Volterra Integral Equations of the First Kind with Jump Discontinuous Kernels

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1 Volterra Integral Equations of the First Kind with Jump Discontinuous Kernels Denis Sidorov Energy Systems Institute, Russian Academy of Sciences INV Follow-up Meeting Isaac Newton Institute for Mathematical Sciences Cambridge, UK, February 10 14, 2014 Denis N. Sidorov Integral Equations with Jump Discontinuous Kernels 1 / 36

2 Outline Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 2 / 36

3 Outline Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 2 / 36

4 Outline Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 2 / 36

5 Outline Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 2 / 36

6 Outline Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 2 / 36

7 Outline Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 2 / 36

8 VIEq with piecewise continuous kernel t 0 K(t, s)x(s)ds = f (t), 0 s t T, f (0) = 0, (1) K 1 (t, s), t, s m 1 K(t, s) = K n (t, s), t, s m n m i = {t, s αi 1 (t) < s < α i (t)}, α 0 (t) = 0, α n (t) = t, i = 1, n α i (t), f (t) C 1 [0,T ], K i (t, s) have continuous derivatives w.r.t. t for t, s m i, K n(t, t) 0, α i (0) = 0, 0 < α 1(t) < α 2(t) < < α n 1(t) < t, α 1(t),..., α n 1(t) increase at least in the small neighborhood 0 t τ, 0 < α 1(0) α n 1(0) < 1 Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 1 / 36

9 VIEq with piecewise continuous kernel t 0 K(t, s)x(s)ds = f (t), 0 s t T, f (0) = 0, (1) K 1 (t, s), t, s m 1 K(t, s) = K n (t, s), t, s m n m i = {t, s αi 1 (t) < s < α i (t)}, α 0 (t) = 0, α n (t) = t, i = 1, n α i (t), f (t) C 1 [0,T ], K i (t, s) have continuous derivatives w.r.t. t for t, s m i, K n(t, t) 0, α i (0) = 0, 0 < α 1(t) < α 2(t) < < α n 1(t) < t, α 1(t),..., α n 1(t) increase at least in the small neighborhood 0 t τ, 0 < α 1(0) α n 1(0) < 1 Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 1 / 36

10 Objectives & Methods Objective Our objective is to construct the solution x(t) C (0,T ] Applications Mathematical models of evolving dynamical systems: vintage capital models, optimal replacement of equipment under technological change, rational harvesting of biological populations. Methods We employ the theory of functional equations a, power-logarithmic asymptotic expansions, the method of steps from delay ODE theory and conventional successive approximations method. For numerical solution the quadrature methods are used. a Gelfond A.O. The Calculus of Finite Differences. 5th Edt, Moscow: URSS Publ., 2012 Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 2 / 36

11 Previous Results t α(t) K(t, s)x(s) ds = f (t), 0 s t T Denisov A. and Lorenzi A. (1995) On a special Volterra integral equation of the first kind. Boll. Un. Mat. Ital. B (7) 9, Apartsyn A. (2003) Nonclassical linear Volterra equations of the first kind. Inverse and Ill-Posed Problems Series. Germany: Walter de Gruyter Publ. Hritonenko N. and Yatsenko Yu. (2009) Integral equation of optimal replacement: analysis and algorithms. Applied Mathematical Modeling Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 3 / 36

12 New α 1(t) 0 K 1 (t, s)x(s) ds+ α 2(t) α 1(t) K 2 (t, s)x(s) ds + t α n 1(t) K n (t, s)x(s) ds = f (t) Sidorov D. (2013) Solution to VIE with piecewise smooth kernels. Differential Equations Vol. 49 (2). P Sidorov D. (2011) VIE with discontinuous kernels in the theory of evolving systems control. Studia Informatica Universalis. Vol. 9 (3). P Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 4 / 36

13 Example 1 x { 1, 0 s < t/2 K(t, s) = 1, t/2 s t t/2 t x(s)ds x(s)ds = t 0 t/2 ( t 2 ) x(t) = 1, x(t) = c ln t ln 2 Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 5 / 36

