Chapter 13

Integral Equations

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Section 13.1Examples of Problems with Solutions[4/100%]
Problem 13.1

Construct the resolvent of the integral equation

x(t)=4∫0tet−sx(s)ds+e3t,t∈[0,T] x(t) = 4 \int _{0}^{t} e^{t-s} x(s) d s+e^{3 t}, t \in [0, T]

and apply it to solve the equation.

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Problem 13.2

Solve in the space L2([0,1])L_{2}(\left[0,1\right]) the integral equation with a degenerate kernel

x(t)=λ∫01(28t5s2−15t3s)x(s)ds+1t3,λ∈C x(t) = \lambda \int _{0}^{1}\left(28 t^{5} s^{2}-15 t^{3} s\right) x(s) d s+\frac{1}{\sqrt[3]{t}}, \lambda \in \mathbb {C}
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Problem 13.3

Find all values of the parameters p,q,rp, q, r, for which the integral equation x(t)=λ∫−11(1+2t+3st)x(s)ds+pt2+qt+r,t∈[−1,1]x(t) = \lambda \int_{-1}^{1}(1+2 t+3 s t) x(s) d s+p t^{2}+q t+r, t \in [-1,1], has a solution in the space L2([−1,1])L_{2}(\left[-1,1\right]) for every λ∈C\lambda \in \mathbb {C}.

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Problem 13.4

Find the characteristic numbers and corresponding normalized eigenfunctions and solve the integral equation with a symmetric kernel

x(t)=λ∫02π(cos⁡(t−s)+2sin⁡2tsin⁡2s)x(s)ds+cos⁡t,t∈[0,2π] x(t) = \lambda \int _{0}^{2 \pi }(\cos (t-s)+2 \sin 2 t \sin 2 s) x(s) d s+\cos t, t \in [0,2 \pi ]
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Section 13.2Problems to Solve[27/93%]
Problem 13.5

Solve the integral equations by the method of successive approximations, taking x0(t)=0x_{0}(t) = 0 :

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(1)

x(t)=∫0tx(s)1+s2ds+arctan⁡t,t∈[0,T],T>0;x(t) = \int_{0}^{t} \frac{x(s)}{1+s^{2}} d s+\arctan t, t \in [0, T], T > 0 ;

(2)

x(t)=∫0te−(t−s)x(s)ds+1,t∈[0,1];x(t) = \int_{0}^{t} e^{-(t-s)} x(s) d s+1, t \in [0,1] ;

(3)

x(t)=∫0t(t−s)x(s)ds+1,t∈[0,1];x(t) = \int_{0}^{t}(t-s) x(s) d s+1, t \in [0,1] ;

(4)

x(t)=∫01e−t−sx(s)ds+1,t∈[0,1];x(t) = \int_{0}^{1} e^{-t-s} x(s) d s+1, t \in [0,1] ;

(5)

x(t)=∫01x(s)1+s2ds+arctan⁡t,t∈[0,1];x(t) = \int_{0}^{1} \frac{x(s)}{1+s^{2}} d s+\arctan t, t \in [0,1] ;

(6)

x(t)=∫01tets2(et−1)x(s)ds+12,t∈[0,1];x(t) = \int_{0}^{1} \frac{t e^{t s}}{2\left(e^{t}-1\right)} x(s) d s+\frac{1}{2}, t \in [0,1] ;

(7)

x(t)=∫01sx(s)ds+1,t∈[0,1];x(t) = \int_{0}^{1} s x(s) d s+1, t \in [0,1] ;

(8)

x(t)=∫12t2sx(s)ds+1,t∈[1,2]x(t) = \int_{1}^{2} \frac{t}{2 s} x(s) d s+1, t \in [1,2].

Problem 13.6

Construct the first two successive approximations of the integral equation, taking x0(t)=0x_{0}(t) = 0 :

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(1)

x(t)=∫01x(s)(1+s+t)2ds+1,t∈[0,1];x(t) = \int_{0}^{1} \frac{x(s)}{(1+s+t)^{2}} d s+1, t \in [0,1] ;

(2)

x(t)=∫0113(ts+s2)x(s)ds+t,t∈[0,1];x(t) = \int_{0}^{1} \frac{1}{3}\left(t s+s^{2}\right) x(s) d s+t, t \in [0,1] ;

(3)

x(t)=∫0π4110arctan⁡(ts)x(s)ds+1,t∈[0,π4]x(t) = \int_{0}^{\frac{\pi }{4}} \frac{1}{10} \arctan (t s) x(s) d s+1, t \in \left[0, \frac{\pi }{4}\right].

Problem 13.7

Construct the resolvent of an integral equation

x(t)=λ∫0tK(t,s)x(s)ds+y(t),t∈[0,T] x(t) = \lambda \int _{0}^{t} K(t, s) x(s) d s+y(t), t \in [0, T]

and solve the equation.

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(1)

K(t,s)=et2−s2,λ=2,y(t)=et2+2tK(t, s) = e^{t^{2}-s^{2}}, \lambda = 2, y(t) = e^{t^{2}+2 t};

(2)

K(t,s)=et−s,λ=2,y(t)=sin⁡tK(t, s) = e^{t-s}, \lambda = 2, y(t) = \sin t;

(3)

K(t,s)=1+t21+s2,λ=1,y(t)=1+t2K(t, s) = \frac{1+t^{2}}{1+s^{2}}, \lambda = 1, y(t) = 1+t^{2};

(4)

K(t,s)=3t−s,λ=−1,y(t)=t3tK(t, s) = 3^{t-s}, \lambda = -1, y(t) = t 3^{t};

(5)

K(t,s)=2+cos⁡t2+cos⁡s,λ=1,y(t)=etsin⁡tK(t, s) = \frac{2+\cos t}{2+\cos s}, \lambda = 1, y(t) = e^{t} \sin t;

(6)

K(t,s)=cosh⁡tcosh⁡s,λ=−1,y(t)=tcosh⁡tK(t, s) = \frac{\cosh t}{\cosh s}, \lambda = -1, y(t) = t \cosh t;

(7)

K(t,s)=es−t,λ=1,y(t)=tet22K(t, s) = e^{s-t}, \lambda = 1, y(t) = t e^{\frac{t^{2}}{2}}.

Problem 13.8

Construct the resolvent of an integral equation

x(t)=λ∫abK(t,s)x(s)ds+y(t),t∈[a,b] x(t) = \lambda \int _{a}^{b} K(t, s) x(s) d s+y(t), t \in [a, b]

and solve the equation.

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(1)

K(t,s)=et+s,[a,b]=[0,1],λ=1e2−1,y(t)=sin⁡πtK(t, s) = e^{t+s},\left[a, b\right] = \left[0,1\right], \lambda = \frac{1}{e^{2}-1}, y(t) = \sin \pi t;

(2)

K(t,s)=sin⁡tcos⁡s,[a,b]=[0,π2],λ=−1,y(t)=cos⁡tK(t, s) = \sin t \cos s,\left[a, b\right] = \left[0, \frac{\pi }{2}\right], \lambda = -1, y(t) = \cos t;

(3)

K(t,s)=tes,[a,b]=[−1,1],λ=e3,y(t)=(3t2+1)e−tK(t, s) = t e^{s},\left[a, b\right] = \left[-1,1\right], \lambda = \frac{e}{3}, y(t) = \left(3 t^{2}+1\right) e^{-t};

(4)

K(t,s)=t2s2,[a,b]=[−1,1],λ=2,y(t)=etK(t, s) = t^{2} s^{2},\left[a, b\right] = \left[-1,1\right], \lambda = 2, y(t) = e^{t};

(5)

K(t,s)=(1+t)(1−s),[a,b]=[−1,0],λ=1,y(t)=πcos⁡πtK(t, s) = (1+t)(1-s),\left[a, b\right] = \left[-1,0\right], \lambda = 1, y(t) = \pi \cos \pi t;

(6)

K(t,s)=ts+t2s2,[a,b]=[−1,1],λ=1,y(t)=3(t+1)K(t, s) = t s+t^{2} s^{2},\left[a, b\right] = \left[-1,1\right], \lambda = 1, y(t) = 3(t+1);

(7)

K(t,s)=1+(2t−1)(2s−1),[a,b]=[0,1],λ=12,y(t)=3t2K(t, s) = 1+(2 t-1)(2 s-1),\left[a, b\right] = \left[0,1\right], \lambda = \frac{1}{2}, y(t) = 3 t^{2};

(8)

K(t,s)=sin⁡tcos⁡s+cos⁡2tsin⁡2s,[a,b]=[0,2π],λ=−1π,y(t)=cos⁡t+sin⁡tK(t, s) = \sin t \cos s+\cos 2 t \sin 2 s,\left[a, b\right] = \left[0,2 \pi \right], \lambda = -\frac{1}{\pi }, y(t) = \cos t+\sin t

(9)

K(t,s)=sin⁡tsin⁡s+cos⁡2tcos⁡2s,[a,b]=[0,2π],λ=12π,y(t)=cos⁡2t;K(t, s) = \sin t \sin s+\cos 2 t \cos 2 s,\left[a, b\right] = \left[0,2 \pi \right], \lambda = \frac{1}{2 \pi }, \quad y(t) = \cos 2 t ;

(10)

K(t,s)=cos⁡tcos⁡s+3sin⁡2tsin⁡2s,[a,b]=[0,2π],λ=16π,y(t)=5cos⁡tK(t, s) = \cos t \cos s+3 \sin 2 t \sin 2 s,\left[a, b\right] = \left[0,2 \pi \right], \lambda = \frac{1}{6 \pi }, y(t) = 5 \cos t;

(11)

K(t,s)=2+cos⁡tcos⁡s,[a,b]=[0,2π],λ=−18π,y(t)=2tK(t, s) = 2+\cos t \cos s,\left[a, b\right] = \left[0,2 \pi \right], \lambda = -\frac{1}{8 \pi }, y(t) = 2 t;

(12)

K(t,s)=etcos⁡s,[a,b]=[0,π],λ=1eπ+1,y(t)=t4K(t, s) = e^{t} \cos s,\left[a, b\right] = \left[0, \pi \right], \lambda = \frac{1}{e^{\pi }+1}, y(t) = \frac{t}{4}.

Problem 13.9

Let RλR_{\lambda } be the resolvent of an integral equation

x(t)=λ∫abK(t,s)x(s)ds+y(t),t∈[a,b] x(t) = \lambda \int _{a}^{b} K(t, s) x(s) d s+y(t), t \in [a, b]

with a kernel K∈C([a,b]2),M=max⁡a≤t,s≤b∣K(t,s)∣>0K \in C\left(\left[a, b\right]^{2}\right), M = \max_{a \leq t, s \leq b}\left|K(t, s)\right| > 0. Show that for λ\lambda such that ∣λ∣<(M(b−a))−1\left|\lambda \right| < (M(b-a))^{-1} and for all t,s∈[a,b]t, s \in [a, b] the function RλR_{\lambda } satisfies the following relations:

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(1)

Rλ(t,s)=λ∫abK(t,u)Rλ(u,s)du+K(t,s)R_{\lambda }(t, s) = \lambda \int_{a}^{b} K(t, u) R_{\lambda }(u, s) d u+K(t, s);

(2)

Rλ(t,s)=λ∫abK(u,s)Rλ(t,u)du+K(t,s);R_{\lambda }(t, s) = \lambda \int_{a}^{b} K(u, s) R_{\lambda }(t, u) d u+K(t, s) ;

(3)

∂Rλ(t,s)∂λ=∫abRλ(t,u)Rλ(u,s)du\frac{\partial R_{\lambda }(t, s)}{\partial \lambda } = \int_{a}^{b} R_{\lambda }(t, u) R_{\lambda }(u, s) d u.

