Solution of Fredholm integral equation

Question

I am trying to determine the $C^2$ solution of the Fredholm integral equation $$u(x)=x+\int_0^1|x-\xi|u(\xi)\, d \xi , \ \ 0\leq x\leq 1$$

I hve done the following:

Let's start with $u_0=1$ then we get $$u_1(x)=Tu_0=x+\int_0^1|x-\xi|u_0(\xi)\, d \xi=x+\int_0^1|x-\xi|\, d \xi=x^2+\frac{1}{2}$$ At the next step we get $$u_2(x)+Tu_1(x)=x+\int_0^1|x-\xi|u_1(\xi)\, d \xi=...=\frac{1}{6}x^2+\frac{1}{2}x^2+\frac{x}{6}+\frac{1}{2}$$
Next we get $$u_3(x)+Tu_2(x)=x+\int_0^1|x-\xi|u_2(\xi)\, d \xi=...=\frac{ x^6}{30}+\frac{x^4}{12}+\frac{x^3}{18}+\frac{x^2}{2}+\frac{3x}{20}+\frac{37}{72}$$

I don't see any pattern.

Are my calculations wrong? Or my intial function $u_0$? Or is the way I am doing that wrong?

Answer 1

An Initial value problem is always converted into a Volterra integral equation and a Boundary value problem is always converted into a Fredholm integral equation and vice-versa.

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By the definition, $$|x-\xi| = \begin{cases} x-\xi \quad \text{if} \quad x \gt \xi\\ \xi - x \quad \text{if} \quad\ x \lt \xi \\ \quad 0 \quad \text{if} \quad\ x = \xi \end{cases}$$

Here for $\ \ 0\leq x\leq 1 \\$ the Fredholm integral equation is

$u(x)=x+\int_0^1|x-\xi|u(\xi)\, d \xi $

$\implies u(x)=x+\int_0^x (x-\xi)u(\xi)\, d \xi + \int_x^1 (\xi-x)u(\xi)\, d \xi\qquad . . . . (1)$

$\implies u'(x)=1+\int_0^x u(\xi)\, d \xi -\int_x^1 u(\xi)\, d \xi\qquad . . . . (2)$ (differentiating with respect $x$ using Leibniz Integral Rule)

$\implies u''(x)=u(x)+u(x)$ (again differentiating with respect $x$ using Leibniz Integral Rule)

$\implies u''(x)-2 u(x)=0 $.

$u(x)=A e^{\sqrt 2x}+Be^{-\sqrt 2x}$, where $A,B$ are integrating constants.

Now at $x=0\quad \text{from $(1)$ and $(2)$}\qquad u(0)=\int_0^1 \xi u(\xi) d\xi,\qquad u'(0)=1-\int_0^1 u(\xi)d\xi$

and at $x=1\quad \text{from $(1)$ and $(2)$}\qquad u(1)=1+\int_0^1 (1-\xi) u(\xi) d\xi,\qquad u'(1)=1+\int_0^1 u(\xi)d\xi$

Therefore the boundary conditions are

$u(0)+u(1)=u'(1) \qquad \text{and} \qquad u'(0)+u'(1)=2$

From these two boundary condition, you can found the value of $A $ and $ B$.

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Leibniz Integral Rule (Differentiation under the integral sign):

Let $f(x, t)$ be a function of $x$ and $t$ such that both $f(x, t)$ and its partial derivative $\frac{\partial f}{\partial x}$ are continuous in $t$ and $x$ in some region of the $(x, t)$-plane, including $a(x) ≤ t ≤ b(x)$, and $ x_0 ≤ x ≤ x_1$. Also suppose that the functions $a(x)$ and $b(x)$ are both continuous and both have continuous derivatives for $x_0 ≤ x ≤ x_1$. Then, for $x_0 ≤ x ≤ x_1$, $$\frac{d}{dx}(\int_{a(x)}^{b(x)} f(x,t) dt)=\int_{a(x)}^{b(x)} \frac{\partial }{\partial x}f(x,t) dt +f( x, b(x)) \frac{db}{dx}-f( x, a(x)) \frac{da}{dx}$$