Equations Whose Kernels Contain Trigonometric Functions

Linear Equations of the Second Kind With Variable Limit of Integration

2.5. Equations Whose Kernels Contain Trigonometric Functions

2.5-1. Kernels Containing Cosine 1. y(x)–A

_x

a

cos(λx)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =Acos(λx) andh(t) = 1.

Solution:

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

a

cos(λx) exp A

λ

sin(λx)–sin(λt) f(t)dt.

2. y(x)–A x

a

cos(λt)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =Aandh(t) = cos(λt).

Solution:

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

a

cos(λt) exp A

λ

sin(λx)–sin(λt) f(t)dt.

3. y(x) +A x

a

cos[λ(x–t)]y(t)dt=f(x).

This is a special case of equation 2.9.34 with g(t) = A. Therefore, solving this integral equation is reduced to solving the following second-order linear nonhomogeneous ordinary differential equation with constant coefficients:

y_xx+Ay_x+λ²y=f_xx +λ²f, f =f(x), with the initial conditions

y(a) =f(a), y_x(a) =f_x(a)–Af(a).

1^◦. Solution with|A|> 2|λ|:

y(x) =f(x) + _x

a

R(x–t)f(t)dt, R(x) = exp

–¹₂AxA²

2k sinh(kx)–Acosh(kx)

, k= ¹₄A²–λ². 2^◦. Solution with|A|< 2|λ|:

y(x) =f(x) + _x

a

R(x–t)f(t)dt, R(x) = exp

–¹₂AxA²

2k sin(kx)–Acos(kx)

, k= λ²– ¹₄A². 3^◦. Solution withλ=±¹₂A:

y(x) =f(x) + x

a

R(x–t)f(t)dt, R(x) = exp

–¹₂Ax₁

2A²x–A .

4. y(x) + _x

a

n k=1

A_kcos[λ_k(x–t)]

y(t)dt=f(x).

This integral equation is reduced to a linear nonhomogeneous ordinary differential equation of order 2nwith constant coefficients. Set

I_k(x) = _x

a

cos[λ_k(x–t)]y(t)dt. (1)

Differentiating (1) with respect toxtwice yields I_k =y(x)–λk

x a

sin[λk(x–t)]y(t)dt, I_k=y_x(x)–λ²_k

x a

cos[λ_k(x–t)]y(t)dt,

(2)

where the primes stand for differentiation with respect tox. Comparing (1) and (2), we see that

I_k=y_x(x)–λ²_kIk, Ik =Ik(x). (3)

With the aid of (1), the integral equation can be rewritten in the form y(x) +

n k=1

AkIk =f(x). (4)

Differentiating (4) with respect toxtwice taking into account (3) yields y_xx (x) +σny_x(x)–

n k=1

Akλ²_kIk=f_xx(x), σn= n

k=1

Ak. (5)

Eliminating the integralI_nfrom (4) and (5), we obtain y_xx(x) +σ_ny_x(x) +λ²_ny(x) +

n–1 k=1

A_k(λ²_n–λ²_k)I_k =f_xx (x) +λ²_nf(x). (6)

Differentiating (6) with respect toxtwice followed by eliminatingI_n–1 from the resulting expression with the aid of (6) yields a similar equation whose left-hand side is a fourth- order differential operator (acting on y) with constant coefficients plus the sum

n–2

k=1

BkIk. Successively eliminating the termsI_n–2,I_n–3,. . . using double differentiation and formula (3), wefinally arrive at a linear nonhomogeneous ordinary differential equation of order 2nwith constant coefficients.

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

5. y(x)–A x

a

cos(λx)

cos(λt)y(t)dt=f(x).

Solution:

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

a

e^A(x–t)cos(λx) cos(λt)f(t)dt.

6. y(x)–A x

a

cos(λt)

cos(λx)y(t)dt=f(x).

Solution:

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

a

e^A(x–t) cos(λt) cos(λx)f(t)dt.

7. y(x)–A _x

a

cos^k(λx) cos^m(µt)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =Acos^k(λx) andh(t) = cos^m(µt).

8. y(x) +A x

a

tcos[λ(x–t)]y(t)dt=f(x).

This is a special case of equation 2.9.34 withg(t) =At.

