=0}\)

            \(y=\dst{\cos(x-a)\int_{x}^{b} F(t)\sin(t-a)\,dt +\sin(x-a)\int_{a}^{x} F(t)\cos(t-a)\,dt}\)

              \(+c_{1}\cos(x-a)\) with \(c_{1}\) arbitrary

12. \(y''-y=F(x)\),  \(y(a)=0\),  \(y(b)=0\)

    Show answer

    \(y=\dst{\frac{\sinh(x-a)}{\sinh(b-a)}\int_{x}^{b}F(t)\sinh(t-b)\,dt +\frac{\sinh(x-b)}{\sinh(b-a)}\int_{a}^{x}F(t)\sinh(t-a)\,dt}\)

13. \(y''-y=F(x)\),  \(y(a)=0\),  \(y'(b)=0\)

    Show answer

    \(y=\dst{-\frac{\sinh(x-a)}{\cosh(b-a)}\int_{x}^{b}F(t)\cosh(t-b)\,dt -\frac{\cosh(x-b)}{\cosh(b-a)}\int_{a}^{x}F(t)\sinh(t-a)\,dt}\)

14. \(y''-y=F(x)\),  \(y'(a)=0\),  \(y'(b)=0\)

    Show answer

    \(y=-\dst{\frac{\cosh(x-a)}{\sinh(b-a)}\int_{x}^{b}F(t)\cosh(t-b)\,dt -\frac{\cosh(x-b)}{\sinh(b-a)}\int_{a}^{x}F(t)\cosh(t-a)\,dt}\)

15. \(y''-y=F(x)\),  \(y(a)-y'(a)=0\),  \(y(b)+y'(b)=0\)

    Show answer

    \(y=-\dst{\frac{1}{2}\left( e^{x}\int_{x}^{b}e^{-t}F(t)\,dt+e^{-x}\int_{a}^{x}e^{t}F(t)\,dt\right)}\)

16. \(y''+ \omega^{2}y=F(x)\),  \(y(0)=0\),  \(y(\pi)=0\)

    Show answer

    If \(\omega\) isn’t a positive integer, then

           \(y=\dst{\frac{1}{\omega\sin\omega\pi} \left(\sin\omega x\int_{x}^{\pi}F(t)\sin\omega(t-\pi)\,dt +\sin\omega(x-\pi)\int_{0}^{x}F(t)\sin\omega t\,dt\right)}\).

         If \(\omega=n\) (positive integer), then \(\dst{\int_{0}^{\pi}F(t)\sin nt\,dt=0}\) is necessary for existence

         of a solution. In this case,

         \(y=-\dst{\frac{1}{n}\left(\sin nx\int_{x}^{\pi}F(t)\cos nt\,dt +\cos nx\int_{0}^{x} F(t)\sin nt\,dt\right) +c_{1}\sin nx}\)

         with \(c_{1}\) arbitrary.

17. \(y''+ \omega^{2}y=F(x)\),  \(y(0)=0\),  \(y'(\pi)=0\)

    Show answer

    If \(\omega\ne n+1/2\)  (\(n=\) integer), then

    \[ y=-\frac{\sin\omega x}{\omega\cos\omega \pi}\int_{x}^{\pi}F(t)\cos\omega(t-\pi)\,dt -\frac{\cos\omega(x-\pi)}{\omega\cos\omega\pi}\int_{0}^{x}F(t)\sin\omega t\,dt. \]

         If \(\omega= n+1/2\) (\(n=\) integer), then \(\dst{\int_{0}^{\pi}F(t)\sin(n+1/2)t\,dt=0}\) is necessary

         for existence of a solution. In this case,

         \(y=-\dst{\frac{\sin(n+1/2)x}{n+1/2}\int_{x}^{\pi}F(t)\cos(n+1/2)t\,dt}\)

               \(-\dst{\frac{\cos(n+1/2)x}{n+/2}\int_{0}^{x}F(t)\sin(n+1/2)t\,dt +c_{1}\sin(n+1/2)x}\)

         with \(c_{1}\) arbitrary,

18. \(y''+ \omega^{2}y=F(x)\),  \(y'(0)=0\),  \(y(\pi)=0\)

    Show answer

    If \(\omega\ne n+1/2\)  (\(n=\) integer), then

    \[ y=\frac{\cos\omega x}{\omega\cos\omega \pi}\int_{x}^{\pi}F(t)\sin\omega(t-\pi)\,dt +\frac{\sin\omega(x-\pi)}{\omega\cos\omega\pi}\int_{0}^{x}F(t)\cos\omega t\,dt. \]

         If \(\omega= n+1/2\) (\(n=\) integer), then \(\dst{\int_{0}^{\pi}F(t)\cos(n+1/2)t\,dt=0}\) is necessary

         for existence of a solution. In this case,

         \(y=\dst{\frac{\cos(n+1/2)x}{n+1/2}\int_{x}^{\pi}F(t)\sin(n+1/2)t\,dt}\)

               \(+\dst{\frac{\sin(n+1/2)x}{n+/2}\int_{0}^{x}F(t)\cos(n+1/2)t\,dt +c_{1}\cos(n+1/2)x}\)

         with \(c_{1}\) arbitrary.

