James V. Herod*

Page maintained with additions by Evans M. Harrell, II, harrell@math.gatech.edu.

**SECTION 6 THE SECOND ALTERNATIVE**

We now discuss the problems where the Second Alternative holds. The
supposition is that there is a nontrivial solution for L(y) = 0, B_{1}(y) =
B_{2}(y) = 0. The
Fredholm Theorems assure us that, if f is continuous, then there is a
solution for L(y) = f, with B_{1}(y) = B_{2}(y) = 0 provided

for all solutions w of the equation
L*(w) = 0, B_{1}*(w) = B_{2}*(w) = 0. As before,
we will construct Green functions G such that, in case f satisfies the above
requirement, then

provides a solution for L(y) = f.

In this second alternative, there may be many solutions for the equation L(y) = f. Consequently, we expect there may be many Green functions. In the technique developed below, G( ,t) is always in

M= {y: B_{0}(y) = B_{1}(y) = 0 }.

This is not necessarily true for Green functions constructed by other methods: see for example the construction found by Don Jones while a graduate research assistant at GEORGIA TECH and given in an appendix.

We again divide the problems into three cases according to the nature of the boundary conditions. We shall illustrate methods of construction.

The first case to consider is where the boundary conditions arise as
__initial__ __conditions__. This case is not pertinent for the initial
value problem has a unique solution. Thus, case one is always in the first
alternative.

**EXAMPLE**:(Second Alternative,__unmixed, two point boundary
conditions__)

Suppose that L(y) = y'' + y' - 2y, B_{1}(y) = y(0) - y'(0), and
B_{2}(y) =
y(1) - y'(1). It is the purpose of this example to show that there is no
function G such that L(G(.,t))(x) = d(x,t). Note that L*(z) = z'' - z' -2z
and M* = {y: 2z(1) = z'(1), 2z(0) = z'(0)}. Nontrivial functions in the
nullspace of {L*, B_{1}*, B_{2}*} are multiples of e^{2t}.^{
}Hence, we are in the second alternative. The Fredholm Alternative
theorem suggests that there will be no function G^{ }such that, if t is
in (0,1), then the distribution equation L(G( ,t)) (x)^{ }= d(x,t)
holds unless

Of course, the value of this integral is not zero.

For this situation, we must modify the construction of the Green function.

**CONSTRUCTION OF G IN THE SECOND ALTERNATIVE, n ^{th }ORDER**

Step (1) Find the nullspace of { L*, M*}

Step (2) Find an orthonormal basis for this nullspace. Call this basis
v_{1},v_{2},...v_{m}, m __<__ n.

Step (3) Construct up such that L(u_{p}) = v_{p}, p = 1, 2, ...n.

Step (4) Construct G such that

**THEOREM** If 0 < t < 1, then there is G(.,t) such that

INDICATION OF PROOF. By the Fredholm Alternative Theorems, there will be such a function G provided

for all w in the nullspace of L*. This can be verified by writing w in terms
of this orthonormal basis ,

and
evaluating the dot product.

**HOW TO CONSTRUCT G SUCH THAT **

First, find linearly independent solutions y_{p}, p=1,...,n, of
the homogeneous equation L(y) = 0. Then, find solutions
u_{p}, p=1,..., m < n, for the equations L(u_{p}, p=1,...,)(x) = v_{p}, p=1,...,(x). It is
not required that these solutions should satisfy any special boundary
conditions.

The problem of finding G is now a problem of finding constants
C_{p} and D_{p} such that

The constants C_{p} and D_{p} are determined by
these 2n equations:

CONTINUATION OF THE PREVIOUS EXAMPLE

Recall that L(y) = y'' +y' -2y, B_{1}(y) = y(0) - y'(0), and B_{2}(y) = y(1) -
y'(1). Linearly independent solutions for L(y) = 0 are e^{-2x} and
e^{x}. A normalized basis for the one-dimensional nullspace of
{L*,B_{1}*,B_{2}*} is \alpha e^{2x} where \alpha is the positive number
given by

A solution u for the equation y'' +y' -2y = \alpha e^{2x} is
u(x) = \alpha e^{2x}/4.

Now, G is given by:

The four constants - A,B,C, and D - can be solved by these four equations:

(1) 0 = B_{1}(G( ,t)) = G(0,t) - G_{x}(0,t) = 3A
+ [[alpha]]^{2}e^{t}/4

(2) 0 = B_{2}(G( ,t)) = G(1,t) - G_{x}(1,t) =
3Ce^{-2} + a^{2}e^{2(1+t)}/4,

(3) 0 = G(t^{+},t) - G(t^{-},t) = (C-A)e^{-2t} +
(D-B)e^{t}, and^{}

^{}(4) 1 = G_{x}(t^{+},t) - G_{x}(t^{-},t)= -2(C-A)e^{-2t} +
(D-B)e^{t}.

Upon solving this system of four equations and four unknowns, an infinity of solutions will be found determined by these three equations:

A = - \alpha^{2}e^{2t}/12, C = Ae^{4}, D-B =
e^{-t}/3.

QED

**EXERCISE**: FOR EACH OF THE FOLLOWING, GIVE L*,B_{1}*, B_{2}*, and G.

(a) L(y) = y'' +y' -2y, B_{1}(y) = y(0)-y'(0),B_{2}(y) = y(1) - y'(1).

(b) L(y) =4y'' - y, B_{1}(y) = y(0) - 2y'(0), B_{2}(y) = y(1) - 2y'(1).

