Uniform convergence of matrix integral sequence

Question

I was given recursively defined: $$ M_k(t)=I+\int_{t_0}^tA(s)M_{k-1}(s)~ds $$ and $M_0=I$ and that $A(t)$ is a matrix with entries that are continuous functions on $t_0\leq t\leq t_1$.

By induction we can get (I think) $$M_n=\sum\limits_{r=0}^n\left(\int_{t_0}^tA(s)~ds\right)^{\,r}.$$

Now, show that $M_n$ converges uniformly (componentwise) to some M (on the given interval). I really have no clue how to do this, any help is appreciated!

To me this seems to boil down to showing that $\int_{t_0}^tA(s)~ds$ is sufficiently small, $<1$ or something to use cauchy, but again, I don't see what given properties I could use!

Answer 1

Unfortunately this formula $$ M_n(t)=I+\sum_{j=0}^n\left(\int_{t_0}^t A(s)\,ds\right)^{\!j} $$ is not in general true.

It is true if $A(s)A(t)=A(t)A(s)$, for all $s,t$.

In order to show that $M_n$ converges uniformly to some $M$, you basically need to follow the steps of the proof of Picard-Lindelöf.

So, subtracting $$ M_n(t)=I+\int_{t_0}^t A(s)M_{n-1}(s)\,ds$$and$$ M_{n+1}(t)=I+\int_{t_0}^t A(s)M_n(s)\,ds\tag{1} $$ we obtain $$ \| M_{n+1}(t)-M_n(t)\| \le \int_{t_0}^t \|A(s)\|\|M_{n}(s)-M_{n-1}(s)\|\,ds\le a\int_{t_0}^t \|M_{n}(s)-M_{n-1}(s)\|\,ds, $$ where $a=\max_{t\in I} \|A(t)\|$, and using the fact that $$ \|M_1(t)-M_0(t)\|\le \int_{t_0}^t \|A(s)\|\,ds\le a(t-t_0), $$ you inductively obtain that $$ \| M_{n+1}(t)-M_n(t)\|\le \frac { a^{n+1}(t-t_0)^{n+1}} {(n+1)!}\le \frac { a^{n+1}(t_1-t_0)^{n+1}} {(n+1)!}, $$ which by virtue of Weierstrass criterion implies that $M_n(t)$ converges usiformly to say $M(t)$, and its limit satisfies the IVP $$ X'=A(t)X, \quad X(t_0)=I. $$ Due to the following fact, letting $n\to\infty$ in $(1)$ we obtain $$ M(t)=I+\int_0^t A(s)M(s)\,ds. $$ Then $M(0)=I$, and $M'(t)=A(t)M(t)$.

Answer 2

This is the same as Yiorgos' answer, but I was writing mine when I discovered his...

Choose any submultiplicative norm on the space of matrices (i.e. $\Vert AB\Vert\leq\Vert A\Vert\,\Vert B\Vert$). Also, for any matrix-valued function $\Phi$ on $[t_0,t_1]$, set $\Vert\Phi\Vert_\infty:=\sup \{ \Vert \Phi(t)\Vert;\, t\in [t_0,t_1]\}$.

Since $A$ is continuous, it is bounded on $[t_0,t_1]$, say $\Vert A\Vert_\infty=M<\infty$.

Let us show by induction that $$\Vert M_{n+1}-M_n(t)\Vert \leq \frac{M^{n+1}}{(n+1)!} (t-t_0)^{n+1}$$ for every $n\geq 0$ and all $t\in [t_0,t_1]$.

For $n=0$ we have $$\Vert M_1(t)-M_0(t)\Vert =\left\Vert \int_{t_0}^t A(s)\, ds\right\Vert\leq M\, (t-t_0)\, . $$

Assume the inequality has been proved for some $n$. Then \begin{eqnarray} \Vert M_{n+2}(t)-M_{n+1}(t)\Vert&=&\left\Vert \int_{t_0}^t A(s)(M_{n+1}(s)-M_n(s))\, ds \right\Vert\\ &\leq&\int_{t_0}^t M\times \frac{M^{n+1}}{(n+1)!} (s-t_0)^{n+1}\, ds\\ &=&\frac{M^{n+2}}{(n+2)!}\, (t-t_0)^{n+2}\cdot \end{eqnarray}

It follows that $\Vert M_{n+1}-M_n\Vert_\infty\leq \frac{M^{n+1}}{(n+1)!} (t_1-t_0)^{n+1}$ for all $n$. Hence, the series $\sum\Vert M_{n+1}-M_n\Vert_\infty$ is convergent, and so the sequence $(M_n)$ is uniformly convergent on $[t_0,t_1]$.