The solution of this equation is given by the formula ˆys = fˆs 1– ˆQ s.. 16.4.6.21 Under the application of this analytical method of solution, the following technical difficulties can
Trang 1On applying the Mellin transform to equation (16.4.6.15) and taking into account the fact that the integral with such a kernel is transformed into the product by the rule (see Subsection 11.3.2)
M
∞
0
1
t Q
x
t
y (t) dt
= ˆQ (s) ˆy(s),
we obtain the following equation for the transform ˆy(s):
ˆy(s) – ˆ Q (s) ˆy(s) = ˆ f (s).
The solution of this equation is given by the formula
ˆy(s) = fˆ(s)
1– ˆQ (s). (16.4.6.18)
On applying the Mellin inversion formula to equation (16.4.6.18) we obtain the solution of the original integral equation
y (x) = 1
2πi
c+i∞
c–i∞
ˆ
f (s)
1– ˆQ (s) x
–s ds. (16.4.6.19)
This solution can also be represented via the resolvent in the form
y (x) = f (x) +
∞
0
1
t N
x
t
f (t) dt, (16.4.6.20) where we have used the notation
N (x) =M– 1{Nˆ (s)}, Nˆ(s) = Qˆ(s)
1– ˆQ (s). (16.4.6.21) Under the application of this analytical method of solution, the following technical
difficulties can occur: (a) in the calculation of the transform for a given kernel K(x) and (b) in the calculation of the solution for the known transform ˆy(s) To find the corresponding
integrals, tables of direct and inverse Mellin transforms are applied (e.g., see Sections T3.5 and T3.6) In many cases, the relationship between the Mellin transform and the Fourier and Laplace transforms is first used:
M{f (x), s}=F{f (e x ), is}=L{f (e x ), –s}+L{f (e–x ), s}, (16.4.6.22) and then tables of direct and inverse Fourier transforms and Laplace transforms are applied (see Sections T3.1–T3.4)
16.4.6-3 Equation with the kernel K(x, t) = t β Q (xt) on the semiaxis.
Consider the following equation on the semiaxis:
y (x) –
∞
0 t
β Q (xt)y(t) dt = f (x). (16.4.6.23)
Trang 2To solve this equation, we apply the Mellin transform On multiplying equation (16.4.6.23)
by x s–1and integrating with respect to x from zero to infinity, we obtain
∞
0 y (x)x
s–1dx– ∞
0 y (t)t
β dt ∞
0 Q (xt)x
s–1dx= ∞
0 f (x)x
s–1dx. (16.4.6.24)
Let us make the change of variables z = xt We finally obtain
ˆy(s) – ˆ Q (s)
∞
0 y (t)t
β–s dt= ˆf (s). (16.4.6.25)
Taking into account the relation
∞
0 y (t)t
β–s dt = ˆy(1 + β – s),
we rewrite equation (16.4.6.25) in the form
ˆy(s) – ˆ Q (s) ˆy(1 + β – s) = ˆ f (s). (16.4.6.26)
On replacing s by1+ β – s in equation (16.4.6.26), we obtain
ˆy(1 + β – s) – ˆ Q(1+ β – s) ˆy(s) = ˆ f(1+ β – s). (16.4.6.27)
Let us eliminate ˆy(1 + β – s) from (16.4.6.26) and (16.4.6.27), and then solve the resulting equation for ˆy(s) We thus find the transform of the solution:
ˆy(s) = fˆ(s) + ˆ Q (s) ˆ f(1+ β – s)
1– ˆQ (s) ˆ Q(1+ β – s) .
On applying the Mellin inversion formula, we obtain the solution of the integral equa-tion (16.4.6.23) in the form
y (x) = 1
2πi
c+i∞
c–i∞
ˆ
f (s) + ˆ Q (s) ˆ f(1+ β – s)
1– ˆQ (s) ˆ Q(1+ β – s) x
–s ds.
16.4.7 Method of Approximating a Kernel by a Degenerate One
16.4.7-1 Approximation of the kernel
For the approximate solution of the Fredholm integral equation of the second kind
y (x) –
b
a K (x, t)y(t) dt = f (x), a≤x≤b, (16.4.7.1)
where, for simplicity, the functions f (x) and K(x, t) are assumed to be continuous, it is useful to replace the kernel K(x, t) by a close degenerate kernel
K(n) (x, t) =
n
k=0
g k (x)h k (t). (16.4.7.2)
Trang 3Let us indicate several ways to perform such a change If the kernel K(x, t) is dif-ferentiable with respect to x on [a, b] sufficiently many times, then, for a degenerate kernel K(n) (x, t), we can take a finite segment of the Taylor series:
K(n) (x, t) =
n
m=0
(x – x0)m
(m)
x (x0, t), (16.4.7.3)
where x0[a, b] A similar trick can be applied for the case in which K(x, t) is differentiable with respect to t on [a, b] sufficiently many times.
To construct a degenerate kernel, a finite segment of the double Fourier series can be used:
K(n) (x, t) =
n
p=0
n
q=0
a pq (x – x0)p (t – t0)q, (16.4.7.4) where
a pq = 1
p ! q!
