An Introduction to HUM

We give here an example of how the Hilbert Uniqueness Method applies.

The major reference is [Lions 1], and the original paper is [Lions 2].

1Problem

Consider the wave equation$$\{\begin{array}{lc}{y}^{\prime \prime }-△y=0& (x,t)\in \Omega \times (0,T)\\ y(0)=y^0,y^{\prime }(0)=y^1& x\in \Omega \\ y=\{\begin{array}{l}v\\ 0\end{array}& \begin{array}{rl}(x,t)& \in \Sigma \\ (x,t)& \in (\partial \Omega \times (0,T))\backslash \Sigma \end{array}\end{array}.$$Here

•	$\Omega$ is an open set of ${\mathbb{R}}^n$, with $C^2$ boundary.
•	$\Sigma$ is a nonempty open subset of $\partial \Omega \times (0,T)$.
•	${}^{\prime }$ denotes $\frac{\partial }{\partial t}$, and the second line means $$y(x,0)=y^0(x),\frac{\partial y}{\partial t}(x,0)=y^1(x),x\in \Omega .$$
•	the initial data $\{y^0,y^1\}$, which will be chosen in some Hilbert space, is given.
•	the control function $v$, in some suitable Hilbert space also yet to be determined, acts on the system via part of the boundary, namely $\Sigma$, and is up to our choice.

Our goal is:

Problem. Given $T>0$ and $\{y^0,y^1\}$ in an appropriate space, find a control $v$ on $\Sigma$, such that $$y(T)=0,y^{\prime }(T)=0,x\in \Omega .$$

Again, the expression above means$$y(x,T)=0,\frac{\partial y}{\partial t}(x,T)=0,x\in \Omega .$$What concerns us are

•	the time $T>0$ when the system arrives at the balanced state.
•	the space that $\{y^0,y^1\}$ lies in.
•	the space that $v$ lies in.

2Solution

We take the following steps.

•	(Preparation: $\varphi$ with properties of use) Start with the wave equation$$\{\begin{array}{lc}{\varphi }^{\prime \prime }-△\varphi =0,& (x,t)\in \Omega \times (0,T)\\ \varphi (0)={\varphi }^0,{\varphi }^{\prime }(0)={\varphi }^1& x\in \Omega \\ \varphi =0,& (x,t)\in \partial \Omega \times (0,T)\end{array},$$(1)which has a unique solution. Three proofs to such result are provided in [Lions 1] Chapter 1 section 3.2: separation of variables, Hille–Yosida, and Lax–Milgram.
•	(The “Inverse” Problem) Then solve$$\{\begin{array}{lc}{\psi }^{\prime \prime }-△\psi =0,& (x,t)\in \Omega \times (0,T)\\ \psi (T)=0,{\psi }^{\prime }(T)=0& x\in \Omega \\ \psi =\{\begin{array}{l}\frac{\partial \varphi }{\partial \nu }\\ 0\end{array}& \begin{array}{rl}(x,t)& \in \Sigma \\ (x,t)& \in (\partial \Omega \times (0,T))\backslash \Sigma \end{array}\end{array},$$(2)where $\varphi$ is from (1), and $\nu$ is the unit outward normal vector of $\Omega$. Be aware that $\psi$ is simply another name for $y$ of the problem. This is a non-homogeneous boundary value problem, which also admits a weak solution $\psi$. In terms of this result, [Lions 2] directs us to [Li & Ma].
•	($\Lambda$: from (1) to (2)) We therefore define a linear operator$$\begin{array}{rl}\Lambda :& C_c^{\infty }(\Omega )\times C_c^{\infty }(\Omega )\to E,\\ & \{{\varphi }^0,{\varphi }^1\}\mapsto \{{\psi }^{\prime }(0),-\psi (0)\},\end{array}$$where the space $E$ depends on what we know about the regularity of $\{{\psi }^{\prime }(0),-\psi (0)\}$ (cf. [Lions 1] Chapter 1 section 4.2).
•	(Calculations) Compute the bilinear form below, where $⟨-,-⟩$ is the $L^2(\Omega )\times L^2(\Omega )$ inner product.$$\begin{array}{rl}⟨\Lambda \{{\varphi }^0,{\varphi }^1\},\{{\varphi }^0,{\varphi }^1\}⟩& =⟨\{{\psi }^{\prime }(0),-\psi (0)\},\{\varphi (0),{\varphi }^{\prime }(0)\}⟩\\ & ={\int }_{\Omega }{\psi }^{\prime }(0)\varphi (0)-\psi (0){\varphi }^{\prime }(0)\\ & ={\int }_{\Sigma }(\frac{\partial \varphi }{\partial \nu }{)}^2.\end{array}$$(3) How to get the last equality. How to get the last equality. Applying Green’s formula in the space (not space–time), we get $${\int }_{\Omega \times (0,T)}\psi △\varphi -\varphi △\psi ={\int }_{\partial \Omega \times (0,T)}\psi \frac{\partial \varphi }{\partial \nu }-\varphi \frac{\partial \psi }{\partial \nu }={\int }_{\partial \Omega \times (0,T)}\psi \frac{\partial \varphi }{\partial \nu }={\int }_{\Sigma }(\frac{\partial \varphi }{\partial \nu }{)}^2.$$Plus, one–dimensional calculus on the $t$-axis yields $${\int }_{\Omega \times (0,T)}(\psi {\varphi }^{\prime \prime }-\varphi {\psi }^{\prime \prime })={\int }_{\Omega \times T}(\psi {\varphi }^{\prime }-\varphi {\psi }^{\prime })-{\int }_{\Omega \times 0}(\psi {\varphi }^{\prime }-\varphi {\psi }^{\prime })={\int }_{\Omega \times 0}(\varphi {\psi }^{\prime }-\psi {\varphi }^{\prime }).$$The left sides of the two lines coincide by the wave equation, thus so do the right sides. $\square$
•	(Key step: A Supposition) The expression above is a seminorm on $C_c^{\infty }(\Omega )\times C_c^{\infty }(\Omega )$, and that it is a norm is equivalent to the following uniqueness theorem (though at the moment we do not know whether or not it is true). Uniqueness Theorem. For $\{{\varphi }^0,{\varphi }^1\}\in C_c^{\infty }(\Omega )\times C_c^{\infty }(\Omega )$, if the solution $\varphi$ to (1) satisfies $$\frac{\partial \varphi }{\partial \nu }=0\text{on }\Sigma ,$$then $\varphi =0$ in $\Omega \times (0,T)$. $\boxed{\text{One not really interested in mathematical strictness may now simply move on to the final step.}}$
•	(Completion of Initial Space) Suppose that the uniqueness theorem holds, then (3) defines a norm on $C_c^{\infty }(\Omega )\times C_c^{\infty }(\Omega )$, written $\Vert {\Vert }_F$. Let the Hilbert space $F$ be the completion of $C_c^{\infty }(\Omega )\times C_c^{\infty }(\Omega )$ with respect to $\Vert {\Vert }_F$. (Note that the construction of $F$ guarantees that $\frac{\partial \varphi }{\partial \nu }$ is always in $L^2(\Sigma )$.)
•	($\Lambda$: invertible now) Then extend $\Lambda$ to be a linear operator$$\Lambda :F\to F^{\prime }$$through the diagram below, where $F^{\prime }$ is the dual space of $F$ with respect to the $L^2(\Omega )\times L^2(\Omega )$ inner product, that is, the space consisting of bounded linear funtionals on $F$ identified with functions (distributions) on $\Omega$ via the pairing between $F^{\prime }$ and $F$ (i.e. $⟨-,-⟩$ in the diagram) that extends the $L^2(\Omega )\times L^2(\Omega )$ inner product on $C_c^{\infty }(\Omega )\times C_c^{\infty }(\Omega )$.$$⟨\Lambda \{{\varphi }^0,{\varphi }^1\},\{{\zeta }^0,{\zeta }^1\}⟩=⟨\{{\varphi }^0,{\varphi }^1\},\{{\zeta }^0,{\zeta }^1\}{⟩}_F.$$$$F\times F\mathbb{R}{F}^{\prime }\times F⟨-,-{⟩}_F\Lambda ⟨-,-⟩$$Conversely, by the Riesz representation theorem, each funtional $f\in F^{\prime }$ (actually $f=\Lambda \{{\varphi }^0,{\varphi }^1\}$) induces exactly one $\{{\varphi }^0,{\varphi }^1\}\in F$. That is to say, $\Lambda$ is an invertible operator.
•	(${\Lambda }^{-1}$: from (2) to (1)) $\Lambda$ being invertible permits us to find, as long as the initial data $\{y^1,-y^0\}\in F^{\prime }$, the unique pair $\{{\varphi }^0,{\varphi }^1\}\in F$ such that$$\Lambda \{{\varphi }^0,{\varphi }^1\}=\{y^1,-y^0\}.$$
•	(The Final Strike: finding $v$) Let$$v=\frac{\partial \varphi }{\partial \nu }\in L^2(\Sigma ).$$

