Sometimes, differentiation possibly multiple of a nonlinear integral equation with subse-quent elimination of the integral terms by means of the original equation makes it possible to re
Trang 1Example 1 Consider the integral equation
x
0
y(x – t)y(t) dt = Ax m, m> – 1 Applying the Laplace transform to the equation under consideration with regard to the relation L {x m} =
Γ(m +1)p–m–1, we obtain
2y2(p) = A Γ(m +1)p–m–1, whereΓ(m) is the gamma function On extracting the square root of both sides of the equation, we obtain
2y(p) = A Γ(m +1)p– m+21 Applying the Laplace inversion formula, we obtain two solutions to the original integral equation:
y1(x) = –
√
A Γ(m +1 )
Γm2+1 x
m–1
2 , y2(x) =
√
A Γ(m +1 )
Γm2+1 x
m–1
2
16.5.2-2 Method of differentiation for integral equations
Sometimes, differentiation (possibly multiple) of a nonlinear integral equation with subse-quent elimination of the integral terms by means of the original equation makes it possible
to reduce this equation to a nonlinear ordinary differential equation Below we briefly list some equations of this type
1◦ The equation
y (x) +
x
a f t , y(t)
can be reduced by differentiation to the nonlinear first-order equation
y
x + f (x, y) – g x (x) =0
with the initial condition y(a) = g(a).
2◦ The equation
y (x) +
x
a (x – t)f t , y(t)
can be reduced by double differentiation (with the subsequent elimination of the integral term by using the original equation) to the nonlinear second-order equation:
y
xx + f (x, y) – g xx (x) =0 (16.5.2.5)
The initial conditions for the function y = y(x) have the form
y (a) = g(a), y
x (a) = g x (a). (16.5.2.6)
3◦ The equation
y (x) +
x
a e
λ(x–t) f t , y(t)
can be reduced by differentiation to the nonlinear first-order equation
y
x + f (x, y) – λy + λg(x) – g x (x) =0 (16.5.2.8)
The desired function y = y(x) must satisfy the initial condition y(a) = g(a).
Trang 24◦ Equations of the form
y (x) +
x
a cosh
λ (x – t)
f t , y(t)
dt = g(x), (16.5.2.9)
y (x) +
x
a sinh
λ (x – t)
f t , y(t)
dt = g(x), (16.5.2.10)
y (x) +
x
a cos
λ (x – t)
f t , y(t)
dt = g(x), (16.5.2.11)
y (x) +
x
a sin
λ (x – t)
f t , y(t)
dt = g(x) (16.5.2.12)
can also be reduced to second-order ordinary differential equations by double differentiation For these equations, see the book by Polyanin and Manzhirov (1998)
16.5.2-3 Successive approximation method
In many cases, the successive approximation method can be applied successfully to solve various types of integral equations The principles of constructing the iteration process are the same as in the case of linear equations For Volterra equations of the form
y (x) –
x
a K x , t, y(t)
dt = f (x), a≤x≤b, (16.5.2.13) the corresponding recurrent expression has the form
y k+1(x) = f (x) +
x
a K x , t, y k (t)
dt, k=0,1,2, (16.5.2.14)
It is customary to take the initial approximation either in the form y0(x)≡ 0or in the form
y0(x) = f (x).
In contrast to the case of linear equations, the successive approximation method has
a smaller domain of convergence Let us present the convergence conditions for the iteration process (16.5.2.14) that are simultaneously the existence conditions for a solution
of equation (16.5.2.13) To be definite, we assume that y0(x) = f (x).
If for any z1and z2we have the relation
|K (x, t, z1) – K(x, t, z2)| ≤ϕ (x, t)|z1– z2|
a x K x , t, f (t)
dt
≤ψ (x)
a ψ
2(t) dt≤N2, b
a
x
a ϕ
2(x, t) dt dx≤M2,
for some constants N and M , then the successive approximations converge to a unique
solution of equation (16.5.2.13) almost everywhere absolutely and uniformly
Trang 3Example 2 Let us apply the successive approximation method to solve the equation
y(x) =
x
0
1+ y2(t)
1+ t2 dt.