∂ p+q
∂x p ∂t q K (x, t)
x=x0
t=t0
, a≤x0 ≤b, a≤t0≤b
A continuous kernel K(x, t) admits an approximation by a trigonometric polynomial of
period2l , where l = b – a.
For instance, we can set
K(n) (x, t) = 1
2a0(t) +
n
k=1
a k (t) cos
kπx l
, (16.4.7.5)
where the a k (t) (k =0,1,2, ) are the Fourier coefficients
a k (t) = 2
l
b
a K (x, t) cos
pπx l
dx (16.4.7.6)
A similar decomposition can be obtained by interchanging the roles of the variables
x and t A finite segment of the double Fourier series can also be applied by setting, for
instance,
a k (t)≈ 12a k0+
n
m=1
a kmcos
mπt l
, k=0,1, , n, (16.4.7.7) and it follows from formulas (16.4.7.5)–(16.4.7.7) that
K(n) (x, t) = 1
4a00+
1 2
n
k=1
a k0cos
kπx l
+ 1 2
n
m=1
a0mcos
mπt l
+
n
k=1
n
m=1
a kmcos
kπx l
cos
mπt l
,
where
a km= 4
l2
b
a
b
a K (x, t) cos
kπx l
cos
mπt l
One can also use other methods of interpolating and approximating the kernel K(x, t).
Trang 416.4.7-2 Approximate solution.
If K(n) (x, t) is an approximate degenerate kernel for a given exact kernel K(x, t) and if a function f n (x) is close to f (x), then the solution y n (x) of the integral equation
y n (x) –
b
a K(n) (x, t)y n (t) dt = f n (x) (16.4.7.9)
can be regarded as an approximation to the solution y(x) of equation (16.4.7.1).
Assume that the following error estimates hold:
b
a |K (x, t) – K(n) (x, t)| dt≤ε, |f (x) – f n (x)| ≤ δ
Next, let the resolvent R n (x, t) of equation (16.4.7.9) satisfy the relation
b
a |R n (x, t)| dt≤M n for a≤x≤b Finally, assume that the following inequality holds:
q = ε(1 + M n) <1
In this case, equation (16.4.7.1) has a unique solution y(x) and
|y (x) – y n (x)| ≤ ε N(1+ M n)2
1– q + δ, N = maxa x b|f (x)|. (16.4.7.10)
Example Let us find an approximate solution of the equation
y (x) –
1/2
0 e
–x2t2y (t) dt =1 (16 4 7 11 )
Applying the expansion in a double Taylor series, we replace the kernel K(x, t) = e–x2t2with the degenerate kernel
K( 2 )(x, t) =1– x2t2+12x4t4 Hence, instead of equation (16.4.7.11) we obtain
y2(x) =1 +
1/2
0 1– x2t2+12x4t4
y2(t) dt. (16 4 7 12 ) Therefore,
y2(x) =1+ A1+ A2x2+ A3x4, (16 4 7 13 ) where
A1=
1/2
0
y2(x) dx, A2= –
1/2
0
x2y2(x) dx, A3= 1
2
1/2
0
x4y2(x) dx. (16 4 7 14 ) From (16.4.7.13) and (16.4.7.14) we obtain a system of three equations with three unknowns; to the fourth decimal place, the solution is
A1 = 0 9930 , A2 = – 0 0833 , A3 = 0 0007 Hence,
y (x)≈y2(x) =1 9930 – 0 0833x2+ 0 0007x4, 0 ≤x≤ 1
2 (16 4 7 15 )
An error estimate for the approximate solution (16.4.7.15) can be performed by formula (16.4.7.10).
Trang 516.4.8 Collocation Method
16.4.8-1 General remarks
Let us rewrite the Fredholm integral equation of the second kind in the form
ε [y(x)]≡y (x) – λ
b
a K (x, t)y(t) dt – f (x) =0 (16.4.8.1) Let us seek an approximate solution of equation (16.4.8.1) in the special form
Y n (x) = Φ(x, A1, , A n) (16.4.8.2)
with free parameters A1, , A n(undetermined coefficients) On substituting the expres-sion (16.4.8.2) into equation (16.4.8.1), we obtain the residual
ε [Y n (x)] = Y n (x) – λ
b
a K (x, t)Y n (t) dt – f (x). (16.4.8.3)
If y(x) is an exact solution, then, clearly, the residual ε[y(x)] is zero Therefore, one tries
to choose the parameters A1, , A n so that, in a sense, the residual ε[Y n (x)] is as small as possible The residual ε[Y n (x)] can be minimized in several ways Usually, to simplify the calculations, a function Y n (x) linearly depending on the parameters A1, , A nis taken
On finding the parameters A1, , A n, we obtain an approximate solution (16.4.8.2) If
lim
n→∞ Y n (x) = y(x), (16.4.8.4)
then, by taking a sufficiently large number of parameters A1, , A n, we find that the
solution y(x) can be found with an arbitrary prescribed precision.
