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11.16-2. The Construction of Eigenfunctions
The method of least squares can also be applied for the approximate construction of characteristic
values and eigenfunctions of the kernel K(x, s), similarly to the way in which it can be done in
the collocation method. Namely, by setting f(x) ≡ 0 and ϕ 0 (x) ≡ 0, which implies ψ 0 (x) ≡ 0, we
determine approximate values of the characteristic values from the algebraic equation
det[c ij (λ)] = 0. (9)
After this, approximate eigenfunctions can be found from the homogeneous system of the form (8),
where, instead of λ, the corresponding approximate value is substituted.
Example. Let us find an approximate solution of the equation
1
2
y(x)= x + sinh(x + t)y(t) dt (10)
–1
by the method of least squares.
2
For the form of an approximate solution we take Y 2 (x)= x + A 2 x + A 1 . This implies
2
ϕ 1 (x)=1, ϕ 2 (x)= x, ϕ 0 (x)= x .
Taking into account the relations
1 1 1
2
sinh(x + t) dt = a sinh x, t sinh(x + t) dt = b sinh x, t sinh(x + t) dt = c sinh x,
–1 –1 –1
a = 2 sinh 1 = 2.3504, b =2e –1 = 0.7358, c = 6 sinh 1 – 4 cosh 1 = 0.8788,
on the basis of formulas (7) of Subsection 11.15-2 we have
ψ 1 =1 – a sinh x, ψ 2 = x – b cosh x, ψ 0 = –c sinh x.
Furthermore, we see that (to the fourth decimal place)
c 11 =2 + a 2 1 sinh 2 – 1 = 6.4935, c 22 = 2 + b 2 1 sinh 2 + 1 = 2.1896,
2 3 2
–1
c 12 = –4(ae –1 + b sinh 1) = –8e sinh 1 = –3.4586, c 10 = ac 1 sinh 2 – 1 = 1.6800, c 20 = –2ce –1 = –0.6466,
2
and obtain the following system for the coefficients A 1 and A 2 :
6.4935A 1 – 3.4586A 2 = –1.6800,
–3.4586A 1 + 2.1896A 2 = 0.6466.
Hence, we have A 1 = –0.5423 and A 2 = –0.5613. Thus,
2
Y 2 (x)= x – 0.5613x – 0.5423. (11)
Since the kernel
K(x, t) = sinh(x + t) = sinh x cosh t + cosh x sinh t
of Eq. (10) is degenerate, we can readily obtain the exact solution
2
y(x)= x + α sinh x + β cosh x, (12)
6 sinh 1 – 4 cosh 1
α = = –0.6821, β = α 1 sinh 2 – 1 = –0.5548.
2 2
2 – 1 sinh 2
2
On comparing formulas (11) and (12) we conclude that the approximate solution Y 2 (x) is close to the exact solution y(x)if
|x| is small. At the endpoints x = ±1, the discrepancy |y(x) – Y 2 (x)| is rather significant.
•
References for Section 11.16: L. V. Kantorovich and V. I. Krylov (1958), B. P. Demidovich, I. A. Maron, and
E. Z. Shuvalova (1963), M. L. Krasnov, A. I. Kiselev, and G. I. Makarenko (1971).
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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