14 Existence and Uniqueness of the Local Solution F (x) def n 1 { } = K n (t, t)x(t)+ α i(t) K i (t, α i (t)) K i+1 (t, α i (t)) x(α i (t))+ i=1 + n α i (t) i=1 α i 1 (t) K i (t, s) x(s)ds f (t) = 0 (2) t Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 6 / 36

15 Reduction to the VIE of the 2nd kind x(t) + Ax + Qx = ˆf (t) (3) A is the functional perturbation operator: Ax def n 1 { } = Kn 1 (t, t) α i(t) K i (t, α i (t)) K i+1 (t, α i (t)) x(α i (t)) i=1 B is the Volterra operator: Qx := t 0 Q(t, s)x(s)ds def = n α i (t) i=1 α i 1 (t) n (t, s) K i(t, s) x(s)ds t K 1 ˆf (t) def = K 1 n (t, t)f (t), K n (t, t) 0 Objective: To obtain the sufficient conditions for existence of local solution, i.e. A + Q q, q < 1, t (0, τ] Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 7 / 36

16 D(t) def = n 1 i=1 α i (t)k 1 n (t, t) Ki (t, α i (t)) K i+1 (t, α i (t)) The idea is to use an equivalent norm x l := sup 0<t<τ e lt x(t), so that the Volterra integral operator Q becomes contractive. Qx l q(l) x l, lim q(l) = 0, l If D(0) < 1, then q 1 < 1 τ > 0 D(t) q 1, t [0, τ]. Then Ax l q 1 x l Ax l + Qx l (q 1 + q(l)) x l q x l, 0 < q < q 1 < 1 l l(q), t [0, τ] Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 8 / 36

17 Existence & Uniqueness of Local Solution Theorem 1: Sufficient Conditions of Existence & Uniqueness of Local Solution Let for t [0, T ] the following conditions be fulfilled: continuous K i (t, s), i = 1, n, α i (t) and f (t) have continuous derivatives wrt t, K n (t, t) 0, 0 = α 0 (t) < α 1 (t) < < α n 1 (t) < α n (t) = t for t (0, T ], α i (0) = 0, f (0) = 0, D(0) < 1, then τ > 0 such as eq. (1) has a unique local solution in C [0,τ]. back to example 1: t/2 x(s)ds t x(s)ds = t, D(t) 1 no unique solution! 0 t/2 Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 9 / 36

18 Existence & Uniqueness of Global Solution Theorem 2: Sufficient Conditions of Existence & Uniqueness of the Global Solution* Let the conditions of the Theorem 1 are fulfilled, and moreover let min τ t T (t α n 1(t)) = h > 0. Then eq. (1) has unique global solution in C [0,T ]. *D.N.Sidorov, E.V.Markova. One the Volterra integral model of evolving dynamical system. Automation and Remote Control (to appear) Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 10 / 36

19 min (t α n 1(t)) = h > 0 τ t T Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 11 / 36

20 Lemma 1 Lemma Let α : [0, T ] R +, α(t) C[0,T 1 ], α(0) = 0, 0 α(1) (0) < 1. Let h (0, T ), I 0 := [0, h],, I k := [(1 + (k 1)ε)h, (1 + kε)h], ε > 0, [0, T ] = m I k. If 0 < α(t) < t, α (1) (t) 1 1+ε for 0 t T then k=0 α : I j j 1 k=0 I k, j = 1, m. (4) Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 12 / 36

21 Proof of the Lemma 1 Idea: proof by induction and mean value theorem. Since α : I j j 1 k=0 to proof by induction that I k α(t) (1 + (j 1)ε)h t I j, then it s enough α(t) t Ij (1 + (j 1)ε)h, j = 1, m. (5) j = 1 : Let a I 1, b = I 1 I0 = h, then a b εh. ξ [0, h], ξ 1 [h, a] : α(a) = α(b) + α (ξ)(a b) = α(0) + α (ξ 1 )h + α (ξ)(a b) 1 1+ε h εεh = h. Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 13 / 36

