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Let $A\in M_{n\times n} (\mathbb{R})$. Use Picard iteration for $\dot x = Ax$, $x(0) = x_0$, to find the solution of this initial-value problem.

I know that the solution must be $$ x = x_0 e^{At} = x_0 \sum_{k=0}^\infty \frac{(At)^k}{k!} $$ I started doing the partial integrals, but I have not been able to show the above. A friend of mine said that it could be shown by induction. Any hints suggestions or full answers to how this can be done?

The Piccard iteration is defined as follows $$x_{n+1}=x_0+\int_0^t Ax_n\,\mathrm{d}t$$ By inserting we see that $$x_1=x_0+\int_0^t Ax_0\,\mathrm{d}t=x_0+A\cdot tx_0$$ and $$x_2=x_0+\int_0^t Ax_1\,ds=x_0+\int_0^tA(x_0+Atx_0)\,\mathrm{d}t$$ but by doing more of these, I do not see an obvious pattern and the integration gets messier and messier. Could anyone lend me a hand? =)

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PiCCard? $ $ $ $ –  Did Nov 1 '12 at 13:18
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1 Answer

up vote 2 down vote accepted

Let's continue : \begin{align} x_1&=x_0+\int_0^t A x_0\,\mathrm{d}t=\left(I+At\right) x_0\\ x_2&=x_0+\int_0^t A x_1\,ds=x_0+\int_0^tA((I+At) x_0)\,\mathrm{d}t=\left(I+A\frac t{1!}+A^2\frac{t^2}{2!}\right)x_0\\ \end{align}

So that you have indeed just to suppose that :

$$x_n=\left(I+A\frac t{1!}+A^2\frac{t^2}{2!}+\cdots +A^n\frac{t^n}{n!}\right)x_0$$ and deduce the $\ x_{n+1}$ term.

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