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Consider the system of linear ODEs $\dot{x}(t)=Ax(t)$, $x(0)=x_0\in\mathbb{R}^n$. Does anyone know a simple proof showing that the solutions are unique that does not require resorting to more general existence/uniqueness results (e.g., those relating to the Picard iteration) nor solving for the solutions explicitly?

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Note that having an explicit solution only shows existence. –  copper.hat Jan 22 '13 at 16:03

3 Answers 3

up vote 3 down vote accepted

If you just need uniqueness, use the Bellman-Gronwall lemma. Suppose $x,y$ are two solutions, then $x(0) = y(0) = x_0$, and both $x,y$ satisfy the ODE. In particular, if we let $\delta = x-y$, we have $\delta(t) = \int_0^t A \delta(\tau) d\tau$, and hence $\|\delta(t)\| \leq \int_0^t \|A\| |\delta(\tau)\| d\tau$, from which it follows (using the integral form of the Bellman-Gronwall lemma) that $\|\delta(t)\| = 0$ for all $t \geq 0$.

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Thanks - that works fine. I'm not accepting it yet though because I'm looking for one to hand out the some undergrad engineering students who are not familiar with matrix norms. –  jkn Jan 22 '13 at 7:19

Since the students are engineers, why don't you want to show them explicit solutions, which surely they'd need to see anyway? If we knew about a matrix exponential $e^{At}$, then to show $x(t) = e^{At}x_0$ let's look at the $t$-derivative of $e^{-At}x(t)$, which is $$ e^{-At}x'(t) + (-Ae^{-At})x(t) = e^{-At}Ax(t) - Ae^{-At}x(t). $$ From the series definition of the matrix exponential, $A$ and $e^{Bt}$ commute if $A$ and $B$ commute, so $A$ and $e^{-At}$ commute. Thus $$ (e^{-At}x(t))' = e^{-At}Ax(t) - Ae^{-At}x(t) = Ae^{-At}x(t) - Ae^{-At}x(t) = 0. $$ Therefore $e^{-At}x(t)$ is a constant vector, and setting $t = 0$ tells us this constant vector has to be $x(0) = x_0$. Thus $e^{-At}x(t) = x_0$, so $x(t) = e^{At}x_0$ if we know that $e^{At}$ and $e^{-At}$ are inverses of each other.

Note that this solution can be thought of as a higher-dimensional version of the integration-free proof that the only solution of the 1-dim. ODE $y'(t) = ay(t)$ with $y(0) = y_0$ is $y_0e^{at}$: if $y(t)$ is a solution then the derivative of $e^{-at}y(t)$ is $$ e^{-at}y'(t) - ae^{-at}y(t) = e^{-at}(ay(t)) - ae^{-at}y(t) = 0. $$ Thus $e^{-at}y(t)$ is a constant function, and at $t = 0$ we see the value is $y(0) = y_0$, so $e^{-at}y(t) = y_0$. Thus $y(t) = y_0e^{at}$. In higher dimensions we just need to be more careful about the order of multiplication (e.g., the way the product rule is formulated for matrix-valued functions).

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If a vector function $y(t)$ has $t$-derivative 0 then it is a constant vector (I assume the $t$-domain is an interval). So, granting the properties of matrix exponentials, the argument implies that the only possible solution is $e^{At}x(0) = e^{At}x_0$. Why do you think the argument doesn't prove $e^{At}x_0$ is the only solution? –  KCd Jan 23 '13 at 20:59
    
Sorry, I completely misread. The initial idea was getting them to prove $\frac{d}{dt}e^{At} = Ae^{At}$, using that to verify that $x(t)=e^{At}x_0$ is indeed a solution, i.e., that $\dot{x}(t)=Ax(t)$ and $x(0)=x_0$ and hopefully finishing it of with a brief uniqueness proof. The only extra effort the your suggestion requires is showing that $A$ and $e^{Bt}$ commute if $A$ and $B$ commute, which is probably the best I could hope for. Thank you for the answer. –  jkn Jan 23 '13 at 22:47
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I'd suggest for the students that you place side-by-side the one-dimensional argument and then the higher-dimensional one, so that they see it is essentially the same idea except for being a bit fussier about the order of multiplication. In fact the way I came up with the solution I posted is by starting with the solution in the one-dimensional case and turning scalars there into either vectors or matrices as needed. –  KCd Jan 23 '13 at 23:15
    
I did, thanks for the suggestion. –  jkn Jan 28 '13 at 10:55

The general solution of the initial value problem $X'=AX$ with $X(0)=X_0$ is $$X(t)=e^{At}X_0$$ Let $Y(t)$ be an another solution satisfying $Y(0)=X_0$. I want to show that $X(t)=Y(t)$. Define $$Z(t)=exp(-tA)\cdot Y(t)$$ $$\qquad\frac{dZ} {dt}=-\exp(-tA)\cdot Y(t)+\exp(-tA)\cdot Y(t) =0$$ therefore $Z(t)$ is a constant. Setting $t=0$ we have $Z(t)=X_0$ so $Y(t)=\exp(At)X_0$ with $Y(0)=X_0$ this completes the proof.

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Ok. There are a lot of misunderstandings about formulas here. This is a main problem, Maisam. –  Babak S. Jan 24 '13 at 19:16
    
i modify it isit true? –  Maisam Hedyelloo Jan 24 '13 at 20:20

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