HANDBOOK OFINTEGRAL EQUATIONS phần 5 ppsx

This is a special case of equation 4.9.1 withgx = cosh kβx and ht = cosmµt.. This is a special case of equation 4.9.1 withgx = cos mµx and ht = coshkβt.. This is a special case of equati

k a

whereC1andC2are arbitrary constants

Forλ(2A + λ) = k2 > 0, the general solution of equation (1) is given by

y(x) = C1cos(kx) + C2sin(kx) + f (x) – 2Aλ

k

 x a

sin[k(x – t)] f (t) dt (4)

Forλ = 2A, the general solution of equation (1) is given by

y(x) = C1+C2x + f (x) + 4A2

 x a

This is a special case of equation 4.9.39 withg(t) = At The solution of the integral equation

can be written via the Bessel functions (or modified Bessel functions) of order 1/3

sin(λk|x – t|)y(t) dt =

 x a

sin[λk(x – t)]y(t) dt +

 b x

sin[λk(x – t)]y(t) dt – λ2k

 b x

sin[λk(t – x)]y(t) dt,

(2)

where the primes denote the derivatives with respect tox By comparing formulas (1) and (2),

we find the relation betweenI k andIk:

Differentiating (6) with respect tox twice and eliminating In–1from the resulting equation

with the aid of (6), we obtain a similar equation whose left-hand side is a second-order linear

differential operator (acting ony) with constant coefficients plus the sum

n–2

k=1

B k I k If we

successively eliminateI n–2,I n–3, , with the aid of double differentiation, then we finally

arrive at a linear nonhomogeneous ordinary differential equation of order 2n with constant

coefficients

3◦ The boundary conditions fory(x) can be found by setting x = a in the integral equation

and all its derivatives (Alternatively, these conditions can be found by settingx = a and x = b

in the integral equation and all its derivatives obtained by means of double differentiation.)

sin(x – t)

x – t f (t) dt, λ≠

2

π.

• Reference: F D Gakhov and Yu I Cherskii (1978).

4.5-3 Kernels Containing Tangent

This is a special case of equation 4.9.1 withg(x) = tan(βx) and h(t) = 1

This is a special case of equation 4.9.1 withg(x) = 1

tan(βx) andh(t) = tan(βt).

 b

a

tank(βx) tan m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = tan k(βx) and h(t) = tanm(µt)

This is a special case of equation 4.9.10 withh(x) = tan(βx).

4.5-4 Kernels Containing Cotangent

This is a special case of equation 4.9.1 withg(x) = cot(βx) and h(t) = 1

This is a special case of equation 4.9.1 withg(x) = 1

cot(βx) andh(t) = cot(βt).

 b

a

cotk(βx) cot m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = cot k(βx) and h(t) = cotm(µt)

This is a special case of equation 4.9.10 withh(x) = cot(βx).

4.5-5 Kernels Containing Combinations of Trigonometric Functions

 b

a

cosk(βx) sin m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = cos k(βx) and h(t) = sinm(µt)

 b

a

[A sin(αx) cos(βt) + B sin(γx) cos(δt)]y(t) dt = f (x).

This is a special case of equation 4.9.18 withg1(x) = sin(αx), h1(t) = A cos(βt), g2(x) = sin(γx),

andh2(t) = B cos(δt)

 b

a

tank(γx) cot m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = tan k(γx) and h(t) = cotm(µt)

 b

a

[A tan(αx) cot(βt) + B tan(γx) cot(δt)]y(t) dt = f (x).

This is a special case of equation 4.9.18 withg1(x) = tan(αx), h1(t) = A cot(βt), g2(x) = tan(γx),

andh2(t) = B cot(δt).

2π 0 2

Here the integral is understood in the sense of the Cauchy principal value Without loss of

generality we may assume thatA2+B2 = 1

Solution:

y(x) = Af (x) + B

2π

 2π 0cot

t – x2



f (t) dt + B

22πA

 2π 0

This is a special case of equation 4.9.1 withg(x) = 1

arccos(βx) andh(t) = arccos(βt).

 b

a

arccosk(βx) arccos m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = arccos k(βx) and h(t) = arccosm(µt)

This is a special case of equation 4.9.10 withh(x) = arccos(βx).

 b

a

[A + B(x – t) arccos(βt)]y(t) dt = f (x).

This is a special case of equation 4.9.8 withh(t) = arccos(βt).

4.6-2 Kernels Containing Arcsine

This is a special case of equation 4.9.1 withg(x) = 1

arcsin(βx) andh(t) = arcsin (βt).

 b

a

arcsink(βx) arcsin m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = arcsin k(βx) and h(t) = arcsinm(µt)

This is a special case of equation 4.9.1 withg(x) = 1

arctan(βx) andh(t) = arctan(βt).

 b

a

arctank(βx) arctan m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = arctan k(βx) and h(t) = arctanm(µt)

This is a special case of equation 4.9.1 withg(x) = 1

arccot(βx) andh(t) = arccot(βt).

 b

a

arccotk(βx) arccot m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = arccot k(βx) and h(t) = arccotm(µt)

4.7-1 Kernels Containing Exponential and Hyperbolic Functions

 b

a

e µ(x–t)cosh[β(x – t)]y(t) dt = f (x).

