0387986995Basic TheoryC

Let 4(x) and O(z) be two solutions of the homogeneous linear differential equation

The first basic result of this section is the following theorem, which concerns the existence and uniqueness of solution of (BP).

Theorem VI-2-1. If the two solutions 4(x) and ip(x) of (VI.2.1) are linearly independent, then problem (BP) has one and only one solutwn on the interval

I(a, b).

Proof

Using the method of variation of parameters (cf. Remark IV-7-2), write the general solution y(x) of the differential equation of (BP) and its derivative y'(x)

where cl and c2 are arbitrary constants and W (x) denotes the Wronskian of

Now, using (VI.2.3) and (VI.2.4), it can be shown that solution

(VI.2.5) satisfies the boundary conditions of (BP) if and only if cl = 0 and c2 =

0. 0

Observation VI-2-2. It can be shown easily that p(x)W(x) is independent of x.

Observation VI-2-3. Under the assumption that the two solutions 40(x) and tfi(x) of (VI.2.1) are linearly independent, the unique solution of (BP) is given by

rb

150 VI. BOUNDARY-VALUE PROBLEMS

Setting

The function G(x,l;) is called Green's function of problem (BP). The following theorem gives the characterization of Green's function.

Theorem VI-2-4. The function G(x, C) given by (VI. 2.7) satisfies the following conditions:

(i) G(x, l:) is continuous with respect to (x, e) on the region D = {(x,) a < x < b, a< £<b},

(iv)as a function of x, G(x, l:) satisfies homogeneous linear differential equation

(VI. 2. 1) if x 0 4,

(v)as a function of x, G(x, 4) satisfies the boundary conditions of (BP), i. e.,

G(b,Z:)cosf3-p(b)8G(b,{)sinf3=0

for a:5 < b.

Furthermore, the function G(x, f) is uniquely determined by these five conditions.

Proof.

It is evident that the function G(x, l;) satisfies these five conditions. Suppose H(x,t) satisfies the five conditions. Conditions (iv) and (v) imply that

for (x,l;') E D, where Ci(1;) and C2(C) must be determined by conditions (i), (ii),

0 and

-PW

Observation VI-2-6. In Theorem VI.2.1, it was assumed that the two solutions ¢(x) and '(x) are linearly independent. Consider the case when this assumption is not satisfied. So, assume that ¢(x) and ?P(x) are linearly dependent. This means that

(f, 0) = (C [y], 0) = (y, x[01) = 0,

if there exists a solution y of problem (VI.2.9). The converse is also true, as shown in the following theorem.

Theorem VI-2-7. Assume that O(x) satisfies condition (VI.2.8). Then, if a real- valued continuous function f (x) on T (a, b) satisfies the condition

problem (BP) has solutions depending on an arbitrary constant.

Proof.

Using the method of variation of parameters, write the general solution y(x) of the differential equation of (BP) and its derivative y(x) respectively in the following form:

(VI.2.11)

where u(x) is a solution of the linear homogeneous differential equation (VI.2.1) such that 0 and p are linearly independent, two quantities ca and c2 are arbitrary constants, and W(x) denotes the Wronskian of {0(x),p(x)}. Note that p(x)W(z) is independent of x and that

The condition that y E V(a,b) and (VI.2.12) imply that ca is arbitrary and c2 =

0. 0

Remark VI-2-8. The materials of this section are also found in (CL, Chapters 7 and 11] and [Har2, Chapter XI].

VI-3. Eigenvalue problems

In this section, we consider the eigenvalue problem

(P(x) dx ) + u(x)y = Ay,

under the assumptions:

(i) the quantities a, b, a, and 0 are real numbers such that a < b,

(ii) u(x) is a real-valued and continuous function on the interval Z(a, b) = {x :

a<x<b},

(iii) the function p(x) is real-valued and continuously differentiable, and p(x) > 0 on I(a, b).

The quantity A is the eigenvalue parameter.

Let 6(x,,\) and ti(x, A) be two solutions of the homogeneous linear differential equation

respectively. Then,

-O(a, A) cos a - p(a)6'(a, A) sin a = 0,

(VI.3.3)

tp(b, A) cos /3 - p(b) 0'(b, A) sin /3 = 0.

These two solutions 6(x, A) and tp(x, A) are analytic in A everywhere in C. Also, they are linearly independent if and only if

Therefore, we obtain the following result.

Theorem VI-3-1. In order that A be an eigenvalue of (EP), it is necessary and sufficient OW ,\ satisfies the equation

or, equivalently,

+'(a, A) cos a - p(a)0'(a, A) sin a = 0.

operator C[y] = ((x)) + u(x)y and a vector space U(a, b) over the complex

number field C by

U(a,b) _ If E C2 (a, b; C) : f (a) cos a - p(a) f'(a)sin a = 0,

f (b) cos a - p(b) f'(b) sin,3 = 0},

where C2(a, b; C) denotes the set of all complex-valued functions which are twice continuously differentiable on the interval I(a, b) = {x : a < x < b}. Also, define an inner product (f, g) for two complex-valued continuous functions f and g on 2(a, b) by

b

(f, 9) = Jn f (09(041

where g(1;) denotes the complex conjugate of g(f ). Since

= P(b)f(b)9'(b) - p(a)f(a)9'(a) + e [-Pwfv)9'(o + uh)fw9(-j <'

J

for f E U(a, b) and g E U(a, b), we obtain

(f, C(91) - (C[f ], 9)

= {p(b)f(b)9'(b) - p(b)9(b)f'(b)} - {p(a)f(a)g(a) - p(a)9(a)f'(a)}.

Also, since

Thus, we extended Theorem VI-2-5 as follows.

Theorem VI-3-4 (self-adjointness). The operator C has the following property:

Theorem VI-3-4 implies the following conclusion.

on T(a, b). 0

Remark VI-3-9. The proof given above also showed that if pi (x) > p2(x) > 0 on

Z(a, b), then we obtain

(a) 02(x) > 61(x) on I(a,b) if 02(a) > 01(a),

02(x) > 01(x) for a < x < b if 02(a) > 01(a) and g2(x) > g1(x) on a < x < b, ( y) in the case when g2(x) > g1(x) on a < x < b, if 01(a) = 0 and 61(b) = 7r and

if 0<02(a)<rr,then 02(b)>a.

Also, y=rsin(0)>0 if and only if 0<0<r(mod 2rr)and p(x) = 1 > 0 at

0 = 0 (mod jr). Theorem VI-1-1 follows from (y).

Setting 9(x,,\) = u(x) - A, apply Lemma VI-3-8 to problem (EP).

Lemma VI-3-10. Let 0(x, A) be the unique solution to the initial-value problem

on the interval Z(a, b), where u(x) is continuous on Z(a, b), p(x) is continuous and positive on I(a, b), A is a real parameter, and at is a fixed real number such that 0 < a < rr. Then, for every c E 2(a, b) such that c > a,

(i) 0(c, A) is a continuous and strictly decreasing function of A for -oo < A <

+00,

(ii)lira 0(c, A) = +oo,

We prove this lemma in three steps.