Equations Whose Kernels Contain Exponential Functions

Linear Equations of the Second Kind With Variable Limit of Integration

2.2. Equations Whose Kernels Contain Exponential Functions

2.2. Equations Whose Kernels Contain Exponential

In this case, the original integral equation has the solution y(x) =f(x) +

_x

a

E₁e^µ1(x–t)+E₂e^µ2(x–t) f(t)dt, where

E₁=A µ₁ µ2–µ1

+B µ₁–λ µ2–µ1

, E₂=A µ₂ µ1–µ2

+B µ₂–λ µ1–µ2

. 3^◦. IfD< 0, then equation (2) has the complex conjugate roots

µ₁=σ+iβ, µ₂=σ–iβ, σ= ¹₂(λ–A–B), β = ¹₂√ –D.

In this case, the original integral equation has the solution y(x) =f(x) +

_x

a

E₁e^σ(x–t)cos[β(x–t)] +E₂e^σ(x–t)sin[β(x–t)]

f(t)dt, where

E₁=–A–B, E₂ = 1

β(–Aσ–Bσ+Bλ).

5. y(x) +A x

a

(e^λx–e^λt)y(t)dt=f(x).

This is a special case of equation 2.9.5 withg(x) =Ae^λx. Solution:

y(x) =f(x) + 1 W

x a

u₁(x)u₂(t)–u₂(x)u₁(t) f(t)dt,

where the primes denote differentiation with respect to the argument specified in the paren- theses, and u₁(x),u₂(x) is a fundamental system of solutions of the second-order linear homogeneous ordinary differential equationu_xx+Aλe^λxu= 0; the functionsu1(x) andu2(x) are expressed in terms of Bessel functions or modified Bessel functions, depending on the sign ofA:

ForAλ> 0, W = λ

π, u₁(x) =J₀ 2√

Aλ λ e^λx/2

, u₂(x) =Y₀ 2√

Aλ λ e^λx/2

, ForAλ< 0,

W =–λ

2, u1(x) =I0

2√

|Aλ| λ e^λx/2

, u2(x) =K0

2√

|Aλ| λ e^λx/2

.

6. y(x) + _x

a

Ae^λx+Be^λt

y(t)dt=f(x).

ForB=–A, see equation 2.2.5. This is a special case of equation 2.9.6 withg(x) =Ae^λxand h(t) =Be^λt.

Differentiating the original integral equation followed by substitutingY(x) = x

a

y(t)dt yields the second-order linear ordinary differential equation

Y_xx + (A+B)e^λxY_x+Aλe^λxY =f_x(x) (1)

under the initial conditions

Y(a) = 0, Y_x(a) =f(a). (2)

A fundamental system of solutions of the homogeneous equation (1) withf ≡0 has the form

Y₁(x) =Φ A

m, 1; –m λe^λx

, Y₂(x) =Ψ A

m, 1;–m λe^λx

, m=A+B, whereΦ

α,β;x andΨ

α,β;x

are degenerate hypergeometric functions.

Solving the homogeneous equation (1) under conditions (2) for an arbitrary function f =f(x) and taking into account the relationy(x) =Y_x(x), we thus obtain the solution of the integral equation in the form

y(x) =f(x)– _x

a

R(x,t)f(t)dt, R(x,t) = Γ(A/m)

λ

∂²

∂x∂t

exp m

λe^λt

Y₁(x)Y₂(t)–Y₂(x)Y₁(t) .

7. y(x) +A _x

a

e^λ(x+t)–e^2λt

y(t)dt=f(x).

The transformationz=e^λx,τ =e^λtleads to an equation of the form 2.1.4.

1^◦. Solution withAλ> 0:

y(x) =f(x)–λk _x

a

e^λtsin

k(e^λx–e^λt)

f(t)dt, k= A/λ.

2^◦. Solution withAλ< 0:

y(x) =f(x) +λk x

a

e^λtsinh

k(e^λx–e^λt)

f(t)dt, k=

|A/λ|.

8. y(x) +A _x

a

e^λx+µt –e^(λ+µ)t

y(t)dt=f(x).

The transformationz=e^µx,τ =e^µt,Y(z) =y(x) leads to an equation of the form 2.1.52:

Y(z) + A µ

z b

(z^k–τ^k)Y(τ)dτ =F(z), F(z) =f(x), wherek=λ/µ,b=e^µa.

9. y(x) +A _x

a

λe^λx+µt–(λ+µ)e^(λ+µ)x

y(t)dt=f(x).

