Maybe you can help here. There is kind of a lengthy setup to understand what the question is asking. There is a paper I'm reading, and in one section of it I can't make heads or tails of the result. The reference is "Global Carleman Estimates for Waves and Applications" by Baudouin, Buhan, Ervedoza.
The setup (taken from the paper) : Suppose $p \in L^{\infty}(\Omega \times (-T,T))$. Given initial data $(y_0^{-T},y_1^{-T}) \in L^2(\Omega)\times H^{-1}(\Omega)$, find a function $u \in L^2(\Gamma_0 \times (-T,T))$ such that the solution $y$ of
\begin{eqnarray}
\partial_t^2y -\Delta y + py = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ in } \Omega \times(-T,T) \\
y = u|_{\Gamma_0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ on } \partial \Omega \times (-T,T)\\
y(-T) = y_0^{-T}, \partial_ty(-T) = y_1^{-T} \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ in } \Omega
\end{eqnarray}
solves $y(T) = \partial_ty(T) = 0$.
There is a claim that we can get an explicit form for $u$ and $y$. Let $\phi = e^{\lambda \psi}$, where $\psi(x,t) = |x-x_0|^2 – \beta t^2 +C$. For $s$ a parameter, define the functional
$$K_{s,p}(z) = \frac{1}{2s}\int_{-T}^{T}\int_{\Omega}e^{2s\phi}|\partial_t^2z – \Delta z + pz|^2 dx \ dt + \frac{1}{2}\int_{-T}^{T}\int_{\Gamma_0}e^{2s\phi}|\partial_{\nu}z|^2 d \sigma dt $$ $$+<(y_0^{-T},y_1^{-T}),(z(-T), \partial_t z(-T))>_{(L^2 \times H^{-1}) \times (H_0^1 \times L^2)}$$
Here, $<(y_0^{-T},y_1^{-T}),(z(-T), \partial_t z(-T))>_{(L^2 \times H^{-1}) \times (H_0^1 \times L^2)} = \int_{\Omega}{y_0^{-T}z_1^{-T} dx} – <y_1^{-T},z_0^{-T}>_{H^{-1} \times H_0^1}$,
and
$<y_1^{-T},z_0^{-T}>_{H^{-1} \times H_0^1} = \int_{\Omega} \nabla(-\Delta_d)^{-1}y_1^{-T}\cdot \nabla z_0^{-T} dx$
where $\Delta_d$ is the Laplace operator with Dirichlet boundary conditions.
Part of the paper shows that $K_{s,p}$ has a unique minimizer $Z[s,p]$, for each $s,p$.
The setup is above. Now comes the two parts I don't get.
(1). The paper claims that the Euler-Lagrange equation given by the minimization of $K_{s,p}$ is
$$\frac{1}{s}\int_{-T}^{T}\int_{\Omega}e^{2s\phi}(\partial_t^2z – \Delta z + pz)(\partial_t^2Z -\Delta Z +pZ) dx \ dt + \int_{-T}^{T}\int_{\Gamma_0}e^{2s\phi}\partial_{\nu}z \partial_{\nu}Z d\sigma dt$$ $$+ <(y_0^{-T},y_1^{-T}),(z(-T), \partial_t z(-T))>_{(L^2 \times H^{-1}) \times (H_0^1 \times L^2)}$$
I don't understand how this result is obtained. From what I know, the Euler Lagrange equations are as follows (from Evans book). If $I[w] = \int L(Dw(x),w(x),x)$, and we call these variables $p, z, x$ respectively, then the Euler Lagrange equations satisfy $-\sum{({L_{p_i}(Du,u,x)})_{x_i}} + L_z(Du,u,x) = 0$. When I try to do this to $K_{s,p}$, I get a huge mess, because it seems like we need to use the product rule. I don't get how it simplifies to this form, and why the third term $<\cdot,\cdot>$ stays the same.