Problem 13.10

Solve an integral equation

x(t)=λ∫abK(t,s)x(s)ds+y(t),t∈[a,b],λ∈C x(t) = \lambda \int _{a}^{b} K(t, s) x(s) d s+y(t), t \in [a, b], \lambda \in \mathbb {C}

with a degenerate kernel.

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(1)

K(t,s)=1,y(t)=1,[a,b]=[−1,1]K(t, s) = 1, y(t) = 1,\left[a, b\right] = \left[-1,1\right];

(2)

K(t,s)=ts+t2s2,y(t)=t2+t4,[a,b]=[−1,1]K(t, s) = t s+t^{2} s^{2}, y(t) = t^{2}+t^{4},\left[a, b\right] = \left[-1,1\right];

(3)

K(t,s)=sin⁡(t−2s),y(t)=cos⁡2t,[a,b]=[0,π]K(t, s) = \sin (t-2 s), y(t) = \cos 2 t,\left[a, b\right] = \left[0, \pi \right];

(4)

K(t,s)=t3+s3,y(t)=1−6t2,[a,b]=[−1,1]K(t, s) = \sqrt[3]{t}+\sqrt[3]{s}, y(t) = 1-6 t^{2},\left[a, b\right] = \left[-1,1\right];

(5)

K(t,s)=t4+5t3s,y(t)=t2−t4,[a,b]=[−1,1]K(t, s) = t^{4}+5 t^{3} s, y(t) = t^{2}-t^{4},\left[a, b\right] = \left[-1,1\right];

(6)

K(t,s)=2ts3+5t2s2,y(t)=7t4+3,[a,b]=[−1,1]K(t, s) = 2 t s^{3}+5 t^{2} s^{2}, y(t) = 7 t^{4}+3,\left[a, b\right] = \left[-1,1\right];

(7)

K(t,s)=t2−ts,y(t)=t2+t,[a,b]=[−1,1]K(t, s) = t^{2}-t s, y(t) = t^{2}+t,\left[a, b\right] = \left[-1,1\right];

(8)

K(t,s)=sin⁡(2t+s),y(t)=π−2t,[a,b]=[0,π]K(t, s) = \sin (2 t+s), y(t) = \pi -2 t,\left[a, b\right] = \left[0, \pi \right];

(9)

K(t,s)=cos⁡(2t+s),y(t)=sin⁡t,[a,b]=[0,π]K(t, s) = \cos (2 t+s), y(t) = \sin t,\left[a, b\right] = \left[0, \pi \right];

(10)

K(t,s)=sin⁡(3t+s),y(t)=cos⁡t,[a,b]=[0,π]K(t, s) = \sin (3 t+s), y(t) = \cos t,\left[a, b\right] = \left[0, \pi \right];

(11)

K(t,s)=sin⁡s+scos⁡t,y(t)=1−2tπ,[a,b]=[0,π]K(t, s) = \sin s+s \cos t, y(t) = 1-\frac{2 t}{\pi },\left[a, b\right] = \left[0, \pi \right];

(12)

K(t,s)=cos⁡2(t−s),y(t)=1+cos⁡4t,[a,b]=[0,π]K(t, s) = \cos^{2}(t-s), y(t) = 1+\cos 4 t,\left[a, b\right] = \left[0, \pi \right];

(13)

K(t,s)=cos⁡tcos⁡s+cos⁡2tcos⁡2s,y(t)=cos⁡3t,[a,b]=[0,2π]K(t, s) = \cos t \cos s+\cos 2 t \cos 2 s, y(t) = \cos 3 t,\left[a, b\right] = \left[0,2 \pi \right];

(14)

K(t,s)=cos⁡tcos⁡s+2sin⁡2tsin⁡2s,y(t)=cos⁡t,[a,b]=[0,2π]K(t, s) = \cos t \cos s+2 \sin 2 t \sin 2 s, y(t) = \cos t,\left[a, b\right] = \left[0,2 \pi \right];

(15)

K(t,s)=sin⁡tsin⁡s+3cos⁡2tcos⁡2s,y(t)=sin⁡t,[a,b]=[0,2π]K(t, s) = \sin t \sin s+3 \cos 2 t \cos 2 s, y(t) = \sin t,\left[a, b\right] = \left[0,2 \pi \right];

Problem 13.11

Solve an integral equation

x(t)=∫abK(t,s)x(s)ds+y(t),t∈[a,b] x(t) = \int _{a}^{b} K(t, s) x(s) d s+y(t), t \in [a, b]

with a degenerate kernel.