9. y(x) +A _x

a

t^kcos^m(λx)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =–Acos^m(λx) andh(t) =t^k.

10. y(x) +A x

a

x^kcos^m(λt)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =–Ax^kandh(t) = cos^m(λt).

11. y(x)– _x

a

Acos(kx) +B–AB(x–t) cos(kx)

y(t)dt=f(x).

This is a special case of equation 2.9.7 withλ=Bandg(x) =Acos(kx).

Solution:

y(x) =f(x) + _x

a

R(x,t)f(t)dt, R(x,t) = [Acos(kx) +B]G(x)

G(t) + B² G(t)

x t

e^B(x–s)G(s)ds, G(x) = exp A

k sin(kx)

.

12. y(x) + _x

a

Acos(kt) +B+AB(x–t) cos(kt)

y(t)dt =f(x).

This is a special case of equation 2.9.8 withλ=Bandg(t) =Acos(kt).

Solution:

y(x) =f(x) + _x

a

R(x,t)f(t)dt, R(x,t) =–[Acos(kt) +B]G(t)

G(x) + B² G(x)

_x

t

e^B(t–s)G(s)ds, G(x) = exp A

k sin(kx)

.

13. y(x) +A _∞

x

cos λ√

t–x

y(t)dt =f(x).

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

–x .

2.5-2. Kernels Containing Sine

14. y(x)–A _x

a

sin(λx)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =Asin(λx) andh(t) = 1.

Solution:

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

a

sin(λx) exp A

λ

cos(λt)–cos(λx) f(t)dt.

15. y(x)–A _x

a

sin(λt)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =Aandh(t) = sin(λt).

Solution:

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

a

sin(λt) exp A

λ

cos(λt)–cos(λx) f(t)dt.

16. y(x) +A x

a

sin[λ(x–t)]y(t)dt=f(x).

This is a special case of equation 2.9.36 withg(t) =A.

1^◦. Solution withλ(A+λ) > 0:

y(x) =f(x)– Aλ k

x a

sin[k(x–t)]f(t)dt, where k=

λ(A+λ).

2^◦. Solution withλ(A+λ) < 0:

y(x) =f(x)– Aλ k

x a

sinh[k(x–t)]f(t)dt, where k=

–λ(λ+A).

3^◦. Solution withA=–λ:

y(x) =f(x) +λ² x

a

(x–t)f(t)dt.

17. y(x) +A _x

a

sin³[λ(x–t)]y(t)dt=f(x).

Using the formula sin³β=–¹₄ sin 3β+ ³₄sinβ, we arrive at an equation of the form 2.5.18:

y(x) + _x

a

–¹₄Asin[3λ(x–t)] + ³₄Asin[λ(x–t)]

y(t)dt=f(x).

18. y(x) + _x

a

A1sin[λ1(x–t)] +A2sin[λ2(x–t)]

y(t)dt=f(x).

This equation can be solved by the same method as equation 2.3.18, by reducing it to a fourth-order linear ordinary differential equation with constant coefficients.

Consider the characteristic equation

z²+ (λ²₁+λ²₂+A₁λ₁+A₂λ₂)z+λ²₁λ²₂+A₁λ₁λ²₂+A₂λ²₁λ₂= 0, (1) whose roots,z1andz2, determine the solution structure of the integral equation.

Assume that the discriminant of equation (1) is positive:

D≡(A₁λ₁–A₂λ₂+λ²₁–λ²₂)²+ 4A₁A₂λ₁λ₂> 0.

In this case, the quadratic equation (1) has the real (different) roots z₁=–¹₂(λ²₁+λ²₂+A₁λ₁+A₂λ₂) + ¹₂√

D, z₂=–¹₂(λ²₁+λ²₂+A₁λ₁+A₂λ₂)– ¹₂√ D.

Depending on the signs ofz₁andz₂the following three cases are possible.

Case 1. Ifz₁ > 0 andz₂ > 0, then the solution of the integral equation has the form (i= 1, 2):

y(x) =f(x) + _x

a

{B₁sinh[µ₁(x–t)] +B₂sinh

µ₂(x–t)

f(t)dt, µi =√ zi, where the coefficientsB₁andB₂are determined from the following system of linear algebraic equations:

B₁µ₁

λ²₁+µ²₁ + B₂µ₂

λ²₁+µ²₂ –1 = 0, B₁µ₁

λ²₂+µ²₁ + B₂µ₂

λ²₂+µ²₂ –1 = 0.

