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Let $A$ be the infinitesimal generator of a $C_0$-semigroup $(S(t))_{t \geq 0}$. Now, for every $x_0 \in X$ the map $t \mapsto S(t) x_0$ is a mild solution of

\begin{equation}\label{eq:1} \dot{x} = Ax, \quad x(0) = x_0. \qquad \text{(*)} \end{equation}

Now, a continuous function $x: [0, \infty) \to X$ is called a mild solution of $\text{(*)}$ if $\int_0^t x(s) \, ds \in D(A)$ where $D(A)$ is the domain of $A$, $x(0) = x_0$ and

$$x(t) - x(0) = A \int_0^t x(\tau) \, d\tau \text{ for all $t \geq 0$}.$$

Now, I have a proof of this but it uses Hille's theorem, but it is quite involved (needs a few tricks) and Hille's theorem is not elementary, does someone know an elementary proof of the uniqueness?

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Hi Jonas. I just encountered the same question and was about to post it. Did you ever find an elementary solution? –  Glen Wheeler May 9 '11 at 18:25
    
@Glen: I have found a solution. It uses elementary tools (Hille's theorem a few times I believe) but there are some smart tricks... Want to see it? –  Jonas Teuwen May 10 '11 at 14:00
    
Of course! If you wish to write it out here, I think it is valuable. –  Glen Wheeler May 10 '11 at 14:12
    
@Glen: I have looked it up if I could find it in a book. Do you know the (great) book by Engel and Nagel about semigroups? The thick one and the short course on operator semigroups both contain a proof. –  Jonas Teuwen May 11 '11 at 19:01
    
I do not know it, but I see that I can retrieve it from work. I learned semigroups from Lunardi, and I must confess that I never thought of looking there. I am not sure that she does mild solutions actually... thanks for the reference. –  Glen Wheeler May 11 '11 at 19:16

1 Answer 1

up vote 1 down vote accepted

You just need some simple properties of the Banach space-valued integral, namely linearity (the integral commutes with continuous linear operators) substitution for translations and the fundamental theorem of calculus, which says: $\frac{1}{h}\int_0 ^h f(t) d t \to f(0)$ (in the Banach space) for $h \to 0$.

So after writing out all definitions (and using linearity and the semigroup properties), your job is to show that for any $t>0$ we have $$\frac{1}{h}\int_0 ^t (S(\tau+h) x_0 - S(\tau)x_0) d \tau \to S(t)x_0 - x_0$$ for $h\to 0$ from above. We can assume $t>h$. Then split the first integral into $\int_0 ^{t-h}+\int_{t-h} ^t$ and the second one into $\int_0 ^h+ \int_h ^t$. By translation the integrals $\int_0 ^{t-h}$ and $\int_h ^t$ cancel, and you just have to apply the fundamental theorem of calculus.

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Could you maybe elaborate on the "writing out all definitions"? –  Jonas Teuwen Nov 26 '10 at 11:27
    
I'm not sure what you have trouble with. You gave the definition of a mild solution. In addition you need to know what the generator of a semigroup is. I am using the following definition (maybe you have a different one?): $Ay:=\lim_{h\to 0; h>0}\frac{1}{h}(S(h)y-y)$ and the domain is the set of $y$ for which this limit exists. –  Florian Nov 26 '10 at 12:47
    
@Florian: Okay, I follow your derivation. What I don't understand completely is why that implies the mild solution is unique. I probably miss some "iff" in some theorem. –  Jonas Teuwen Nov 27 '10 at 19:24
    
The question is because you seem to use $x(t) = S(t)x_0$ in the derivation but you also conclude that... –  Jonas Teuwen Nov 29 '10 at 20:11
    
Now I see the word "Uniqueness" in the title. I was just looking at the question which said "... is a mild solution. ... Does someone know an elementary proof for this fact?" So your question is actually "why are mild solutions unique?". This needs a different treatment. –  Florian Nov 29 '10 at 21:17

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