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(1)

K(t,s)=∑k=1100sin⁡ktsin⁡ksk,y(t)=sin⁡2t,[a,b]=[0,π]K(t, s) = \sum_{k = 1}^{100} \frac{\sin k t \sin k s}{\sqrt{k}}, y(t) = \sin \sqrt{2} t,\left[a, b\right] = \left[0, \pi \right];

(2)

K(t,s)=s2+t2,y(t)=1,[a,b]=[0,1]K(t, s) = s^{2}+t^{2}, y(t) = 1,\left[a, b\right] = \left[0,1\right];

(3)

K(t,s)=cos⁡(t−s),y(t)=sin⁡t,[a,b]=[0,π2]K(t, s) = \cos (t-s), y(t) = \sin t,\left[a, b\right] = \left[0, \frac{\pi }{2}\right];

(4)

K(t,s)={4ts,0≤s≤1,3t2,1<s≤2,y(t)=1,[a,b]=[0,2]K(t, s) = \begin{cases} 4 t s, 0 \leq s \leq 1, \\ 3 t^{2}, 1 < s \leq 2, \end{cases} y(t) = 1,\left[a, b\right] = \left[0,2\right];

(5)

K(t,s)={3t2−2s,−1≤s≤0,5ts3,0<s≤1,y(t)=t2+1K(t, s) = \begin{cases} 3 t^{2}-2 s, & -1 \leq s \leq 0, \\ 5 t s^{3}, & 0 < s \leq 1, \end{cases} \quad y(t) = t^{2}+1,

[a,b]=[−1,1] \left[a, b\right] = \left[-1,1\right]
Problem 13.12

Determine for which λ∈C\lambda \in \mathbb {C} an integral equation

x(t)=λ∫abK(t,s)x(s)ds+y(t),t∈[a,b] x(t) = \lambda \int _{a}^{b} K(t, s) x(s) d s+y(t), t \in [a, b]

has a solution in the space L2([a,b])L_{2}(\left[a, b\right]).

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(1)

K(t,s)=et−s,y(t)=1,t∈[a,b]K(t, s) = e^{t-s}, y(t) = 1, t \in [a, b];

(2)

K(t,s)=ts2e−s4,y(t)=t,t∈[a,b]K(t, s) = t s^{2} e^{-s^{4}}, y(t) = t, t \in [a, b].

Problem 13.13

For which y∈L2([−1,1])y \in L_{2}(\left[-1,1\right]) does an equation

x(t)=λ∫−11K(t,s)x(s)ds+y(t) x(t) = \lambda \int _{-1}^{1} K(t, s) x(s) d s+y(t)

have a solution in L2([−1,1])L_{2}(\left[-1,1\right]) for arbitrary λ∈C\lambda \in \mathbb {C}, if

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(1)

K(t,s)=ts3K(t, s) = t s^{3},

(2)

K(t,s)=st+2K(t, s) = \frac{s}{t+2},

(3)

K(t,s)=t+1s3K(t, s) = \frac{t+1}{\sqrt[3]{s}}?

For which λ\lambda is the solution unique?

Problem 13.14

Find all values of parameters p,q,rp, q, r, for which an integral equation

x(t)=λ∫abK(t,s)x(s)ds+y(t),t∈[a,b] x(t) = \lambda \int _{a}^{b} K(t, s) x(s) d s+y(t), t \in [a, b]

has a solution in the space L2([a,b])L_{2}(\left[a, b\right]) for every λ∈C\lambda \in \mathbb {C}.

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(1)

K(t,s)=ts5+t4s2,y(t)=pt2+qt+r,[a,b]=[−1,1]K(t, s) = t s^{5}+t^{4} s^{2}, y(t) = p t^{2}+q t+r,\left[a, b\right] = \left[-1,1\right];

(2)

K(t,s)=1+t3s,y(t)=pt2+qt+r,[a,b]=[−1,1]K(t, s) = 1+t^{3} s, y(t) = p t^{2}+q t+r,\left[a, b\right] = \left[-1,1\right];

(3)

K(t,s)=t2+ts2,y(t)=pt+q,[a,b]=[−1,1]K(t, s) = t^{2}+t s^{2}, y(t) = p t+q,\left[a, b\right] = \left[-1,1\right];

(4)

K(t,s)=2t5s3+t2s2,y(t)=pt+q,[a,b]=[−1,1]K(t, s) = 2 t^{5} s^{3}+t^{2} s^{2}, y(t) = p t+q,\left[a, b\right] = \left[-1,1\right];

(5)

K(t,s)=its+t2s4,y(t)=pt2+q,[a,b]=[−1,1]K(t, s) = i t s+t^{2} s^{4}, y(t) = p t^{2}+q,\left[a, b\right] = \left[-1,1\right];

(6)

K(t,s)=5t2s+3ts2,y(t)=pt+qt3,[a,b]=[−1,1]K(t, s) = 5 t^{2} s+3 t s^{2}, y(t) = p t+q t^{3},\left[a, b\right] = \left[-1,1\right];

(7)

K(t,s)=t3−s3,y(t)=pt2+qt+r,[a,b]=[−1,1]K(t, s) = \sqrt[3]{t}-\sqrt[3]{s}, y(t) = p t^{2}+q t+r,\left[a, b\right] = \left[-1,1\right];

(8)

K(t,s)=3t+ts−5t2s2,y(t)=pt5,[a,b]=[−1,1]K(t, s) = 3 t+t s-5 t^{2} s^{2}, y(t) = p t^{5},\left[a, b\right] = \left[-1,1\right];

(9)

K(t,s)=2t5s+3it2s2,y(t)=pt3+qt,[a,b]=[−1,1]K(t, s) = 2 t^{5} s+3 i t^{2} s^{2}, y(t) = p t^{3}+q t,\left[a, b\right] = \left[-1,1\right];

(10)

K(t,s)=ssin⁡t+cos⁡s,y(t)=pt+q,[a,b]=[−π2,π2]K(t, s) = s \sin t+\cos s, y(t) = p t+q,\left[a, b\right] = \left[-\frac{\pi }{2}, \frac{\pi }{2}\right];

(11)

K(t,s)=cos⁡(t+s),y(t)=psin⁡t+q,[a,b]=[0,π]K(t, s) = \cos (t+s), y(t) = p \sin t+q,\left[a, b\right] = \left[0, \pi \right];

(12)

K(t,s)=tcos⁡s+sin⁡tsin⁡s,y(t)=p+qcos⁡t+rsin⁡t,[a,b]=[−π,π]K(t, s) = t \cos s+\sin t \sin s, y(t) = p+q \cos t+r \sin t,\left[a, b\right] = \left[-\pi , \pi \right];

(13)

K(t,s)=tsin⁡s+cos⁡t,y(t)=pt+q,[a,b]=[−π,π]K(t, s) = t \sin s+\cos t, y(t) = p t+q,\left[a, b\right] = \left[-\pi , \pi \right].

