CHAPTER 8 Integral Equations 8.1. Introduction Integral equations appears in most applied areas and are as important as differential equations. In fact, as we will see, many problems can be formulated (equivalently) as either a differential or an integral equation. Example 8.1. Examples of integral equations are: Z x (a) y(x) = x − (x −t)y(t)dt. 0 Z x (b) y(x) = f (x) + λ k(x −t)y(t)dt, where f (x) and k(x) are specified functions. 0 Z 1 (c) y(x) = λ k(x,t)y(t)dt, where 0 ( x(1 −t), 0 ≤ x ≤ t, k(x,t) = t(1 − x), t ≤ x ≤ 1. Z 1 (d) y(x) = λ (1 − 3xt)y(t)dt. 0 Z 1 (e) y(x) = f (x) + λ (1 − 3xt)y(t)dt. 0 ♦ A general integral equation for an unknown function y(x) can be written as Z b f (x) = a(x)y(x) + k(x,t)y(t)dt, a where f (x),a(x) and k(x,t) are given functions (the function f (x) corresponds to an external force). The function k(x,t) is called the kernel. There are different types of integral equations. We can classify a given equation in the following three ways. • The equation is said to be of the First kind if the unknown function only appears under the integral sign, i.e. if a(x) ≡ 0, and otherwise of the Second kind. • The equation is said to be a Fredholm equation if the integration limits a and b are constants, and a Volterra equation if a and b are functions of x. • The equation are said to be homogeneous if f (x) ≡ 0 otherwise inhomogeneous. Example 8.2. A Fredholm equation (Ivar Fredholm): Z b k(x,t)y(t)dt + a(x)y(x) = f (x). a 67 68 8. INTEGRAL EQUATIONS ♦ Example 8.3. A Volterra equation (Vito Volterra): Z x k(x,t)y(t)dt + a(x)y(x) = f (x). a ♦ Example 8.4. The storekeeper’s control problem. To use the storage space optimally a storekeeper want to keep the stores stock of goods constant. It can be shown that to manage this there is actually an integral equation that has to be solved. Assume that we have the following definitions: a = number of products in stock at time t = 0, k(t) = remainder of products in stock (in percent) at the time t, u(t) = the velocity (products/time unit) with which new products are purchased, u(τ)∆τ = the amount of purchased products during the time interval ∆τ. The total amount of products at in stock at the time t is then given by: Z t ak(t) + k(t − τ)u(τ)dτ, 0 and the amount of products in stock is constant if, for some constant c0, we have Z t ak(t) + k(t − τ)u(τ)dτ = c0. 0 To find out how fast we need to purchase new products (i.e. u(t)) to keep the stock constant we thus need to solve the above Volterra equation of the first kind. ♦ Example 8.5. (Potential) Let V(x,y,z) be the potential in the point (x,y,z) coming from a mass distribution ρ(ξ,η,ζ) in Ω (see Fig. 8.1.1). Then ZZZ ρ(ξ,η,ζ) V(x,y,z) = −G dξdηdζ. Ω r The inverse problem, to determine ρ from a given potential V, gives rise to an integrated equation. Furthermore ρ and V are related via Poisson’s equation ∇2V = 4πGρ. FIGURE 8.1.1. A potential from a mass distribution Ω r (ξ,η,ζ) (x,y,z) 8.2. INTEGRAL EQUATIONS OF CONVOLUTION TYPE 69 ♦ 8.2. Integral Equations of Convolution Type We will now consider integral equations of the following type: Z x y(x) = f (x) + k(x −t)y(t)dt = f (x) + k ? y(x), 0 where k ? y(x) is the convolution product of k and y (see p. 45). The most important technique when working with convolutions is the Laplace transform (see sec. 6.2). Example 8.6. Solve the equation Z x y(x) = x − (x −t)y(t)dt. 0 Solution: The equation is of convolution type with f (x) = x and k(x) = x. We observe that 1 L(x) = and Laplace transforming the equation gives us s2 1 1 1 1 L [y] = − L [x ? y] = − L [x]L [y] = − L [y], i.e. s2 s2 s2 s2 1 L [y] = , 1 + s2 1 and thus y(x) = L−1 = sinx. 1 + s2 Answer: y(x) = sinx. ♦ Example 8.7. Solve the equation Z x y(x) = f (x) + λ k(x −t)y(t)dt, 0 where f (x) and k(x) are fixed, given functions. Solution: The equation is of convolution type, and applying the Laplace transform yields L [y] = L [ f ] + λL [k]L [y], i.e. L [ f ] L [y] = . 