Handbook of mathematics for engineers and scienteists part 129 docx

16.5.3.10 To different roots of this system, there are different corresponding solutions of the nonlinear integral equation.. The number of solutions of the integral equation is defined

For the case in whichΦ(t, y) is a polynomial in y, i.e.,

Φ(t, y) = p0(t) + p1(t)y + · · · + p n (t)y n, (16.5.3.7)

where p0(t), , p n (t) are, for instance, continuous functions of t on the interval [a, b], system (16.5.3.6) becomes a system of nonlinear algebraic equations for A1, , A m The number of solutions of the integral equation (16.5.3.3) is equal to the number of solu-tions of system (16.5.3.6) Each solution of system (16.5.3.6) generates a solution (16.5.3.5)

of the integral equation

2◦ Consider the Urysohn equation with a degenerate kernel of the special form

y (x) +

 b

a

n

k=1

g k (x)f k t , y(t)

dt = h(x). (16.5.3.8)

Its solution has the form

y (x) = h(x) +

n



k=1

λ k g k (x), (16.5.3.9)

where the constants λ kcan be defined by solving the algebraic (or transcendental) system

of equations

λ m+

 b

a f m



t , h(t) +

n



k=1

λ k g k (t)



dt=0, m=1, , n. (16.5.3.10)

To different roots of this system, there are different corresponding solutions of the nonlinear integral equation It may happen that (real) solutions are absent

A solution of the Urysohn equation with a degenerate kernel in the general form

f x , y(x)

+

 b

a

n

k=1

g k x , y(x)

h k t , y(t)

dt=0 (16.5.3.11) can be represented in the implicit form

f x , y(x)

+

n



k=1

λ k g k x , y(x)

where the parameters λ kare determined from the system of algebraic (or transcendental) equations:

λ k – H k (λ) =0, k=1, , n,

H k (λ) =

 b

a h k t , y(t)

dt,  λ={λ1, , λ n} (16.5.3.13)

Into system (16.5.3.13), we must substitute the function y(x) = y(x, λ), which can be

obtained by solving equation (16.5.3.12)

The number of solutions of the integral equation is defined by the number of solutions obtained from (16.5.3.12) and (16.5.3.13) It can happen that there is no solution

Example 1 Let us solve the integral equation

y (x) = λ

 1

0 xty

3(t) dt (16 5 3 14 )

with parameter λ We write

A=

 1

0

ty3(t) dt. (16 5 3 15 )

In this case, it follows from (16.5.3.14) that

y (x) = λAx. (16 5 3 16 )

On substituting y(x) in the form (16.5.3.16) into relation (16.5.3.15), we obtain

A=

 1

0

tλ3A3t3dt Hence,

For λ >0 , equation (16.5.3.17) has three solutions:

A1= 0 , A2=  5

λ3

1/2

, A3= –  5

λ3

1/2

.

Hence, the integral equation (16.5.3.14) also has three solutions for any λ >0 :

y1(x) ≡ 0 , y2(x) =

 5

λ3

1/2

x, y3(x) = –

 5

λ3

1/2

x.

For λ≤ 0, equation (16.5.3.17) has only the trivial solution y(x)≡ 0

16.5.3-2 Method of integral transforms

1◦ Consider the following nonlinear integral equation with quadratic nonlinearity on a

semiaxis:

μy (x) – λ

 ∞

0

1

t y

x

t



y (t) dt = f (x). (16.5.3.18)

To solve this equation, the Mellin transform can be applied, which, with regard to the convolution theorem (see Section 11.3), leads to a quadratic equation for the transform

ˆy(s) =M{y (x)}:

μ ˆy(s) – λ ˆy2(s) = ˆ f (s).

This implies

ˆy(s) = μ

μ2–4λ ˆ f (s)

The inverse transform y(x) =M–1{ˆy(s)}obtained by means of the Mellin inversion formula (if it exists) is a solution of equation (16.5.3.18) To different signs in the formula for the images (16.5.3.19), there are two corresponding solutions of the original equation

2◦ By applying the Mellin transform, one can solve nonlinear integral equations of the

form

y (x) – λ

 ∞

β y (xt)y(t) dt = f (x). (16.5.3.20) The Mellin transform (see Table 11.3 in Section 11.3) reduces (16.5.3.20) to the following

functional equation for the transform ˆy(s) =M{y (x)}:

