1
$\begingroup$

In Theorem 2.1 of Prof. Lawrie notes (which can be found here), he provided without proof the following result:

Let $(f,g)\in H^s\times H^{s-1}$ and $h\in L^1([0,T], H^{s-1})$. Then the linear wave equation $$\begin{cases} \Box u = h \\ u(0,x) = f(x)\\ \partial_tu(0,x) = g(x) \end{cases}$$ has a unique solution $u\in C([0,T],H^s)\cap C^1([0,T],H^{s-1})$ which satisfies, for each $0\le t\le T$, the following energy estimates: $$ \|u(t,\cdot)\|_{H^s} + \|\partial_t u(t,\cdot)\|_{H^{s-1}} \le C(1+t)\left(\|f\|_{H^s} + \|g\|_{H^{s-1}} + \int_0^t\|h(\tau,\cdot)\|_{H^{s-1}} d\tau\right) $$

To this theorem, I have three questions:

  1. I am not too worried about the uniqueness but I have no idea how to proof the inequality. In particular, how should I bound $\|u(t,\cdot)\|_{H^s}$? I've tried the energy method with $E(t):=\frac{1}{2}\int_{\mathbb{R}^n}|u(t,x)|^2\,dx$ but I didn't have much success with this. How do I make the $(1+t)$ appear when bounding this term?

  2. When bounding $\|\partial_t u(t,\cdot)\|_{H^{s-1}}$, I used integration by parts and simply discarded the boundary term; am I allowed to do that? The working for that portion looks like this \begin{align} \int_{\mathbb{R}^n} (\partial u)\cdot(\partial_t\partial u)dx &= \int_{\mathbb{R}^n} \left[(\partial_t u)(\partial_t^2 u) + \sum_{i=1}^n (\partial_i u)(\partial_i \partial_t u)\right]dx \\ &= \int_{\mathbb{R}^n} \left[(\partial_t u)(\partial_t^2 u) - \sum_{i=1}^n (\partial_i^2 u)(\partial_t u)\right]dx \end{align}

  3. What is going on with all the spaces here? What is $h\in L^1([0,T], H^{s-1})$? Isn't the function $h$ suppose to be $\mathbb{R}^{1+n}\to\mathbb{R}$? Then how can it be a map of $[0,T]\to H^{s-1}(\mathbb{R^n})$? And what does $u\in C([0,T],H^s)\cap C^1([0,T],H^{s-1})$ mean? How do you even intersect these two spaces? I'm really perplexed by this so please help shed some light here. (I know this post, to some extent, discuss this but whatever is going on there makes no sense to me as well)

Any help would be appreciated; thanks in advance.

$\endgroup$
5
  • $\begingroup$ I took a look at the note. Prof. Lawrie did prove Theorem 2.1 in his note. It may help. $\endgroup$
    – Pleroma
    Commented Apr 11, 2022 at 3:36
  • $\begingroup$ Yea, I saw it after making this post. If my understanding is correct, he used Fourier transform which resulted in $u$ being a distribution. But shouldn't $u$ be a function since $u\in C(...)\cap C^1(...)$. I was wondering if this proof can be done without invoking distribution (I don't mind a non-optimum $C$ if necessary). $\endgroup$
    – Tham
    Commented Apr 11, 2022 at 7:13
  • $\begingroup$ Also, if possible, do you have any idea what's going on with all the spaces (as per point 3 of my question)? Prof Lawrie did not give the definition of his notation. I presume it's a common knowledge so he omitted it but I'm new to this field so it'll be good for me to see the definition at least once. $\endgroup$
    – Tham
    Commented Apr 11, 2022 at 7:24
  • $\begingroup$ I answered point 3 of your question below. Unfortunately, I am not familiar with the wave equations. I will answer point 1 and 2 if I have any ideas. $\endgroup$
    – Pleroma
    Commented Apr 11, 2022 at 8:39
  • $\begingroup$ Thank you, this helps. I'll try to figure out the other two points on my own. $\endgroup$
    – Tham
    Commented Apr 12, 2022 at 7:20

1 Answer 1

2
$\begingroup$

For the first part of the third question, although the function $h=h(t,x)$ is from $ [0,T]\times \mathbb{R}^n$ to $\mathbb{R}$. One can fix a $t\in [0,T]$ to regard $h(t,\cdot)$ as a function of $x\in\mathbb{R}$. Varying $t\in [0,T]$, we get a map from $[0,T]$ to $H^{s-1}(\mathbb{R}^n)$. So basically we can regard the function $h$ as a map:$[0,T]\to H^{s-1}(\mathbb{R}^n)$.

In this way, one can define the space $L^1([0,T];H^{s-1}(\mathbb{R}^n))$ by $$ \| h \|_{L^1([0,T];H^{s-1}(\mathbb{R}^n))}:= \int^T_0 \| h(t,\cdot) \|_{H^{s-1}(\mathbb{R}^n)}\,dt <\infty. $$

For the second part, similarly, $u \in C([0,T],H^s)$ means that we regard $u$ as a map $t\to u(t,\cdot)$ and this map is continuous and lies in the space $C([0,T],H^s)$ with $$ \| u \|_{C([0,T];H^{s}(\mathbb{R}^n))}:= \max_{0\leq t \leq T} \| u(t,\cdot) \|_{H^{s}(\mathbb{R}^n)} <\infty. $$ To define the space $C^1([0,T],H^{s-1})$, it requires the map $t\to u(t,\cdot)$ is continuously differentiable. The norm of $C^1([0,T],H^{s-1}(\mathbb{R}^n))$ is therefore given by $$ \| u\|_{C^1([0,T],H^{s-1}(\mathbb{R}^n))}:= \| u \|_{C([0,T],H^{s-1}(\mathbb{R}^n))}+\| u'\|_{C([0,T],H^{s-1}(\mathbb{R}^n))}, $$ where $u'$ denotes the Fréchet derivative of the map $t \to u(t,\cdot)$. Of course, $u\in C([0,T],H^s)\cap C^1([0,T],H^{s-1})$ means $u\in C([0,T],H^s)$ and $u\in C^1([0,T],H^{s-1})$.

$\endgroup$
1
  • 1
    $\begingroup$ I saw that you self-deleted your answer to a poor-quality question. Thank you for helping to maintain site quality! $\endgroup$
    – user21820
    Commented Apr 11, 2022 at 15:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .