This theorem locates, for certain one-term EVPs, a finite interval containing at
least one eigenvalue:
Theorem 2.9 (Inclusion). For an EVP of the form (2.67) let the
following conditions apply: The EVP is self-adjoint,
R b
a uKudx [ 0 for
u2u TF , g n has fixed sign on x2[a;b], and
ðÀ1Þ
n
Z b
a
u g n v
ðnÞ
h
i ðnÞ
dx ¼
Z b
a
g n u
ðnÞ v
ðnÞ dx for u; v 2 u TF :
ð2:68Þ
Then define two functions u 0 (x)2C
2n [a;b] and u 1 (x)2u TF so that
Ku 1 = Lu 0 , and a third function U(x) through:
UðxÞ ¼ u
ðnÞ
0
.
u
ðnÞ
1 :
ð2:69Þ
If U(x) remains between finite and positive limits U min and U max for x2[a;
b], then the interval [U min ;U max ] contains at least one eigenvalue.
The theorem is applied in three steps: 1) Choose a test function u 1 (x); 2) calculate u 0 (x) as the solution of Ku 1 = Lu 0 ; 3) determine the minimum and maximum
of U(x) on [a;b].
The function u 0 is not required to be a test function (satisfying all boundary
conditions). However, a closer interval (U max − U min ) is obtained if u 0 satisfies
some or all boundary conditions, since u 0 then resembles more of an eigenfunction.
Thus, it is advisable to leave some free constants in u 1 that allows u 0 to be matched
to the boundary conditions.
Example 2.17 Avibrating rod having length l = 1 obeys the differential equation
−c
2 u′′ = x
2 u and boundary conditions u(0) = u′(1) = 0. The EVP is self-adjoint
(cf. Example 2.8). Further,
R b
a uKudx ¼
R 1
0 uðÀc
2 u
00
Þdx ¼ c
2
R 1
0 ðu
0
Þ
2 dx [ 0 for
u2u TF , and g 0 (x) = 1 has fixed sign on [0;1]. Thus, the inclusion theorem is
applicable.
For applying the theorem we choose a polynomial u 1 (x) = a 0 + a 1 x + a 2 x
2 +
a 3 x
3 + a 4 x
4 , and solve Ku 1 = Lu 0 for u 0 to yield u 0 (x) = −c
2 u 1 ′′ = −c
2 (2a 2 + 6
a 3 x + 12x
2 ). To be a test function u 1 must satisfy the boundary conditions, giving
a 0 = 0 and a 1 + 2a 2 + 3a 3 = −4. Though not strictly necessary, we may use the
remaining free constants to force also u 0 to satisfy the boundary conditions, giving
a 2 = 0 and a 3 = −4. Consequently u 0 (x) = 12c
2 x(2 − x) and u 1 (x) = x(8 −
4x
2 + x
3 ), so that the function U(x) becomes:
2.7 Properties of Eigenvalues and Eigenfunctions
73
least one eigenvalue:
Theorem 2.9 (Inclusion). For an EVP of the form (2.67) let the
following conditions apply: The EVP is self-adjoint,
R b
a uKudx [ 0 for
u2u TF , g n has fixed sign on x2[a;b], and
ðÀ1Þ
n
Z b
a
u g n v
ðnÞ
h
i ðnÞ
dx ¼
Z b
a
g n u
ðnÞ v
ðnÞ dx for u; v 2 u TF :
ð2:68Þ
Then define two functions u 0 (x)2C
2n [a;b] and u 1 (x)2u TF so that
Ku 1 = Lu 0 , and a third function U(x) through:
UðxÞ ¼ u
ðnÞ
0
.
u
ðnÞ
1 :
ð2:69Þ
If U(x) remains between finite and positive limits U min and U max for x2[a;
b], then the interval [U min ;U max ] contains at least one eigenvalue.
The theorem is applied in three steps: 1) Choose a test function u 1 (x); 2) calculate u 0 (x) as the solution of Ku 1 = Lu 0 ; 3) determine the minimum and maximum
of U(x) on [a;b].
The function u 0 is not required to be a test function (satisfying all boundary
conditions). However, a closer interval (U max − U min ) is obtained if u 0 satisfies
some or all boundary conditions, since u 0 then resembles more of an eigenfunction.
Thus, it is advisable to leave some free constants in u 1 that allows u 0 to be matched
to the boundary conditions.
Example 2.17 Avibrating rod having length l = 1 obeys the differential equation
−c
2 u′′ = x
2 u and boundary conditions u(0) = u′(1) = 0. The EVP is self-adjoint
(cf. Example 2.8). Further,
R b
a uKudx ¼
R 1
0 uðÀc
2 u
00
Þdx ¼ c
2
R 1
0 ðu
0
Þ
2 dx [ 0 for
u2u TF , and g 0 (x) = 1 has fixed sign on [0;1]. Thus, the inclusion theorem is
applicable.
For applying the theorem we choose a polynomial u 1 (x) = a 0 + a 1 x + a 2 x
2 +
a 3 x
3 + a 4 x
4 , and solve Ku 1 = Lu 0 for u 0 to yield u 0 (x) = −c
2 u 1 ′′ = −c
2 (2a 2 + 6
a 3 x + 12x
2 ). To be a test function u 1 must satisfy the boundary conditions, giving
a 0 = 0 and a 1 + 2a 2 + 3a 3 = −4. Though not strictly necessary, we may use the
remaining free constants to force also u 0 to satisfy the boundary conditions, giving
a 2 = 0 and a 3 = −4. Consequently u 0 (x) = 12c
2 x(2 − x) and u 1 (x) = x(8 −
4x
2 + x
3 ), so that the function U(x) becomes:
2.7 Properties of Eigenvalues and Eigenfunctions
73