Case 2. Ifz₁< 0 andz₂< 0, then the solution of the integral equation has the form y(x) =f(x) +

x a

{B1sin[µ1(x–t)] +B2sin

µ2(x–t)

f(t)dt, µi=

|zi|,

whereB₁andB₂are determined from the system B1µ1

λ²₁–µ²₁ + B2µ2

λ²₁–µ²₂ –1 = 0, B1µ1

λ²₂–µ²₁ + B2µ2

λ²₂–µ²₂ –1 = 0.

Case 3. Ifz1> 0 andz2< 0, then the solution of the integral equation has the form y(x) =f(x) +

x a

{B₁sinh[µ₁(x–t)] +B₂sin

µ₂(x–t)

f(t)dt, µ_i=

|z_i|,

whereB1andB2are determined from the system B₁µ₁

λ²₁+µ²₁ + B₂µ₂

λ²₁–µ²₂ –1 = 0, B₁µ₁

λ²₂+µ²₁ + B₂µ₂

λ²₂–µ²₂ –1 = 0.

Remark. The solution of the original integral equation can be obtained from the solution of equation 2.3.18 by performing the following change of parameters:

λk→iλk, µk→iµk, Ak →–iAk, Bk →–iBk, i²=–1 (k= 1, 2).

19. y(x) + _x

a

n k=1

A_ksin[λ_k(x–t)]

y(t)dt=f(x).

1^◦. This integral equation can be reduced to a linear nonhomogeneous ordinary differential equation of order 2nwith constant coefficients. Set

Ik(x) = x

a

sin[λk(x–t)]y(t)dt. (1)

Differentiating (1) with respect toxtwice yields I_k =λ_k

_x

a

cos[λ_k(x–t)]y(t)dt, I_k=λ_ky(x)–λ²_{k x}

a

sin[λ_k(x–t)]y(t)dt, (2) where the primes stand for differentiation with respect tox. Comparing (1) and (2), we see that

I_k=λ_ky(x)–λ²_kI_k, I_k =I_k(x). (3) With aid of (1), the integral equation can be rewritten in the form

y(x) + n

k=1

AkIk =f(x). (4)

Differentiating (4) with respect toxtwice taking into account (3) yields y_xx(x) +σ_ny(x)–

n k=1

A_kλ²_kI_k =f_xx (x), σ_n= n

k=1

A_kλ_k. (5)

Eliminating the integralI_nfrom (4) and (5), we obtain y_xx (x) + (σ_n+λ²_n)y(x) +

n–1 k=1

A_k(λ²_n–λ²_k)I_k=f_xx(x) +λ²_nf(x). (6) Differentiating (6) with respect toxtwice followed by eliminatingIn–1 from the resulting expression with the aid of (6) yields a similar equation whose left-hand side is a fourth- order differential operator (acting on y) with constant coefficients plus the sum

n–2

k=1

B_kI_k. Successively eliminating the termsI_n–2,I_n–3,. . . using double differentiation and formula (3), wefinally arrive at a linear nonhomogeneous ordinary differential equation of order 2nwith constant coefficients.

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

2^◦. Let usfind the rootsz_kof the algebraic equation n

k=1

λkAk

z+λ²_k + 1 = 0. (7)

By reducing it to a common denominator, we arrive at the problem of determining the roots of annth-degree characteristic polynomial.

Assume that allz_kare real, different, and nonzero. Let us divide the roots into two groups z1 > 0, z2> 0, . . ., zs > 0 (positive roots);

zs+1< 0, zs+2< 0, . . ., zn< 0 (negative roots).

Then the solution of the integral equation can be written in the form y(x) =f(x)+

x a

s k=1

Bksinh

µk(x–t) +

n k=s+1

Cksin

µk(x–t)

f(t)dt, µk=

|zk|. (8) The coefficients Bk andCk are determined from the following system of linear algebraic equations:

s k=0

Bkµk

λ²_m+µ²_k + n k=s+1

Ckµk

λ²_m–µ²_k –1 = 0, µk =

|zk| m= 1, 2,. . .,n. (9) In the case of a nonzero rootz_s = 0, we can introduce the new constantD= B_sµ_s and proceed to the limitµ_s →0. As a result, the termD(x–t) appears in solution (8) instead of Bssinh

µs(x–t)

and the corresponding termsDλ^–2_mappear in system (9).

