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--- hamburgminicourse2017:the_abstract_cauchy_problem_in_banach_space [2017/04/02 13:53] – created markus
+++ hamburgminicourse2017:the_abstract_cauchy_problem_in_banach_space [2017/04/02 15:57] (current) – markus
@@ Line 18: / Line 18: @@
 with domain $D(A)$, i.e. all $x\in X$ for which the limit exists.
-Theorems that show the existence of such a semigroup by demanding specific properties from its generator are called //generation theorems//. An example is Stone's theorem in the Hilbert space case and the theorems of Hille-Yoshida and Lumer-Phillips in the more general setting of Banach spaces. The important relation between the semigroup and the Cauchy problem above is given by the following theorem with two different types of solutions (discussed later).
+Theorems that show the existence of such a semigroup by demanding specific properties from its generator are called //generation theorems//. An example is [[Stone's theorem]] in the Hilbert space case and the theorems of Hille--Yoshida and Lumer--Phillips in the more general setting of Banach spaces. The important relation between the semigroup and the Cauchy problem above is given by the following theorem with two different types of solutions (discussed later).
 **Theorem.** Let $T(t)$ be a $\mathcal{C}^0$ semigroup and $A$ its infinitesimal generator. Then it holds for $t>0$ that
@@ Line 26: / Line 26: @@
 **Proof.** (1.) Take $h >0$, then
 \begin{align*}
-\frac{T(h)-\id}{h} \int_0^t T(s) x \,ds &= \frac{1}{h} \int_0^t (T(s+h)x-T(s)x) \,ds \\
+\frac{T(h)-\mathrm{id}}{h} \int_0^t T(s) x \,ds &= \frac{1}{h} \int_0^t (T(s+h)x-T(s)x) \,ds \\
 &= \frac{1}{h} \int_t^{t+h} T(s) x \,ds - \frac{1}{h} \int_0^h T(s) x \,ds
 \end{align*}
@@ Line 33: / Line 33: @@
 (2.) Now because of boundedness of $T(t)$ we have as $h \searrow 0$
 \[
-\frac{T(h)-\id}{h} T(t) x = T(t) \frac{T(h)-\id}{h} x \longrightarrow T(t) A x
+\frac{T(h)-\mathrm{id}}{h} T(t) x = T(t) \frac{T(h)-\mathrm{id}}{h} x \longrightarrow T(t) A x
 \]
 thus $T(t)x \in D(A)$ and $AT(t)x = T(t)Ax$ as well as the right derivative of $T(t)x$ fulfilling
@@ Line 44: / Line 44: @@
 &= \lim_{h \searrow 0} T(t-h) \left( \frac{T(h)x-x}{h} - Ax \right) + \lim_{h \searrow 0} (T(t-h)Ax - T(t)Ax)
 \end{align*}
-Both limit terms vanish, the first due to $x \in D(A)$ and boundedness of $T(t-h)$, the second by strong continuity of $T(t)$.
+Both limit terms vanish, the first due to $x \in D(A)$ and boundedness of $T(t-h)$, the second by strong continuity of $T(t)$. $\Box$
+That such a semigroup really always is linked to a usefully defined generator is secured by the next theorem.
+**Theorem.** The generator of a $\mathcal{C}^0$ semigroup is always densely defined and closed.
+**Proof.**  Like before it holds
+\[
+x = \lim_{h \searrow 0}\frac{1}{h} \int_0^h T(s) x \,ds
+\]
+and by (1.) above the argument of the limit is in $D(A)$. Hence any $x \in X$ can be approached by a limit sequence and $D(A)$ is dense in $X$. Assume now such a sequence $x_n \in D(A), x_n \rightarrow x$ and $Ax_n \rightarrow y$. Then integrating out part (2.) over the time interval $[0,h]$ we can write
+\[
+T(h)x_n-x_n = \int_0^h T(s) A x_n \,ds.
+\]
+In the limit $n \rightarrow \infty$ this yields
+\[
+T(h)x - x = \int_0^h T(s) y \,ds.
+\]
+Finally we divide by $h$ and let $h \searrow 0$ and get
+\[
+Ax = \lim_{h \searrow 0} \frac{1}{h} \int_0^h T(s) y \,ds = y.
+\]
+This shows closedness. $\Box$
+We can find two **types of solutions** to the Cauchy problem formulated here:
+  - $u \in (0,\infty) \rightarrow D(A)$ continuous in $X$ and continuously differentiable that really solve $\frac{d}{dt}u(t)=Au(t)$ with $\lim_{t\rightarrow 0}u(t)=x\in X$ are called //classical// or //strong solutions//.
+  - $u(t) = T(t)x$ (also continuous in $X$) just defined by application of the corresponding evolution semigroup are called //generalised solutions// and can be defined for any initial state $x\in X$. They solve a weaker form of the Cauchy problem from (2.) in the theorem above:
+\[
+A\int_0^t u(s)\,ds = u(t)-x.
+\]
+(Such solutions of an integral equation are called //mild//.)
+ >> [[Stone's theorem]]