This is a special case of equation 4.9.18 withg1(x) = eµxcosh(βx), h1(t) = e –µtcosh(βt),

g2(x) = eµxsinh(βx), and h2(t) = –e –µtsinh(βt)

 b

a

e µ(x–t)sinh[β(x – t)]y(t) dt = f (x).

This is a special case of equation 4.9.18 withg1(x) = eµxsinh(βx), h1(t) = e–µtcosh(βt),

g2(x) = eµxcosh(βx), and h2(t) = –e–µtsinh(βt)

 b

a

te µ(x–t)sinh[β(x – t)]y(t) dt = f (x).

This is a special case of equation 4.9.18 withg1(x) = eµxsinh(βx), h1(t) = te–µtcosh(βt),

g2(x) = eµxcosh(βx), and h2(t) = –te–µtsinh(βt)

4.7-2 Kernels Containing Exponential and Logarithmic Functions

Solution witha > 0, b > 0, and x > 0:

y(x) = f (x) + a

2–b2

2b

 ∞0

1

t exp

–blnx t



f (t) dt.

• Reference: F D Gakhov and Yu I Cherskii (1978).

4.7-3 Kernels Containing Exponential and Trigonometric Functions

This is a special case of equation 4.9.18 with g1(x) = e µxcos(βx), h1(t) = e–µtcos(βt),

g2(x) = e µxsin(βx), and h2(t) = e–µtsin(βt)

e µ(x–t)sin(xt)f (t) dt, λ ≠ ±2/π

This is a special case of equation 4.9.18 with g1(x) = e sin(βx), h1(t) = e cos(βt),

g2(x) = e µxcos(βx), and h2(t) = –e–µtsin(βt)

This is a special case of equation 4.9.18 with g1(x) = e µxsin(βx), h1(t) = te–µtcos(βt),

g2(x) = e µxcos(βx), and h2(t) = –te–µtsin(βt)

 b

a

xe µ(x–t)sin[β(x – t)]y(t) dt = f (x).

This is a special case of equation 4.9.18 with g1(x) = xeµxsin(βx), h1(t) = e–µtcos(βt),

g2(x) = xeµxcos(βx), and h2(t) = –e–µtsin(βt)

This is a special case of equation 4.9.18 withg1(x) = eµxtan(βx), h1(t) = e –µt,g2(x) = eµx,

andh2(t) = –e –µttan(βt)

This is a special case of equation 4.9.1 withg(x) = e µxandh(t) = cot(βt).

4.7-4 Kernels Containing Hyperbolic and Logarithmic Functions

 b

coshk(βx) ln m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = ln m(µx) and h(t) = coshk(βt).

This is a special case of equation 4.9.1 withg(x) = ln m(µx) and h(t) = cothk(βt)

4.7-5 Kernels Containing Hyperbolic and Trigonometric Functions

 b

a

coshk(βx) cos m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = cosh k(βx) and h(t) = cosm(µt)

 b

a

coshk(βt) cos m(µx)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = cos m(µx) and h(t) = coshk(βt)

 b

a

coshk(βx) sin m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = cosh k(βx) and h(t) = sinm(µt)

This is a special case of equation 4.9.1 withg(x) = sin m(µx) and h(t) = coshk(βt).

 b

a

sinhk(βx) cos m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = sinh k(βx) and h(t) = cosm(µt)

 b

a

sinhk(βt) cos m(µx)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = cos m(µx) and h(t) = sinhk(βt)

 b

a

sinhk(βx) sin m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = sinh k(βx) and h(t) = sinm(µt)

 b

a

sinhk(βt) sin m(µx)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = sin m(µx) and h(t) = sinhk(βt)

 b

a

tanhk(βx) cos m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = tanh k(βx) and h(t) = cosm(µt)

 b

a

tanhk(βt) cos m(µx)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = cos m(µx) and h(t) = tanhk(βt)

 b

a

tanhk(βx) sin m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = tanh k(βx) and h(t) = sinm(µt)

 b

a

tanhk(βt) sin m(µx)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = sin m(µx) and h(t) = tanhk(βt)

4.7-6 Kernels Containing Logarithmic and Trigonometric Functions

 b

a

cosk(βx) ln m(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 withg(x) = ln (µx) and h(t) = cosk(βt).

This is a special case of equation 4.9.1 withg(x) = ln m(µx) and h(t) = cotk(βt)

4.8 Equations Whose Kernels Contain Special

tJ ν(xt)f (t) dt, λ ≠ ±1.