If y0(x)≡ 0 , then
y1(x) =
x
0
dt
1+ t2 = arctan x,
y2(x) =
x
0
1 + arctan2t
1+ t2 dt = arctan x +
1
3 arctan3x,
y3(x) =
x
0
1+ arctan t + 13arctan3t
1+ t2 dt = arctan x +
1
3 arctan3x+3⋅52 arctan5x+7⋅91 arctan7x.
On continuing this process, we can observe that y k (x) → tan(arctan x) = x as k → ∞, i.e., y(x) = x The
substitution of this result into the original equation shows the validity of the result.
16.5.2-4 Newton–Kantorovich method
A merit of the iteration methods when applied to Volterra linear equations of the second kind
is their unconditional convergence under weak restrictions on the kernel and the right-hand side When solving nonlinear equations, the applicability domain of the method of simple iterations is smaller, and if the process is still convergent, then, in many cases, the rate of convergence can be very low An effective method that makes it possible to overcome the indicated complications is the Newton–Kantorovich method
Let us apply the Newton–Kantorovich method to solve a nonlinear Volterra equation of the form
y (x) = f (x) +
x
a K x , t, y(t)
dt (16.5.2.15)
We obtain the following iteration process:
y k (x) = y k–1(x) + ϕ k–1(x), k=1,2, , (16.5.2.16)
ϕ k–1(x) = ε k–1(x) +
x
y x , t, y k–1(t)
ϕ k–1(t) dt, (16.5.2.17)
ε k–1(x) = f (x) +
x
a K x , t, y k–1(t)
dt – y k–1(x). (16.5.2.18)
The algorithm is based on the solution of the linear integral equation (16.5.2.17) for
the correction ϕ k–1(x) with the kernel and right-hand side that vary from step to step.
This process has a high rate of convergence, but it is rather complicated because we must solve a new equation at each step of iteration To simplify the problem, we can replace equation (16.5.2.17) with the equation
ϕ k–1(x) = ε k–1(x) +
x
y x , t, y0(t)
ϕ k–1(t) dt (16.5.2.19)
or with the equation
ϕ k–1(x) = ε k–1(x) +
x
y x , t, y m (t)
ϕ k–1(t) dt, (16.5.2.20)
whose kernels do not vary In equation (16.5.2.20), m is fixed and satisfies the condition
m < k –1
Trang 4It is reasonable to apply equation (16.5.2.19) with an appropriately chosen initial
ap-proximation Otherwise we can stop at some mth approximation and, beginning with this
approximation, apply the simplified equation (16.5.2.20) The iteration process thus ob-tained is the modified Newton–Kantorovich method In principle, it converges somewhat slower than the original process (16.5.2.16)–(16.5.2.18); however, it is not so cumbersome
in the calculations
Example 3 Let us apply the Newton–Kantorovich method to solve the equation
y(x) =
x
0
[ty2(t) –1] dt.
The derivative of the integrand with respect to y has the form
K y t, y(t)
= 2ty(t).
For the zero approximation we take y0(x)≡ 0 According to (16.5.2.17) and (16.5.2.18) we obtain ϕ0(x) = –x and y1 (x) = –x Furthermore, y2 (x) = y1 (x) + ϕ1(x) By (16.5.2.18) we have
ε1(x) =
x
0
[t(–t)2– 1] dt + x = 14x4 The equation for the correction has the form
ϕ1(x) = –2
x
0
t2ϕ1(t) dt + 1
4x4
and can be solved by any of the known methods for Volterra linear equations of the second kind In the case under consideration, we apply the successive approximation method, which leads to the following results (the number of the step is indicated in the superscript):
ϕ(10)= 14x4,
ϕ(11)= 14x4– 2
x
0
1
4t6dt= 14x4–141x7,
ϕ(12)= 14x4– 2
x
0
t2 14t4–141 x7
dt= 14x4– 141x7+701 x10.