This equation can be obtained by differentiating an equation of the form 1.2.22:

_x

a

1 +Ae^λx(e^µt–e^µx)

y(t)dt=F(x), F(x) = _x

a

f(t)dt.

Solution:

y(x) = d dx

e^λxΦ(x) _x

a

F(t) e^λt

t

dt Φ(t)

, Φ(x) = exp Aµ

λ+µe^(λ+µ)x

.

10. y(x) + x

a

A1e^λ1(x–t)+A2e^λ2(x–t)

y(t)dt=f(x).

1^◦. Introduce the notation I1=

x a

e^λ1(x–t)y(t)dt, I2= x

a

e^λ2(x–t)y(t)dt.

Differentiating the integral equation twice yields (thefirst line is the original equation) y+A1I1+A2I2=f, f =f(x), (1) y_x+ (A₁+A₂)y+A₁λ₁I₁+A₂λ₂I₂=f_x, (2) y_xx+ (A₁+A₂)y_x + (A₁λ₁+A₂λ₂)y+A₁λ²₁I₁+A₂λ²₂I₂=f_xx . (3) EliminatingI1andI2, we arrive at the second-order linear ordinary differential equation with constant coefficients

y_xx + (A₁+A₂–λ₁–λ₂)y_x+ (λ₁λ₂–A₁λ₂–A₂λ₁)y=f_xx –(λ₁+λ₂)f_x +λ₁λ₂f. (4) Substitutingx=ainto (1) and (2) yields the initial conditions

y(a) =f(a), y_x(a) =f_x(a)–(A1+A2)f(a). (5) Solving the differential equation (4) under conditions (5), we can find the solution of the integral equation.

2^◦. Consider the characteristic equation

µ²+ (A1+A2–λ1–λ2)µ+λ1λ2–A1λ2–A2λ1= 0 (6) which corresponds to the homogeneous differential equation (4) (withf(x)≡0). The structure of the solution of the integral equation depends on the sign of the discriminant

D≡(A₁–A₂–λ₁+λ₂)²+ 4A₁A₂ of the quadratic equation (6).

IfD> 0, the quadratic equation (6) has the real different roots µ1= ¹₂(λ1+λ2–A1–A2) + ¹₂√

D, µ2= ¹₂(λ1+λ2–A1–A2)– ¹₂√ D.

In this case, the solution of the original integral equation has the form y(x) =f(x) +

_x

a

B₁e^µ1(x–t)+B₂e^µ2(x–t) f(t)dt, where

B₁=A₁µ₁–λ₂

µ₂–µ₁ +A₂µ₁–λ₁

µ₂–µ₁, B₂=A₁µ₂–λ₂

µ₁–µ₂ +A₂µ₂–λ₁ µ₁–µ₂. IfD< 0, the quadratic equation (6) has the complex conjugate roots

µ1=σ+iβ, µ2=σ–iβ, σ= ¹₂(λ1+λ2–A1–A2), β= ¹₂√ –D.

In this case, the solution of the original integral equation has the form y(x) =f(x) +

_x

a

B₁e^σ(x–t)cos[β(x–t)] +B₂e^σ(x–t)sin[β(x–t)]

f(t)dt.

where

B1=–A1–A2, B2= 1 β

A1(λ2–σ) +A2(λ1–σ) .

11. y(x) + x

a

Ae^λ(x+t)–Ae^2λt+Be^λt

y(t)dt=f(x).

The transformationz=e^λx,τ =e^λt,Y(z) =y(x) leads to an equation of the form 2.1.5:

Y(z) + _z

b

B₁(z–τ) +A₁

Y(τ)dτ =F(z), F(z) =f(x), whereA₁=B/λ,B₁ =A/λ,b=e^λa.

12. y(x) + _x

a

Ae^λ(x+t)+Be^2λt+Ce^λt

y(t)dt=f(x).

The transformationz=e^λx,τ =e^λt,Y(z) =y(x) leads to an equation of the form 2.1.6:

Y(z)– z

b

(A1z+B1τ+C1)Y(τ)dτ =F(z), F(z) =f(x), whereA1=–A/λ,B1=–B/λ,C1=–C/λ,b=e^λa.

13. y(x) + x

a

λe^λ(x–t)+A

µe^µx+λt–λe^λx+µt

y(t)dt=f(x).

This is a special case of equation 2.9.23 withh(t) =A.