ˆy(s) – λ ˆy(s) ˆy(1– s + β) = ˆ f (s). (16.5.3.21)

On replacing s by1– s + β in (16.5.3.21), we obtain the relationship

ˆy(1– s + β) – λ ˆy(s) ˆy(1– s + β) = ˆ f(1– s + β). (16.5.3.22)

On eliminating the quadratic term from (16.5.3.21) and (16.5.3.22), we obtain

ˆy(s) – ˆ f (s) = ˆy(1– s + β) – ˆ f(1– s + β). (16.5.3.23)

We express ˆy(1– s + β) from this relation and substitute it into (16.5.3.21) We arrive at the

quadratic equation

λ ˆy2(s) –

1+ ˆf (s) – ˆ f(1– s + β)

ˆy(s) + ˆ f (s) =0 (16.5.3.24)

On solving (16.5.3.24) for ˆy(s), by means of the Mellin inversion formula we can find a

solution of the original integral equation (16.5.3.20)

16.5.3-3 Method of differentiating for integral equations

1◦ Consider the equation

y (x) +

 b

a |x – t|f t , y(t)

Let us remove the modulus in the integrand:

y (x) +

 x

a (x – t)f t , y(t)

dt+

 b

x (t – x)f t , y(t)

dt = g(x). (16.5.3.26)

Differentiating (16.5.3.26) with respect to x yields

y 

x (x) +

 x

a f t , y(t)

dt–

 b

x f t , y(t)

dt = g  x (x). (16.5.3.27) Differentiating (16.5.3.27), we arrive at a second-order ordinary differential equation for

y = y(x):

y 

xx+2f (x, y) = g  xx (x). (16.5.3.28) Let us derive the boundary conditions for equation (16.5.3.28) We assume that –∞ < a < b < ∞ By setting x = a and x = b in (16.5.3.26), we obtain the relations

y (a) +

 b

a (t – a)f t , y(t)

dt = g(a),

y (b) +

 b

a (b – t)f t , y(t)

dt = g(b).

(16.5.3.29)

Let us solve equation (16.5.3.28) for f (x, y) and substitute the result into (16.5.3.29) Integrating by parts yields the desired boundary conditions for y(x):

y (a) + y(b) + (b – a)

g 

x (b) – y x  (b)

= g(a) + g(b),

y (a) + y(b) + (a – b)

g 

x (a) – y  x (a)



= g(a) + g(b). (16.5.3.30) Let us point out a useful consequence of (16.5.3.30):

y 

x (a) + y x  (b) = g  x (a) + g  x (b),

which can be used together with one of conditions (16.5.3.30)

Equation (16.5.3.28) under the boundary conditions (16.5.3.30) determines the solution

of the original integral equation (there may be several solutions)

2◦ The equations

y (x) +

 b

a e

λ|x–t| f t , y(t)

dt = g(x),

y (x) +

 b

a sinh λ|x – t|f t , y(t)

dt = g(x),

y (x) +

 b

a sin λ|x – t|f t , y(t)

dt = g(x),

can also be reduced to second-order ordinary differential equations by means of the differ-entiation For these equations, see the book by Polyanin and Manzhirov (1998)

16.5.3-4 Successive approximation method

Consider the nonlinear Urysohn integral equation in the canonical form

y (x) =

 b

a K x , t, y(t)

dt, a≤x≤b (16.5.3.31) The iteration process for this equation is constructed by the formula

y k (x) =

 b

a K x , t, y k–1 (t)

dt, k=1,2, (16.5.3.32)

If the functionK(x, t, y) is jointly continuous together with the derivative K 

y (x, t, y) (with respect to the variables x, t, and ρ, a≤x≤b , a≤t≤b, and|y| ≤ρ) and if

 b

a supy |K(x, t, y)|dt≤ρ,

 b

a supy |K 

y (x, t, y)|dt≤β <1, (16.5.3.33)

then for any continuous function y0(x) of the initial approximation from the domain

{|y| ≤ρ , a ≤ x ≤ b}, the successive approximations (16.5.3.32) converge to a

continu-ous solution y ∗ (x), which lies in the same domain and is unique in this domain The rate of

convergence is defined by the inequality

|y ∗ (x) – y

k (x)| ≤ 1β k

– β supx |y1(x) – y0(x)|, a≤x≤b, (16.5.3.34)

which gives an a priori estimate for the error of the kth approximation The a posteriori

estimate (which is, in general, more precise) has the form

|y ∗ (x) – y

k (x)| ≤ 1β

– β supx |y k (x) – y k–1 (x)|, a≤x≤b (16.5.3.35)

