Towards Optimal Control of Amyloid Fibrillation

Abstract

Epigallocatechin-3-gallate, as a representative amyloid inhibitors, has shown a promising ability against A$\beta$ fibrillation by directly degradating the mature fibrils. Most previous studies have been focusing on its functional mechanisms, meanwhile its optimal dosage has been seldom considered. To solve this critical issue, we refer to the generalized Logistic model for amyloid fibrillation and inhibition and adopt the optimal control theory to balance the effectiveness and cost (or toxicity) of inhibitors. The optimal control trajectory of inhibitors is analytically solved, based on which the influence of model parameters, the difference between the optimal control strategy and several other traditional drug dosing strategies are systematically compared and validated through experiments. It is found that the strategy of multiple-times adding is more suitable for a long-term disease treatment, while single high-dose therapy is preferred for a short-term treatment. We hope our findings can shed light on the rational usage of amyloid inhibitors in clinic.

Appendices

Appendix A: Proof of Main Theorems

1.1 A.1 Some Related Lemmas on Optimal Control

The Bolza problem of optimal control reads

$$\begin{aligned} \begin{array}{l} \inf _{u(t)} J[u(t)]=\Phi \left[ X(T)\right] +\int _{0}^{T} L \left[ X\left( t\right) , u\left( t\right) , t\right] d t \\ \text {subject to} \\ \dot{X}(t)=f(X(t),u(t),t),\quad t \in {[0,T]},\quad X(0)=x_0. \end{array} \end{aligned}$$

(15)

The Pontryagin’s Maximum Principle (PMP) provides a necessary condition, whose concrete form is stated as follows.

Lemma 1

(Pontryagin 1987) (PMP) Let $u^*(t) \in \Theta \subset \mathbb {R}^{m}$ be a bounded, measurable and admissible control that optimizes Eq. (15), with $\Theta$ be the control set, and $X^*$ be its corresponding state trajectory. Define a Hamiltonian

$$\begin{aligned} H(X,u,\lambda ,t)=\lambda ^T f(X,u,t)-L(X,u,t), \end{aligned}$$

where $\lambda \in \mathbb {R}^d$. Then there exists an absolutely continuous process $\lambda (t)$ such that

$$\begin{aligned}{} & {} \dot{X^*}(t)=\frac{\partial H(X^*(t),u^*(t),\lambda ^*(t),t)}{\partial \lambda },\quad X^*(0)=x_0 \end{aligned}$$

(16)

$$\begin{aligned}{} & {} \dot{\lambda }^*(t)=-\frac{\partial H(X^*(t),u^*(t),\lambda ^*(t),t)}{\partial X},\quad \lambda ^*(T)=-\frac{\partial \Phi (X^*(T))}{\partial X}\nonumber \\{} & {} H(X^*(t),u^*(t),\lambda ^*(t),t) \ge H(X^*(t),u(t),\lambda ^*(t),t),\quad \nonumber \\{} & {} \forall u\in \Theta \quad \text {and} \quad a.e. ~t\in [0,T] \end{aligned}$$

(17)

Here Eq. (16) reduces to the state equation under the optimal control, while the co-state $\lambda ^*$ evolves backward according to Eq. (17) with a fixed terminal state $\lambda ^*(T)$.

Lemma 2

(Lenhart and Workman 2007) Suppose that $-f(X,u,t)$ and L(X, u, t) are both continuously difierentiable functions with respect to their three arguments and convex in X and u. Define a Hamiltonian

$$\begin{aligned} H(X,u,\lambda ,t)=\lambda ^T f(X,u,t)-L(X,u,t), \end{aligned}$$

where $\lambda \in \mathbb {R}^d$. Suppose $u^*(t)$ is a control, with associated state $X^*(t)$, and $\lambda ^*(t)$, a piecewise difierentiable function, such that $u^*(t), X^*(t)$, and $\lambda ^*(t)$, together satisfy on $0 \le t \le T$:

$$\begin{aligned}{} & {} \dot{X^*}(t)=\frac{\partial H(X^*(t),u^*(t),\lambda ^*(t),t)}{\partial \lambda },\quad X^*(0)=x_0 \end{aligned}$$

(18)

$$\begin{aligned}{} & {} \dot{\lambda }^*(t)=-\frac{\partial H(X^*(t),u^*(t),\lambda ^*(t),t)}{\partial X},\quad \lambda ^*(T)=0 \end{aligned}$$

(19)

$$\begin{aligned}{} & {} \lambda ^*(t) \le 0 \end{aligned}$$

(20)

$$\begin{aligned}{} & {} H_{u}(X^*(t),u^*(t),\lambda ^*(t),t) = 0,\quad \nonumber \\{} & {} \forall u\in \Theta \quad \text {and} \quad a.e. ~t\in [0,T] \end{aligned}$$