Solution:

y(x) = 1 e^λx

d dx

Φ(x)

x a

F(t) e^λt

t

e^2λt Φ(t)dt

, Φ(x) = exp

Aλ–µ

λ+µe^(λ+µ)x

, F(x) = _x

a

f(t)dt.

14. y(x)– x

a

λe^–λ(x–t)+A

µe^λx+µt–λe^µx+λt

y(t)dt=f(x).

This is a special case of equation 2.9.24 withh(x) =A.

Assume thatf(a) = 0. Solution:

y(x) = _x

a

w(t)dt, w(x) =e^–λx d dx

e^2λx Φ(x)

_x

a

f(t) e^λt

t

Φ(t)dt

, Φ(x) = exp

Aλ–µ

λ+µe^(λ+µ)x

.

15. y(x) + _x

a

λe^λ(x–t)+Ae^βt

µe^µx+λt–λe^λx+µt

y(t)dt=f(x).

This is a special case of equation 2.9.23 withh(t) =Ae^βt. Solution:

y(x) =e^–(λ+β)x d dx

Φ(x)

_x

a

F(t) e^λt

t

e^(2λ+β)t Φ(t) dt

, Φ(x) = exp

A λ–µ

λ+µ+βe^(λ+µ+β)x

, F(x) = x

a

f(t)dt.

16. y(x)– x

a

λe^–λ(x–t)+Ae^βx

µe^λx+µt–λe^µx+λt

y(t)dt=f(x).

This is a special case of equation 2.9.24 withh(x) =Ae^βx. Assume thatf(a) = 0. Solution:

y(x) = x

a

w(t)dt, w(x) =e^–λx d dx

e^(2λ+β)x Φ(x)

x a

f(t) e^(λ+β)t

t

Φ(t)dt

, Φ(x) = exp

A λ–µ

λ+µ+βe^(λ+µ+β)x

.

17. y(x) + x

a

ABe^(λ+1)x+t–ABe^λx+2t–Ae^λx+t–Be^t

y(t)dt=f(x).

The transformationz=e^x,τ =e^t,Y(z) =y(x) leads to an equation of the form 2.1.56:

Y(z) + z

b

ABz^λ+1–ABz^λτ–Az^λ–B

Y(τ)dτ =F(z), whereF(z) =f(x) andb=e^a.

18. y(x) + x

a

ABe^x+λt–ABe^(λ+1)t+Ae^λt+Be^t

y(t)dt=f(x).

The transformationz =e^x,τ =e^t,Y(z) =y(x) leads to an equation of the form 2.1.57 (in whichλis substituted byλ–1):

Y(z) + z

b

ABzτ^λ–1–ABτ^λ+Aτ^λ–1+B

Y(τ)dτ =F(z), whereF(z) =f(x) andb=e^a.

19. y(x) + x

a

n k=1

A_ke^λk(x–t)

y(t)dt=f(x).

1^◦. This integral equation can be reduced to annth-order linear nonhomogeneous ordinary differential equation with constant coefficients. Set

Ik(x) = x

a

e^λk(x–t)y(t)dt. (1)

Differentiating (1) with respect toxyields I_k =y(x) +λk

_x

a

e^λk(x–t)y(t)dt, (2)

where the prime stands for differentiation with respect to x. From the comparison of (1) with (2) we see that

I_k =y(x) +λkIk, Ik=Ik(x). (3) The integral equation can be written in terms ofI_k(x) as follows:

y(x) + n

k=1

A_kI_k =f(x). (4)

Differentiating (4) with respect toxand taking account of (3), we obtain y_x(x) +σny(x) +

n k=1

AkλkIk=f_x(x), σn= n

k=1

Ak. (5)

Eliminating the integralInfrom (4) and (5), wefind that

y_x(x) +

σn–λn)y(x) + n–1

k=1

Ak(λk–λn)Ik =f_x(x)–λnf(x). (6) Differentiating (6) with respect to xand eliminatingI_n–1 from 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

A¹_kIk. If we proceed with successively eliminatingI_n–2,I_n–3,. . .,I₁with the aid of differentiation and formula (3), then we will finally arrive at an nth-order linear nonhomogeneous ordinary differential equation with constant coefficients.

The initial conditions fory(x) can be obtained by settingx=ain the integral equation and all its derivative equations.

2^◦. The solution of the equation can be represented in the form y(x) =f(x) +

x a

n k=1

Bke^µk(x–t)

f(t)dt. (7)

The unknown constantsµ_k are the roots of the algebraic equation n

k=1

Ak

z–λk

+ 1 = 0, (8)

which is reduced (by separating the numerator) to the problem offinding the roots of an nth-order characteristic polynomial.