We restrict ourselves to the second approximation and obtain
y2(x) = –x + 14x4–141x7+ 701x10
and then pass to the third iteration step of the Newton–Kantorovich method:
y3(x) = y2(x) + ϕ2(x),
ε2(x) = 1601 x10–18201 x13– 78401 x16+93401 x19+1078001 x22,
ϕ2(x) = ε2(x) +2
x
0
t –t + 14t4– 141t7+ 701t10
ϕ2(t) dt.
When solving the last equation, we restrict ourselves to the zero approximation and obtain
y3(x) = –x + 14x4–141x7+11223x10– 18201 x13–78401 x16+93401 x19+1078001 x22.
The application of the successive approximation method to the original equation leads to the same result at the fourth step.
As usual, in the numerical solution the integral is replaced by a quadrature formula The main difficulty of the implementation of the method in this case is in evaluating the derivative of the kernel The problem can be simplified if the kernel is given as an analytic expression that can be differentiated in the analytic form However, if the kernel is given
by a table, then the evaluation must be performed numerically
Trang 516.5.2-5 Collocation method.
When applied to the solution of a nonlinear Volterra equation
x
a K x , t, y(t)
dt = f (x), a≤x≤b, (16.5.2.21)
the collocation method is as follows The interval [a, b] is divided into N parts on each of
which the desired solution can be presented by a function of a certain form
2y(x) = Φ(x, A1, , A m), (16.5.2.22)
involving free parameters A i , i =1, , m.
On the (k +1)st part x k≤x≤x k+1, where k =0,1, , N –1, the solution can be written
x k
K x , t, 2y(t) dt = f (x) –Ψk (x), (16.5.2.23) where the integral
Ψk (x) =
x k
a K x , t, 2y(t) dt (16.5.2.24) can always be calculated for the approximate solution2y(x), which is known on the interval
a≤x≤x k and was previously obtained for k –1parts The initial value y(a) of the desired
solution can be found by an auxiliary method or is assumed to be given
To solve equation (16.5.2.23), representation (16.5.2.22) is applied, and the free
param-eters A i (i =1, , m) can be defined from the condition that the residuals vanish:
ε (A i , x k,j) =
x k,j
x k
K x k,j , t, Φ(t, A1, , A m)
dt – f (x k,j) –Ψk (x k,j), (16.5.2.25)
where x k,j (j = 1, , m) are the nodes that correspond to the partition of the interval [x k , x k+1] into m parts (subintervals) System (16.5.2.25) is a system of m equations for
A1, , A m.
For convenience of the calculations, it is reasonable to present the desired solution on any part as a polynomial
2y(x) =
m
i=1
A i ϕ i (x), (16.5.2.26)
where ϕ i (x) are linearly independent coordinate functions For the functions ϕ i (x), power and trigonometric polynomials are frequently used; for instance, ϕ i (x) = x i–1
In applications, the concrete form of the functions ϕ i (x) in formula (16.5.2.26), as well
as the form of the functionsΦ in (16.5.2.20), can sometimes be given on the basis of physical
reasoning or defined by the structure of the solution of a simpler model equation
16.5.2-6 Quadrature method
To solve a nonlinear Volterra equation, we can apply the method based on the use of quadrature formulas The procedure of constructing the approximate system of equations
is the same as in the linear case (see Subsection 16.2.7)
Trang 61◦ We consider the nonlinear Volterra equation of the form
y (x) –
x
a K x , t, y(t)
on an interval a≤x≤b Assume that K x , t, y(t)
and f (x) are continuous functions From equation (16.5.2.27) we find that y(a) = f (a) Let us choose a constant integration step h and consider the discrete set of points x i = a + h(i –1), where i =1, , n For
