Dear virtual lecturers,

once more a late comment, this time on the proof of Theorem 11.2.1. (It occurred to me in connection with the preparation for project M.) It seems to me that the proof is unnecessarily complicated, once one has proved Lemma 11.2.2.

In Lemma 11.2.2 it is not explicitly mentioned that the function T is also holomorphic on $C_{Re>-{\rho }_0}$. (This is used anyway in the proof in the Lecture Notes.) This means that $T$ is a material law with $s_b\left(T\right)\le -{\rho }_0$. Hence $T\left({\partial }_{t,\nu }\right)\in L\left(L_{2,\nu }\right(R;H\left)\right)$ for all $\nu >-{\rho }_0$. Also, $S_{\nu }=T\left({\partial }_{t,\nu }\right)$ for $\nu >0$ (with $S_{\nu }$ from Theorem 11.2.1, i.e., from Picard's theorem). Now, from Theorem 5.3.5 we know that $T\left({\partial }_{t,\nu }\right)$ and $T\left({\partial }_{t,\mu }\right)$ coincide on the intersection of $L_{2,\nu }(R;H)$ and $L_{2,\mu }(R;H)$, for all $\mu ,\nu >s_b\left(T\right)$. This implies that, for $\nu >0$, $\rho \in [0,{\rho }_0)$, $f\in L_{2,\nu }(R;H)\cap L_{2,-\rho }(R;H)$, one has $$S_{\nu }f=T\left({\partial }_{t,-\rho }\right)f\in L_{2,-\rho }(R;H).$$

This shows the asserted exponential stability.

Best wishes, Jürgen

Dear virtual lecturers,

once more concerning the proof of Prop. 11.12: It might well be that the inclusion $$S_{\nu }{\partial }_{t,\nu }\subseteq {\partial }_{t,\nu }{S}_{\nu }$$ (stated on line 3 of the proof) will not be immediately clear to all readers. Here is the argument: If $S$ is the material law for $S_{\nu }$, then $${\partial }_{t,\nu }^{-1}{S}_{\nu }=S_{\nu }{\partial }_{t,\nu }^{-1},$$ because both sides correspond to the material law $\frac{1}{z}S\left(z\right)$. This implies that the range of the right hand side is contained in $dom\left({\partial }_{t,\nu }\right)$, and $$S_{\nu }={\partial }_{t,\nu }{S}_{\nu }{\partial }_{t,\nu }^{-1}.$$ Composing on the right by ${\partial }_{t,\nu }$ one obtains $$S_{\nu }{\partial }_{t,\nu }={\partial }_{t,\nu }{S}_{\nu }{\partial }_{t,\nu }^{-1}{\partial }_{t,\nu }={\partial }_{t,\nu }{S}_{\nu }{\mid }_{dom\left({\partial }_{t,\nu }\right)}\subseteq {\partial }_{t,\nu }{S}_{\nu }.$$

Best wishes, Jürgen

Dear virtual lecturers,

concerning the proof of Theorem 11.2.1: I think that one cannot understand the part beginning with “Let $T(\cdot )$ be as in Lemma 11.2.2.”, if one has not in mind the proof of Theorem 6.3.1 (Picard). As I understand, you want to use that the operator $S_{\nu }$, written out in the formula lines 3, 4, and 5 on p. 139, is just $N\left({\partial }_{t,\nu }\right)$, where $N\left(z\right):=T(z{)}^{-1}$ (and where $T\left(z\right)$ corresponds to $B\left(z\right)$ in the proof of Theorem 6.3.1). Without this information, I wouldn't know how to derive the formula line 5 on this page.

Concerning the idea of the proof, I suggest to show that $S_{\nu }\in L(L_{2,\nu }\cap L_{2,-\rho }(R\left)\right)$, by showing that $S_{\nu }$ is bounded on $\{u\in L_{2,-\rho }(R);sptu\text{bounded below}\}$, and using that the latter space is dense in $L_{2,\nu }\cap L_{2,-\rho }\left(R\right)$. (In the proof of the boundedness your manipulations with the functions $U_n$ would be needed!) Would you agree that this would also work?

Best wishes, Jürgen

Dear all,

just two comments.

Concerning the proof of Prop. 11.1.2: To see that $S\nu =S\left({\partial }_{t,\nu }\right)$ for some material law, one might quote Theorem 8.2.1. However, as Marcus pointed out to me, it also is part of the proof of Picard's theorem (Thm 6.2.1) that ${\partial }_{t,\nu }$ corresponds to the material law $N$ on p. 76.

Concerning the proof of Lemma 11.2.2: I was a bit puzzled by the expression on the first line of (11.2), coming out of thin air. I would like to offer a way to understand better what is happening here.

Writing out the relation $T\left(z\right)(u,v)=(f,g)$ one obtains two equations $$\begin{array}{cc}z{M}_{0}u-C^{\ast }v& =f,\\ Cu+M_1\left(z\right)v& =g.\end{array}$$ In order to eliminate $v$ I multiply the first line by $(C^{\ast }{)}^{-1}$, the second by $M_1(z{)}^{-1}$: $$\begin{array}{cc}z(C^{\ast }{)}^{-1}{M}_{0}u-v& =(C^{\ast }{)}^{-1}f,\\ M_1(z{)}^{-1}Cu+v& =M_1(z{)}^{-1}g.\end{array}$$ Adding these two equations and retaining the second equation in the first system above I get an equivalent system $$\begin{array}{cc}\left(z\right(C^{\ast }{)}^{-1}{M}_0{C}^{-1}+M_1\left(z{)}^{-1}\right)Cu& =(C^{\ast }{)}^{-1}f+M_1(z{)}^{-1}g,\\ M_1\left(z\right)v& =g-Cu\end{array}$$ and this is (11.2). Now one shows - as in the middle of p. 138 - that $S\left(z\right)$ is invertible in $L\left(H_1\right)$ and then gets that $(u,v)$ is uniquely determined as on the lines beginning by $u:=$ and $v:=$ in the Lecture Notes. (Note: Instead of retaining the second equation one also could have retained the first equation $z{M}_{0}u-C^{\ast }v=f$ in the first system. Then one would have obtained $v:=\left(C^{\ast }{)}^{-1}\right(z{M}_{0}u-f)$, and it would have become clear that $v\in dom\left(C^{\ast }\right)$.)

Best wishes, Jürgen

Dear all,

there is a small misprint in Exercise 11.4. In the second equality of (11.7) there is a minus sign missing, so the correct expression should be $$q=-(1-k\ast ){grad}_0\theta .$$

Best regards

Sascha