Problem 13.15

For which functions y∈C([0,π])y \in C(\left[0, \pi \right]) does the integral equation

x(t)=∫0πsin⁡(t−s)x(s)ds+y(t),t∈[0,π] x(t) = \int _{0}^{\pi } \sin (t-s) x(s) d s+y(t), t \in [0, \pi ]

have a solution in the space C([0,π])C(\left[0, \pi \right])?

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Problem 13.16

Find all real values of the parameter pp, for which the integral equation x(t)=λ∫01(pt−s)x(s)ds+y(t),t∈[0,1]x(t) = \lambda \int_{0}^{1}(p t-s) x(s) d s+y(t), \quad t \in [0,1], has a solution for all λ∈R\lambda \in \mathbb {R} and all y∈L2([0,1])y \in L_{2}(\left[0,1\right]).

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Problem 13.17

Find all values of λ\lambda, for which an integral equation

x(t)=λ∫abK(t,s)x(s)ds+y(t),t∈[a,b] x(t) = \lambda \int _{a}^{b} K(t, s) x(s) d s+y(t), t \in [a, b]

has a unique solution for every y∈C([a,b])y \in C(\left[a, b\right]).

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(1)

K(t,s)=12+cos⁡2(t+s),[a,b]=[0,2π]K(t, s) = \frac{1}{2}+\cos^{2}(t+s),\left[a, b\right] = \left[0,2 \pi \right];

(2)

K(t,s)=sin⁡(t−s),[a,b]=[0,2π]K(t, s) = \sin (t-s),\left[a, b\right] = \left[0,2 \pi \right];

(3)

K(t,s)=45t2s2−2,[a,b]=[0,1]K(t, s) = 45 t^{2} s^{2}-2,\left[a, b\right] = \left[0,1\right].

Problem 13.18

Find all values of λ\lambda, for which an integral equation

x(t)=λ∫01K(t,s)x(s)ds+y(t),t∈[0,1] x(t) = \lambda \int _{0}^{1} K(t, s) x(s) d s+y(t), t \in [0,1]

has a solution for every y∈Lp([0,1])y \in L_{p}(\left[0,1\right]).

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(1)

K(t,s)=(ts)25+(st)25,p=2K(t, s) = \left(\frac{t}{s}\right)^{\frac{2}{5}}+\left(\frac{s}{t}\right)^{\frac{2}{5}}, p = 2;

(2)

K(t,s)=s1−t+1−st,p=32K(t, s) = \frac{s}{\sqrt{1-t}}+\frac{1-s}{\sqrt{t}}, p = \frac{3}{2};

(3)

K(t,s)=st4+ts4,p=3K(t, s) = \frac{\sqrt{s}}{\sqrt[4]{t}}+\frac{\sqrt{t}}{\sqrt[4]{s}}, p = 3.

(4)

K(t,s)=t71−s2,p=+∞K(t, s) = \frac{t^{7}}{\sqrt{1-s^{2}}}, p = +\infty.

Problem 13.19

Let f,a1,…,an∈C([0,T]),T>0f, a_{1}, \ldots , a_{n} \in C(\left[0, T\right]), T > 0, and c0,c1,…,cn−1∈Cc_{0}, c_{1}, \ldots , c_{n-1} \in \mathbb {C}. Prove that the function x∈Cn([0,T])x \in C^{n}(\left[0, T\right]), for which x(k)(0)=ck,k=0,1,…,n−1x^{(k)}(0) = c_{k}, k = 0,1, \ldots , n-1, is a solution of the differential equation

x(n)(t)+a1(t)x(n−1)(t)+…+an(t)x(t)=f(t),t∈[0,T] x^{(n)}(t)+a_{1}(t) x^{(n-1)}(t)+\ldots +a_{n}(t) x(t) = f(t), t \in [0, T]

if and only if the function y=x(n)y = x^{(n)} is a solution of the Volterra integral equation of the second kind y(t)=−∫0tK(t,s)y(s)ds+F(t),t∈[0,T]y(t) = -\int_{0}^{t} K(t, s) y(s) d s+F(t), t \in [0, T], where

K(t,s)=∑k=1nak(t)(t−s)k−1(k−1)!,F(t)=f(t)−∑k=1n(∑j=0k−1cn−k+jtjj!)ak(t). K(t, s) = \sum _{k = 1}^{n} a_{k}(t) \frac{(t-s)^{k-1}}{(k-1)!}, \quad F(t) = f(t)-\sum _{k = 1}^{n}\left(\sum _{j = 0}^{k-1} c_{n-k+j} \frac{t^{j}}{j!}\right) a_{k}(t).
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Problem 13.20

Find the eigenvalues and corresponding normalized eigenfunctions of an integral equation x(t)=λ∫abK(t,s)x(s)ds,t∈[a,b]x(t) = \lambda \int_{a}^{b} K(t, s) x(s) d s, t \in [a, b].