(c) L(y) = y'' - 2y' - 3y, B_{1}(y) = 3y(0) - y'(0), B_{2}(y) = 3y(1) - y'(1).

**EXAMPLE**(Second Alternative, __mixed, two point boundary
conditions__.)

Suppose that L(y) = y'', B_{1}(y) = y(0) + y(1), B_{2}(y) = y'(0) - y'(1). Then
L*(z) = z'', B_{1}*(z) = z(0)- z(1), B_{2}*(z) = z'(0) + z'(1). All solutions of
{L,B_{1}, B_{2}} are multiples of 2x - 1 . A nontrivial solution of [L*,B_{1}*, B_{2}*} is
the constant function 1. Also, the function V(x) = 1 forms a basis for the
null space of 0 = L*(z) in M*. The function u(x) = x^{2}/2 satisfies
L(u) = 1. Thus

G(x,t) = BLC{(A(A + Bx - x^{2}/2 if x < t,C + Dx -
x^{2}/2 if t < x.))

We have four unknowns; we have the following four equations:

(1) 0 = G(0,t) +G(1,t) = A + C + D - 1/2

(2) 0 = dG(x,t)/dx|x=0 - dG(x,t)/dx|x=1 = B - (D-1)

(3) 0 = G(t^{+},t) - G(t^{-},t) = C-A + (D-B)t

(4) 1 = dG(x,t)/dx|x=t+ - dG(x,t)/dx|x=t- = D-B.

As expected, there is an infinity of solutions to these equations which may be found by choosing D and then

B = D - 1

2C = -(t+D) + 1/2

A = C + t.

QED

**EXERCISE:** I Verify that each of these problems is second alternative and
find L*,B_{1}*, B_{2}*,and G.

(1) L(y) = y'', B_{1}(y) = y(0) - y(1), B_{2}(y) = y'(0) - y'(1),

(2) L(y) = y'' + 9[[pi]]^{2}y, B_{1}(y) = y(0) - y(1), B_{2}(y) = y'(0) +
y'(1),

(3) L(y) = y'' + y' - 2y, B_{1}(y) = e y(0) - y(1), B_{2}(y) = e y'(0) - y'(1).

II. Construct L*, B* and G for each of the following L's and with periodic boundary conditions y(0) = y(1), y'(0) = y'(1):

(a) L(y) = y'',

(b) L(y) = y'' + \pi^{2}y

(c) L(y) = 2y'' + y' - y,

** NONHOMOGENEOUS BOUNDARY CONDITIONS**

To solve the equations L(y) = f, B_{1}(y) = \alpha, B_{2}(y) = \beta, first
construct G for the problem L(y) = f, B_{1}(y)= 0, B_{2}(y) = 0. Then construct
functions z_{1} and z_{2} such that B_{1}(z_{1}) = 0, B_{2}(z_{1}) != 0, and B_{1}(z_{2}) != 0, B_{2}(z_{2})
= 0. The solution for the original problem is

QED

A COMPENDIUM OF PROBLEMS

1. Find a formula for u if u'' = f and

(a) u(0) = u(1) = 0.

(b) u(0) = u'(0) = 0.

(c) u(0) = 3, u(1) = 5,

(d) u'(0) = 3u(0), u'(1) = 5 u(1),

(e) u(0) = u(1), u'(0) = u'(1),

(f) u(0) = 3, u'(0) = 5,

(g) u'(0) = 3, u(0) = 5,

(h) u(0) = 0, I(0,1, ) u(x) dx = 0.

(answers)

2. Find a formula for u if u'' + 9u = f and

(a) u(0) = 3, u'(1) = 5.

(b) u(0) - u'(0) = 3, u(1) = 5.

3. Find a formula for u if (x u'(x))' = f and u(1) = 0, u(2) = 5.

4. (a) Find conditions on f in order that u'' + 4 \pi^{2} u = f,
u(0)=u(1), u'(0) = u'(1) should have a solution.

(b) Give the Green function for this problem.

(c) By finding the Green function for the problem L(y) = y'', y(0) = y(1), y'(0) = y'(1), re-write this equation as an integral equation such as was studied in the previous chapter.

5. Here is a linear differential operator with boundary conditions:

L(y)(x) = (e^{x} y' )' and B_{1}(y) = y(0), B_{2}(y) =
y'(0).

(a) Show that (e^{x} y')' z - y
(e^{x}z')' = [ e^{x} (z y' -
z' y)]'.

(b) Give L* and B*.

(c) Give the Green function for the problem L(y) = f with
B_{1}(y) = B_{2}(y) = 0.

(d) Rewrite the problem (e^{x} y')' + sin(x) y(x) =
f(x) , y(0) = y'(0) = 0 as an integral equation in the form

y = **K**(y) + F.

Be sure to identify **K** and F carefully.

6. Consider the differential equation: f is continuous on [0, \pi ] and

(sin(x) y'(x))' + 2 sin(x) y(x) = f(x)

y(0) = 0 = y( \pi ).

(a) In the context of this course, what is the appropriate space and linear operator L?

(b) What is the adjoint of L in this space? Explain your answer.

(c) Is this problem 1^{st} or 2^{nd} alternative?

(d) If possible, solve this problem with f(x) = x. If it is not possible, explain why not.

Back to Section 2.5

Return to Table of Contents (Green Track)

Return to Evans Harrell's