3Remark

In summary, with a specific control $\frac{\partial \varphi }{\partial \nu }$ on $\Sigma$, we get the initial state that will evolve into the balanced state at $T$; when given an initial state in an attempt to find a control forcing it into the balanced state at $T$, the control is obtained by reversing the process. Or in other words, operating $\Lambda$ is what we have already been able to do, and ${\Lambda }^{-1}$ solves the problem.

The fact that $\Lambda$ is invertible is a consequence of introducing the Hilbert structure $F$ on the space of initial data. The difficulty, then, lies in establishing the uniqueness theorem. Finally, two concluding remarks are in order.

Q	Does the time $T$ play any role in the problem?
A	Yes. Recall that as a hyperbolic system, the effect of boundary data (here the control $v$) is of finite propagation speed. So $T$ has to be taken sufficiently large. ($T$ is not big enough in the picture, as the shaded cone of dependence at $(x^0,T)$ does not touch $\partial \Omega \times (0,T)$.)
Q	Are the spaces $F,F^{\prime }$ necessarily abstract?
A	Surely not. In some cases $F$ can be characterized explicitly. For example, if $$T>2\underset{x\in \Omega }{\sup }|x-x^0|,\Sigma =\{x\in \partial \Omega \mid (x-x^0)\cdot \nu (x)>0\}\times (0,T)$$for an arbitrary $x^0\in {\mathbb{R}}^n$, then as is shown in [Lions 1] Chapter 1 corollary 5.1, the uniqueness theorem holds, and $$F=H_0^1(\Omega )\times L^2(\Omega ),F^{\prime }=H^{-1}(\Omega )\times L^2(\Omega ).$$Hence the problem is completely solved in this case.

References

[Lions 1]	J.L. Lions. 分布系统的精确能控性、摄动和镇定 第一卷: 精确能控性. 法兰西数学精品译丛. 严金海等译. 高等教育出版社, 2012.
[Lions 2]	J.L. Lions. Exact Controllability, Stabilization and Perturbations for Distributed Systems. SIAM Review, Vol. 30, No. 1 (Mar., 1988), pp.1–68. jstor: 2031306.
[Li & Ma]	J.L. Lions and E. Magnenes. Non-Homogeneous Boundary Value Problems and Applications. Vol I-II. Grundlehren der mathematischen Wissenschaften 181-182. Springer, 1972.