x = x i, equation (16.5.2.27) becomes
y (x i) –
x i
a K x i , t, y(t)
dt = f (x i) (16.5.2.28)
Applying the quadrature formula (see Subsection 16.1.5) to the integral in (16.5.2.28),
choosing x j (j =1, , i) to be the nodes in t, and neglecting the truncation error, we arrive
at the following system of nonlinear algebraic (or transcendental) equations:
y1= f1, y i–
i
j=1
A ij K ij (y j ) = f i i=2, , n, (16.5.2.29)
where A ij are the coefficients of the quadrature formula on the interval [a, x i ]; y i are
the approximate values of the solution y(x) at the nodes x i ; f i = f (x i ); and K ij (y j) =
K (x i , t j , y j)
Relations (16.5.2.29) can be rewritten as a sequence of recurrent nonlinear equations,
y1= f1, y i – A ii K ii (y i ) = f i+
i–1
j=1
A ij K ij (y j), i=2, , n, (16.5.2.30)
for the approximate values of the desired solution at the nodes
2◦ When applied to the Volterra equation of the second kind in the Hammerstein form
y (x) –
x
a Q (x, t)Φ t , y(t)
dt = f (x), (16.5.2.31)
the main relations of the quadrature method have the form (x1 = a)
y1= f1, y i–
i
j=1
A ij Q ijΦj (y j ) = f i i=2, , n, (16.5.2.32)
where Q ij = Q(x i , t j) and Φj (y j) = Φ(t j , y j) These relations lead to the sequence of nonlinear recurrent equations
y1 = f1, y i – A ii Q iiΦi (y i ) = f i+
i–1
j=1
A ij Q ijΦj (y j), i=2, , n, (16.5.2.33)
whose solutions give approximate values of the desired function
Trang 7Example 4 In the solution of the equation
y(x) –
x
0 e
–(x–t)y2(t) dt = e–x, 0 ≤x≤ 0 1 ,
where Q(x, t) = e–(x–t), Φ t, y(t)
= y2(t), and f (x) = e–x, the approximate expression has the form
y(x i) –
x i
0 e
–(xi–t) 2(t) dt = e–xi.
On applying the trapezoidal rule to evaluate the integral (with step h =0 02 ) and finding the solution at the nodes
x i= 0 , 0.02, 0.04, 0.06, 0.08, 0.1, we obtain, according to (16.5.2.33), the following system of computational relations:
y1= f1, y i– 0 01Q ii y2= f i+
i–1
j=1
0 02Q ij y2j, i= 2, ,6
Thus, to find an approximate solution, we must solve a quadratic equation for each value y i, which makes it possible to write out the answer
y i= 50 50
1 – 0 04f i+
i–1
j=1
0 02Q ij y j2
1/2 , i= 2, ,6
16.5.3 Equations with Constant Integration Limits
16.5.3-1 Nonlinear equations with degenerate kernels
1◦ Consider a Urysohn equation of the form
y (x) =
b
a Q (x, t)Φ t , y(t)
dt, (16.5.3.1)
where Q(x, t) and Φ(t, y) are given functions and y(x) is the unknown function.
Let the kernel Q(x, t) be degenerate, i.e.,
Q (x, t) =
m
k=1
g k (x)h k (t). (16.5.3.2)
In this case equation (16.5.3.1) becomes
y (x) =
m
k=1
g k (x)
b
a h k (t)Φ t , y(t)
dt (16.5.3.3)
We write
A k=
b
a h k (t)Φ t , y(t)
dt, k=1, , m, (16.5.3.4)
where the constants A kare yet unknown Then it follows from (16.5.3.3) that
y (x) =
m
k=1
A k g k (x). (16.5.3.5)
On substituting the expression (16.5.3.5) for y(x) into relations (16.5.3.4), we obtain (in the general case) m transcendental equations of the form
A k =Ψk (A1, , A m), k=1, , m, (16.5.3.6)
which contain m unknown numbers A1, , A m
... solve various types of integral equations The principles of constructing the iteration process are the same as in the case of linear equations For Volterra equations of the formy (x)... the concrete form of the functions ϕ i (x) in formula (16.5.2.26), as well
as the form of the functionsΦ in (16.5.2.20), can sometimes be given on the basis of physical
reasoning... dx≤M2,
for some constants N and M , then the successive approximations converge to a unique
solution of equation (16.5.2.13) almost everywhere absolutely and uniformly