(21)

Then for all controls u, we have

$$\begin{aligned} J(u^{*}) \le J(u). \end{aligned}$$

Lemma 3

(Michel Petrovitch 1901) (Comparison Theorem) Let both functions f(x, y) and F(x, y) be continuous in the plane region G and satisfy the inequality

$$\begin{aligned} f(x,y)<F(x,y),\qquad (x,y) \in G. \end{aligned}$$

Let $y=\phi (x)$ and $y=\Phi (x)$ be solutions to the initial value problems

$$\begin{aligned} \frac{d y}{d x}=f(x,y),\qquad y(x_{0})=y_{0} \end{aligned}$$

and

$$\begin{aligned} \frac{d y}{d x}=F(x,y),\qquad y(x_{0})=y_{0} \end{aligned}$$

respectively on the interval $a<x<b$, where $(x_{0},y_{0})\in G$. Then we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \phi (x)<\Phi (x),&{}\qquad x_{0}<x<b,\\ \phi (x)>\Phi (x),&{}\qquad a<x<x_{0}. \end{array}\right. } \end{aligned}$$

By utilizing above lemmas, we can prove that the following boundary value problem has a solution.

Theorem 2

For $t \in [0,T]$, the boundary value problem,

$$\begin{aligned} \left\{ \begin{array}{cl} \dot{M}(t)=&{}k_{a}M(t)(1-\frac{M(t)}{M_{max}})-k_{d} u(t) M(t),\\ \dot{u}(t)=&{}\frac{k_{a}}{M_{max}}u(t)M(t) - \frac{k_{d}}{w^2}{M(t)}^{2},~t< T,\\ M(0)=&{}{m_0}>0, \quad u(T)=0, \end{array}\right. \end{aligned}$$

(22)

admits a solution.

Proof

By utilizing Lemma 3, we can deduce that

$$\begin{aligned} 0 \le M(t)<M_{max} \text { and } 0 \le u(t)<k_{d}M_{max}^2T/w^2. \end{aligned}$$

Let $h_{1}(x, y)=k_{a}x(1-\frac{x}{M_{max}})-k_{d} x y$ and $h_{2}(x, y)=\frac{k_{a}}{M_{max}}xy - \frac{k_{d}}{w^2}{x}^{2}$. Clearly, both $h_{1}$ and $h_{2}$ have continuous partial derivatives on $\mathbb {R}^2$. By employing the existence and uniqueness theorem for ordinary differential equations and the continuous dependence on initial values, the initial value problem

$$\begin{aligned} \left\{ \begin{array}{cl} \dot{M}(t)=&{}k_{a}M(t)(1-\frac{M(t)}{M_{max}})-k_{d} u(t) M(t), \\ \dot{u}(t)=&{}\frac{k_{a}}{M_{max}}u(t)M(t) - \frac{k_{d}}{w^2}{M(t)}^{2}, ~t< T, \\ M(0)=&{}{m_{0}}>0, \quad u(0)=u_{0}, \end{array}\right. \end{aligned}$$

(23)

has a unique solution on the interval [0, T] , and the solution $u=\phi (t, u_{0})$ with respect to $u_{0}$ is continuous.

Based on the comparison theorem, it is evident that $u=\phi (T, k_{d}M_{max}^2T/w^2)>0$ and $u=\phi (T, 0)<0$. Hence, there exists a $u_{0} \in (0, k_{d}M_{max}^2T/w^2)$ such that $u=\phi (T, u_{0})=0$. Therefore, the boundary value problem has a solution. $\square$

Theorem 3

The analytical solution to the boundary value problem (22) is given by:

$$\begin{aligned} M= & {} \frac{\phi ^{2}(t)-C_{1}}{2\sqrt{A}\phi (t)+B}, \end{aligned}$$

(24)

$$\begin{aligned} u= & {} \frac{1}{k_{d}}\left[ \left( \sqrt{A}-\frac{k_{a}}{M_{max}}\right) M-\phi (t)+k_{a}\right] , \end{aligned}$$

(25)

where

$$\begin{aligned} \phi (t)=\frac{2\sqrt{C_{1}}}{1-e^{\sqrt{C_{1}}t+C_{2}}}-\sqrt{C_{1}}, \end{aligned}$$

(26)

with $C_{1}$ and $C_{2}$ being constants determined by the conditions $M(0)={m_{0}}$ and $u(T)=0$. Furthermore, $A=\frac{k_{d}^{2}}{w^{2}}+\frac{k_{a}^{2}}{M_{max}^{2}}$ and $B=-\frac{2k_{a}^{2}}{M_{max}}$.

Proof

Introduce function $s(t)=\ln (M(t))$. The boundary value problem in (22) can be transformed into:

$$\begin{aligned}{} & {} \dot{s}(t)=k_{a}\left( 1-\frac{e^{s(t)}}{M_{max}}\right) -k_{d}u(t), \end{aligned}$$

(27)

$$\begin{aligned}{} & {} \dot{u}(t)=\frac{k_{a}}{M_{max}}u(t)e^{s(t)} - \frac{k_{d}}{w^2}e^{2s(t)}, \end{aligned}$$

(28)

$$\begin{aligned}{} & {} s(0)=\ln (m_{0}), \quad u(T)=0. \end{aligned}$$

(29)

Next, differentiate equation (27) with respect to t and substitute equations (28) and (27) into it. Now, we obtain

$$\begin{aligned} \ddot{s}(t)=Ae^{2s}+\frac{B}{2}e^{s}, \end{aligned}$$

where $A=\frac{k_{d}^{2}}{w^{2}}+\frac{k_{a}^{2}}{M_{max}^{2}}$ and $B=-\frac{2k_{a}^{2}}{M_{max}}$. Integrating above formula once, we have:

$$\begin{aligned} {\dot{s}}^{2}(t)= Ae^{2s}+Be^{s}+C_{1}, \end{aligned}$$

(30)

where $C_{1}$ is a constant.

By performing separation of variables and Euler’s substitution on both sides, we arrive at

$$\begin{aligned} \dot{s}(t)=\phi (t)-\sqrt{A}e^{s}, \end{aligned}$$

(31)

where

$$\begin{aligned} \phi (t)=\frac{2\sqrt{C_{1}}}{1-e^{\sqrt{C_{1}}t+C_{2}}}-\sqrt{C_{1}}, \end{aligned}$$

and $C_{1}$ and $C_{2}$ are two constants. Considering equation (30), now we have

$$\begin{aligned} (\phi (t)-\sqrt{A}e^{s})^2= Ae^{2s}+Be^{s}+C_{1}. \end{aligned}$$

Its solution gives

$$\begin{aligned} M(t)= & {} \frac{\phi ^{2}(t)-C_{1}}{2\sqrt{A}\phi (t)+B}, \end{aligned}$$

(32)

$$\begin{aligned} u(t)= & {} \frac{1}{k_{d}}\left[ \left( \sqrt{A}-\frac{k_{a}}{M_{max}}\right) M-\phi (t)+k_{a}\right] . \end{aligned}$$

(33)

$\square$

1.2 A.2 Proof of Theorem 1

The optimal control problem satisfying the constraint (6) is given by:

$$\begin{aligned} \begin{array}{l} \inf _{u(t)\in [0,u_{max}]} J[u(\cdot )]=\int _{0}^{T}[M(t)^2+w^2u(t)^2]d t \\ \text {subject to} \\ \dot{M(t)}=k_{a}M(t)\left( 1-\frac{M(t)}{M_{max}}\right) -k_{d} u(t) M(t),\quad M^*(0)={m_0}>0, \end{array} \end{aligned}$$

(34)

where $M_{max} \ge M(t) \ge 0, \forall t \in [0,T]$.

According to Lemma 1, an optimal solution to (34) should satisfy the following equations,

$$\begin{aligned}{} & {} \dot{M^*}(t)=k_{a}M^*-k_{a}(M^{*})^2/M_{max}-k_{d} u^* M^*, \end{aligned}$$

(35)

$$\begin{aligned}{} & {} \dot{\lambda ^*}(t)=-k_{a}\lambda ^*+2k_{a}\lambda ^*M^*/M_{max}+k_{d} u^*\lambda ^* +2M^*,~t<T, \end{aligned}$$

(36)

$$\begin{aligned}{} & {} u^*(t) \in {\arg \max }_{u\in [0,+\infty ]} \left[ \lambda \left( k_{a}M^*-k_{a}(M^*)^2/M_{max}-k_{d} u M^*\right) -(M^*)^2-w^2u^2 \right] , \nonumber \\{} & {} \end{aligned}$$

(37)

$$\begin{aligned}{} & {} M^*(0)={m_0}, \qquad \lambda ^* (T)=0. \end{aligned}$$

(38)

We get $u^*(t)=-k_{d}\lambda ^*M^*/2w^2$ by Eq. (37). Further, we can get

$$\begin{aligned} \left\{ \begin{array}{cl} \dot{M^*}(t)=&{}k_{a}M^*(t)(1-\frac{M^*(t)}{M_{max}})-k_{d} u^*(t) M^*(t),\\ \dot{u^*}(t)=&{}\frac{k_{a}}{M_{max}}u^*(t)M^*(t) - \frac{k_{d}}{w^2}{M^*(t)}^{2},~t< T, \\ M^*(0)=&{}{m_0}, \quad u^* (T)=0. \end{array}\right. \end{aligned}$$

(39)

From Eq. (39), it can be asserted that $u^* \ge 0$, otherwise it contradicts with $u^*(T)=0$. By Lemma 3, we can further get $u^*(t)<k_{d}M_{max}^2T/w^2$. To sum up, $0 \le u^*<k_{d}M_{max}^2T/w^2$. We have demonstrated that if the optimal control problem has a solution, it must satisfy Eq. (39).