(2) Let $Y = \frac{1}{s}e^{2s\phi}(\partial_t^2 – \Delta + p)Z[s,p]$, and let $U[s,p] = e^{2s\phi}\partial_{\nu}Z[s,p]|_{\Gamma_0}$.
Then, we get
$$\int_{-T}^{T}\int_{\Omega}e^{2s\phi}(\partial_t^2z – \Delta z + pz)Y dx \ dt + \int_{-T}^{T}\int_{\Gamma_0}e^{2s\phi}\partial_{\nu}z U d\sigma dt$$ $$+ <(y_0^{-T},y_1^{-T}),(z(-T), \partial_t z(-T))>_{(L^2 \times H^{-1}) \times (H_0^1 \times L^2)} = 0$$
The paper claims that this is the dual formulation of the problem. What does this mean exactly, and how does this help us show that Y,U works as a solution?
Any help is greatly appreciated. Thanks in advance
Best Answer
Derivation of Euler-Lagrange equation: If $z$ minimizes $K_{s,p}(z)$, then any small perturbation on $z$ will make this functional bigger. Hence we want: $\newcommand{\e}{\epsilon}$ $$ \lim_{\e\to 0}\frac{d}{d\e} K_{s,p}(z+\e v) = 0.\tag{$\dagger$} $$ This means the perturbation $\e v$ in the test function space will drive the functional away from its local minimum (just like calculus).
$K_{s,p}(z+\e v)$ in (1) reads: $\newcommand{\lsub}[2]{{\vphantom{#2}}_{#1}{#2}}$ $$ \begin{aligned}K_{s,p}(z+\e v) &= \frac{1}{2s}\int_{-T}^{T}\int_{\Omega}e^{2s\phi}|\partial_t^2 (z+\e v) - \Delta (z+\e v) + p(z+\e v)|^2 dx \,dt \\ &\quad + \frac{1}{2}\int_{-T}^{T}\int_{\Gamma_0}e^{2s\phi}|\partial_{\nu}(z+\e v)|^2 d \sigma\, dt \\ &\quad + \lsub{L^2 \times H^{-1}}{\Big\langle (y_0^{-T},y_1^{-T}),((z+\e v)(-T), \partial_t (z+\e v)(-T))\Big\rangle}_{H_0^1 \times L^2}. \end{aligned}\tag{1}$$
Let the first term in (1) be $I_1$, first for the integrand: $$ \begin{aligned} & |\partial_t^2 (z+\e v) - \Delta (z+\e v) + p(z+\e v)|^2 \\ =& \left|(\partial_t^2 z - \Delta z + pz) + \e (\partial_t^2 v - \Delta v + pv)\right|^2 \\ =& |\partial_t^2 z - \Delta z + pz|^2 + \e^2 |\partial_t^2 v - \Delta v + pv|^2 \\ &\quad +2\e (\partial_t^2 z - \Delta z + pz)(\partial_t^2 v - \Delta v + pv). \end{aligned}$$Taking derivative of $\e$, first term gone, let $\e \to 0$, second term gone, what is left is the cross term with a factor of 2, hence: $$ \lim_{\e\to 0}\frac{d}{d\e} I_1 = \frac{1}{s}\int_{-T}^{T}\int_{\Omega}e^{2s\phi}(\partial_t^2 z - \Delta z + pz)(\partial_t^2 v - \Delta v + pv)dx \, dt. \tag{2} $$
Second term in (1), say $I_2$, expand the integrand: $$ |\partial_{\nu}(z+\e v)|^2 = |\partial_{\nu}z|^2 + \e^2|\partial_{\nu}v|^2 + 2\e \,\partial_{\nu}z \,\partial_{\nu}v, $$ Similar argument as above: $$ \lim_{\e\to 0}\frac{d}{d\e} I_2 = \int_{-T}^{T}\int_{\Gamma_0}e^{2s\phi}\partial_{\nu}z \,\partial_{\nu}v \,d\sigma\, dt. \tag{3} $$