22 Proof of the Lemma 1 Let (5) be fulfilled for j = 2,... m 1. Then a I m and b = I m Im 1 = (1 + (m 1)ε)h : α(a) = α(b) + α (ξ)(a b) (1 + (m 2)ε)h ε εh 1 + (m 2)ε + ε < 1 + (m 1)ε 1 + ε Therefore a I m : α(a) (1 + (m 1)ε)h. Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 14 / 36

23 Proof of the Th. 2 The local solution exists on [0, τ] due to conditions of the Theorem 1. Let > 0, I 0 = [0, τ],..., I k = [τ + (k 1), τ + k ], k = 1, N, [0, T ] N I k. α i (t) : [0, τ] [0, τ i ], τ i < τ, i = 1, n 1. For t [τ, T ] k=0 due to the Lemma s conditions we have α i (t) t h. It allows us to select h, such as α i (t) : I k k 1 j=0 I j. (6) Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 15 / 36

24 Proof of the Th. 2 Inclusion (6) allows us to extend the local solution x 0 (t) onto the whole [τ, T ] with step using the method of steps. Indeed, let us construct the desired continuation of the solution on section I 1. For that objective we solve the VIE of the 2nd kind t x(t) + τ τ Q(t, s)x(s)ds = Ax 0 t [τ, τ + ] = I 1, 0 Q(t, s)x 0 (s)ds + ˆf (t), (7) where in the right hand side we already have the known x 0 (t). Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 16 / 36

25 Proof of the Th. 2 Continuous function x 1 (t), which satisfy eq. (7) for t I 1 is continuous extension of the solution x 0 (t) onto ([τ, τ + ]. Let us fix q (0, 1), let sup Q(t, s) = c < and min h, q ). Then operator s,t c t τ Q(t, s)x(s)ds in C [τ,τ+ ] will be contracting, and q is contraction coefficient. Hence eq. (7) has the unique solution x 1 (t) C [τ,τ+ ]. Therefore the function { x0 (t), 0 < t τ, ˆx 1 = x 1 (t), τ t < τ +, is constructed and it continuously extends the solution on [τ, τ + ]. Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 17 / 36

26 Proof of the Th. 2 The next continuation x 2 (t) C [τ+,τ+2 ]. Now we have to solve the VIEq with known ˆx 1 : x(t)+ t τ+ Q(t, s)x(s)ds = Aˆx 1 τ+ Finally, we will construct the function ˆx 2 (t) = 0 Q(t, s)ˆx 1 (s)ds, t [τ+, τ+2 ]. { x0 (t), 0 < t τ, x 1 (t), τ t τ +, x 2 (t), τ + t τ + 2. We can continue this process and within finite number of steps on (0, T ] the desired x(t) C (0,T ], will be constructed. Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 18 / 36

27 Supplementary Condition of Local Smoothness So, in case of D(0) < 1 the global solution exists and unique: local solution is constructed by successive approximations and continued by method of steps with successive approximations on each step. Therefore theorerical instest is to study the case D(0) 1 A. Exists polynomial P i (t, s) = M K iνµ t ν s µ, i = 1, n,, ν=1 ν+µ=0 f M (t) = M f ν t ν, αi M (t) = M α iν t ν, i = 1, n 1, where ν=1 0 < α 11 < α 12 < < α n 1,n < 1 such as for t +0, s +0 the following estimates hold: K i (t, s) P i (t, s) = O((t + s) M+1 ), i = 1, n, f (t) f M (t) = O(t M+1 ), α i (t) α M i (t) = O(t M+1 ), i = 1, n 1. Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 19 / 36

28 Supplementary Condition of Local Smoothness Let 0 α i (0) < 1, α i(0) = 0, i = 1, n 1. Then ε (0, 1) τ (0, T ] : max α α n 1 (t) i (t) ε, sup ε. i=1,n 1, t [0,τ] t (0,τ] t B. For fixed q (0, 1), τ (0, T ], 0 < ε < 1 max t [0,τ] εm D(t) q < 1 Estimate is true for big enough M Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 20 / 36