After theµkhave been calculated, the coefficientsBk can be found from the following linear system of algebraic equations:

n k=1

B_k λm–µk

+ 1 = 0, m= 1,. . .,n. (9)

Another way of determining theBkis presented in item 3^◦below.

If all the rootsµ_kof equation (8) are real and different, then the solution of the original integral equation can be calculated by formula (7).

To a pair of complex conjugate rootsµk,k+1=α±iβof the characteristic polynomial (8) there corresponds a pair of complex conjugate coefficientsBk,k+1in equation (9). In this case, the corresponding termsBke^µk(x–t)+Bk+1e^µk+1(x–t)in solution (7) can be written in the form Bke^α(x–t)

cosβ(x–t)

+Bk+1e^α(x–t)

sinβ(x–t)

, whereBkandBk+1are real coefficients.

3^◦. Fora= 0, the solution of the original integral equation is given by y(x) =f(x)–

_x

0

R(x–t)f(t)dt, R(x) =L^–1 R(p)

, (10)

whereL^–1 R(p)

is the inverse Laplace transform of the function R(p) = K(p)

1 +K(p), K(p) = n

k=1

Ak

p–λk

. (11)

The transform R(p) of the resolvent R(x) can be represented as a regular fractional function:

R(p) = Q(p)

P(p), P(p) = (p–µ1)(p–µ2). . .(p–µn),

whereQ(p) is a polynomial inpof degree <n. The rootsµkof the polynomialP(p) coincide with the roots of equation (8). If all µk are real and different, then the resolvent can be determined by the formula

R(x) = n

k=1

Bke^µkx, Bk = Q(µk) P(µ_k), where the prime stands for differentiation.

2.2-2. Kernels Containing Power-Law and Exponential Functions

20. y(x) +A x

a

xe^λ(x–t)y(t)dt =f(x).

Solution:

y(x) =f(x)–A _x

a

xexp₁

2A(t²–x²) +λ(x–t) f(t)dt.

21. y(x) +A _x

a

te^λ(x–t)y(t)dt=f(x).

Solution:

y(x) =f(x)–A x

a

texp₁

2A(t²–x²) +λ(x–t) f(t)dt.

22. y(x) +A x

a

(x–t)e^λty(t)dt=f(x).

This is a special case of equation 2.9.4 withg(t) =Ae^λt. Solution:

y(x) =f(x) + A W

_x

a

u₁(x)u₂(t)–u₂(x)u₁(t)

e^λtf(t)dt,

whereu₁(x),u₂(x) is a fundamental system of solutions of the second-order linear homo- geneous ordinary differential equationu_xx+Ae^λxu= 0; the functionsu₁(x) andu₂(x) are expressed in terms of Bessel functions or modified Bessel functions, depending on signA:

W = λ

π, u1(x) =J0

2√ A λ e^λx/2

, u2(x) =Y0

2√ A λ e^λx/2

for A> 0, W =–λ

2, u1(x) =I0

2√

|A|

λ e^λx/2

, u2(x) =K0

2√

|A|

λ e^λx/2

for A< 0.

23. y(x) +A x

a

(x–t)e^λ(x–t)y(t)dt=f(x).

1^◦. Solution withA> 0:

y(x) =f(x)–k x

a

e^λ(x–t)sin[k(x–t)]f(t)dt, k=√ A.

2^◦. Solution withA< 0:

y(x) =f(x) +k _x

a

e^λ(x–t)sinh[k(x–t)]f(t)dt, k=√ –A.

24. y(x) +A x

a

(x–t)e^λx+µty(t)dt=f(x).

The substitutionu(x) =e^–λxy(x) leads to an equation of the form 2.2.22:

u(x) +A _x

a

(x–t)e^(λ+µ)tu(t)dt=f(x)e^–λx.

25. y(x)– x

a

(Ax+Bt+C)e^λ(x–t)y(t)dt=f(x).

The substitutionu(x) =e^–λxy(x) leads to an equation of the form 2.1.6:

u(x)–A x

a

(Ax+Bt+C)u(t)dt=f(x)e^–λx.

26. y(x) +A _x

a

x²e^λ(x–t)y(t)dt =f(x).

Solution:

y(x) =f(x)–A _x

a

x²exp₁

3A(t³–x³) +λ(x–t) f(t)dt.

27. y(x) +A x

a

xte^λ(x–t)y(t)dt=f(x).

