# Help with Initial value problem : $y'= x^2+ xy^2, y(0) = 0$; Picard–Lindelöf Approximation.

i need solve this: $$y'=x^2+xy^2 , y(0)= y(t_0)= 0$$

a) Compute, starting from the constant function $u_0=0$ the successive approximations $u_1,u_2,u_3$ (in the sense of the theorem of Picard–Lindelöf)

b) show that the sequence $u_n$ on an interval $[-\frac{1}{2},\frac{1}{2}]$ converges uniformly to the solution of the problem.

My solution:

1. $$u_{n+1}(t)=y(x_{0})+\int_{x_0}^{t}f(t,u_{n}(t))dx ; t\in[x_0,x_0+\epsilon]$$

it follow:

$$u_{1}(t)= 0 + \int_{0}^{x}f(t,u_0)dt = \int_{0}^{x} t^2 dt = \frac{x^3}{3}$$

$$u_{2}(t) = 0+\int_{0}^{x} f(t,u_1)dt = \int_{0}^{x} t^2+ t \frac{x^6}{9}dt = \frac{x^3}{3} + \frac{x^8}{18}$$

$$u_{3}(t) = 0 + \int_{0}^{x} f(t,u_2)dt = \int_{0}^{x} t^2+t (\frac{x^3}{3}+\frac{x^8}{18})^2 dt = \frac{1}{648}x^3(x^{15}+12x^{10}+36x^5 + 216)$$

1. How I can show this and find $u_n$? I can not solve this differential equation (wolframalpha too can not solve!)

Show me, please, how I can finish this exercise!

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In part 1, you should write $u_1(x)$, etc instead of $u_1(t)$. Also, you could simplify the computations a bit by integrating $t^2$ once and for all: $$u_{n+1}(x) = \frac{x^3}{3}+\int_0^x t u_{n}(t)\,dt$$ In part 2, you should prove that the sequence $(u_n)$ is Cauchy in the space $C[-1/2,1/2]$. For this it suffices (why?) to show that for all (sufficiently large) $n$ we have $\|u_{n+1}-u_n\|\le \lambda\|u_n-u_{n-1}\|$ with some constant $\lambda<1$ independent on $n$. The norm here is the supremum norm on $[-1/2,1/2]$. By the integral triangle inequality, $$|u_{n+1}(x)-u_{n}(x)| = \left|\int_0^x t (u_{n}(t)-u_{n-1}(t))\,dt\right| \le \|u_n-u_{n-1}\| \left|\int_0^x t \,dt\right| = \|u_n-u_{n-1}\| \frac{x^2}{2}$$ hence $$\| u_{n+1}-u_{n}\| \le \|u_n-u_{n-1}\| \sup_{x\in[-1/2,1/2]}\frac{x^2}{2}$$ which gives us the desired $\lambda<1$.
Now that we know that the sequence converges uniformly, it is not difficult to prove that its limit $u$ satisfies $$u(x) = \frac{x^3}{3}+\int_0^x t u(t)\,dt$$ which means $u$ solves the initial value problem.