29 Approximation and Regularization of the Solution Let condition D(0) < 1 is not fuilfilled. C. Select N such as lim t 0 Theorem* 3 ( t K(t,s) x(s) ds f (t) 0 ) t N = 0 Let x(t) be the known function such as the discrepancy C is true for N M. Then eq. (1) has the solution x(t) = x(t) + t N u(t), where u(t) C [0,T ] is unique and can be constructed by means of the successive approximations method. *D.Sidorov. Solution to VIE with piecewise smooth kernels. Differential Equations Vol. 49 (2). P Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 21 / 36

30 Regularization For u(t) we have the equation t 0 K(t, s)s N u(s) ds = g(t) ( ) t g(t) := K(t, s) x(s) ds + f (t), 0 g(t) C (1) [0,T ], g(0) = 0, g (t) = o(t N ), as t +0 Definition The equation (*) has unique solution and we call it as regularization of the equation (1). Function x(t) is approximation of solution to eq. (1) Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 22 / 36

31 Question Remains: How to Construct an Approximation x(t) to meet the Condition C? Idea: To construct x(t) as power-logarithmic asymptotic expansion. We would need an additional smoothness of all the gived functions according to the condition A since we need the Taylor coefficients. No need in condition D(0) < 1 Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 23 / 36

32 Relaxation of the condition D(0) < 1 Theorem 4 (Relaxed Sufficient Condition for Existence & Uniqueness) Let conditions (B) and (C) be fulfilled, and B(j) := K n (0, 0) + n 1 (α i (0))1+j (K i (0, 0) K i+1 (0, 0)) 0 for i=1 j N {0}. Then eq. (1) has unique solution x(t) = x M (t) + t N u(t) in C [0,T ], M N. Moreover, for t +0 polynomial x(t) x M (t) = M x i t i is an Mth order asymptotic approximation of such solution. i=0 Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 24 / 36

33 Proof Since B(j) 0, j N {0} then all the coefficients in the solution x M (t) = M x i t i can be determined. We can construct the solution in the i=0 form x(t) = x M (t) + t N u(t), M N. We get the integral-functional equation satisfying the contraction mapping principle on (0, τ] and the function u(t) C (0,T ] can be uniquely determined with successive approximations. Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 25 / 36

34 Construction of the Power-Logarithmic Asymptotic of Solution We search for the approximation of solution as x(t) = N x j t j. To find x j using method of undetermined coefficients we must solve the reccurent sequence of linear algebraic equations B(j)x j = M j (x 0,, x j 1 ), M 0 = f (0) Characteristic Equation B(j) K n (0, 0) + n 1 (α i (0))1+j (K i (0, 0) K i+1 (0, 0)) = 0 i=1 B(j) 0 for j N {0} If B(j ) = 0 then coefficient x j of power-logarithmic expansion is solution to the difference eq.: K n(0, 0)x j (z) + n 1 (α i (0)) 1+j {K i (0, 0) K i+1 (0, 0)} x j (z + ln α i (0)) = M j (x 0(z),, x j 1(z)), z := ln t i=1 j=0 Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 26 / 36

35 Construction of the Power-Logarithmic Asymptotic of Solution We search for the approximation of solution as x(t) = N x j t j. To find x j using method of undetermined coefficients we must solve the reccurent sequence of linear algebraic equations B(j)x j = M j (x 0,, x j 1 ), M 0 = f (0) Characteristic Equation B(j) K n (0, 0) + n 1 (α i (0))1+j (K i (0, 0) K i+1 (0, 0)) = 0 i=1 B(j) 0 for j N {0} If B(j ) = 0 then coefficient x j of power-logarithmic expansion is solution to the difference eq.: K n(0, 0)x j (z) + n 1 (α i (0)) 1+j {K i (0, 0) K i+1 (0, 0)} x j (z + ln α i (0)) = M j (x 0(z),, x j 1(z)), z := ln t i=1 j=0 Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 26 / 36