Solution:

y(x) =f(x)–A x

a

xtexp₁

3A(t³–x³) +λ(x–t) f(t)dt.

28. y(x) +A x

a

t²e^λ(x–t)y(t)dt=f(x).

Solution:

y(x) =f(x)–A x

a

t²exp₁

3A(t³–x³) +λ(x–t) f(t)dt.

29. y(x) +A _x

a

(x–t)²e^λ(x–t)y(t)dt=f(x).

Solution:

y(x) =f(x)– _x

a

R(x–t)f(t)dt, R(x) = ²₃ke^(λ–2k)x– ²₃ke^(λ+k)x

cos√ 3kx

–√ 3 sin√

3kx

, k=₁

4A1/3

.

30. y(x) +A x

0

(x²–t²)e^λ(x–t)y(t)dt=f(x).

The substitutionu(x) =e^–λxy(x) leads to an equation of the form 2.1.11:

u(x) +A _x

0

(x²–t²)u(t)dt=f(x)e^–λx.

31. y(x) +A x

a

(x–t)ⁿe^λ(x–t)y(t)dt=f(x), n= 1, 2,. . . Solution:

y(x) =f(x) + _x

a

R(x–t)f(t)dt, R(x) = 1

n+ 1e^λx n

k=0

exp(σ_kx)

σ_kcos(β_kx)–β_ksin(β_kx) , where

σk=|An!|ⁿ⁺¹¹ cos 2πk

n+ 1

, βk=|An!|ⁿ⁺¹¹ sin 2πk

n+ 1

for A< 0, σ_k=|An!|ⁿ⁺¹¹ cos

2πk+π n+ 1

, β_k=|An!|ⁿ⁺¹¹ sin

2πk+π n+ 1

for A> 0.

32. y(x) +b x

a

exp[λ(x–t)]

√x–t y(t)dt =f(x).

Solution:

y(x) =e^λx

F(x) +πb² x

a

exp[πb²(x–t)]F(t)dt

, where

F(x) =e^–λxf(x)–b _x

a

e^–λtf(t)

√x–t dt.

33. y(x) +A x

a

(x–t)t^ke^λ(x–t)y(t)dt=f(x).

The substitutionu(x) =e^–λxy(x) leads to an equation of the form 2.1.49:

u(x) +A _x

a

(x–t)t^ku(t)dt=f(x)e^–λx.

34. y(x) +A _x

a

(x^k–t^k)e^λ(x–t)y(t)dt=f(x).

The substitutionu(x) =e^–λxy(x) leads to an equation of the form 2.1.52:

u(x) +A x

a

(x^k–t^k)u(t)dt=f(x)e^–λx.

35. y(x)–λ _x

0

e^µ(x–t)

(x–t)^αy(t)dt=f(x), 0 <α< 1.

Solution:

y(x) =f(x) + x

0

R(x–t)f(t)dt, where R(x) =e^µx ∞

n=1

λΓ(1–α)x^1–αn

xΓ

n(1–α) .

36. y(x) +A x

a

exp

λ(x²–t²)

y(t)dt=f(x).

Solution:

y(x) =f(x)–A _x

a

exp

λ(x²–t²)–A(x–t) f(t)dt.

37. y(x) +A x

a

exp

λx²+βt²

y(t)dt=f(x).

In the case β = –λ, see equation 2.2.36. This is a special case of equation 2.9.2 with g(x) =–Aexp

λx²) andh(t) = exp βt²

.

38. y(x) +A _∞

x

exp –λ√

t–x

y(t)dt=f(x).

This is a special case of equation 2.9.62 withK(x) =Aexp –λ√

–x .

39. y(x) +A x

a

exp

λ(x^µ–t^µ)

y(t)dt=f(x), µ> 0.

This is a special case of equation 2.9.2 withg(x) =–Aexp λx^µ

andh(t) = exp –λt^µ

. Solution:

y(x) =f(x)–A x

a

exp

λ(x^µ–t^µ)–A(x–t) f(t)dt.

40. y(x) +k _x

0

1 x exp

–λt

x

y(t)dt=g(x).

This is a special case of equation 2.9.71 withf(z) =ke^–λz. For a polynomial right-hand side,g(x) =

N n=0

Anxⁿ, a solution is given by

y(x) = N n=0

A_n 1 +kBn

xⁿ, Bn = n!

λⁿ⁺¹ –e^–λ n

k=0

n!

k!

1 λ^n–k+1.

2.3. Equations Whose Kernels Contain Hyperbolic