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

  • $\begingroup$ Do you really have $z$ in the ELE in question number 1) in the last term or rather $Z$? $\endgroup$ – gerw May 26 '13 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). $\endgroup$ – Euler....IS_ALIVE May 27 '13 at 21:39

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 $u$.

  • $\begingroup$ 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. $\endgroup$ – Euler....IS_ALIVE Jun 2 '13 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 '13 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 '13 at 19:10

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 $v$.

  • $\begingroup$ 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. $\endgroup$ – Euler....IS_ALIVE May 26 '13 at 4:33
  • $\begingroup$ also taking this $\epsilon$ derivative doesn't seem to give anything close to the stated result? $\endgroup$ – Euler....IS_ALIVE May 28 '13 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 '13 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 '13 at 7:11
  • 1
    $\begingroup$ 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 '13 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.