36 Construction of the Power-Logarithmic Asymptotic of Solution We search for the approximation of solution as x(t) = N x j t j. To find x j using method of undetermined coefficients we must solve the reccurent sequence of linear algebraic equations B(j)x j = M j (x 0,, x j 1 ), M 0 = f (0) Characteristic Equation B(j) K n (0, 0) + n 1 (α i (0))1+j (K i (0, 0) K i+1 (0, 0)) = 0 i=1 B(j) 0 for j N {0} If B(j ) = 0 then coefficient x j of power-logarithmic expansion is solution to the difference eq.: K n(0, 0)x j (z) + n 1 (α i (0)) 1+j {K i (0, 0) K i+1 (0, 0)} x j (z + ln α i (0)) = M j (x 0(z),, x j 1(z)), z := ln t i=1 j=0 Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 26 / 36

37 Example 1 t/2 Back to example 1: x(s)ds t 0 t/2 x(s)ds = t Characteristic eq. B(j) 1 + (1/2) j = 0, root : j = 0 ( ) t x x(t) = 1, 2 Solution: x(t) = C 1 + C 2 ln t ( C 1 + C 2 ln t ) C 1 C 2 ln t = 1 C 2 = 1 2 ln 2 x(t) = ln t ln 2 + const. Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 27 / 36

38 Example 1 t/2 Back to example 1: x(s)ds t 0 t/2 x(s)ds = t Characteristic eq. B(j) 1 + (1/2) j = 0, root : j = 0 ( ) t x x(t) = 1, 2 Solution: x(t) = C 1 + C 2 ln t ( C 1 + C 2 ln t ) C 1 C 2 ln t = 1 C 2 = 1 2 ln 2 x(t) = ln t ln 2 + const. Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 27 / 36

39 Example 1 t/2 Back to example 1: x(s)ds t 0 t/2 x(s)ds = t Characteristic eq. B(j) 1 + (1/2) j = 0, root : j = 0 ( ) t x x(t) = 1, 2 Solution: x(t) = C 1 + C 2 ln t ( C 1 + C 2 ln t ) C 1 C 2 ln t = 1 C 2 = 1 2 ln 2 x(t) = ln t ln 2 + const. Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 27 / 36

40 Example 1 t/2 Back to example 1: x(s)ds t 0 t/2 x(s)ds = t Characteristic eq. B(j) 1 + (1/2) j = 0, root : j = 0 ( ) t x x(t) = 1, 2 Solution: x(t) = C 1 + C 2 ln t ( C 1 + C 2 ln t ) C 1 C 2 ln t = 1 C 2 = 1 2 ln 2 x(t) = ln t ln 2 + const. Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 27 / 36

41 1. Generalization to the Nonlinear VIE t 0 K(t, s, x(s)) ds, 0 s t T, f (0) = 0 (8) K 1 (t, s)g 1 (s, x(s)), t, s m 1, K(t, s, x(s)) = K 1 (t, s)g 1 (s, x(s)), t, s m n, m i = {t, s α i 1 (t) < s < α i (t)}, α 0(t) = 0, α n(t) = t, i = 1, n, K i (t, s), f (t), α i (t) have continuous derivatives w.r.t. t for t, s m i, K n (t, t) 0, α i (0) = 0, 0 < α 1 (t) < α 2 (t) < < α n 1 (t) < t, α 1 (t),..., α n 1 (t) increase at least in the small neighborhood 0 t τ. Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 28 / 36

