Method of successive approximation to solve integral equation.

Question

Use the method of successive approximations to solve the integral equation: $$u(t)=1+\int_{0}^{t}(t-s)u(s)ds \space,u(t_0)=1$$

Solution: I used the Picard's iteration: $$u_1=1+\int_{t_0}^{t}(1+\int_{0}^t ((t-s)u(s)ds))ds$$ $$=1+(t-t_0)+\int_{t_0}^{t}\int_{0}^t ((t-s)u(s)ds))ds$$

Is this the right approach to the problem ? should I have to go to $n$ iterations ?

Answer 1

Picard's iteration would be $u_0= 1$,

$u_1= 1+ \int_0^t (s- t)(1)ds= 1+ \left[s^2/2- st\right]_0^t= 1- t^2/2$,

$u_2= 1+ \int_0^t (s- t)(1- s^2/2) ds= 1+ \int_0^t t- s+ ts^2/2- s^3/2 ds= 1+ \left[ts- s^2/2+ ts^3/6- s^4/8\right]_0^t= 1+ t^2/2+ t^4/24$,

etc.

Answer 2

$$u(t)=1+\int_{0}^{t}(t-s)u(s)ds \tag 1$$ First differentiation with respect to $t$ : $$u'=(t-t)u(t)+\int_{0}^{t}u(s)ds=\int_{0}^{t}u(s)ds$$ Second differentiation with respect to $t$ : $$u''=u(t)$$ $$u(t)=c_1 e^t+c_2 e^{-t}$$ The constants are not arbitrary since $u(t)$ has to satisfy Eq.$1$. $$c_1 e^t+c_2 e^{-t}=1+\int_{0}^{t}(t-s)\left( c_1 e^s+c_2 e^{-s}\right)ds$$ $$c_1 e^t+c_2 e^{-t}=1+c_1 e^t+c_2 e^{-t}-c_1t+c_2t-c_1-c_2$$ After simplification : $$(c_2-c_1)t-c_1-c_2=0$$ Valid any $t$ implies $$c_1=c_2=\frac12$$ $$u(t)=\frac12(e^t+e^{-t})=\cosh(t)$$ The solution of equation $(1)$ is $u(t)=\cosh(t)$

Specifying the condition $u(t_0)=1$ is excessive. Either $t_0=0$ and the condition is consistent with the integral equation, or $t_0\neq 0$ which is contradictory and there is no solution.

This might be the cause of the difficulty to use the method of successive approximations : Without a-priori setting $t_0=0$ no solution can be found.