By settingx = 2z ,t = 2τ ,y(x) = Y (z), and f (x) = F (z), we arrive at an equation of the

form 4.8.3:

Y (z) + λ

 ∞0

[AYµ(αx) + BYν(βt)]y(t) dt = f (x).

This is a special case of equation 4.9.5 withg(x) = AYµ(αx) and h(t) = BYν(βt).

 b

a

[AYµ(x)Yµ(t) + BYν(x)Yν(t)]y(t) dt = f (x).

This is a special case of equation 4.9.14 withg(x) = Yµ(x) and h(t) = Yν(t)

 b

[AY µ(x)Y ν(t) + BY ν(x)Y µ(t)]y(t) dt = f (x).

[AIµ(αx) + BIν(βt)]y(t) dt = f (x).

This is a special case of equation 4.9.5 withg(x) = AI µ(αx) and h(t) = BIν(βt)

 b

a

[AIµ(x)Iµ(t) + BIν(x)Iν(t)]y(t) dt = f (x).

This is a special case of equation 4.9.14 withg(x) = I µ(x) and h(t) = Iν(t)

 b

a

[AIµ(x)Iν(t) + BIν(x)Iµ(t)]y(t) dt = f (x).

This is a special case of equation 4.9.17 withg(x) = I µ(x) and h(t) = I ν(t)

This is a special case of equation 4.9.5 withg(x) = AKµ(αx) and h(t) = BKν(βt).

This is a special case of equation 4.9.17 withg(x) = Kµ(x) and h(t) = Kν(t).

4.9 Equations Whose Kernels Contain Arbitrary

1◦ Assume thatλ≠ b

a g(t)h(t) dt

–1.Solution:

y(x) = f (x) + λkg(x), where k = 1 –λ

 b a g(t)h(t) dt

–1 b a h(t)f (t) dt.

2◦ Assume thatλ =

 b

a g(t)h(t) dt

–1.For

The limits of integration may take the valuesa = – ∞ and/or b = ∞, provided that the

corresponding improper integral converges

where the constantsA1andA2are given by

A1= f1–λ[f1g1– (b – a)f2]

[g2– (b – a)g2]λ2– 2g1λ + 1, A2 =

f2–λ(f2g1–f1g2)[g2– (b – a)g2]λ2– 2g1λ + 1,

2◦ Solution withλ = λ1≠ λ2andf1=f2= 0:

y(x) = f (x) + Cy1(x), y1(x) = g(x) +



g2

b – a,

whereC is an arbitrary constant and y1(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ1.

3◦ Solution withλ = λ2 ≠ λ1andf1=f2= 0:

y(x) = f (x) + Cy2(x), y2(x) = g(x) –



g2

b – a,

whereC is an arbitrary constant and y2(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ2

4◦ The equation has no multiple characteristic values

g1=

 b a g(x) dx, g2=

 b a

g2(x) dx

1◦ Solution withλ ≠ λ1,2:

y(x) = f (x) + λ[A1g(x) + A2],where the constantsA1andA2are given by

A1= f1+λ[f1g1– (b – a)f2][(b – a)g2–g2]λ2+ 1 , A2=

– 2+λ(f2g1–f1g2)[(b – a)g2–g2]λ2+ 1,

f1=

 b a

f (x) dx, f2 =

 b a

f (x)g(x) dx.

2◦ Solution withλ = λ1≠ λ2andf1=f2= 0:

y(x) = f (x) + Cy1(x), y1(x) = g(x) + 1 –λ1g1

λ1(b – a),whereC is an arbitrary constant and y1(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ1.

3◦ The solution withλ = λ2 ≠ λ1 andf1 =f2 = 0 is given by the formulas of item 2◦ in

which one must replaceλ1andy1(x) by λ2andy2(x), respectively

4◦ The equation has no multiple characteristic values

The characteristic values of the equation:

λ1,2= (A + B)g1±(A – B)2g2+ 4AB(b – a)g2

2AB[g2– (b – a)g2] ,where

g1=

 b a g(x) dx, g2=

 b a

g2(x) dx

1◦ Solution withλ ≠ λ1,2:

y(x) = f (x) + λ[A1g(x) + A2],where the constantsA1andA2are given by

whereC is an arbitrary constant and y1(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ1.

3◦ The solution withλ = λ2 ≠ λ1 andf1 =f2 = 0 is given by the formulas of item 2◦ in

which one must replaceλ1andy1(x) by λ2andy2(x), respectively

4◦ Solution withλ = λ1,2=λ ∗andf1=f2= 0, where the characteristic valueλ ∗= 2

(A + B)g1

is double:

y(x) = f (x) + Cy ∗(x), y ∗(x) = g(x) – (A – B)g1

2A(b – a).HereC is an arbitrary constant and y ∗(x) is an eigenfunction of the equation corresponding

where the constantsA1andA2are given by

A1= f1–λ[f1s3– (b – a)f2]

[s1s3– (b – a)s2]λ 2– (s1+s3)λ + 1, A2 =

f2–λ(f2s1–f1s2)[s1s3– (b – a)s2]λ 2– (s1+s3)λ + 1,

2◦ Solution withλ = λ1≠ λ2andf1=f2= 0:

y(x) = f (x) + Cy1(x), y1(x) = g(x) + 1 –λ1s1

λ1(b – a),

whereC is an arbitrary constant and y1(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ1.