42 (D) Lipschitz cond.: G i (s, x 1) G i (s, x 2) (x 1 x 2) q i x 1 x 2, x 1, x 2 R 1 (E) q n + n 1 α i (0) Kn 1 (0, 0)(K i (0, 0) K i+1 (0, 0)) (1 + q i ) < 1 i=1 Theorem 2 (Sufficient Condition for Existence & Uniqueness for NLVIEq) Let for t [0, T ] the following conditions be fulfilled: K i (t, s), G i (s, x(s)), i = 1, n are continuous, α i (t) and f (t) have continuous derivatives wrt t, K n (t, t) 0, 0 = α 0 (t) < α 1 (t) < < α n 1 (t) < α n (t) = t, for t (0, T ], α i (0) = 0, f (0) = 0, (D) and (E). Then τ > 0 such as eq. (8) posseses unique local solution. Moreover if min τ t T (t α n 1(t)) = h > 0 then such local solution is continuously extendable on the entire [τ, T ] using the method of steps and successive approximations. Therefore exists unique global solution in C [0,T ]. Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 29 / 36

43 2. : Systems of linear VIEs with piecewise continuous kernels: Denis Sidorov. Solvability of system of the VIEs of the 1st kind with piecewise continuous kernels. Russian Mathematics Vol. 57(1). P Abstract integral-operator Volterra equations in Banach spaces: Alfredo Lorenzi. Integro-differential operator equations of the first kind of degenerate type in Banach spaces and applications to integro-differential PDEs. Eurasian Journal of Mathematical and Computer Applications (to appear) K i (t, s) L(E 1 E 2 ), G i : E 1 E 1 Nikolay Sidorov and Denis Sidorov. Volterra integral equation with piece-wisely defined kenel in Banach Spaces. In Nonclassic Equations of Mathematical Physics, Edt. Prof. Kojanov. Sobolev Institute of Mathematics SB RAS, P arxiv: Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 30 / 36

44 α(t) 0 K 1 (t, s)x(s)ds + t α(t) K 2 (t, s)x(s)ds = f (t), t [0, T ], (9) 0 < α(t) < t t (0, T ], α(0) = 0, K 1 (t, s), K 2 (t, s), f (t) are continuous and smooth sufficiently, f (0) = 0, K 2 (t, ] t) 0 t [0, T ]. t i = t 0 + ih, i = 1, n, t 0 = 0, nh = T, l = + 1 [ α(ti ) h l 1 h K 1 (t i, t j )x h (t j ) + (α(t i ) t l 1 )K 1 (t i, α(t i ))x h (α(t i )) + (10) j=1 i +(t l α(t i ))K 2 (t i, t l )x h (t l ) + h K 2 (t i, t j )x h (t j ) = f (t i ) j=l+1 i = 1, n Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 31 / 36

45 x h (α(t 1 )) = f (t 1 ) (α(t 1 ) t 0 )K 1 (t 1, α(t 1 )) + (t 1 α(t 1 ))K 2 (t 1, α(t 1 )) (11) x(0) = If D(0) < 1 (Theorem 2) then f (0) α (0)[K 1 (0, 0) K 2 (0, 0)] + K 2 (0, 0), (12) α (0)[K 1 (0, 0) K 2 (0, 0)] + K 2 (0, 0) 0 Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 32 / 36

46 1st Example t 3 0 t (1 + t s)x(s)ds t 3 x(s)ds = t t3, t [0, 2], 81 x(t) = t 2. ε 1 = max x(t i) x1 h(t i), ε 2 = max x(t i) x2 h(t i), 1 i n 0 i n x1 h is solution computed with (11) in a(t 1) x2 h is solution computed with (12), (11) and extrapolation in the 1st node. h ε 1 ε 2 1/32 0, , /64 0, , /128 0, , Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 33 / 36

47 2nd Example. 2 sin( t 2 ) 0 t 1 t x(s)ds x(s)ds = 3 sin3 2 + t3, t [0, 2π], 3 sin( t 2 ) x(t) = t 2 h ε 1 π/128 0, π/256 0, π/512 0, Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 34 / 36

48 T H A N K Y O U F O R Y O U R A T T E N T I O N Denis N. Sidorov contact.dns@gmail.com Integral Equations with Jump Discontinuous Kernels 35 / 36

49 Special Session: Optimization in Inverse Problems Denis N. Sidorov Integral Equations with Jump Discontinuous Kernels 36 / 36