# Turnpike property for functionals involving $L^1$−norm

Authors: - 25 October 2017

### Introduction

We introduce the following notation: $L^2=L^2\left(\Omega\right)$, $L^2_T=L^2\left(\Omega\times \left(0,T\right)\right)$,

and the correspondent norms $\norm{\cdot}=\langle \cdot,\cdot\rangle$ and $\norm{\cdot}_T=\langle \cdot,\cdot\rangle_T$. Moreover, we define the norms

We want to study the following optimal control problem:

where $L: \ L^2_T \to L^2_T$ is defined by

and $y$ is the solution of the PDE given by

Notice that, by integration by parts, $L^*\mu=B^\ast p$, where $\varphi$ is solution of the adjoint equation:

### The stationary problem

#### Numerical algorithm

In order to compute a numerical solution of problem

after a discretization by finite differences, we use a prox-prox splitting: first write the state as $y=A^{-1}Bu$, then

• Proximal-point step:
• $$\begin{equation*} \begin{split} \tilde{u}_{k} & =argmin_{u\in L^2} \left\{\frac{\beta}{2}\norm{u}^2 + \frac{\gamma}{2}\norm{A^{-1}Bu-z}^2+ \frac{1}{2\lambda_k}\norm{u-u_k}^2\right\}\\ & = \left[\left(\beta+\frac{1}{\lambda_k}\right)I+\gamma B^*A^{-*}A^{-1}B\right]^{-1}\left(\frac{1}{\lambda_k}u_k+\gamma B^*A^{-*}z\right). \end{split} \end{equation*}$$
• Proximal-point step:
• $$\begin{equation*} \begin{split} u_{k+1}& =argmin_{u\in L^2} \left\{\alpha_c \norm{u}_{1,T} + \frac{1}{2\lambda_k}\norm{u-\tilde{u}_k}_{T}^2\right\}\\ & = shrink(\tilde{u}_k,\alpha_c \lambda_k). \end{split} \end{equation*}$$

Remark: Notice that, when $\alpha_s=0$, the solution of $\left(\mathcal{P}^c_s\right)$ is simply given by

### Evolutionary problem

#### Optimality conditions

Define the classical Lagrangian

By integration by parts, we have

Deriving with respect to the three variables $\left(u,y,p\right)$, we obtain the optimality system:

where the relation between the optimal control and the dual state is given by

The latter is equivalent to

where the operator of $soft-shrinkage$ is defined by

Finally,

#### Numerical algorithm

In order to compute a numerical solution of problem $\left(\mathcal{P}_c\right)$, after a discretization by finite differences, we use a grad-prox splitting:

• $$\begin{equation*} \begin{split} \tilde{u}_{k} & = u_k-\lambda_k \nabla_u \left[\frac{\beta}{2}\norm{u}^2_{T} + \frac{\gamma}{2}\norm{Lu-z}_{T}^2 \right]\left(u_k\right) \\ & = u_k-\lambda_k \left[\beta u_k+\gamma L^*\left(Lu_k-z\right)\right]\\ & = u_k - \lambda_k\left[\beta u_k+\gamma B^*p_k\right], \end{split} \end{equation*}$$ where $$\begin{equation*} \begin{cases} y_k'+Ay_k=Bu_k & \left(\Omega \times \left(0,T\right)\right)\\ y_k=0 & \left(\partial\Omega\times \left(0,T\right)\right)\\ y_k(0)=0 & \left(\Omega\right) \end{cases} \end{equation*}$$ and $$\begin{equation*} \begin{cases} -p_k'+A^* p_k =y_k-z & \left(\Omega \times \left(0,T\right)\right)\\ p_k=0 & \left(\partial\Omega\times \left(0,T\right)\right)\\ p_k(T)=0 & \left(\Omega\right). \end{cases} \end{equation*}$$
• Proximal-point step:
• $$\begin{equation*} \begin{split} u_{k+1}& =argmin_{u\in L^2_T} \left\{\alpha_c \norm{u}_{1,T} + \frac{1}{2\lambda_k}\norm{u-\tilde{u}_k}_{T}^2\right\}\\ & = shrink(\tilde{u}_k,\alpha_c \lambda_k). \end{split} \end{equation*}$$