A solution of an equation of the form (16.5.3.31) with an additional term f (x) on the

right-hand side can be constructed in a similar manner

Example 2 Let us apply the successive approximation method to solve the equation

y (x) =

 1

0

xty2(t) dt – 125x+ 1 The recurrent formula has the form

y k (x) =

 1

0 xty

2

k–1(t) dt – 125x+ 1 , k= 1 , 2, For the initial approximation we take y0(x) =1 The calculation yields

y1(x) =1 + 0 083x,

y8(x) =1 + 0 27x,

y16(x) = 1 + 0 318x,

y2(x) =1 + 0 14x,

y9(x) =1 + 0 26x,

y17(x) = 1 + 0 321x,

y3(x) =1 + 0 18x,

y10(x) =1 + 0 29x,

y18(x) = 1 + 0 323x,

.

Thus, the approximations tend to the exact solution y(x) =1 +13x We see that the rate of convergence

of the iteration process is fairly small Note that in Paragraph 16.5.3-5, the equation in question is solved by a more efficient method.

16.5.3-5 Newton–Kantorovich method

We consider the Newton–Kantorovich method in connection with the Urysohn equation in the canonical form (16.5.3.31) The iteration process is constructed as follows:

y k (x) = y k–1 (x) + ϕ k–1 (x), k=1,2, , (16.5.3.36)

ϕ k–1 (x) = ε k–1 (x) +

 b

a K  y x , t, y k–1 (t)

ϕ k–1 (t) dt, (16.5.3.37)

ε k–1 (x) =

 b

a K x , t, y k–1 (t)

dt – y k–1 (x). (16.5.3.38)

At each step of the algorithm, a linear integral equation for the correction ϕ k–1 (x) is solved.

Under some conditions, the process (16.5.3.36) has a high rate of convergence; however, it is rather complicated because at each iteration we must obtain the new kernelK 

y x , t, y k–1 (t)

for equations (16.5.3.37)

The algorithm can be simplified by using the equation

ϕ k–1 (x) = ε k–1 (x) +

 b

a K  y x , t, y0(t)

ϕ k–1 (t) dt (16.5.3.39)

instead of (16.5.3.37) If the initial approximation is chosen successfully, then the differ-ence between the integral operators in (16.5.3.37) and (16.5.3.39) is small, and the kernel

in (16.5.3.39) remains the same in the course of the solution

The successive approximation method that consists of the application of formulas

(16.5.3.36), (16.5.3.38), and (16.5.3.39) is called the modified Newton–Kantorovich method.

In principle, its rate of convergence is less than that of the original (unmodified) method; however, this version of the method is less complicated in calculations, and therefore it is frequently preferable

Let the functionK(x, t, y) be jointly continuous together with the derivatives K 

y (x, t, y)

andK 

yy (x, t, y) with respect to the variables x, t, y, where a≤x≤b and a≤t≤b, and let the following conditions hold:

1◦ For the initial approximation y0(x), the resolvent R(x, t) of the linear integral

equa-tion (16.5.3.37) with the kernelK 

y x , t, y0(t)

satisfies the condition

 b

a |R(x, t)|dt≤A<∞, a≤x≤b

2◦ The residual ε0(x) of equation (16.5.3.38) for the approximation y0(x) satisfies the

inequality

|ε0(x)|=

a b K x , t, y0(t)

dt – y0(x)

≤B <∞.

3◦ In the domain|y (x) – y0(x)| ≤ 2(1+ A)B, the following relation holds:

 b

a supy K 

yy (x, t, y)dt≤D<∞.

4◦ The constants A, B, and D satisfy the condition

H= (1+ A)2BD≤ 1

2.

In this case, under assumptions1◦–4◦ , the process (16.5.3.36) converges to a solution y ∗ (x)

of equation (16.5.3.31) in the domain

|y (x) – y0(x)| ≤(1–√

1–2H )H–1(1– A)B, a≤x≤b This solution is unique in the domain

|y (x) – y0(x)| ≤ 2(1+ A)B, a≤x≤b The rate of convergence is determined by the estimate

|y ∗ (x) – y

k (x)| ≤ 21–k(2H)2k 1(1– A)B, a≤x≤b Thus, the above conditions establish the convergence of the algorithm and the existence, the position, and the uniqueness domain of a solution of the nonlinear equation (16.5.3.31)

These conditions impose certain restrictions on the initial approximation y0(x) whose choice

is an important independent problem that has no unified approach As usual, the initial

approximation is determined either by more detailed a priori analysis of the equation under

consideration or by physical reasoning implied by the essence of the problem described

by this equation Under a successful choice of the initial approximation, the Newton– Kantorovich method provides a high rate of convergence of the iteration process to obtain

an approximate solution with given accuracy

Remark. Let the right-hand side of equation (16.5.3.31) contain an additional term f (x) Then such an equation can be represented in the form (16.5.3.31), where the integrand is K x , t, y(t)

+ (b – a)–1f (x).

Example 3 Let us apply the Newton–Kantorovich method to solve the equation

y (x) =

 1

0

xty2(t) dt – 125x+ 1 (16 5 3 40 )

For the initial approximation we take y0(x) =1 According to (16.5.3.38), we find the residual

ε0(x) =

 1

0 xty

2(t) dt – 125x+ 1– y0(x) = x

 1

0 t dt– 125x+ 1 – 1 =121x.

The y-derivative of the kernel K(x, t, y) = xty2(t), which is needed in the calculations, has the form K 

y (x, t, y) =

2xty (t) According to (16.5.3.37), we form the following equation for ϕ0 (x):

ϕ0(x) = 121x+ 2x

 1

0

ty0(t)ϕ0(t) dt,

where the kernel turns out to be degenerate, which makes it possible to obtain the solution ϕ0(x) =14xdirectly Now we define the first approximation to the desired function:

y1(x) = y0(x) + ϕ0 (x) =1 +14x.

We continue the iteration process and obtain

ε1(x) =

 1

0

xt 1 +14t

dt+ 1 –125 x

– 1 +14x

= 641x.

The equation for ϕ1(x) has the form

ϕ1(x) = 641x+ 2x

 1

0 t 1 +14t

dt+ 1 –125x

– 1 +14x

,

and the solution is ϕ1(x) = 403x Hence, y2(x) =1 +14x+403x= 1 + 0 325x The maximal difference between

the exact solution y(x) =1 +13x and the approximate solution y2(x) is observed at x =1 and is less than 0.5%.

This solution is not unique The other solution can be obtained by taking the function y0(x) =1 + 0 8x

for the initial approximation In this case we can repeat the above sequence of approximations and obtain the

following results (the numerical coefficient of x is rounded):

y1(x) =1 + 0 82x, y2(x) = 1 + 1 13x, y3(x) = 1 + 0 98x, .,

and the subsequent approximations tend to the exact solution y(x) =1+ x.

We see that the rate of convergence of the iteration process performed by the Newton–Kantorovich method

is significantly higher than that performed by the method of successive approximations (see Example 2 in Paragraph 16.5.3-4).

To estimate the rate of convergence of the performed iteration process, we can compare the above results with the realization of the modified Newton–Kantorovich method In connection with the latter, for the above versions of the approximations we can obtain

y (x) =1+ k n x; k0 k1 k2 k3 k4 k5 k6 k7 k8 .

0 0 25 0 69 0 60 0 51 0 44 0 38 0 36 0 345 . .

The iteration process converges to the exact solution y(x) =1 +13x.

We see that the modified Newton–Kantorovich method is less efficient than the Newton–Kantorovich method, but more efficient than the method of successive approximations (see Example 2).

16.5.3-6 Quadrature method

To solve an arbitrary nonlinear equation, we can apply the method based on the application

of quadrature formulas The procedure of composing the approximating system of equations

is the same as in the linear case (see Subsection 16.4.11) We consider this procedure for

an example of the Urysohn equation

y (x) –

 b

a K x , t, y(t)

dt = f (x), a≤x≤b (16.5.3.41)

We set x = x i (i =1, , n) Then we obtain

y (x i) –

 b

a K x i , t, y(t)

dt = f (x i) i=1, , n. (16.5.3.42)

solution of the original integral equation (16.5.3.20)

16.5.3-3 Method of differentiating for integral... remains the same in the course of the solution

The successive approximation method that consists of the application of formulas

(16.5.3.36), (16.5.3.38), and (16.5.3.39) is called