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(1)

K(t,s)=ts,[a,b]=[1,2]K(t, s) = t s,\left[a, b\right] = \left[1,2\right];

(2)

K(t,s)=∑n=1∞sin⁡ntsin⁡nsn2+n,[a,b]=[0,π]K(t, s) = \sum_{n = 1}^{\infty } \frac{\sin n t \sin n s}{n^{2}+n},\left[a, b\right] = \left[0, \pi \right];

(3)

K(t,s)=5ts−2t−2s+1,[a,b]=[0,1]K(t, s) = 5 t s-2 t-2 s+1,\left[a, b\right] = \left[0,1\right];

(4)

K(t,s)=6i(t−s),[a,b]=[0,1]K(t, s) = 6 i(t-s),\left[a, b\right] = \left[0,1\right];

(5)

K(t,s)=sin⁡(t+s),[a,b]=[−π,π]K(t, s) = \sin (t+s),\left[a, b\right] = \left[-\pi , \pi \right];

(6)

K(t,s)=11−αei(t−s),0<α<1,[a,b]=[−π,π]K(t, s) = \frac{1}{1-\alpha e^{i(t-s)}}, 0 < \alpha < 1,\left[a, b\right] = \left[-\pi , \pi \right].

Problem 13.21

Find the eigenvalues and corresponding normalized eigenfunctions of an integral equation with a symmetric kernel

x(t)=λ∫abK(t,s)x(s)ds+y(t),t∈[a,b] x(t) = \lambda \int _{a}^{b} K(t, s) x(s) d s+y(t), t \in [a, b]

and solve the equation for each λ∈C\lambda \in \mathbb {C}.

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(1)

K(t,s)=sin⁡tsin⁡s,[a,b]=[0,2π],y(t)=cos⁡t+sin⁡tK(t, s) = \sin t \sin s,\left[a, b\right] = \left[0,2 \pi \right], y(t) = \cos t+\sin t;

(2)

K(t,s)=cos⁡2(t−s),[a,b]=[−π,π],y(t)=sin⁡2tK(t, s) = \cos^{2}(t-s),\left[a, b\right] = \left[-\pi , \pi \right], y(t) = \sin 2 t;

(3)

K(t,s)={t(1−s),t≤s,s(1−t),s≤t,[a,b]=[0,1],y(t)=14sin⁡2πt+19sin⁡3πt;K(t, s) = \begin{cases} t(1-s), & t \leq s, \\ s(1-t), & s \leq t, \end{cases} \quad [a, b] = \left[0,1\right], \quad y(t) = \frac{1}{4} \sin 2 \pi t+\frac{1}{9} \sin 3 \pi t ;

(4)

K(t,s)={(t+1)s,t≤s,t(s+1),s≤t,[a,b]=[0,1],y(t)=sin⁡πt+πcos⁡πt;K(t, s) = \begin{cases} (t+1) s, & t \leq s, \\ t(s+1), & s \leq t, \end{cases} \quad [a, b] = \left[0,1\right], \quad y(t) = \sin \pi t+\pi \cos \pi t ;

(5)

K(t,s)={(t+1)(1−s),t≤s,(1−t)(s+1),s≤t,[a,b]=[−1,1],y(t)=1K(t, s) = \begin{cases} (t+1)(1-s), & t \leq s, \\ (1-t)(s+1), & s \leq t, \end{cases} \quad [a, b] = \left[-1,1\right], \quad y(t) = 1;

(6)

K(t,s)={(et−e−t)(es+e2−s),t≤s,(es−e−s)(et+e2−t),s≤t,[a,b]=[0,1],y(t)=tK(t, s) = \begin{cases} \left(e^{t}-e^{-t}\right)\left(e^{s}+e^{2-s}\right), & t \leq s, \\ \left(e^{s}-e^{-s}\right)\left(e^{t}+e^{2-t}\right), & s \leq t, \end{cases} \quad [a, b] = \left[0,1\right], y(t) = t;

(7)

K(t,s)={sin⁡tcos⁡s,t≤s,sin⁡scos⁡t,s≤t,[a,b]=[0,π2],y(t)=cos⁡tK(t, s) = \begin{cases} \sin t \cos s, & t \leq s, \\ \sin s \cos t, & s \leq t, \end{cases} \quad [a, b] = \left[0, \frac{\pi }{2}\right], y(t) = \cos t;

(8)

K(t,s)={sin⁡tcos⁡s,t≤s,sin⁡scos⁡t,s≤t,[a,b]=[0,π],y(t)=sin⁡32tK(t, s) = \begin{cases} \sin t \cos s, & t \leq s, \\ \sin s \cos t, & s \leq t, \end{cases} \quad [a, b] = \left[0, \pi \right], y(t) = \sin \frac{3}{2} t;

(9)

K(t,s)={cos⁡tsin⁡s,t≤s,cos⁡ssin⁡t,s≤t,[a,b]=[0,π],y(t)=t;K(t, s) = \begin{cases} \cos t \sin s, & t \leq s, \\ \cos s \sin t, & s \leq t, \end{cases} \quad [a, b] = \left[0, \pi \right], \quad y(t) = t ;

(10)

K(t,s)={sin⁡tsin⁡(1−s),t≤s,sin⁡(1−t)sin⁡s,s≤t,[a,b]=[0,1],y(t)=cos⁡πt;K(t, s) = \begin{cases} \sin t \sin (1-s), & t \leq s, \\ \sin (1-t) \sin s, & s \leq t, \end{cases} \quad [a, b] = \left[0,1\right], y(t) = \cos \pi t ;

(11)

K(t,s)=min⁡(t,s),[a,b]=[0,1],y(t)=sin⁡πtK(t, s) = \min (t, s),\left[a, b\right] = \left[0,1\right], y(t) = \sin \pi t.