3◦ The solution withλ = λ2 ≠ λ1 andf1 =f2 = 0 is given by the formulas of item 2◦ in

which one must replaceλ1andy1(x) by λ2andy2(x), respectively

4◦ Solution withλ = λ1,2=λ ∗andf1=f2= 0, where the characteristic valueλ ∗= 2

s1+s3

is double:

y(x) = f (x) + Cy ∗(x), y ∗(x) = g(x) – s1–s3

2(b – a).HereC is an arbitrary constant and y ∗(x) is an eigenfunction of the equation corresponding

s0=

 b

a h(x) dx, s1=

 b

a g(x)h(x) dx, s2=

A1= Af1–ABλ(f1s1–f2s0)

AB(s2

1–s0s2)λ 2– (A + B)s1λ + 1, A2 =

Bf2–ABλ(f2s1–f1s2) AB(s2

1–s0s2)λ 2– (A + B)s1λ + 1,

f1 =

 b a

f (x)h(x) dx, f2 =

 b a

f (x)g(x)h(x) dx.

2◦ Solution withλ = λ1 ≠ λ2andf1=f2= 0:

y(x) = f (x) + Cy1(x), y1(x) = g(x) + 1 –λ1As1

λ1As0 ,

whereC is an arbitrary constant and y1(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ1.

4 Solution withλ = λ1,2=λ ∗andf1=f2= 0, where the characteristic valueλ ∗=

(A + B)s1

is double:

y(x) = f (x) + Cy ∗(x),whereC is an arbitrary constant and y ∗(x) is an eigenfunction of the equation corresponding

toλ ∗ Two cases are possible

(a) IfA ≠ B, then 4ABs0s2= –(A – B)2s2and

y ∗(x) = g(x) – (A – B)s1

2As0 .(b) IfA = B, then, in view of 4ABs0s2= –(A – B)2s2= 0, we have

 b a g(x)h(x) dx, s2=

 b a

g2(x)h(x) dx

1◦ Solution withλ ≠ λ1,2:

y(x) = f (x) + λ[A1g(x) + A2],

where the constantsA1andA2are given by

f (x)h(x) dx, f2 =

 b a

f (x)g(x)h(x) dx.

2◦ Solution withλ = λ1 ≠ λ2andf1=f2= 0:

y(x) = f (x) +  Cy1(x), y1(x) = g(x) + 1 –λ1As1

λ1As0 ,

(A + B)s1+Cs0 is double:

y(x) = f (x) +  Cy ∗(x),where C is an arbitrary constant and y ∗(x) is an eigenfunction of the equation corresponding

toλ ∗ Two cases are possible

(a) IfAs1– (Bs1+Cs0)≠ 0, then 4As0(Bs2+Cs1) = –[(A – B)s1–Cs0]2and

 b

a

[A + B(x – t)h(t)]y(t) dt = f (x).

The characteristic values of the equation:

λ1,2= A(b – a) ±[A(b – a) – 2Bh1]2+ 2Bh0[A(b2–a2) – 2Bh2]

B

A(b – a)[2h1– (b + a)h0] – 2B(h2

1–h0h2) ,where

h0=

 b a h(x) dx, h1=

 b a xh(x) dx, h2=

 b a

to the characteristic valueλ1.

4 Solution withλ = λ1,2=λ ∗andf1=f2= 0, where the characteristic valueλ ∗=

A(b – a)

(A≠ 0) is double:

y(x) = f (x) + Cy ∗(x),whereC is an arbitrary constant, and y ∗(x) is an eigenfunction of the equation corresponding

toλ ∗ Two cases are possible

(a) IfA(b – a) – 2Bh1≠ 0, then

1 for h0≠ 0 and A(b2–a2) = 2Bh2,

x for h0= 0 andA(b2–a2)≠ 2Bh2,

C1+C2x for h0= 0 andA(b2–a2) = 2Bh2,whereC1andC2are arbitrary constants

h0=

 b a h(x) dx, h1=

 b a xh(x) dx, h2=

 b a

A(b – a) + (B + C)h1 is double:

y(x) = f (x) +  Cy ∗(x),where C is an arbitrary constant and y ∗(x) is an eigenfunction of the equation corresponding

toλ ∗ Two cases are possible

(a) IfA(b – a) + (C – B)h1≠ 0, then

1 for h0≠ 0 and A(b2–a2) = –2Ch2,

x for h0= 0 andA(b2–a2)≠ –2Ch2,

The characteristic values of the equation:

λ1,2= A(b – a) ±[A(b – a) + 2Bh1] 2– 2Bh0[A(b 2–a2) + 2Bh2]

B{h0[A(b2–a2) + 2Bh2] – 2h1[A(b – a) + Bh1]} ,where

h0=

 b a h(x) dx, h1=

 b a xh(x) dx, h2=

 b a

x2h(x) dx.

