Dear virtual lecturers,

here are some minor remarks from our group meeting in Darmstadt.

p. 2 l. -6: We were not sure whether it is more common to refer to the convolution kernel of the semigroup as the fundamental solution (in the case of the initial value problem)?

Last sentence of Section 1.1 (starting with “We emphasise …”): We didn't get what you want to say here…

Paragraph after the first displayed formula in Section 1.3: The explanation of $M\left({\partial }_t\right)$ is unclear. Maybe it is just a matter of punctuation: Do you mean that $M$ is an analytic function on some open subset of $C$ with values in the bounded operators between Hilbert spaces in which we can plug in ${\partial }_t$ e.g. via its sectorial functional calculus? Then we recommand adding a colon after “is” and removing the comma before $M$ (or rewriting the whole sentence in a better understandable way).

Third line over “Wave Equation”: We recommand to write “spatial” instead of “spacial” for consistency reasons.

Reformulation of the wave equation: It seems to us like you had to add the second (tautological) equation only to ensure that the equation fits into your framework. Is this correct or is there a deeper reason or explanation from a methological point of view that explains this second line?

Exercise 1.1: Just curious: Some of us think they have seen this assertion with merely measurability instead of continuity. Maybe someone has a reference or construction of a counterexample for this modified version at hand and can provide it to us.

Best, Sebastian

Dear all,

I've got a question regarding the wave equation near the bottom of page 6. The wave equation together with the definitions of $v$ and $q$ (which amounts to three equations in total) are transformed into a system of two equations, which means that one equation is lost. In which sense is the wave equation equivalent to the system of two equations?

Best wishes, Hendrik

Since I solved problem 1 differently from the posted solution, then I will post my solution here. If somebody has a comment or spotted a mistake, please reply or leave a comment on this forum. Thank you.

Since $\varphi \left(t\right)\varphi (-t)=1$, then we only need consider $t\ge 0$. First, we claim that $\varphi \left(t\right)\ge 0$ for all $t\ge 0$. Assume that $\varphi \left(t_1\right)<0$ for some $t_1>0$. Since $\varphi \left(0\right)=1$ and $\varphi$ is continuous, then there exist $t_2\in (0,t_1)$ such that $\varphi \left(t_2\right)=0$. However, $$\varphi \left(t_1\right)=\varphi (t_1-t_2+t_2)=\varphi (t_1-t_2)\varphi \left(t_2\right)=0,$$ which contradict with assumption $\varphi \left(t_1\right)<0$, hence $\varphi \left(t\right)\ge 0$ for all $t\ge 0$.

Now, define $$\psi \left(t\right):={\int }_0^t\varphi \left(s\right)ds;.$$ The function $\psi \left(t\right)$ is non decreasing function, differentiable with $\frac{d\psi }{dt}=\varphi \left(t\right)$, and in particular ${lim}_{t\to 0^{+}}\frac{\psi \left(t\right)}{t}=\varphi \left(0\right)=1$. Thus, for some small $t_0>0$, $\psi \left(t_0\right)\ne 0$, then we can write $$\begin{array}{ccc}\varphi \left(t\right)& =& {\psi }^{-1}\left(t_0\right)\psi \left(t_0\right)\varphi \left(t\right)={\psi }^{-1}\left(t_0\right){\int }_0^t\psi (s+t)ds\\ & =& {\psi }^{-1}\left(t_0\right){\int }_t^{t+t_0}\psi \left(s\right)ds\\ & =& {\psi }^{-1}\left(t_0\right)\left[\psi \right(t_0+t)-\psi (t\left)\right].\end{array}$$ Then $\varphi$ is differentiable with derivative, $$\begin{array}{ccc}\frac{d\varphi }{dt}& =& \underset{h\to 0^{+}}{lim}\left[\frac{\varphi \left(h\right)-1}{h}\right]\varphi \left(t\right)\\ & =& \underset{h\to 0^{+}}{lim}{\psi }^{-1}\left(t_0\right)[\frac{\psi (t_0-h)-\psi \left(t_0\right)}{h}-\frac{\psi \left(h\right)}{h}]\varphi \left(t\right)\\ & =& {\psi }^{-1}\left(t_0\right)\left[\varphi \right(t_0)-1]\varphi \left(t\right).\end{array}$$

If we set $\alpha ={\psi }^{-1}\left(t_0\right)\left[\varphi \right(t_0)-1]\in R$, then $\varphi$ satisfies differential equation $$\{\begin{array}{c}\frac{d\varphi }{dt}=\alpha \varphi \\ \varphi \left(0\right)=1\end{array}$$ thus, we have $$\varphi \left(t\right)=e^{\alpha t}.$$

We are happy to announce the opening of our discussion forum and await your comments and questions.