An Adaptive Newton Algorithm for Optimal Control Problems with Application to Optimal Electrode Design

Abstract

In this work, we present an adaptive Newton-type method to solve nonlinear constrained optimization problems, in which the constraint is a system of partial differential equations discretized by the finite element method. The adaptive strategy is based on a goal-oriented a posteriori error estimation for the discretization and for the iteration error. The iteration error stems from an inexact solution of the nonlinear system of first-order optimality conditions by the Newton-type method. This strategy allows one to balance the two errors and to derive effective stopping criteria for the Newton iterations. The algorithm proceeds with the search of the optimal point on coarse grids, which are refined only if the discretization error becomes dominant. Using computable error indicators, the mesh is refined locally leading to a highly efficient solution process. The performance of the algorithm is shown with several examples and in particular with an application in the neurosciences: the optimal electrode design for the study of neuronal networks.

Appendix

In Appendix, we derive the flux function g, the admissible set $Q_{\text {ad}}$ and their dependence on the sizes $s_k$ and positions $m_k$ of the k-th hole, $k=1\ldots p$. The fixed size of the hole at the tip is denoted by $s_0$. The remaining holes are numbered from bottom (tip) to the top. In order to derive the flux function g, we assume that the holes do not overlap and that they remain above the tip of the pipette and below the upper boundary of the simulation domain. Moreover, due to fabrication reasons a minimum distance $\delta _1>0$ has to remain between them. In formulas, this means that

$$\begin{aligned} y_{\text {tip}} + \delta _1&\le m_1 - \frac{s_1}{2},\quad m_k+ \frac{s_k}{2} + \delta _1 \le m_{k+1} -\frac{s_{k+1}}{2} \quad k=1,\dots ,p-1,\quad m_p + \frac{s_p}{2} \le y_{\text {top}} \end{aligned}$$

(43)

where $y_{\text {tip}}=0$ and $y_{\text {top}}$ denote the vertical coordinate of the tip and the upper boundary of the simulation domain, respectively. Moreover, we assume that the sizes of the holes are bounded below by a constant $\delta _2>0$

$$\begin{aligned} s_k \ge \delta _2\quad \text {for } k=1,\dots ,p, \end{aligned}$$

(44)

which is again due to fabrication limitations. The admissible set is defined by

$$\begin{aligned} Q_{\text {ad}} := \left\{ q:=(m_1,\dots ,m_p,s_1,\dots ,s_p) \in {\mathbb {R}}^{2p}\; \big | \; q \text { fulfills } (43) \text { and } (44) \right\} . \end{aligned}$$

(45)

Note that by this definition, $Q_{\text {ad}}$ is closed and bounded. A rough bound for the sizes $s_k$ is given by $0 \le s_k < y_{\text {top}}-y_{\text {tip}}$ and for the positions $m_k$ by $y_{\text {tip}}< m_k < y_{\text {top}},\, i=1,\ldots ,p$.

Fig. 12

Scheme of the electric circuit around the micropipette

A scheme of the electric circuit is shown in Fig. 12. For simplicity, we present only the case of a micropipette with two holes on each side. The derivation of corresponding formulas for a different number of holes is analogous. We assume that a fixed current ${\overline{I}}$ is applied at the top of the micropipette and calculate the current $I_k$ that flows out of the micropipette at the holes $k=0,\dots ,2$. The flux function g at hole k is then given by the current density $J_k$

$$\begin{aligned} g|_{\varGamma _k}= J_k = \frac{I_k}{|\varGamma _k|} \quad \text { on } \varGamma _k. \end{aligned}$$

The micropipette is filled with a conducting liquid. We assign a specific resistance $R_j$, $j=0, \dots , 3$, to each of the parts of the micropipette. The resistances of the conducting liquid in the small holes on the left and right in between the isolating wall are denoted by $R_1$ and $R_2$, the resistances of the parts in the interior of the micropipette by $R_3$ and $R_0$. Denoting the thickness of the wall by d, the resistance of a hole is given by

$$\begin{aligned} R_k = \rho \frac{d}{\pi s_k^2} \quad (k=1,2), \end{aligned}$$

where $\rho = 1/\sigma$ is the electrical resistivity. To calculate the resistance of the conical part below $x_2$, we introduce the notation a(x) for the area inside the micropipette at position x, see Fig. 13. The area $a_0=a(0)$ of $\varGamma _0$ at the tip of the micropipette is given by $a(0)=\pi s_0^2$, the area $a_1=a(m_1)$ of $\varGamma _1$ at the first hole by

$$\begin{aligned} a(m_1)=\pi (s_0 + \tan (\theta )m_1)^2, \end{aligned}$$

where $\theta$ is the inclination angle of the micropipette. The resistance of the conical part below $x_2$ is then given by (see e.g., [54])

$$\begin{aligned} R_0 = \rho \int _0^{m_1} \frac{1}{a(x)^2} \mathrm{d}x = \frac{\rho }{\pi } \mathrm {cot} (\theta ) \left( \frac{1}{s_0} -\frac{1}{s_0 + m_1 \tan (\theta )}\right) . \end{aligned}$$

Fig. 13

Scheme of the tip region of the micropipette including area elements to calculate the resistances