19. \(y''+ \omega^{2}y=F(x)\),  \(y'(0)=0\),  \(y'(\pi)=0\)

    Show answer

    If \(\omega\) isn’t a positive integer, then

        \(y=\dst{\frac{1}{\omega\sin\omega\pi} \left(\cos\omega x\int_{x}^{\pi}F(t)\cos\omega(t-\pi)\,dt +\cos\omega(x-\pi)\int_{0}^{x}F(t)\cos\omega t\,dt\right)}\).

         If \(\omega=n\) (positive integer), then \(\dst{\int_{0}^{\pi}F(t)\cos nt\,dt=0}\) is necessary for existence

         of a solution. In this case,

         \(y=-\dst{\frac{1}{n}\left(\cos nx\int_{x}^{\pi}F(t)\sin nt\,dt +\sin nx\int_{0}^{x} F(t)\cos nt\,dt\right) +c_{1}\cos nx}\)

         with \(c_{1}\) arbitrary.

20. Let \(\{z_{1},z_{2}\}\) be a fundamental set of solutions of \(Ly=0\). Given that the homogeneous boundary value problem

    \[ Ly=0, \quad B_{1}(y)=0,\quad B_{2}(y)=0 \]

    has a nontrivial solution, express it explicity in terms of \(z_{1}\) and \(z_{2}\).

    Show answer

    \(y_{1}=B_{1}(z_{2})z_{1}-B_{1}(z_{1})z_{2}\)

21. If the boundary value problem has a solution for every continuous \(F\), then find the Green’s function for the problem and use it to write an explicit formula for the solution. Otherwise, if the boundary value problem does not have a solution for every continuous \(F\), find a necessary and sufficient condition on \(F\) for the problem to have a solution, and find all solutions. Assume that \(a<b\).

    1. \(y''=F(x)\),   \(y(a)=0\),  \(y(b)=0\)

    2. \(y''=F(x)\),   \(y(a)=0\),  \(y'(b)=0\)

    3. \(y''=F(x)\),   \(y'(a)=0\),  \(y(b)=0\)

    4. \(y''=F(x)\),   \(y'(a)=0\),  \(y'(b)=0\)

    Show answer

    (a) \(G(x,t)= \begin{cases} \dst{\frac{(t-a)(x-b)}{b-a}} &a\le t\le x,\\[.11in] \dst{\frac{(x-a)(t-b)}{b-a)}}&x\le t\le b \end{cases}\)

         \(y=\dst{\frac{1}{b-a}\left((x-a)\int_{x}^{b}(t-b)F(t)\,dt+(x-b)\int_{a}^{x}(t-a)F(t)\,dt\right)}\)

       (b)  \(G(x,t)= \begin{cases} a-t & a\le t\le x\\ a-x &x\le t\le b \end{cases}\)   \(y=\dst{(a-x)\int_{x}^{b}F(t)\,dt+\int_{a}^{x}(a-t)F(t)\,dt}\)

       (c)  \(G(x,t)= \begin{cases} x-b & a\le t\le x\\ t-b &x\le t\le b \end{cases}\)   \(y=\dst{\int_{x}^{b}(t-b)F(t)\,dt+(x-b)\int_{a}^{x}F(t)\,dt}\)

       (d)  \(\dst{\int_{a}^{b}F(t)\,dt=0}\) is a necessary condition for existence of a solution. Then

            \(y=\dst{\int_{x}^{b}tF(t)\,dt+x\int_{a}^{x}F(t)\,dt+c_{1}}\) with \(c_{1}\) arbitrary.

22. Find the Green’s function for the boundary value problem

    \begin{equation} y''=F(x), \quad y(0)-2y'(0)=0, \quad y(1)+2y'(1)=0. \tag{A} \end{equation}

    Then use the Green’s function to solve (A) with (a) \(F(x)=1\), (b) \(F(x)=x\), and (c) \(F(x)=x^{2}\).

