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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

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  • $\begingroup$ Do you really have $z$ in the ELE in question number 1) in the last term or rather $Z$? $\endgroup$
    – gerw
    May 26, 2013 at 17:52
  • $\begingroup$ @gerw according to the paper it's $z$. It could be a typo of course, (there are other typos in the paper). $\endgroup$ May 27, 2013 at 21:39

2 Answers 2

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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 $u$.

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  • $\begingroup$ Thank you for your answer. I believe the first part was already answered above, and you confirmed it. For the second part, normally when I think of a weak solution, there should be some integration by parts in the formulation. I don't see that in their derivation? I'm not sure how we get the boundary data for the weak formulation either; that is, translating this boundary data into the inner product form. $\endgroup$ Jun 2, 2013 at 18:52
  • $\begingroup$ @Euler....IS_ALIVE For second order problem (original pde has a $\Delta$ in space, like Poisson equation and your equation), there are two approaches (and more) to get the weak form, one is multiplying with a test function $v$ and integration by parts like you said. The other is making use the fact that the solution to $-\Delta u = f$, is also the minimizer of the quadratic functional $$ \mathcal{F}[u] = \frac{1}{2}\int_{\Omega} |\nabla u|^2 -\int_{\Omega} fu$$, then using calculus of variation we could get a weak form same with the first approach. $\endgroup$
    – Shuhao Cao
    Jun 2, 2013 at 19:02
  • $\begingroup$ @Euler....IS_ALIVE The functional in my previous comments is Galerkin type, the functional in your question is least square type, if we have $\mathcal{L}u = f$ in $\Omega$ and $\mathcal{R}u = g$ on $\partial \Omega$, then you can also use the least square functional to deduce the weak formulation: $$J[u] = \|\mathcal{L}u - f\|^2 + \|\mathcal{R}u - g\|^2 $$. $\endgroup$
    – Shuhao Cao
    Jun 2, 2013 at 19:10
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Consider $g(\epsilon)_{z,v}=K_{s,p}(z+\epsilon v)$ for $v$ an arbitrary function in the appropriate space. Now take the derivative in $\epsilon$ and evaluate at $\epsilon =0$. This is the directional derivative of the functional along the line given by $v$.

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  • $\begingroup$ I'm sorry.. I don't understand how this helps us. Could you please elaborate? For example, doing this procedure still doesn't explain how we get the $Z$ in there, and also it doesn't answer the second question. $\endgroup$ May 26, 2013 at 4:33
  • $\begingroup$ also taking this $\epsilon$ derivative doesn't seem to give anything close to the stated result? $\endgroup$ May 28, 2013 at 5:49
  • $\begingroup$ $$\tilde{K}_{s,p}(z+\epsilon Z) = \frac{1}{2s}\int_{-T}^{T}\int_{\Omega}e^{2s\phi}|\partial_t^2 (z+\epsilon Z) - \Delta (z+\epsilon Z) + p(z+\epsilon Z)|^2 dx \ dt + \frac{1}{2}\int_{-T}^{T}\int_{\Gamma_0}e^{2s\phi}|\partial_{\nu}(z+\epsilon Z)|^2 d \sigma dt $$ Now take the derivative in epsilon and evaluate at epsilon=0: $$d\tilde{K}_{s,p}(z,Z) = \frac{1}{s}\int_{-T}^{T}\int_{\Omega}e^{2s\phi}\left(\partial_t^2-\Delta+ p\right)z\left(\partial_t^2-\Delta +p\right)Z dx \ dt + \int_{-T}^{T}\int_{\Gamma_0}e^{2s\phi}\partial_{\nu}z \partial_{\nu}Z d \sigma dt $$, right? $\endgroup$
    – guacho
    May 28, 2013 at 7:10
  • $\begingroup$ In the other terms I think that there is $Z$, not $z$, but maybe I don't understand it appropriately. I hope this helps you. $\endgroup$
    – guacho
    May 28, 2013 at 7:11
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    $\begingroup$ You have something like $$ |(\partial_t -\Delta+p) (z+\epsilon Z)|^2=|(\partial_t -\Delta+p)z+\epsilon (\partial_t -\Delta+p) Z|^2 $$ $$ =[(\partial_t -\Delta+p)z]^2+\epsilon^2[ (\partial_t -\Delta+p) Z)]^2+2[(\partial_t -\Delta+p)z]\epsilon[(\partial_t -\Delta+p) Z)] $$ $\endgroup$
    – guacho
    May 28, 2013 at 12:59

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