1◦ Solution withλ ≠ λ1,2:

y(x) = f (x) + λ AE1+ (BE1x + E2)h(x)

,where the constantsE1andE2are given by

 b a

f (x) dx, f2=

 b a

xf (x) dx.

2◦ Solution withλ = λ1≠ λ2andf1=f2= 0:

y(x) = f (x) + Cy1(x), y1(x) = A + Bxh(x) + 1 –λ1[A(b – a) + Bh1]

whereC is an arbitrary constant and y1(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ1.

4 Solution withλ = λ1,2=λ ∗andf1=f2= 0, where the characteristic valueλ ∗=

A(b – a)

(A≠ 0) is double:

y(x) = f (x) + Cy ∗(x),whereC is an arbitrary constant and y ∗(x) is an eigenfunction of the equation corresponding

toλ ∗ Two cases are possible

(a) IfA(b – a) ≠ –2Bh1, then

A + Bxh(x) for h0≠ 0 and A(b2–a2) = –2Bh2,

h(x) for h0= 0 andA(b2–a2)≠ –2Bh2,

C1[A + Bxh(x)] + C2h(x) for h0= 0 andA(b2–a2) = –2Bh2,whereC1andC2are arbitrary constants

D = [A(b – a) + (B – C)h1]2+ 2Ch0[A(b2–a2) + 2Bh2],where

h0=

 b a h(x) dx, h1=

 b a xh(x) dx, h2=

 b a

x2h(x) dx.

1◦ Solution withλ ≠ λ1,2:

y(x) = f (x) + λ AE1+ (BE1x + E2)h(x)

,where the constantsE1andE2are given by

A(b – a) + (B + C)h1 (A≠ 0) is double:

y(x) = f (x) +  Cy ∗(x),where C is an arbitrary constant and y ∗(x) is an eigenfunction of the equation corresponding

toλ ∗ Two cases are possible

(a) IfA(b – a) + Bh1≠ Ch1, then

A + Bxh(x) for h0≠ 0 and A(b2–a2) = –2Bh2,

h(x) for h0= 0 andA(b2–a2)≠ –2Bh2,

22(s1s3–s2) , λ2 =

s1+s3–

(s1–s3)2+ 4s2

22(s1s3–s2) ,where

 b

a

h2(x) dx

1◦ Solution withλ ≠ λ1,2:

y(x) = f (x) + λ[A1g(x) + A2h(x)],

where the constantsA1andA2are given by

f (x)g(x) dx, f2=

 b a

f (x)h(x) dx.

2◦ Solution withλ = λ1 ≠ λ2andf1=f2= 0:

y(x) = f (x) + Cy1(x), y1(x) = g(x) + 1 –λ1s1

λ1s2 h(x),

whereC is an arbitrary constant and y1(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ1.

3◦ The solution withλ = λ2 ≠ λ1 andf1 =f2 = 0 is given by the formulas of item 2◦ in

which one must replaceλ1andy1(x) by λ2andy2(x), respectively.

4◦ Solution withλ = λ1,2=λ ∗andf1=f2= 0, where the characteristic valueλ ∗= 1/s1is

double:

y(x) = f (x) +  C1g(x) +  C2h(x),

where C1and C2are arbitrary constants

The characteristic values of the equation:

λ1= s1–s3+

(s1+s3)2– 4s2

22(s2

2–s1s3) , λ2 =

s1–s3–

(s1+s3)2– 4s2

22(s2

2–s1s3) ,where

f (x)g(x) dx, f2=

 b a

f (x)h(x) dx.

2◦ Solution withλ = λ1 ≠ λ2andf1=f2= 0:

y(x) = f (x) + Cy1(x), y1(x) = g(x) + 1 –λ1s1

λ1s2 h(x),

whereC is an arbitrary constant and y1(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ1

3◦ The solution withλ = λ2 ≠ λ1 andf1 =f2 = 0 is given by the formulas of item 2◦ in

which one must replaceλ1andy1(x) by λ2andy2(x), respectively.

4◦ Solution withλ = λ1,2=λ ∗andf1=f2= 0, where the characteristic valueλ ∗= 2

s1–s3

is double:

y(x) = f (x) + Cy ∗(x), y ∗(x) = g(x) – s1+s3

2s2 h(x),whereC is an arbitrary constant and y ∗(x) is an eigenfunction of the equation corresponding

s1=

 b a

g2(x) dx, s2=

 b a g(x)h(x) dx, s3=

 b a

h2(x) dx

1◦ Solution withλ ≠ λ1,2:

whereC is an arbitrary constant and y1(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ1.

3◦ The solution withλ = λ2 ≠ λ1 andf1 =f2 = 0 is given by the formulas of item 2◦ in

which one must replaceλ1andy1(x) by λ2andy2(x), respectively.