Remarks: Another possibility is to include the term $\frac{\beta}{2}\norm{u}^2_{T}$ in the proximal step. Notice that, for

then $\nabla f$ is Lipschitz continuous. Indeed, for $u_i\in L^2_T$ ($i=1,2$), then

where

and

By linearity $\delta y=y_2-y_1$ and $\delta p=p_2-p_1$ solve the same equations with right-hand-sides $B(u_2-u_1)$ and $\delta y$, respectively. Then

where we defined

In order the prox-grad method to converge, the restriction on the step size is given by

### The stationary problem

#### Optimality conditions

Finally, we obtain a single equation in the dual variable $p$:

#### Numerical algorithm

In order to compute a numerical solution of problem $\left(\mathcal{P}_s\right)$, after a discretization by finite differences, we use a prox-prox splitting on the Augmented Energy: first write the state as $y=A^{-1}Bu$, then

• Proximal-point step:
• $$\begin{equation*} \begin{split} u_{k+1} & =argmin_{u\in L^2} \left\{\frac{\beta}{2}\norm{u}^2 + \frac{\gamma}{2}\norm{A^{-1}Bu-z}^2+\frac{\delta}{2\lambda_k}\norm{A^{-1}Bu-y_k}^2 +\frac{1}{2\lambda_k}\norm{u-u_k}^2\right\}\\ & = \left[\left(\beta+\frac{1}{\lambda_k}\right)I+\left(\gamma+\frac{\delta}{\lambda_k}\right) B^*A^{-*}A^{-1}B\right]^{-1}\left[\frac{1}{\lambda_k}u_k+ B^*A^{-*}\left(\gamma z+\frac{\delta}{\lambda_k}y_k\right)\right]. \end{split} \end{equation*}$$
• Proximal-point step:
• $$\begin{equation*} \begin{split} y_{k+1}& =argmin_{y\in L^2} \left\{ \alpha_s \norm{y}_{1} + \frac{\delta}{2\lambda_k}\norm{y-A^{-1}Bu_{k+1}}^2+\frac{1}{2\lambda_k}\norm{y-y_k}_{T}^2\right\}\\ & = shrink(\tilde{y}_k,\tilde{\lambda}_k), \end{split} \end{equation*}$$ where we defined $$\begin{eqnarray*} \tilde{y}_k & = & \frac{y_k+\delta A^{-1}B u_{k+1}}{1+\delta};\\ \tilde{\lambda}_k & = & \frac{\alpha_s \lambda_k}{1+\delta}. \end{eqnarray*}$$

Remark: Notice that again, when $\alpha_s=0$, the solution of $\left(\mathcal{P}_s\right)$ is simply given by

### Evolutionary problem

#### Optimality conditions

Define the classical Lagrangian

By integration by parts, we have

Deriving with respect to the three variables $\left(u,y,p\right)$, we obtain the optimality system:

where the relation between the optimal control and the dual state is given by

The adjoint equation is equivalent to

Finally,

#### Numerical algorithm

In order to compute a numerical solution of problem $\left(\mathcal{P}_s\right)$, after a discretization by finite differences, we use a grad-prox splitting on the following Augmented Energy:

Then,

• $$\begin{equation*} \begin{split} u_{k+1} & = u_k-\lambda_k \nabla_u \left[\frac{\beta}{2}\norm{u}^2_{T}+ \frac{\gamma}{2}\norm{Lu-z}_{T}^2+\frac{\delta}{2\lambda}\norm{Lu-y}_{T}^2 \right]\left(u_k\right) \\ & = u_k-\lambda_k \left[\beta u_k+\gamma L^*\left(Lu_k-z\right)+\frac{\delta}{\lambda_k}L^*\left(Lu_k-y_k\right)\right]\\ & = \left(1-\beta\lambda_k\right)u_k - B^*p_k, \end{split} \end{equation*}$$ where $$\begin{equation*} \begin{cases} y_{u_k}'+Ay_{u_k}=Bu_k & \left(\Omega \times \left(0,T\right)\right)\\ y_{u_k}=0 & \left(\partial\Omega\times \left(0,T\right)\right)\\ y_{u_k}(0)=0 & \left(\Omega\right) \end{cases} \end{equation*}$$ and $$\begin{equation*} \begin{cases} -p_k'+A^* p_k =\left(\gamma \lambda_k + \delta\right)y_{u_k}-\gamma \lambda_k z - \delta y_k & \left(\Omega \times \left(0,T\right)\right)\\ p_k=0 & \left(\partial\Omega\times \left(0,T\right)\right)\\ p_k(T)=0 & \left(\Omega\right). \end{cases} \end{equation*}$$
• Proximal-point step:
• $$\begin{equation*} \begin{split} y_{k+1}& =argmin_{y\in L^2_T} \left\{\alpha_s \norm{y}_{1,T} + \frac{\delta}{2\lambda_k}\norm{y-Lu_{k+1}}_T^2+\frac{1}{2\lambda_k}\norm{y-y_k}_{T}^2\right\}\\ & = shrink(\tilde{y}_k,\tilde{\lambda}_k), \end{split} \end{equation*}$$ where we defined $$\begin{eqnarray*} \tilde{y}_k & = & \frac{y_k+\delta L u_{k+1}}{1+\delta};\\ \tilde{\lambda}_k & = & \frac{\alpha_s \lambda_k}{1+\delta}. \end{eqnarray*}$$
Remark: Another possibility is to consider $$\begin{equation*} \mathcal{L}_{\lambda}\left(u,y\right)=\frac{\beta}{2}\norm{u}^2_{T}+\alpha_s \norm{y}_{1,T}+ \frac{\gamma}{2}\norm{y-z}_{T}^2+\frac{\delta}{2\lambda}\norm{Lu-y}_{T}^2. \end{equation*}$$

### Computational experiments

In the following, we present the setting for the numerical experiments.
• Spacial domain: $\Omega=\left(0,1\right)$;
• Time interval: $\left[0,T\right]$, with $T=1$;
• Weight-parameters: $\alpha_c=[0, 0.01]$, $\alpha_s=[0, 0.65]$, $\beta=0.0001$ and $\gamma=1$;
• Trajectory target: $$z(x)=\mathcal{I}_{\left[x_a,x_b\right]},$$ where $x_a=1.7/3$, $x_b=3.5/4$;
• Control operator: for $x_1=1/7$ and $x_2=4/5$, $$B=\mathcal{I}_{\left[x_1,x_2\right]};$$
• $A$ is the finite difference discretization of $-\Delta$;
• Numerical grid: $N_x=300$ in space, $N_t=100$ in time.

### Stationary solutions Figure 1: $\alpha_c=\alpha_s=0$. Figure 2: $\alpha_c=0.01$, $\alpha_s=0$. Figure 3: $\alpha_c=0$, $\alpha_s=0.65$.

### Evolutionary problem

#### Optimal control Figure 4a: Optimal control for $\alpha_c=\alpha_s=0$. In red, the controllable subdomain; in blue, the stationary optimal controls. Figure 4b: Optimal control for $\alpha_c=0.01$, $\alpha_s=0$. In red, the controllable subdomain; in blue, the stationary optimal controls. Figure 4c: Optimal control for $\alpha_c=0$, $\alpha_s=0.65$. In red, the controllable subdomain; in blue, the stationary optimal controls.

#### Optimal state Figure 5a: Optimal state for $\alpha_c=\alpha_s=0$ . In red, the target $z$; in blue, the stationary optimal states. Figure 5b: Optimal state for $\alpha_c=0.01$, $\alpha_s=0$. In red, the target $z$; in blue, the stationary optimal states. Figure 5c: Optimal state for $\alpha_c=0$, $\alpha_s=0.65$. In red, the target $z$; in blue, the stationary optimal states. Figure 6a: Optimal adjoint for $\alpha_c=\alpha_s=0$. Figure 6b: Optimal adjoint for $\alpha_c=0.01$, $\alpha_s=0$. Figure 6c: Optimal adjoint for $\alpha_c=0$, $\alpha_s=0.65$.