Problem 13.22

Transform an integral equation μx(t)=∫abK(t,s)x(s)ds+y(t),t∈[a,b]\mu x(t) = \int_{a}^{b} K(t, s) x(s) d s+y(t), t \in [a, b], into a differential one and solve it:

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(1)

y(t)=t2,K(t,s)={π232t(2−s),t≤s,π232s(2−t),s<t,[a,b]=[0,2],μ=1y(t) = \frac{t}{2}, K(t, s) = \begin{cases} \frac{\pi^{2}}{32} t(2-s), & t \leq s, \\ \frac{\pi^{2}}{32} s(2-t), & s < t, \end{cases} \quad [a, b] = \left[0,2\right], \mu = 1;

(2)

y(t)=tet,K(t,s)={−1sinh⁡1sinh⁡tsinh⁡(s−1),t≤s,−1sinh⁡1sinh⁡ssinh⁡(t−1),s<t,[a,b]=[0,1],μ=1;y(t) = t e^{t}, \quad K(t, s) = \begin{cases} -\frac{1}{\sinh 1} \sinh t \sinh (s-1), & t \leq s, \\ -\frac{1}{\sinh 1} \sinh s \sinh (t-1), & s < t, \end{cases} \quad [a, b] = \left[0,1\right], \mu = 1 ;

(3)

y(t)=sin⁡3πt,K(t,s)={(s−1)t,0≤t≤s,(t−1)s,s<t≤1,[a,b]=[0,1],μ=0;y(t) = \sin^{3} \pi t, K(t, s) = \begin{cases} (s-1) t, & 0 \leq t \leq s, \\ (t-1) s, & s < t \leq 1, \end{cases} \quad [a, b] = \left[0,1\right], \quad \mu = 0 ;

(4)

y(t)=2t2−3t+52,K(t,s)=∣t−s∣+t2,[a,b]=[1,2],μ=0y(t) = 2 t^{2}-3 t+\frac{5}{2}, K(t, s) = \left|t-s\right|+t^{2},\left[a, b\right] = \left[1,2\right], \mu = 0.

Problem 13.23

By differentiation, transform a Fredholm integral equation of the first kind ∫abK(t,s)x(s)ds=y(t),t∈[a,b]\int_{a}^{b} K(t, s) x(s) d s = y(t), t \in [a, b], into an equation of the second kind:

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(1)

K(t,s)=1∣t−s∣+4,[a,b]=[0,1]K(t, s) = \frac{1}{\left|t-s\right|+4},\left[a, b\right] = \left[0,1\right];

(2)

K(t,s)=sin⁡∣t−s∣,[a,b]=[1,2]K(t, s) = \sin \left|t-s\right|,\left[a, b\right] = \left[1,2\right];

(3)

K(t,s)=ei∣t−s∣1+∣t−s∣,[a,b]=[0,1]K(t, s) = \frac{e^{i\left|t-s\right|}}{1+\left|t-s\right|},\left[a, b\right] = \left[0,1\right];

(4)

K(t,s)=t+s+2sign⁡(t−s),[a,b]=[0,1]K(t, s) = t+s+2 \operatorname {sign}(t-s),\left[a, b\right] = \left[0,1\right].

Problem 13.24

By differentiation, transform a Volterra integral equation of the first kind ∫atK(t,s)x(s)ds=y(t),t∈[a,b]\int_{a}^{t} K(t, s) x(s) d s = y(t), t \in [a, b], into an equation of the second kind:

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(1)

K(t,s)=t−s+1,y(t)=t,[a,b]=[0,1]K(t, s) = t-s+1, y(t) = t,\left[a, b\right] = \left[0,1\right];

(2)

K(t,s)=ets,y(t)=et−1,[a,b]=[0,1]K(t, s) = e^{t s}, y(t) = e^{t}-1,\left[a, b\right] = \left[0,1\right];

(3)

K(t,s)=sin⁡(t−s),y(t)=cos⁡t−1,[a,b]=[0,π]K(t, s) = \sin (t-s), y(t) = \cos t-1,\left[a, b\right] = \left[0, \pi \right].

Problem 13.25

By change of variable and differentiation, transform a Volterra integral equation of the first kind

∫aφ(t)K(t,s)x(s)ds=y(t),t∈[a,b] \int _{a}^{\varphi (t)} K(t, s) x(s) d s = y(t), t \in [a, b]

into an equation of the second kind:

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(1)

K(t,s)=1−t2+s,y(t)=t4,φ(t)=t2,[a,b]=[0,1]K(t, s) = 1-t^{2}+s, y(t) = t^{4}, \varphi (t) = t^{2},\left[a, b\right] = \left[0,1\right];

(2)

K(t,s)=t+s,y(t)=t4+t62,φ(t)=t3,[a,b]=[1,2]K(t, s) = t+s, y(t) = t^{4}+\frac{t^{6}}{2}, \varphi (t) = t^{3},\left[a, b\right] = \left[1,2\right];

(3)

K(t,s)=1+t+s,y(t)=t2−12+tt−13,φ(t)=t,[a,b]=[0,1]K(t, s) = 1+t+s, y(t) = \frac{t^{2}-1}{2}+\frac{t \sqrt{t}-1}{3}, \varphi (t) = \sqrt{t},\left[a, b\right] = \left[0,1\right].

Problem 13.26

Let KK be a 2π2 \pi-periodic function such that K(t)=∣t∣K(t) = \left|t\right| for t∈[−π,π]t \in [-\pi , \pi ]. Find the eigenvalues and normalized eigenfunctions of an integral equation

x(t)=λ∫−ππK(t+s)x(s)ds+y(t),t∈[−π,π] x(t) = \lambda \int _{-\pi }^{\pi } K(t+s) x(s) d s+y(t), t \in [-\pi , \pi ]

and solve the equation for each λ∈C\lambda \in \mathbb {C}.