4◦ Solution withλ = λ1,2=λ ∗andf1=f2= 0, where the characteristic valueλ ∗= 2

As1+Bs3

is double:

y(x) = f (x) + Cy ∗(x),whereC is an arbitrary constant and y ∗(x) is an eigenfunction of the equation corresponding

toλ ∗ Two cases are possible

(a) IfAs1 ≠ Bs3, then 4ABs2

 b a

h2(x) dx, s3=

 b a

1–s2s3)λ2– 2s1λ + 1, A2 =

f2–λ(f2s1–f1s3)(s2

2◦ Solution withλ = λ1 ≠ λ2andf1=f2= 0:

y(x) = f (x) + Cy1(x), y1(x) = g(x) +



s3

s2 h(x),

whereC is an arbitrary constant and y1(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ1.

whereC is an arbitrary constant and y2(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ2

4◦ The equation has no multiple characteristic values

 b a

h2(x) dx, s3=

 b a

– 2+λ(f2s1–f1s3)

(s2s3–s2)λ2+ 1 ,

f1=

 b a

f (x)h(x) dx, f2=

 b a

whereC is an arbitrary constant and y1(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ1.

3◦ Solution withλ = λ2 ≠ λ1andf1=f2= 0:

y(x) = f (x) + Cy2(x), y2(x) = g(x) –



s2–s2s3+s1

whereC is an arbitrary constant and y2(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ2.

The characteristic values of the equation:

 b a

f (x)h(x) dx, f2=

 b a

f (x)g(x) dx.

2◦ Solution withλ = λ1≠ λ2andf1=f2= 0:

y(x) = f (x) + Cy1(x), y1(x) = g(x) + 1 –λ1As1

λ1As2 h(x),

whereC is an arbitrary constant and y1(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ1.

3◦ The solution withλ = λ2 ≠ λ1 andf1 =f2 = 0 is given by the formulas of item 2◦ in

which one must replaceλ1andy1(x) by λ2andy2(x), respectively

4◦ Solution withλ = λ1,2=λ ∗andf1=f2= 0, where the characteristic valueλ ∗= 2

 b a h2(x)g1(x) dx, s22=

 b a h2(x)g2(x) dx

are convergent

where the constantsA1andA2are given by

A1= f1–λ(f1s22–f2s12)

(s11s22–s12s21)λ2– (s11+s22)λ + 1, A2 =

f2–λ(f2s11–f1s21)(s11s22–s12s21)λ2– (s11+s22)λ + 1,

whereC is an arbitrary constant and y1(x) is an eigenfunction of the equation corresponding

to the characteristic valueλ1:

y1(x) = g1(x) + 1 –λ1s11

λ1s12 g2(x) = g1(x) +

λ1s21

1 –λ1s22g2(x).

3◦ The solution withλ = λ2 ≠ λ1 andf1 =f2 = 0 is given by the formulas of item 2◦ in

which one must replaceλ1andy1(x) by λ2andy2(x), respectively

4◦ Solution withλ = λ1,2=λ ∗andf1=f2= 0, where the characteristic valueλ ∗= 2

s11+s22

is double (there is no double characteristic value provided thats11 = –s22):

y(x) = f (x) + Cy ∗(x),whereC is an arbitrary constant and y ∗(x) is an eigenfunction of the equation corresponding

toλ ∗ Two cases are possible

(a) Ifs11≠ s22, thens12 = –14(s11–s22)2/s21,s21≠ 0, and

y ∗(x) = g1(x) – s11–s22

2s12 g2(x).

Note that in this case,s12ands21have opposite signs

(b) Ifs11=s22, then, in view ofs12s21 = –14(s11–s22)2= 0, we have

The characteristic values of the integral equation (counting the multiplicity, we have exactly

n of them) are the roots of the algebraic equation

smk=

 b a hm(x)gk(x) dx; m, k = 1, , n,

are assumed to be convergent

Solution with regularλ:

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅–λsn1 · · · –λsnk–1 fn –λsnk+1 · · · 1 – λsnn









For solutions of the equation in the case in whichλ is a characteristic value, see

cos(λx) + BIc–AIs

I2+I2 ssin(λx),

Ic= 1 +

 ∞–∞K(z) cos(λz) dz, Is=

 ∞–∞K(z) sin(λz) dz.

–iuxdx is the Fourier transform of K(x) In this case, the

equation has a unique solution, which is given by

y(x) = f (x) +

 ∞–∞R(x – t)f (t) dt,

R(x) = √1

2π

 ∞–∞

The Wiener–Hopf equation of the second kind.*

Here 0≤ x < ∞, K(x) ∈ L1(–∞, ∞), f(x) ∈ L1(0, ∞), and y(x) ∈ L1(0, ∞).