Third term $I_3$ in (1): $$ \begin{aligned} & \lsub{L^2 \times H^{-1}}{\Big\langle (y_0^{-T},y_1^{-T}),((z+\e v)(-T), \partial_t (z+\e v)(-T))\Big\rangle}_{H_0^1 \times L^2} \\ =& \int_{\Omega}{y_0^{-T}(z+\e v)(-T) dx} - \lsub{H^{-1}}{\big\langle y_1^{-T},\partial_t(z+\e v)(-T)\big\rangle}_{H_0^1} \\ =& \int_{\Omega}{y_0^{-T}z(-T) dx} - \lsub{H^{-1}}{\big\langle y_1^{-T},\partial_t z(-T)\big\rangle}_{H_0^1} \\ &\quad + \e \left(\int_{\Omega}{y_0^{-T}v(-T) dx} - \lsub{H^{-1}}{\big\langle y_1^{-T},\partial_tv(-T)\big\rangle}_{H_0^1}\right). \end{aligned} $$ Taking derivative makes first term gone: $$ \begin{aligned} \lim_{\e\to 0}\frac{d}{d\e} I_3 &= \int_{\Omega}{y_0^{-T}v(-T) dx} - \lsub{H^{-1}}{\big\langle y_1^{-T},\partial_t v(-T)\big\rangle}_{H_0^1} \\ &=\lsub{L^2 \times H^{-1}}{\Big\langle (y_0^{-T},y_1^{-T}),(v(-T), \partial_t v(-T))\Big\rangle}_{H_0^1 \times L^2}. \end{aligned}\tag{4} $$
Now (2)+(3)+(4) yields the expression of $(\dagger)$: $$ \begin{aligned} &\frac{1}{s}\int_{-T}^{T}\int_{\Omega}e^{2s\phi}(\partial_t^2 z - \Delta z + pz)(\partial_t^2 v - \Delta v + pv)dx \, dt \\ &+ \int_{-T}^{T}\int_{\Gamma_0}e^{2s\phi}\partial_{\nu}z \,\partial_{\nu}v \,d\sigma\, dt \\ &+ \lsub{L^2 \times H^{-1}}{\Big\langle (y_0^{-T},y_1^{-T}),(v(-T), \partial_t v(-T))\Big\rangle}_{H_0^1 \times L^2} = 0.\end{aligned}\tag{$\ddagger$} $$ I am using $v$ instead of $z$ as the test function, and the minimizer $Z[s,p]$ is my $z$ (replacing $z$ with $Z$, $v$ with $z$ leads to your equation). And $(\ddagger)$ is the Euler-Lagrange equation.
What this paper claims is that there exists a unique $Z$ depending on the choice of $s$ and $p$, such that $(\ddagger)$ for any $v$ (in the paper author uses $z$), then $Z$ minimizes the functional $K_{s,p}$, he should show the existence and uniqueness of $(\ddagger)$ subject to certain boundary conditions somewhere in the paper.
Now onto the second question: $Y$,$U$ does not work as the solution, what the author essentially does is just a change of notation. It is using $Y$ to represent some expression of the minimizer $Z$ (solution) in the interior, and $U$ to represent some other expression of the minimizer $Z$ (solution) on the boundary. The author stating "dual formulation" is with respect to the original PDE: the minimizer $z= Z$ of the functional $K_{s,p}$ satisfying $(\ddagger)$ for any $v$, and at the same time, serves as the weak solution to the original PDE: $$\left\{\begin{aligned} &\partial_t^2 z -\Delta z + p z= 0 \quad\text{ in } \Omega \times(-T,T), \\ &z = u|_{\Gamma} \quad \text{ on } \partial \Omega \times (-T,T),\\ &z(-T) = y_0^{-T},\; \partial_t z(-T) = y_1^{-T} \quad \text{ in } \Omega, \end{aligned}\right.$$ with appropriately chosen boundary data $u$.