Theorem 4

The solution to problem (34) exists and satisfies the system of equations in (39).

Proof

From the above, we only need to show the existence of a solution for the problem (34).

Introduce the function $s(t)=\ln (M(t))$ and consider the equivalent problem obtained by the variable substitution, which will be referred as OC’ problem. The OC’ problem consists of three main components:

• (i) The control constraint set is the same as the problem (34).

• (ii) The control equation is:

  $$\begin{aligned} \frac{d}{d t} s(t) =k_{a}(1-\frac{e^{s}}{M_{max}}) - k_{d} u, \qquad s(0)=\ln (m_{0}). \end{aligned}$$

• (iii) The objective functional is:

  $$\begin{aligned} J[u(\cdot )] =\int _{0}^{T}e^{2s}+w^2 u(t)^2d t. \end{aligned}$$

The OC’ problem aims to minimize the above objective functional, and its equivalence to the problem (34) is evident.

Next, we verify the conditions of Lemma 2 for the existence of the OC’ problem. By Theorem 2, the boundary value problem for $t \in [0, T]$ is given by:

$$\begin{aligned} \left\{ \begin{array}{cl} \dot{s}(t)=&{}k_{a}(1-\frac{e^{s(t)}}{M_{max}})-k_{d} u(t), \\ \dot{u}(t)=&{}\frac{k_{a}}{M_{max}}u(t)e^{s(t)} - \frac{k_{d}}{w^2}e^{2s(t)}, ~t< T, \\ s(0)=&{}\ln (m_{0}), \quad u(T)=0, \end{array}\right. \end{aligned}$$

(40)

and its solution exists.

By setting $p=-\frac{2w^2}{k_{d}}u$, we have:

$$\begin{aligned} -k_{d}p-2w^2u=0, \end{aligned}$$

(41)

and

$$\begin{aligned} \dot{p}(t)=\frac{k_{a}}{M_{max}}e^{s(t)}p(t)+2e^{2s(t)}, \quad p(T)=0, \end{aligned}$$

(42)

which implies

$$\begin{aligned} p(t)\le 0, \end{aligned}$$

(43)

otherwise, it contradicts $p(T)=0$.

Now, we only need to check that $-f(x, u)=-k_{a}(1-\frac{e^{x}}{M_{max}}) + k_{d} u$ and $L(x, u)=e^{2x}+w^2 u(t)^2$ are both convex functions. This can be done by directly computing the Hessian matrix. By using Lemma 2, the existence of solutions to the OC’ problem is verfied, which implies the solution to the problem (34) exists too. $\square$

1.3 A.3 A Feedback Control

Equation (7) can be transformed into a feedback control given by

$$\begin{aligned} \left\{ \begin{array}{cl} \dot{M^*}(t)=&{}k_{a}M^*(t)(1-\frac{M^*(t)}{M_{max}})-k_{d}M^*(t)u^*(t),\\ u^*(t)=&{}\frac{k_{a}(M_{max}-M^*(t))}{M_{max}k_{d}}\pm \sqrt{\frac{{M^*(T)}^2 -{M^*(t)}^2}{w^2}+\left( \frac{k_{a}(M_{max}-M^*(t))}{M_{max}k_{d}}\right) ^2},~t< T, \\ M^*(0)=&{}{m_0}. \end{array}\right. \nonumber \\ \end{aligned}$$

(44)

First, we can derive from Eq. (7) that

$$\begin{aligned}{} & {} \left[ k_{a}M^*(t)\left( 1-\frac{M^*(t)}{M_{max}}\right) -k_{d} u^*(t) M^*(t)\right] du^*+\\{} & {} \left[ \frac{k_{d}}{w^2}{M^*(t)}^{2}-\frac{k_{a}}{M_{max}}u^*(t)M^*(t)\right] dM^{*}=0,\\{} & {} \left[ k_{a}\left( 1-\frac{M^*(t)}{M_{max}}\right) -k_{d} u^*(t)\right] du^*+ \left[ \frac{k_{d}}{w^2}{M^*(t)} -\frac{k_{a}}{M_{max}}u^*(t)\right] dM^{*}=0,\\{} & {} d\left[ {M^*}^2-w^2{u^*}^2-\frac{2wk_{a}}{M_{max}k_{d}}M^*u^*+\frac{2wk_{a}}{k_{d}}u^*\right] =0,\\{} & {} {M^*}^2-w^2{u^*}^2-\frac{2wk_{a}}{M_{max}k_{d}}M^*u^*+\frac{2wk_{a}}{k_{d}}u^* =const. \end{aligned}$$