For the integral equation to be solvable, it is necessary and sufficient that

Ω(u) = 1 – ˇ K(u) ≠ 0, –∞ < u < ∞, (1)where ˇK(u) =∞

–∞K(x)e

iux dx is the Fourier transform (in the asymmetric form) of K(x).

In this case, the index of the equation can be introduced,

ν = –ind Ω(u) = – 1

2π argΩ(u)

∞–∞.

1◦ Solution withν = 0:

y(x) = f (x) +

 ∞0

R(x, t)f (t) dt,

and the functionsR+(x) and R–(x) satisfy the conditions R+(x) = 0 and R–(x) = 0 for x < 0

and are uniquely defined by their Fourier transforms as follows:

1 +

 ∞0

R ±(t)e±iut dt = exp

–1

2 lnΩ(u) ∓ 1

2πi

 ∞–∞

lnΩ(t)

t – u dt

Alternatively, R+(x) and R–(x) can be obtained by constructing the solutions of the

equations

R+(x) +

 ∞0

K(x – t)R+(t) dt = K(x), 0≤ x ≤ ∞,

R–(x) +

 ∞0

R ◦(x, t)



f (t) + ν

R(0) + (x – s)R(1)

– (t – s) ds,and the functionsR(0)+ (x) and R(1)

– (x) are uniquely defined by their Fourier transforms:

1 +

 ∞0

R(0)±(t)e±iut dt

,

1 +

 ∞0

R(0)±(t)e±iut dt = exp

–1

2lnΩ

◦(u)∓ 1

2πi

 ∞–∞

lnΩ◦(t)

t – u dt

,

Ω◦(u)(u + i)ν=Ω(u)(u – i) ν

3◦ Forν < 0, the solution exists only if the conditions

 ∞0

f (x)ψm(x) dx = 0, m = 1, 2, , –ν,

are satisfied Hereψ1(x), , ψν(x) is the system of linearly independent solutions of the

transposed homogeneous equation

ψ(x) –

 ∞0

K(t – x)ψ(t) dt = 0.

Then

y(x) = f (x) +

 ∞0

R ∗(x, t)f (t) dt,where

R ∗(x, t) = R(1)+ (x – t) + R(0)– (t – x) +

 ∞0

R(1)+ (x – s)R(0)– (t – s) ds,and the functionsR+(1)(x) and R(0)

– (x) are uniquely defined in item 2◦by their Fourier forms

trans-• References: V I Smirnov (1974), F D Gakhov and Yu I Cherskii (1978), I M Vinogradov (1979).

–iuxdx, K(u) = √1

2π

 ∞–∞K(x)e

where

Ik(λ) =

 ∞–∞K(z)e

(β+λ)zdz, Im(λ) =

 ∞–∞M (z)e

The solution can be obtained with the aid of the inverse Mellin transform as follows:

y(x) = 1

2πi

 c+i ∞ c–i ∞

Eigenfunctions of this integral equation are determined by the roots of the following

tran-scendental (algebraic) equation for the parameterλ:

 ∞0

The general solution is the linear combination (with arbitrary constants) of the

eigenfunc-tions of the homogeneous integral equation

It is assumed that the improper integral is convergent andB ≠ 0 The general solution of

the integral equations is the sum of the above solution and the solution of the homogeneous

HereC = 0.5772 is the Euler constant, ψ(z) = [ln Γ(z)] 

z is the logarithmic derivative of the gamma function, and thes kare the negative roots of the transcendental equationΓ(s k) = 2, whereΓ(z) is the gamma function.

• Reference: M L Krasnov, A I Kisilev, and G I Makarenko (1971).

 b

x g(t)y(t) dt = f x (x) (2)

Differentiating (2), we arrive at a second-order ordinary differential equation fory = y(x),

y(a) +

 b a

(t – a)g(t)y(t) dt = f (a),

y(b) +

 b a

(b – t)g(t)y(t) dt = f (b)

(4)

Let us expressg(x)y from (3) via y  xxandf xxand substitute the result into (4) Integrating

by parts yields the desired boundary conditions fory(x),

y(a) + y(b) + (b – a)[f x (b) – y x(b)] = f (a) + f (b),

y(a) + y(b) + (a – b)[f x (a) – yx (a)] = f (a) + f (b) (5)Note a useful consequence of (5),

y  x(a) + yx (b) = fx (a) + fx (b), (6)which can be used together with one of conditions (5)

Equation (3) under the boundary conditions (5) determines the solution of the original

integral equation Conditions (5) make it possible to calculate the constants of integration

that occur in the solution of the differential equation (3)

e λ(x–t) g(t)y(t) dt +

 b x

e λ(t–x) g(t)y(t) dt = f (x). (1)Differentiating (1) with respect tox twice yields

y xx(x) + 2λg(x)y(x) + λ2

 x a

e λ(x–t) g(t)y(t) dt + λ2

 b x

e λ(t–x) g(t)y(t) dt = f xx (x) (2)Eliminating the integral terms from (1) and (2), we arrive at a second-order ordinary

differential equation fory = y(x),

y xx+ 2λg(x)y – λ2y = f xx (x) – λ2f (x). (3)