Similarly, we get for the resistance $R_3$ of the part between the points $x_1$ and $x_2$

$$\begin{aligned} R_3 = \frac{\rho }{\pi } \mathrm {cot} (\theta ) \left( \left( s_0 + m_1 \mathrm{tan}(\theta )\right) ^{-1} - \left( s_0 + m_2 \mathrm{tan}(\theta )\right) ^{-1}\right) . \end{aligned}$$

The voltage difference between point $x_2$ and a point ${\hat{x}}$ far off the micropipette can be used to derive the following formula by using Ohm’s law (see Fig. 12)

$$\begin{aligned} I_2 \cdot R_2 = I_3 \cdot R_{0,1,3} \quad (= u(x_2) - u({\hat{x}}) ). \end{aligned}$$

(46)

Here, $R_{0,1,3}$ stands for the total resistance of the parts $R_0$, $R_1$ and $R_3$ which is given by

$$\begin{aligned} R_{0,1,3} = R_3 + (R_0^{-1} + 2 R_1^{-1})^{-1}. \end{aligned}$$

Furthermore, by Kirchhoff’s current law the current ${\overline{I}}$ splits at point $x_2$ to

$$\begin{aligned} {\overline{I}} = I_3 + 2 I_2. \end{aligned}$$

(47)

(46) and (47) can be solved for the two unknowns $I_2$ and $I_3$. In the same way, it holds at point $x_1$

$$\begin{aligned} R_0 \cdot I_0 = R_1 \cdot I_1 \end{aligned}$$

(48)

and

$$\begin{aligned} I_3 = 2 I_1 + I_0. \end{aligned}$$

(49)

Given $I_3$, (48) and (49) define $I_0$ and $I_1$. Inserting the formulas for the resistances, a direct calculation results in

$$\begin{aligned} I_0&= {\overline{I}} \frac{R_1 R_2}{\left( R_2 + 2 R_{0,1,3} \right) \left( R_1 + 2 R_0\right) } = {\overline{I}} \frac{\left( s_0\,c + m_1 \right) ^2 \left( s_0\,c + m_2 \right) d^2 \, s_0}{T(m,s)},\nonumber \\ I_1&= {\overline{I}} \frac{R_0 R_2}{\left( R_2 + 2 R_{0,1,3} \right) \left( R_1 + 2 R_0\right) } = {\overline{I}} \frac{\left( s_0\,c + m_1 \right) \left( s_0\,c + m_2 \right) d \, c \, m_1 \, s_1^2}{T(m,s)},\nonumber \\ I_2&= {\overline{I}} \frac{R_{0,1,3}}{R_2 + 2 R_{0,1,3}} = {\overline{I}} \frac{\left( m_2 d s_0^2 c^2 + 2 c m_2 d s_0 m_1+ 2 m_2 s_1^2 c^2 m_1-2 s_1^2 c^2 m_1^2+d m_1^2 m_2 \right) \,c \, s_2^2}{T(m,s)} \end{aligned}$$

(50)

with $c:={cot} (\theta )$ and

$$\begin{aligned} T(m,s)&={{ s_0}}^{4} c^{3}{d}^{2}+2\,{{ s_0}}^{3} c^{2}{d}^{2}{} { m_1}+2\,d{{ s_0}} ^{2} c^{3}{} { m_1}\,{{ s_1 }}^{2}+{{ s_0}}^{3} c ^{2}{ m_2}\,{d}^{2}+2\,{{ s_0}}^{2}c { m_2} \,{d}^{2}{} { m_1}\\&\quad +2\,d{ s_0}\,c^{2}{} { m_2}\,{ m_1}\,{{ s_1}}^{2}+ {{ m_1}}^{2}{{ s_0 }}^{2}c {d}^{2}+2\,d{{ m_1}}^{2}{} { s_0}\, c ^{2}{{ s_1}}^{2}+{{ m_1}}^{2}{} { m_2}\,{d}^{2}{} { s_0} \\&\quad +2\,d{{ m_1}}^{2}{} { m_2}\, c{{ s_1}}^{2}+2 c^{3}{{ s_2}}^{2}{} { m_2}\,d{{ s_0}}^{2} +4\, c ^{2}{{ s_2}}^{2}{} { m_2} \,d{ s_0}\,{ m_1}\\ {}&\quad +\,4\, c^{3}{{ s_2}}^{2}({ m_2}-{ m_1})\,{ m_1}\,{{ s_1}}^{2} +2\,c {{ m_1}}^{2}d{{ s_2}}^{2}{ m_2} . \end{aligned}$$

As we use a two-dimensional setting, the size of $\varGamma _k$ is given by $|\varGamma _k|=s_k$. The flux function at hole k is thus given by

$$\begin{aligned} J_k = \frac{I_k}{s_k} \quad \text { on } \varGamma _k. \end{aligned}$$

(51)

As the parameters $s_0, d$ and c as well as the variables $m_k$ and $s_k$ ($k=1,2$) are strictly positive, the denominator T(m, s) in the definition of $I_k$ is positive for all values of $(m_1, m_2, s_1, s_2) \in {\mathbb {R}}_+^4$ with $m_2-m_1\ge 0$. The denominator ($s_k$) in the definition of $J_k$ (51) cancels out with the nominator in (50), such that $J_k$ has no singularities for $m_2\ge m_1$. Therefore, $J_k(q)$ is a smooth function on $Q_{\text {ad}}$.