As ${M^*(T)}^2=const$ by $u^{*}(T)=0$, we have

$$\begin{aligned} {u^*}^2-2\frac{k_{a}(M_{max}-M^*)}{k_{d}M_{max}}u^*+\frac{{M^*(T)}^2-{M^*}^2}{w^2}=0, \end{aligned}$$

(45)

and

$$\begin{aligned} u^*(t)=\frac{k_{a}(M_{max}-M^*(t))}{M_{max}k_{d}}\pm \sqrt{\frac{{M^*(T)}^2-{M^*(t)}^2}{w^2}+\left( \frac{k_{a}(M_{max}-M^*(t))}{M_{max}k_{d}}\right) ^2}.\nonumber \\ \end{aligned}$$

(46)

It is found that, when $M^*(t) \ll M_{max}$ is satisfied, the positive sign is initially chosen; and then at the intermediate time $t_1$ which satisfies $\frac{{M^*(T)}^2-{M^*(t_1)}^2}{w^2}+\left( \frac{k_{a}(M_{max}-M^*(t_1))}{M_{max}k_{d}}\right) ^2=0$, it changes to the minus sign until T. This choice will cause mass concentration of aggregates to decrease first and then increase. Otherwise, we just choose the minus sign for the entire control process. This choice results in a monotonous increase in the mass concentration of aggregates.

Appendix B: Derivation of the Analytical Solution

The optimal control problem under the linear approximation is given by

$$\begin{aligned} \left\{ \begin{array}{cl} \dot{M^*}(t)=&{}k_{a}M^*-k_{d} u^* M^*,\\ \dot{\lambda ^*}(t)=&{}-k_{a}\lambda ^*+k_{d} u^*\lambda ^* +2M^*,~t<T, \\ M^*(0)=&{}{m_0}, \qquad \lambda ^* (T)=0. \end{array}\right. \end{aligned}$$

(47)

Additionally, based on the optimality condition, $H(M^*(t),u^*(t),\lambda ^*(t),t) \ge H(M^*(t),u(t),\lambda ^*(t),t)$ with $H(M,u,\lambda ,t)=\lambda (k_{a}M-k_{d} u M)-M^2-w^2u^2$,

we have

$$\begin{aligned} u^*(t) \in {\arg \max }_{u\in [0,+\infty ]} \left[ \lambda ^{*} \left( k_{a}M^*-k_{d} u M^*\right) -(M^*)^2-w^2u^2 \right] , \end{aligned}$$

(48)

or equivalently,

$$\begin{aligned} 0={\frac{\partial H(M,u,\lambda ,t)}{\partial u}}\bigg \vert _{M^{*},u^{*},\lambda ^{*}} =-2w^2u^*(t)-k_{d} \lambda ^*(t) M^*(t). \end{aligned}$$

(49)

It is straightforward to solve above algebraic equation, which leads to $u^*(t)=-{k_{d} \lambda ^*(t) M^*(t)}/{2w^2}$ by noticing $w^2>0$. Combining this with the terminal co-state $\lambda ^*(T)=0$ deduces $u^* (T)=0$.

In the next step, by taking the time derivative on both sides of the algebraic equation (49),

$$\begin{aligned} -\left( \frac{2w^2}{k_{d}}\right) \frac{du^*}{dt}=\lambda ^* \frac{dM^*}{dt}+ M^*\frac{d\lambda ^*}{dt}, \end{aligned}$$

and substituting the ODEs of $M^*$ and $\lambda ^*$ in Eq. (47) into the above one, we arrive at

$$\begin{aligned} \frac{du^*}{dt}=-\frac{k_{d}}{w^2}(M^*)^2. \end{aligned}$$

(50)

By multiplying $-2k_{d}M^*/{w^2}$ on both sides of the first equation in (47) and substituting Eq. (50), a decoupled equation for the co-state variable $u^*(t)$ is derived, i.e.

$$\begin{aligned} \frac{d^2u^*}{dt^2}=2k_{a}\frac{du^*}{dt}-2k_{d}u^*\frac{du^*}{dt}, \end{aligned}$$