20. y(x)–A x

a

sin(λx)

sin(λt)y(t)dt=f(x).

Solution:

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

a

e^A(x–t)sin(λx) sin(λt)f(t)dt.

21. y(x)–A x

a

sin(λt)

sin(λx)y(t)dt=f(x).

Solution:

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

a

e^A(x–t)sin(λt) sin(λx)f(t)dt.

22. y(x)–A x

a

sin^k(λx) sin^m(µt)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =Asin^k(λx) andh(t) = sin^m(µt).

23. y(x) +A _x

a

tsin[λ(x–t)]y(t)dt=f(x).

This is a special case of equation 2.9.36 withg(t) =At.

Solution:

y(x) =f(x) + Aλ W

_x

a

t

u₁(x)u₂(t)–u₂(x)u₁(t) f(t)dt,

whereu1(x),u2(x) is a fundamental system of solutions of the second-order linear ordinary differential equationu_xx+λ(Ax+λ)u= 0, andW is the Wronskian.

Depending on the sign ofAλ, the functionsu1(x) andu2(x) are expressed in terms of Bessel functions or modified Bessel functions as follows:

ifAλ> 0, then

u1(x) =ξ^1/2J_1/32

3

√Aλ ξ^3/2

, u2(x) =ξ^1/2Y_1/32

3

√Aλ ξ^3/2 , W = 3/π, ξ=x+ (λ/A);

ifAλ< 0, then

u1(x) =ξ^1/2I_1/32

3

√–Aλ ξ^3/2

, u2(x) =ξ^1/2K_1/32

3

√–Aλ ξ^3/2 , W =–³₂, ξ=x+ (λ/A).

24. y(x) +A _x

a

xsin[λ(x–t)]y(t)dt=f(x).

This is a special case of equation 2.9.37 withg(x) =Axandh(t) = 1.

Solution:

y(x) =f(x) +Aλ W

x a

x

u₁(x)u₂(t)–u₂(x)u₁(t) f(t)dt,

whereu₁(x),u₂(x) is a fundamental system of solutions of the second-order linear ordinary differential equationu_xx+λ(Ax+λ)u= 0, andW is the Wronskian.

The functionsu1(x),u2(x), andW are specified in 2.5.23.

25. y(x) +A _x

a

t^ksin^m(λx)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =–Asin^m(λx) andh(t) =t^k. 26. y(x) +A

x a

x^ksin^m(λt)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =–Ax^kandh(t) = sin^m(λt).

27. y(x)– _x

a

Asin(kx) +B–AB(x–t) sin(kx)

y(t)dt=f(x).

This is a special case of equation 2.9.7 withλ=Bandg(x) =Asin(kx).

Solution:

y(x) =f(x) + x

a

R(x,t)f(t)dt, R(x,t) = [Asin(kx) +B]G(x)

G(t) + B² G(t)

x t

e^B(x–s)G(s)ds, G(x) = exp

–A

k cos(kx)

.

28. y(x) + x

a

Asin(kt) +B+AB(x–t) sin(kt)

y(t)dt=f(x).

This is a special case of equation 2.9.8 withλ=Bandg(t) =Asin(kt).

Solution:

y(x) =f(x) + x

a

R(x,t)f(t)dt, R(x,t) =–[Asin(kt) +B]G(t)

G(x) + B² G(x)

x t

e^B(t–s)G(s)ds, G(x) = exp

–A

k cos(kx)

.

29. y(x) +A _∞

x

sin λ√

t–x

y(t)dt=f(x).

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

–x .

2.5-3. Kernels Containing Tangent

30. y(x)–A _x

a

tan(λx)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =Atan(λx) andh(t) = 1.

Solution:

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

a

tan(λx) cos(λt) cos(λx)

^A/λf(t)dt.

31. y(x)–A x

a

tan(λt)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =Aandh(t) = tan(λt).

Solution:

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

a

tanh(λt) cos(λt) cos(λx)

^A/λf(t)dt.

32. y(x) +A x

a

tan(λx)–tan(λt)

y(t)dt=f(x).