    Show answer

    \(G(x,t)= \begin{cases} \dst{-\frac{(2+t)(3-x)}{5}},& 0\le t\le x,\\[.11in] -\dst{\frac{(2+x)(3-t)}{5}}, &x\le t\le 1 \end{cases}\)   (a) \(y=\dst{\frac{x^{2}-x-2}{2}}\)   (b) \(y=\dst{\frac{5x^{2}-7x-14}{30}}\)

        (c)\(\, y=\dst{\frac{5x^{4}-9x-18}{60}}\)

23. Find the Green’s function for the boundary value problem

    \begin{equation} x^{2}y''+xy'+(x^{2}-1/4)y=F(x), \quad y(\pi/2)=0,\quad y(\pi)=0, \tag{A} \end{equation}

    given that

    \[ y_{1}(x)=\frac{\cos x}{\sqrt{x}} \text{\; and\; \; } y_{2}(x)=\frac{\sin x}{\sqrt{x}} \]

    are solutions of the complementary equation. Then use the Green’s function to solve (A) with (a) \(F(x)=x^{3/2}\) and (b) \(F(x)=x^{5/2}\).

    Show answer

    \(G(x,t)= \begin{cases} \dst{\frac{\cos t\sin x}{t^{3/2}\sqrt{x}}}, &\dst{\frac{\pi}{2}}\le t\le x,\\[.11in] \dst{\frac{\cos x\sin t}{t^{3/2}\sqrt{x}}}, &x\le t\le \pi \end{cases}\)

          (a)  \(y=\dst{\frac{1+\cos x-\sin x}{\sqrt x}}\)   (b)  \(y=\dst{\frac{x+\pi \cos x-\pi/2\sin x}{\sqrt x}}\)

24. Find the Green’s function for the boundary value problem

    \begin{equation} x^{2}y''-2xy'+2y=F(x), \quad y(1)=0,\quad y(2)=0, \tag{A} \end{equation}

    given that \(\{x,x^{2}\}\) is a fundamental set of solutions of the complementary equation. Then use the Green’s function to solve (A) with (a) \(F(x)=2x^{3}\) and (b) \(F(x)~=~6x^{4}\).

    Show answer

    \(G(x,t)= \begin{cases} \dst{\frac{(t-1)x(x-2)}{t^{3}}},\; 1\le t\le x,\\[.11in] \dst{\frac{x(x-1)(t-2)}{t^{3}}},\; x\le t\le 2 \end{cases}\)

        (a) \(y=x(x-1)(x-2)\)  (b) \(y=x(x-1)(x-2)(x+3)\)

25. Find the Green’s function for the boundary value problem

    \begin{equation} x^{2}y''+xy'-y=F(x), \quad y(1)-2y'(1)=0,\quad y'(2)=0, \tag{A} \end{equation}

    given that \(\{x,1/x\}\) is a fundamental set of solutions of the complementary equation. Then use the Green’s function to solve (A) with (a) \(F(x)=1\), (b) \(F(x)=x^{2}\), and (c) \(F(x)=x^{3}\).

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    \(G(x,t)= \begin{cases} -\dst{\frac{1}{22}\left(3+\frac{1}{t^{2}}\right)\left(x+\frac{4}{x}\right)},&\; 1\le x\le t,\\[.11in] -\dst{\frac{1}{22}\left(3x+\frac{1}{x}\right)\left(1+\frac{4}{t^{2}}\right)},&\; x\le t\le 2 \end{cases}\)

        (a) \(y=\dst{\frac{x^{2}-11x+4}{11x}}\)  (b) \(y=\dst{\frac{11x^{3}-45x^{2}-4}{33x}}\)  (c) \(y=\dst{\frac{11x^{4}-139x^{2}-28}{88x}}\)

26. \(y''=F(x)\),   \(\alpha y(0)+\beta y'(0)=0\),   \(\rho y(1)+\delta y'(1)=0\)

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    \(\alpha(\rho+\delta)-\beta\rho\ne0\)   \(G(x,t)= \begin{cases} \dst{\frac{(\beta-\alpha t)(\rho+\delta-\rho x)} {\alpha(\rho+\delta)-\beta\rho}}, \quad 0\le t\le x, \\[.11in] \dst{\frac{(\beta-\alpha x)(\rho+\delta-\rho t)} {\alpha(\rho+\delta)-\beta\rho}},\quad x\le t\le 1 \end{cases}\)

27. \(y''+y=F(x)\),   \(\alpha y(0)+\beta y'(0)=0\),   \(\rho y(\pi)+\delta y'(\pi)=0\)