(51)

which is of the second order. Integration once gives

$$\begin{aligned} \frac{du^*}{dt}=-k_{d}{u^*}^2+2k_{a}u^*-C, \end{aligned}$$

where C is an undetermined constant. The above equation can be rewritten as

$$\begin{aligned}\frac{du^*}{dt}=-k_{d} \left[ \bigg (u^*-\frac{k_{a}}{k_{d}}\bigg )^2+\bigg (\frac{C}{k_{d}}-\frac{k_{a}^2}{k_{d}^2} \bigg )\right] , \end{aligned}$$

whose solution gives the optimal trajectories of $u^*$ and $M^*$,

$$\begin{aligned} u^*(t)=C_1\tan (-k_{d}C_1t+C_2)+\frac{k_{a}}{k_{d}},\quad M^*(t)=wC_1\sec (-k_{d}C_1t+C_2). \end{aligned}$$

The constants $C_1$ and $C_2$ are determined by the boundary conditions $M^*(0)={m_0}$ and $u^* (T)=0$ as

$$\begin{aligned} w^2C_1^2\sec ^2(C_2)= {m_0}^2, \quad C_1^2\tan ^2(-k_{d}C_1T+C_2)=\frac{k_{a}^2}{k_{d}^2}, \end{aligned}$$

which further gives

$$\begin{aligned} u^*(0)=C_1\tan (C_2)+\frac{k_{a}}{k_{d}},\quad M^*(T)=wC_1\sec (-k_{d}C_1T+C_2). \end{aligned}$$

Appendix C: Analytical Solutions for Upper Bounded Cases

The optimal control problem with general upper bound constraints under the linear approximation is given by

$$\begin{aligned} \left\{ \begin{array}{cl} \dot{M^*}(t)=&{}k_{a}M^*-k_{d} u^* M^*,\\ \dot{\lambda ^*}(t)=&{}-k_{a}\lambda ^*+k_{d} u^*\lambda ^* +2M^*,~t<T, \\ M^*(0)=&{}{m_0}, \qquad \lambda ^* (T)=0. \end{array}\right. \end{aligned}$$

(52)

Based on the optimality condition, $H(M^*(t),u^*(t),\lambda ^*(t),t) \ge H(M^*(t),u(t),\lambda ^*(t),t)$ with $H(M,u,\lambda ,t)=\lambda (k_{a}M-k_{d} u M)-M^2-w^2u^2$, we have

$$\begin{aligned} u^*(t) \in {\arg \max }_{u\in [0,u_{max}]} \left[ \lambda ^{*} \left( k_{a}M^*-k_{d} u M^*\right) -(M^*)^2-w^2u^2 \right] , \end{aligned}$$

(53)

or equivalently,

$$\begin{aligned} u^*(t) \in {\arg \max }_{u\in [0,u_{max}]} \left[ -k_{d}M^*\lambda ^{*} u -w^2u^2 \right] . \end{aligned}$$

(54)

Let $Q = -{k_{d}M^{*}\lambda ^{*}}/{(2w^2)}$. Since $dQ/dt = -k_{d}/(2w^2)d(M^{*}\lambda ^{*})/dt = - k_{d}(M^*)^2/w^2 < 0$, the axis of symmetry (also the position of maximal value) of $A[u]:= -k_{d}M^*\lambda ^{*} u -w^2u^2$ decreases from $-{k_{d}m_{0}\lambda ^{*}(0)}/{(2w^2)}$ to 0. Hence, at the initial time, if $-{k_{d}m_{0}\lambda ^{*}(0)}/{(2w^2)}\not \in [0,u_{max}]$, we have $u^{*} = u_{max}$, which persist to time $t_{1}$ satisfying $u^*(t_1)=-{k_{d} \lambda ^*(t_1) M^*(t_1)}/{2w^2}$. Within the interval $[t_{1}, T]$, the optimal control problem reduces to the unbounded case, and results in Appendix B apply.

Appendix D: Objective Functional with $L_{1}$ Norm

The quadratic cost in the main text is adopted to illustrate our conclusions, alternative forms of the objective functional are allowed too, such as the $L_{1}$ norm which reads

$$\begin{aligned} J[u(\cdot )]=\int _{0}^{T}[M(t)+wu(t)]d t. \end{aligned}$$

(55)

To keep the non-negativity of M(t) and u(t), additional constraints are introduced too, that is

$$\begin{aligned} \forall t \ge 0, \qquad M_{max} \ge M(t) \ge 0 \quad \text { and} \quad u_{max}\ge u(t) \ge 0, \end{aligned}$$

(56)

According to the optimality condition, $H(M^*(t),u^*(t),\lambda ^*(t),t) \ge H(M^*(t),u(t),\lambda ^*(t),t)$ with $H(M,u,\lambda ,t)=\lambda (k_{a}M(1-M/M_{max})-k_{d} u M)-M-w u$, we have

$$\begin{aligned} u^*(t) ={\left\{ \begin{array}{ll} u_{max}, \quad -(\lambda ^* M^* k_{d} +w)>0,\\ 0, \quad -(\lambda ^* M^* k_{d} +w)<0. \end{array}\right. } \end{aligned}$$

(57)

The above description actually is a Bang–bang control.