This is a special case of equation 2.9.5 withg(x) =Atan(λx).

Solution:

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

x a

Y₁(x)Y₂(t)–Y₂(x)Y₁(t) f(t)dt,

whereY1(x),Y2(x) is a fundamental system of solutions of the second-order linear ordinary differential equation cos²(λx)Y_xx +AλY = 0,W is the Wronskian, and the primes stand for the differentiation with respect to the argument specified in the parentheses.

As shown in A. D. Polyanin and V. F. Zaitsev (1995, 1996), the functionsY1(x) andY2(x) can be expressed via the hypergeometric function.

33. y(x)–A _x

a

tan(λx)

tan(λt)y(t)dt=f(x).

Solution:

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

a

e^A(x–t)tan(λx) tan(λt)f(t)dt.

34. y(x)–A _x

a

tan(λt)

tan(λx)y(t)dt=f(x).

Solution:

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

a

e^A(x–t) tan(λt) tan(λx)f(t)dt.

35. y(x)–A _x

a

tan^k(λx) tan^m(µt)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =Atan^k(λx) andh(t) = tan^m(µt).

36. y(x) +A x

a

t^ktan^m(λx)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =–Atan^m(λx) andh(t) =t^k. 37. y(x) +A

x a

x^ktan^m(λt)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =–Ax^kandh(t) = tan^m(λt).

38. y(x)– _x

a

Atan(kx) +B–AB(x–t) tan(kx)

y(t)dt=f(x).

This is a special case of equation 2.9.7 withλ=Bandg(x) =Atan(kx).

39. y(x) + _x

a

Atan(kt) +B+AB(x–t) tan(kt)

y(t)dt=f(x).

This is a special case of equation 2.9.8 withλ=Bandg(t) =Atan(kt).

2.5-4. Kernels Containing Cotangent

40. y(x)–A _x

a

cot(λx)y(t)dt =f(x).

This is a special case of equation 2.9.2 withg(x) =Acot(λx) andh(t) = 1.

Solution:

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

a

cot(λx) sin(λx) sin(λt)

^A/λf(t)dt.

41. y(x)–A x

a

cot(λt)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =Aandh(t) = cot(λt).

Solution:

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

a

coth(λt) sin(λx) sin(λt)

^A/λf(t)dt.

42. y(x)–A _x

a

cot(λx)

cot(λt)y(t)dt=f(x).

Solution:

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

a

e^A(x–t)cot(λx) cot(λt)f(t)dt.

43. y(x)–A _x

a

cot(λt)

cot(λx)y(t)dt=f(x).

Solution:

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

a

e^A(x–t) cot(λt) cot(λx)f(t)dt.

44. y(x) +A x

a

t^kcot^m(λx)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =–Acot^m(λx) andh(t) =t^k. 45. y(x) +A

x a

x^kcot^m(λt)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =–Ax^kandh(t) = cot^m(λt).

46. y(x)– _x

a

Acot(kx) +B–AB(x–t) cot(kx)

y(t)dt=f(x).

This is a special case of equation 2.9.7 withλ=Bandg(x) =Acot(kx).

47. y(x) + x

a

Acot(kt) +B+AB(x–t) cot(kt)

y(t)dt=f(x).

This is a special case of equation 2.9.8 withλ=Bandg(t) =Acot(kt).

2.5-5. Kernels Containing Combinations of Trigonometric Functions 48. y(x)–A

_x

a

cos^k(λx) sin^m(µt)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =Acos^k(λx) andh(t) = sin^m(µt).

49. y(x)– _x

a

A+Bcos(λx)–B(x–t)[λsin(λx) +Acos(λx)]

y(t)dt=f(x).

This is a special case of equation 2.9.38 withb=Bandg(x) =A.

50. y(x)– x

a

A+Bsin(λx) +B(x–t)[λcos(λx)–Asin(λx)]

y(t)dt=f(x).

This is a special case of equation 2.9.39 withb=Bandg(x) =A.

51. y(x)–A x

a

tan^k(λx) cot^m(µt)y(t)dt=f(x).

This is a special case of equation 2.9.2 withg(x) =Atan^k(λx) andh(t) = cot^m(µt).

2.6. Equations Whose Kernels Contain Inverse