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(1)

y(t)=1+2sin⁡t−cos⁡ty(t) = 1+2 \sin t-\cos t;

(2)

y(t)=cos⁡2t+sin⁡3ty(t) = \cos 2 t+\sin 3 t;

(3)

y(t)=sign⁡ty(t) = \operatorname {sign} t;

(4)

y(t)=t2y(t) = t^{2}.

Problem 13.27

Let KK be a function from problem 13.26. Find the characteristic numbers and normalized eigenfunctions of an integral equation

x(t)=λ∫−ππK(t−s)x(s)ds+y(t),t∈[−π,π] x(t) = \lambda \int _{-\pi }^{\pi } K(t-s) x(s) d s+y(t), t \in [-\pi , \pi ]

and solve the equation for each λ∈C\lambda \in \mathbb {C}.

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(1)

y(t)=1+sin⁡t−2cos⁡2ty(t) = 1+\sin t-2 \cos 2 t;

(2)

y(t)=sin⁡t+sin⁡2t+sin⁡3ty(t) = \sin t+\sin 2 t+\sin 3 t;

(3)

y(t)=sign⁡ty(t) = \operatorname {sign} t;

(4)

y(t)=t2y(t) = t^{2}.

Problem 13.28

Let K∈C(R)K \in C(\mathbb {R}) be a 2π2 \pi-periodic function such that ∫−ππK(t)eiktdt≠0\int_{-\pi }^{\pi } K(t) e^{i k t} d t \neq 0 for all k∈Zk \in \mathbb {Z}. Prove that μk=(∫−ππK(t)eiktdt)−1\mu_{k} = \left(\int_{-\pi }^{\pi } K(t) e^{i k t} d t\right)^{-1} and φk(t)=12πe−ikt,k∈Z\varphi_{k}(t) = \frac{1}{\sqrt{2 \pi }} e^{-i k t}, k \in \mathbb {Z}, are the characteristic numbers and corresponding normalized eigenfunctions of the integral equation

x(t)=λ∫−ππK(t−s)x(s)ds x(t) = \lambda \int _{-\pi }^{\pi } K(t-s) x(s) d s
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Problem 13.29

Let {μn:n≥1}\left\{ \mu_{n}: n \geq 1\right\} be a sequence of characteristic numbers and {φn:n≥1}\left\{ \varphi_{n}: n \geq 1\right\} be a corresponding orthonormal sequence of eigenfunctions of the integral equation x(t)=λ∫abK(t,s)x(s)ds,t∈[a,b]x(t) = \lambda \int_{a}^{b} K(t, s) x(s) d s, t \in [a, b], with Hermitian kernel K∈C([a,b]2)K \in C\left(\left[a, b\right]^{2}\right). Prove that

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(1)

∑n=1∞∣φn(t)∣2μn2=∫ab∣K(t,s)∣2ds,t∈[a,b]\sum_{n = 1}^{\infty } \frac{\left|\varphi_{n}(t)\right|^{2}}{\mu_{n}^{2}} = \int_{a}^{b}\left|K(t, s)\right|^{2} d s, t \in [a, b];

(2)

∑n=1∞1μn2=∫[a,b]2∣K(t,s)∣2dtds\sum_{n = 1}^{\infty } \frac{1}{\mu_{n}^{2}} = \int_{\left[a, b\right]^{2}}\left|K(t, s)\right|^{2} d t d s;

(3)

(Ax,x)=∑n=1∞∣(x,φn)∣2μn,x∈L2([a,b])(A x, x) = \sum_{n = 1}^{\infty } \frac{\left|\left(x, \varphi_{n}\right)\right|^{2}}{\mu_{n}}, x \in L_{2}(\left[a, b\right]), where AA is the integral operator with kernel KK.

Problem 13.30
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(1)

Find all λ∈R\lambda \in \mathbb {R}, for which the homogeneous integral equation x(t)=λ∫e−∣t−s∣x(s)ds,t∈Rx(t) = \lambda \int e^{-\left|t-s\right|} x(s) d s, t \in \mathbb {R}, has a nonzero solution x∈L∞(R)x \in L_{\infty }(\mathbb {R}).

(2)

Prove that for all λ<12\lambda < \frac{1}{2} and y∈L∞(R)y \in L_{\infty }(\mathbb {R}) the integral equation x(t)=λ∫e−∣t−s∣x(s)ds+y(t),t∈Rx(t) = \lambda \int e^{-\left|t-s\right|} x(s) d s+y(t), t \in \mathbb {R}, has a unique solution in the space L∞(R)L_{\infty }(\mathbb {R}), which is expressed by the formula

x(t)=λ1−2λ∫Re−1−2λ∣t−s∣y(s)ds+y(t),t∈R x(t) = \frac{\lambda }{\sqrt{1-2 \lambda }} \int _{\mathbb {R}} e^{-\sqrt{1-2 \lambda }\left|t-s\right|} y(s) d s+y(t), t \in \mathbb {R}
Problem 13.31

Let y∈C1([0,T]),T>0,y(0)=0y \in C^{1}(\left[0, T\right]), T > 0, y(0) = 0 and 0<α<10 < \alpha < 1. Prove that the function x(t)=sin⁡αππ∫0ty′(s)(t−s)1−αdsx(t) = \frac{\sin \alpha \pi }{\pi } \int_{0}^{t} \frac{y^{\prime }(s)}{(t-s)^{1-\alpha }} d s satisfies the Abel equation

∫0tx(s)(t−s)αds=y(t),t∈[0,T] \int _{0}^{t} \frac{x(s)}{(t-s)^{\alpha }} d s = y(t), t \in [0, T]
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