2◦ Let us derive the boundary conditions for equation (3) We assume that the limits of

integration satisfy the conditions –∞ < a < b < ∞ By setting x = a and x = b in (1), we

obtain two consequences

y(a) + e–λa

 b a

Let us expressg(x)y from (3) via y  xxandf xxand substitute the result into (4) Integrating

by parts yields the conditions

e λb ϕ  x(b) – eλa ϕ  x(a) = λeλa ϕ(a) + λe λb ϕ(b),

e–λbϕ  x(b) – e–λaϕ  x(a) = λe–λaϕ(a) + λe–λbϕ(b), ϕ(x) = y(x) – f (x).

Finally, after some manipulations, we arrive at the desired boundary conditions fory(x):

ϕ  x(a) + λϕ(a) = 0, ϕ  x(b) – λϕ(b) = 0; ϕ(x) = y(x) – f (x). (5)Equation (3) under the boundary conditions (5) determines the solution of the original

integral equation Conditions (5) make it possible to calculate the constants of integration

that occur in solving the differential equation (3)

1 Let us remove the modulus in the integrand:

y xx(x) + 2λg(x)y(x) + λ2

 x a

differential equation fory = y(x),

y xx+ 2λg(x)y – λ2y = f xx (x) – λ2f (x). (3)

2◦ Let us derive the boundary conditions for equation (3) We assume that the limits of

integration satisfy the conditions –∞ < a < b < ∞ By setting x = a and x = b in (1), we

obtain two corollaries

y(a) +

 b a

sinh[λ(t – a)]g(t)y(t) dt = f (a),

Let us expressg(x)y from (3) via y  xxandf xxand substitute the result into (4) Integrating

by parts yields the desired boundary conditions fory(x),

sinh[λ(b – a)]ϕ x(b) – λ cosh[λ(b – a)]ϕ(b) = λϕ(a),

sinh[λ(b – a)]ϕ x(a) + λ cosh[λ(b – a)]ϕ(a) = –λϕ(b); ϕ(x) = y(x) – f (x). (5)

Equation (3) under the boundary conditions (5) determines the solution of the original

integral equation Conditions (5) make it possible to calculate the constants of integration

that occur in solving the differential equation (3)

sin[λ(x – t)]g(t)y(t) dt +

 b x

sin[λ(t – x)]g(t)y(t) dt = f (x) (1)Differentiating (1) with respect tox twice yields

y xx(x) + 2λg(x)y(x) – λ2

 x a

sin[λ(x – t)]g(t)y(t) dt

 b

2◦ Let us derive the boundary conditions for equation (3) We assume that the limits of

integration satisfy the conditions –∞ < a < b < ∞ By setting x = a and x = b in (1), we

obtain two consequences

sin[λ(b – t)]g(t)y(t) dt = f (b)

(4)

Let us expressg(x)y from (3) via y  xxandf xxand substitute the result into (4) Integrating

by parts yields the desired boundary conditions fory(x),

sin[λ(b – a)]ϕ x(b) – λ cos[λ(b – a)]ϕ(b) = λϕ(a),sin[λ(b – a)]ϕ x(a) + λ cos[λ(b – a)]ϕ(a) = –λϕ(b); ϕ(x) = y(x) – f (x). (5)

Equation (3) under the boundary conditions (5) determines the solution of the original

integral equation Conditions (5) make it possible to calculate the constants of integration

that occur in solving the differential equation (3)

4.9-4 Equations of the Formy(x) +b

Eigenfunctions of this integral equation* are determined by the roots of the following

char-acteristic (transcendental or algebraic) equation forµ:

 b a

1◦ For a real (simple) rootµ kof equation (1), there is a corresponding eigenfunction

yk(x) = exp(µkx).

2◦ For a real rootµkof multiplicityr, there are corresponding r eigenfunctions

y k1(x) = exp(µ k x), y k2(x) = x exp(µ k x), , y kr(x) = xr–1exp(µk x).

3◦ For a complex (simple) rootµ k=α k+iβ kof equation (1), there is a corresponding pair

of eigenfunctions

y k(1)(x) = exp(αkx) cos(βkx), y(2)k (x) = exp(αkx) sin(βkx).

4◦ For a complex rootµk =αk+iβk of multiplicityr, there are corresponding r pairs of

The general solution is the linear combination (with arbitrary constants) of the

eigenfunc-tions of the homogeneous integral equation

* In the equations below that containy(x – t) in the integrand, the arguments can have, for example, the domain

(a) –∞ < x < ∞, –∞ < t < ∞ for a = –∞ and b = ∞ or (b) a ≤ t ≤ b, –∞ ≤ x < ∞, for a and b such that –∞ < a < b < ∞.

Case (b) is a special case of (a) iff (t) is nonzero only on the interval a ≤ t ≤ b.