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Let $u:[0,t_{e}]\to\mathcal{D}(A)$ satisfy

$$\begin{cases} \frac{du}{dt}=Au & 0\le t \le t_{e} \\ u(0)=x \end{cases}$$

I want to prove that necessarily $u(t)=T(t)x$.

So it's clear to see that $T(0)x=x=u(0)$. And $\frac{d}{dt}T(t)x=AT(t)x=Au$, if $u(t)=T(t)x$.

How can I complete this proof? I assume, at least, by making the supposition that $u(t)\ne T(t)x$ and then looking for a proof by contradiction.

Pedro
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Jason Born
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1 Answers1

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You proved that $T(t)x$ is a solution. So, it remains to show that the problem has only one solution.

Let $u$ and $v$ be two solutions. Notice that the function $w=u-v$ satisfies $$\left\{\begin{align} &\frac{dw}{dt}=Aw, & 0\le t \le t_{e} \\ &w(0)=0. \end{align}\right.$$ Pick $s\in[0,t_e]$ and define $\phi:[0,s]\to X$ be setting $\phi(t)=T(t)w(s-t)$. Then, $$\begin{align} \frac{d}{dt}\phi(t)&=\frac{d}{dt}T(t)w(s-t)-T(t)\frac{d}{dt}w(s-t)\\ &=AT(t)w(s-t)-T(t)Aw(s-t)\\ &=T(t)Aw(s-t)-T(t)Aw(s-t)\\ &=0. \end{align}$$

It follows that $\phi$ is constant, that is, $\phi(t_1)=\phi(t_2)$ for all $t_1,t_2\in[0,s]$. Taking $t_1=0$ and $t_2=s$, we obtain

$$u(s)-v(s)=w(s)=T(0)w(s-0)=\phi(0)=\phi(s)=T(s)w(s-s)=T(s)0=0.$$

Since $s$ is arbitrary, we conclude that $u=v$ on $[0,t_e]$.

Remark: This argument appears in many books in the proof of the following result: If $A:D(A)\subset X\to X$ is the infinitesimal generator of a $C_0$-semigroup $\{T(t)\}_{t\geq 0}$ on $X$, then, for each $U_0\in D(A)$, the abstract Cauchy problem $U_t=AU$, $U(0)=U_0$ has an unique classical solution. Moreover, this solution is given by $U(t)=T(t)U_0$. See, for example, Goldstein's book (page 83) and Zheng's book (page 34). In fact, as we can see in Pazy's book (page 105) , a similar argument can also be applied to prove uniqueness of classical solution for inhomogeneous Cauchy problems.

Pedro
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  • Since my proposed answer is a counterexample, I'm struggling to figure out the error in your answer. I think you need to show rigorously that $\frac d{dt} \phi(t)$ exists. (You also need to define in what space this derivative is calculated.) – Stephen Montgomery-Smith Aug 30 '15 at 18:07
  • I'm thinking about it even more. How do you justify the product formula? Things that work fine in finite dimensions don't necessarily follow in the infinite dimensional case. – Stephen Montgomery-Smith Aug 30 '15 at 18:46
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    @StephenMontgomery-Smith I don't have a explicit proof for this product rule, but I added some references in the post where it is applied (also, here there is something similar). In fact, my answer is the classical argument that can be found in operator semigroups books. Since, probably, you know these books, I think that I'm missing something. – Pedro Aug 30 '15 at 20:26
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    No, I am probably wrong! – Stephen Montgomery-Smith Aug 30 '15 at 20:29
  • So while I have written papers on semigroup theory, I have read very little of the literature! – Stephen Montgomery-Smith Aug 30 '15 at 22:19
  • @Pedro Isn't this proof essentially the same as if $S={S(t)}{t\ge 0}$ and $T={T(t)}{t\ge 0}$ both have infinitesimal generator $A$, then $S=T$? – Jason Born Aug 31 '15 at 10:08