1 − λL [k] L [ f ] Answer: y(x) = L−1 . 1 − λL [k] ♦ 70 8. INTEGRAL EQUATIONS 8.3. The Connection Between Differential and Integral Equations (First-Order) Example 8.8. Consider the differential equation (initial value problem) ( y0(x) = f (x,y), (8.3.1) y(x0) = y0. By integrating from x0 to x we obtain Z x Z x y0(t)dt = f (t,y(t))dt, x0 x0 i.e. Z x (8.3.2) y(x) = y0 + f (t,y(t))dt. x0 On the other hand, if (8.3.2) holds we see that y(x0) = y0, and y0(x) = f (x,y(x)), which implies that (8.3.1) holds! Thus the problems (8.3.1) and (8.3.2) are equivalent. ♦ In fact, it is possible to formulate many initial and boundary value problems as integral equations and vice versa. In general we have: Initial value problem ⇒ The Volterra equation, Dynamical system Boundary value problem ⇒ The Fredholm equation. Picard’s method (Emile Picard) Problem: Solve the initial value problem ( y0 = f (x,y), y(x0) = A. Or equivalently, solve the integral equation : Z x y(x) = A + f (t,y(t))dt. x0 We will solve this integral equation by constructing a sequence of successive approximations to y(x). First choose an initial approximation, y0(x) (it is common to use y0(x) = y(x0)), then define the sequence: y1(x), y2(x),...,yn(x) by Z x y1(x) = A + f (t,y0(t))dt, x0 Z x y2(x) = A + f (t,y1(t))dt, x0 . Z x yn(x) = A + f (t,yn−1(t))dt. x0 Our hope is now that y(x) ≈ yn(x). 8.3. THE CONNECTION BETWEEN DIFFERENTIAL AND INTEGRAL EQUATIONS (FIRST-ORDER) 71 By a famous theorem (Picard’s theorem) we know that under certain conditions on f (x,y) we have y(x) = lim yn(x). n→∞ Example 8.9. Solve the equation ( y0(x) = 2x(1 + y), y(0) = 0. Solution: (With Picard’s method) We have the integral equation Z x y(x) = 2t(1 + y(t))dt, 0 and as the initial approximation we take y0(x) ≡ 0. We then get Z x Z x Z x 2 y1(x) = 2t(1 + y0(t))dt = 2t(1 + 0)dt = 2tdt = x , 0 0 0 Z x Z x Z x 2 3 2 1 4 y2(x) = 2t(1 + y1(t))dt = 2t(1 +t )dt = 2t + 2t dt = x + x , 0 0 0 2 Z x 6 2 1 4 2 1 4 x y3(x) = 2t(1 +t + t )dt = x + x + , 0 2 2 6 . x4 x6 x2n y (x) = x2 + + + ··· + . n 2 6 n! We see that x2 lim yn(x) = e − 1. n→∞ x2 REMARK 22. Observe that y(x) = e − 1 is the exact solution to the equation. (Show this!) REMARK 23. In case one can can guess a general formula for yn(x) that formula can often be verified by, for example, induction. LEMMA 8.1. If f (x) is continuous for x ≥ a then: Z x Z s Z x f (y)dyds = f (y)(x − y)dy. a a a Z s PROOF. Let F(s) = f (y)dy. Then we see that: a Z x Z s Z x Z x f (y)dyds = F(s)ds = 1 · F(s)ds a a a a Z x x 0 {integration by parts} = [sF(s)]a − sF (s)ds a Z x = xF(x) − aF(a) − s f (s)ds a Z x Z s = x f (y)dy − 0 − y f (y)dy a a Z s = f (y)(x − y)dy. a 72 8. INTEGRAL EQUATIONS 8.4. The Connection Between Differential and Integral Equations (Second-Order) Example 8.10. Assume that we want to solve the initial value problem ( u00(x) + u(x)q(x) = f (x), x > a, (8.4.1) 0 u(a) = u0, u (a) = u1. We integrate the equation from a to x and get Z x 0 u (x) − u1 = [ f (y) − q(y)u(y)]dy, a and another integration yields Z x Z x Z x Z s 0 u (s)ds = u1ds + [ f (y) − q(y)u(y)]dyds. a a a a By Lemma 8.1 we get Z x u(x) − u0 = u1(x − a) + [ f (y) − q(y)u(y)](x − y)dy, a which we can write as Z x Z x u(x) = u0 + u1(x − a) + f (y)(x − y)dy + q(y)(y − x)u(y)dy a a Z x = F(x) + k(x,y)u(y)dy, a where Z x F(x) = u0 + u1(x − a) + f (y)(x − y)dy, and a k(x,y) = q(y)(y − x).