So when do we have $(\lambda ^* M^* k_{d} +w)=0$? It is found that this happens at only one time point. Based on PMP, we have

$$\begin{aligned} \frac{d \lambda ^*(t)}{d t}=-k_{a}\lambda ^*+2k_{a}\lambda ^* M^*/M_{max}+k_{d} u^*\lambda ^* +1, \lambda ^*(T)=0 \end{aligned}$$

and

$$\begin{aligned} \frac{d N(t)}{d t}=(k_{a} N/M_{max}+1)M^*, \end{aligned}$$

(58)

where $N(t)=\lambda ^*(t) M^*(t)$. Since $N(T)=0$, we assert that $N(t)>-M_{max}/k_{a}$. It tells us an important fact that when $w/k_{d}>M_{max}/k_{a}$, there must be $u^*(t)=0, t\in [0,T]$. This states an extreme case when an inhibitor is too toxic and ineffective, we would better not to use it at all. Because of $(k_{a} N/M_{max}+1)M^*, N(t)$ increases monotonically to 0, which means $-(\lambda ^* M^* k_{d} +w)=0$ holds at one time point at most. Solving Eq. (58), we get

$$\begin{aligned} N(t)=\frac{M_{max}}{k_{a}}\left[ \exp (k_{a}/M_{max}\int _{T}^{t}M^*(\tau )d \tau )-1\right] . \end{aligned}$$

If the switch happens at $t_1$, it must satisfy the condition $N(t_{1})=-w/k_{d}$. We have

$$\begin{aligned} \frac{d M^*(t)}{d t}= {\left\{ \begin{array}{ll} k_{a}M^*(t)\left( 1-\frac{M^*(t)}{M_{max}}\right) -k_{d} u_{max} M^*(t), \quad t \in [0, t_1],\\ k_{a}M^*(t)\left( 1-\frac{M^*(t)}{M_{max}}\right) ,\quad t \in (t_{1},T]. \end{array}\right. } \end{aligned}$$

(59)

Through calculations, $t_{1}$ must satisfy

$$\begin{aligned} (n_{1}-1)(\exp (k_{a}(T-t_{1}))-1)+(1/n_2-M_{max}/m_{0})\exp (-k_{a}n_{2}t_{1})-1/n_{2}=0,\nonumber \\ \end{aligned}$$

(60)

where $n_{1}=k_{d}M_{max}/(k_{a}w)$ and $n_{2}=1-k_{d}u_{max}/k_{a}$. To sum up, the Bang–bang control is given by

$$\begin{aligned} u^*(t) ={\left\{ \begin{array}{ll} u_{max}, \quad t \in [0,t_1],\\ 0, \quad t \in (t_1,T]. \end{array}\right. } \end{aligned}$$

(61)

where $t_{1}$ satisfies (60).

We have known that the optimal control trajectory is uniquely determined by the switching time $t_{1}$. How the switching time is affected by parameters $k_a, k_d, w$ and $u_{max}$ is of great importance. As shown in Fig. 8a1, b1, the switching time remains zero (meaning no inhibitor) when the apparent fibril growth rate $k_{a}$ is either too large or too small, or when the rate constant for fibril inhibition $k_{d}$ is very small. In the presence of large fibril inhibition rate $k_{d}$ or small toxicity of inhibitors w, the switching time is close to 1 (meaning using inhibitors at every time), as long as $u_{max}$ remains small.

Fig. 8

Influence of model parameters on the Bang–bang control. Subplots in the first row show how the switching time depends on $k_a, k_d, w$ and $u_{max}$. The corresponding optimal trajectories for the inhibitor concentration and fibril concentration with parameters at five marked positions are illustrated in second and third rows. Default parameters are set as $m_0=2.7 \times 10^{-3}$ $\mu \textbf{ M}$, $k_{a}=5.5 \times 10^{-2} \min ^{-1}$, $M_{max}=1$ $\mu \textbf{ M}$, $k_{d}=k_{a}$, $w=1$ and $u_{max}=10m_{0}$ for subplots (a–c). In subplots d, $k_{d}=k_{a}/m_{0}$ is taken

Appendix E: Strategies for Drug Dosing

The strategies used for drug dosing in the maintext are summarized as follows:

Lump-sum adding All inhibitors will be added at the initial instant as shown in Fig. 7a, b. The corresponding ODEs for M(t) and u(t) are

$$\begin{aligned} \left\{ \begin{array}{cl} \dot{M}(t)=&{}k_{a}M(1-\frac{M}{M_{max}})-k_{d} u M,\\ \dot{u}(t)=&{}-k_{d} u M,~t\le T,\\ M(0)=&{}{m_0}, \quad u (0)=u_{tot}. \end{array}\right. \end{aligned}$$

(62)

Constant adding Inhibitors will be added at a constant rate during the whole procedure from $t=0$ to $t=T$,

$$\begin{aligned} \left\{ \begin{array}{cl} \dot{M}(t)=&{}k_{a}M(1-\frac{M}{M_{max}})-k_{d} u M,\\ \dot{u}(t)=&{}\frac{u_{tot}}{T}-k_{d} u M,~t\le T,\\ M(0)=&{}{m_0}, \quad u (0)=\frac{u_{tot}}{T}. \end{array}\right. \end{aligned}$$

(63)

Periodic adding

During the whole procedure, inhibitors are added intermittently. The adjacent adding of inhibitors and non-adding forms a cycle. Assume that there are $N$ cycles $\{0,1,2,\cdots ,N-1\}$. Each cycle includes a time interval $(T/N-t_1)$ during which inhibitors are added uniformly, and a time interval $t_1$ when no inhibitor is added. Let

$$\begin{aligned} u_{per}(t)=\left\{ \begin{array}{cl} \frac{u_{tot}}{T-N t_1} &{} \quad \frac{kT}{N}\le t<\frac{(k+1)T}{N}-t_1, \quad k=0,1,2,...N-1,\\ 0&{}\quad \frac{(k+1)T }{N}-t_1 \le t <\frac{(k+1)T }{N},\quad k=0,1,2,...N-1. \end{array}\right. \end{aligned}$$

The following ODEs are obtained,

$$\begin{aligned} \left\{ \begin{array}{cl} \dot{M}(t)=&{}k_{a}M(1-\frac{M}{M_{max}})-k_{d} u M,\\ \dot{u }(t)=&{}u_{per}-k_{d} u M,\\ M(0)=&{}{m_0}, \quad u (0)=\frac{u_{per}}{T-N t_1}. \end{array}\right. \end{aligned}$$

(64)

Multiple-times adding

In this strategy, inhibitors are equally divided into N pieces and each piece will be added sequentially into the solution after a constant time interval (see Fig. 7a, b for example). Acccordingly, the optimal control is given by

$$\begin{aligned} \left\{ \begin{array}{cl} \dot{M}(t)=&{}k_{a}M(1-\frac{M}{M_{max}})-k_{d} u M,\\ \dot{u}(t)=&{}\frac{u_{tot}}{N}\sum _{i=1}^N\delta (t-iT/N) -k_{d} u M,~t\le T,\\ M(0)=&{}{m_0}, \quad u (0)=\frac{u_{tot}}{N}. \end{array}\right. \end{aligned}$$

(65)

Appendix F: Experimental Setup

A$\beta 40$ (purity $>98\%$) was purchased from ChinaPeptides. Other chemical reagents were bought from Aladdin. A$\beta 40$ was pre-treated with ammonium hydroxide as previously described (Li et al. 2021). A$\beta 40$ was then solubilized in 6 mM NaOH to $100 \,\mu \textbf{ M}$ while ThT and EGCG were directly solubilized in PBS. For ThT kinetics, the final concentrations of A$\beta 40$ and ThT were $5 \,\mu \textbf{ M}$ and $20 \,\mu \textbf{ M}$, respectively. Corning 3603 96-well plates were used and ThT kinetics was monitored at a Tecan Infinite M200 PRO microplate reader with continuously shaking. Fluorescence intensities at ex = 440 nm and em = 480 nm were recorded every 6 min.

For the “Lump-sum” group, $10 \,\mu \textbf{M}$ EGCG was mixed with A$\beta 40$ and ThT at $t=0$ min.

For the “Twice adding” group, $5~ \,\mu \textbf{ M}$ EGCG was mixed with A$\beta 40$ and ThT at $t=0$ min. Then stock solutions of EGCG ($0.5\,\,m\textbf{M}$) was added to the mixture at $t=30$ min or $t=60$ min to increase the concentration of EGCG to $10~\,\mu \textbf{ M}$ while minimize the changes of A$\beta 40$ concentration.

For the “Four times adding” group, $2.5\,\mu \textbf{ M}$ EGCG was mixed with A$\beta 40$ and ThT at $t=0$ min. Then stock solutions of EGCG ($0.5\,\,m\textbf{M}$) was added to the mixture at $t=30, 60, 90$ min or $t=60, 120, 180$ min to increase the concentration of EGCG to 5, 7.5 and $10~\mu \textbf{ M}$, respectively.