Now let us go to the description of a concrete method of construction of an approximate
solution Y n (x).
16.4.8-2 Approximate solution
We set
Y n (x) = ϕ0(x) +
n
i=1
A i ϕ i (x), (16.4.8.5)
where ϕ0(x), ϕ1(x), , ϕ n (x) are given functions (coordinate functions) and A1, , A n are indeterminate coefficients, and assume that the functions ϕ i (x) (i =1, , n) are linearly
independent Note that, in particular, we can take ϕ0(x) = f (x) or ϕ0(x)≡ 0 On substituting the expression (16.4.8.5) into the left-hand side of equation (16.4.8.1), we obtain the residual
ε [Y n (x)] = ϕ0(x) +
n
i=1
A i ϕ i (x) – f (x) – λ
b
a K (x, t)
ϕ0(t) +
n
i=1
A i ϕ i (t)
dt, or
ε [Y n (x)] = ψ0(x, λ) +
n
i=1
A i ψ i (x, λ), (16.4.8.6)
Trang 6ψ0(x, λ) = ϕ0(x) – f (x) – λ
b
a K (x, t)ϕ0(t) dt,
ψ i (x, λ) = ϕ i (x) – λ
b
a K (x, t)ϕ i (t) dt, i=1, , n
(16.4.8.7)
According to the collocation method, we require that the residual ε[Y n (x)] be zero at the given system of the collocation points x1, , x n on the interval [a, b], i.e., we set
ε [Y n (x j)] =0, j=1, , n, where
a≤x1< x2<· · · < x n–1< x n≤b
It is common practice to set x1= a and x n = b.
This, together with formula (16.4.8.6), implies the linear algebraic system
n
i=1
A i ψ i (x j , λ) = –ψ0(x j , λ), j =1, , n, (16.4.8.8)
for the coefficients A1, , A n If the determinant of system (16.4.8.8) is nonzero,
det[ψ i (x j , λ)] ≠ 0, then system (16.4.8.8) uniquely determines the numbers A1, , A n,
and hence makes it possible to find the approximate solution Y n (x) by formula (16.4.8.5).
16.4.8-3 Eigenfunctions of the equation
On equating the determinant with zero, we obtain the relation
det[ψ i (x j , λ)] =0, which enables us to find approximate values 2λ k (k =1, , n) for the characteristic values
of the kernel K(x, t).
If we set
f (x)≡ 0, ϕ0(x)≡ 0, λ= 2λ k,
then, instead of system (16.4.8.8), we obtain the homogeneous system
n
i=1
2
A(k)
i ψ i (x j, 2λ k) =0, j =1, , n (16.4.8.9)
On finding nonzero solutions 2A(k)
i (i =1, , n) of system (16.4.8.9), we obtain
approx-imate eigenfunctions for the kernel K(x, t):
2
Y(k)
n (x) =
n
i=1
2
A(k)
i ϕ i (x), that correspond to its characteristic value λ k≈2λ k
Trang 7Example Let us solve the equation
y (x) –
1
0
t2y (t)
x2+ t2 dt = x arctan
1
by the collocation method.
We set
Y2(x) = A1+ A2x.
On substituting this expression into equation (16.4.8.10), we obtain the residual
ε [Y2(x)] = –A1xarctan 1
x + A2
x– 1
2+
x2
2 ln
1 + 1
x2
– x arctan1
x.
On choosing the collocation points x1 = 0and x2 = 1 and taking into account the relations
lim
x→0 xarctan 1
x = 0 , lim
x→0 x
2 ln
1 + 1
x2
= 0 ,
we obtain the following system for the coefficients A1and A2 :
0 ×A1–12A2= 0 , –π4A1+12( 1 + ln 2)A2= π4.
This implies A2 = 0and A1 = – 1 Thus,
We can readily verify that the approximate solution (16.4.8.11) thus obtained is exact.
16.4.9 Method of Least Squares
16.4.9-1 Description of the method
By analogy with the collocation method, for the equation
ε [y(x)]≡y (x) – λ
b
a K (x, t)y(t) dt – f (x) =0 (16.4.9.1)
we set
Y n (x) = ϕ0(x) +
n
i=1
A i ϕ i (x), (16.4.9.2)
where ϕ0(x), ϕ1(x), , ϕ n (x) are given functions, A1, , A nare indeterminate
coeffi-cients, and ϕ i (x) (i =1, , n) are linearly independent
On substituting (16.4.9.2) into the left-hand side of equation (16.4.9.1), we obtain the residual
ε [Y n (x)] = ψ0(x, λ) +
n
i=1
A i ψ i (x, λ), (16.4.9.3)
where ψ0(x, λ) and ψ i (x, λ) (i =1, , n) are defined by formulas (16.4.8.7)
According to the method of least squares, the coefficients A i (i =1, , n) can be found from the condition for the minimum of the integral
I =
b
a {ε [Y n (x)]}2dx= b
a
ψ0(x, λ) +
n
i=1
A i ψ i (x, λ)
2
dx (16.4.9.4)