So let $I = (a, b)$ be an interval of the real line, $A$ be an $n \times n$ matrix of continuous real-valued functions, and let $B$ be a $1 \times n$ column of continuous real-valued functions on $I$.

The theorem states that for every $t_{0} \in I$ and $X^* \in \mathbb{R}^n$ there exists a unique column $X$ such that $$X'(t)=A(t)X(t)+B(t), X(t_{0}) = X^*$$ for every $t \in I$.

I've proven that there exists a solution by successive approximations. I defined a sequence $$X_{0}(t) = X^* + \int_{t_{0}}^{t} B(u)du = \widetilde{B}(t),$$ $$X_{n+1}(t)=\widetilde{B}(t)+\int_{t_{0}}^{t}A(u)X_{n}(u)du,$$ and proved that the series $X_{0}(t)+\sum_{k=1}^{\infty}(X_{k}(t)-X_{k-1}(t))$ is uniformly convergent when $t \to t_{1}$ for any $t_{1} \in I$. Therefore, the function $X(t)=\lim_{n \to \infty} X_{n}(t)$ is well defined and is indeed the solution to my equation.

What baffles me here is the uniqueness. I've only seen other proofs of uniqueness with the fixed point theorem, but I'm not aware of any connections between the fixed point theorem and the successive approximations method. Are there any? I tried to prove uniqueness by assuming that there is another function $Y$ that satisfies the equation, and I get stuck at $$F(t)=\int_{t_{0}}^{t} A(u)F(u)du,$$ where $F(t)=Y(t)-X(t)$.

Is this the right way to go? Or is there maybe a connection between the two methods (Banach fixed point and successive approximations) that I don't know about?


Let $X_1,X_2$ be two solutions with the same initial condition $X(0)$. From the integral equations you get $$ |X_1-X_2|\leq\int_0^t |A(s)||X_1-X_2|ds $$ By the Gronwall's inequality it implies that $$ X_1=X_2. $$

Added: and of course successive approximations are just a less abstract way to look at the contracting map principle, by which the solution exists and unique.

  • $\begingroup$ What I meant to ask (not really ask, rather verify that it doesn't) is if the successive approximation method has a theorem related to it that guarantees the uniqueness of the solution akin to the fixed point theorem. $\endgroup$ – Matija Sreckovic Mar 23 '17 at 23:06
  • 1
    $\begingroup$ @MatijaSreckovic I would say that it depends on the problem considered. $\endgroup$ – Artem Mar 23 '17 at 23:16

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