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    \(\alpha\delta-\beta\rho\ne0\)   \(G(x,t)= \begin{cases} \dst{\frac{(\beta\cos t-\alpha\sin t)(\delta \cos x-\rho\sin x)} {\alpha\delta-\beta\rho}},\; &0\le t\le x,\\[.11in] \dst{\frac{(\beta\cos x-\alpha\sin x)(\delta \cos t-\rho\sin t)} {\alpha\delta-\beta\rho}},\; &x\le t\le \pi \end{cases}\)

28. \(y''+y=F(x)\),   \(\alpha y(0)+\beta y'(0)=0\),   \(\rho y(\pi/2)+\delta y'(\pi/2)=0\)

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    \(\alpha\rho+\beta\delta\ne0\)   \(G(x,t)= \begin{cases} \dst{\frac{(\beta\cos t-\alpha\sin t)(\rho\cos x+\delta\sin x)} {\alpha\rho+\beta\delta}}\; & x\le t\le \pi \\[.11in] \dst{\frac{(\beta\cos x-\alpha\sin x)(\rho\cos t+\delta\sin t)} {\alpha\rho+\beta\delta}}\; &0\le t\le x \end{cases}\)

29. \(y''-2y'+2y=F(x)\),   \(\alpha y(0)+\beta y'(0)=0\),   \(\rho y(\pi)+\delta y'(\pi)=0\)

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    \(\alpha\delta-\beta\rho\ne0\)   \(G(x,t)= \begin{cases} \dst{\frac{e^{x-t)}(\beta\cos t-(\alpha+\beta)\sin t) (\delta\cos x-(\rho+\delta)\sin x)}{\alpha\delta-\beta\rho}}\; &0\le t\le x, \\[.11in] \dst{\frac{e^{x-t}(\beta\cos x-(\alpha+\beta)\sin x) (\delta\cos t-(\rho+\delta)\sin t)}{\alpha\delta-\beta\rho}},\; &x\le t\le \pi \end{cases}\)

30. \(y''-2y'+2y=F(x)\),   \(\alpha y(0)+\beta y'(0)=0\),   \(\rho y(\pi/2)+\delta y'(\pi/2)=0\)

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    \(\beta \delta+(\alpha+\beta)(\rho+\delta)\ne0\)   \(G(x,t)= \begin{cases} \dst{\frac{e^{x-t}(\beta\cos t-(\alpha+\beta)\sin t) ((\rho+\delta)\cos x+\delta \sin x)} {\beta \delta+(\alpha+\beta)(\rho+\delta}},& 0\le t\le x, \\[.11in] \dst{\frac{e^{x-t}(\beta\cos x-(\alpha+\beta)\sin x) ((\rho+\delta)\cos t+\delta \sin t)} {\beta \delta+(\alpha+\beta)(\rho+\delta}}, &x\le t\le \pi/2 \end{cases}\)

31. Find necessary and sufficient conditions on \(\alpha\), \(\beta\), \(\rho\), and \(\delta\) for the boundary value problem

    \begin{equation} y''-y=F(x), \quad \alpha y(a)+\beta y'(a)=0, \quad \rho y(b)+\delta y'(b)=0 \tag{A} \end{equation}

    to have a unique solution for every continuous \(F\), and find the Green’s function for (A). Assume that \(a<b\).

    Show answer

    \((\rho+\delta)(\alpha-\beta)e^{(b-a)}-(\rho-\delta)(\alpha+\beta)e^{(a-b)}\ne0\)

        \(G(x,t)= \begin{cases} \dst{\frac{((\alpha-\beta)e^{(t-a)}-(\alpha+\beta)e^{-(t-a)}) ((\rho-\delta)e^{(x-b)}-(\rho+\delta)e^{-(x-b)})} {2[(\rho+\delta)(\alpha-\beta)e^{(b-a)}-(\rho-\delta)(\alpha+\beta)e^{(a-b)}]}},\; &0\le t\le x,\\[.11in] \dst{\frac{((\alpha-\beta)e^{(x-a)}-(\alpha+\beta)e^{-(x-a)}) ((\rho-\delta)e^{(t-b)}-(\rho+\delta)e^{-(t-b)})} {2[(\rho+\delta)(\alpha-\beta)e^{(b-a)}-(\rho-\delta)(\alpha+\beta)e^{(a-b)}]}}\; &x\le t\le \pi \end{cases}\)

32. Show that the assumptions of Theorem 13.1.3 imply that the unique solution of

    \[ Ly=F, \quad B_{1}(y)=k_{1},\quad B_{2}(y)=f_{2} \]

    is

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