solving a nonlinear differential equation

Question

a friend of mine asked a question before, which seemed easy at first glance, but it has occupied my mind afterward. Here is the question: \begin{equation} \dot{X_1}=\dot{U_1}U_2-\dot{f}(t)\\ \dot{X_2}=\dot{U_1}X_1-\dot{g}(t) \end{equation} Where $\dot{f}$ and $\dot{g}$ are bounded known derivatives with zeros as the initial values. Find bounded signals $U_1$ and $U_2$, so that $X_1$ and $X_2$ stabilize (converge to a bounded set or became a bounded function). I have suggested the answer below, for $\dot{X_1}$ equation, but I can't extend it to a general solution: \begin{equation} U_1 = -\cos(f)\\ U_2 = \sin(f) \end{equation} By applying them to the $\dot{X_1}$ equation, we get: \begin{align} \dot{X_1}&=\dot{f}\sin(f)\sin(f)-\dot{f}\\ &=-\dot{f}cos(2f)+\dot{f}-\dot{f}\\ &=-\frac{1}{2}\dot{\sin}(2f)\\ \to X_1&=-\frac{1}{2}\sin(2f) \end{align} Thanks in advance, and any tips would be appreciated.

Answer 1

This may not be a solution in the form you are looking for, but if you allow feedback ($U_1,U_2$ depend on $X_1,X_2$), then $\dot{U_1},U_2$ can be designed as controllers to enforce boundedness of $X_1,X_2$. We have two controllers $\dot{U_1},U_2$ and two outputs $X_1,X_2$. However, when $X_1(t)=0$, we loose controlability of $X_2(t)$ since $\dot{X_2} = -\dot{g}$ in this case no matter what we assign to $\dot{U_1},U_2$. Thus, in addition to the goal of enforcing $X_1,X_2$ to be bounded, it is required to make $|X_1(t)|$ bounded from bellow at least after some time in order to ensure controlability. Under this setting, a possible controller design is the following: first, we cancel the term $-\dot{f}$ to make $X_1$ converge to a desired bounded interval far from the origin, e.g. of the form $|X_1(t)|\in[\beta/2,\beta]$ for all $t\geq t^*$ after some time $t^*>0$ and some $\beta>0$. Then, we design a controller for $X_2$ provided that $X_1$ reaches such interval.

The concrete controller is described in the following:

Controller:

Let $X_1(0)$ be the initial condition for $X_1(t)$. Additionally, choose any $\beta,\delta>0$ and any $\varepsilon>\max\left(|\dot{g}|,\frac{|\dot{g}|(1+\bar{\sigma})}{\underline{\sigma}}\right)$ with $\underline{\sigma} := \frac{|X_1(0)|+\beta/2}{|X_1(0)|+\beta + \delta}>0$ and $\bar{\sigma}:=\frac{|X_1(0)|+\beta}{|X_1(0)|+\beta/2 + \delta}>0$. Moreover, define $\text{sign}(x)=1$ if $x\geq 0$ and $\text{sign}(x)=-1$ otherwise. Hence, the control feedback is:

$$ \dot{U}_1(t) = \frac{\dot{g}(t)-\varepsilon\text{sign}(X_2(t))}{(|X_1(t)|+\delta)\text{sign}(X_1(t))} $$

$$ {U}_2(t) = \frac{[\dot{f}(t)-X_1(t)+X_1(0)+\beta\text{sign}(X_1(0))](|X_1(t)|+\delta)\text{sign}(X_1(t))}{\dot{g}(t)-\varepsilon\text{sign}(X_2(t))} $$

Proof of correctness:

To check that this works, first lets analyze $\dot{X}_1$: $$ \begin{aligned} \dot{X}_1 &= \dot{U}_1U_2 - \dot{f} \\&= \left(\frac{\dot{g}-\varepsilon\text{sign}(X_2)}{(|X_1|+\delta)\text{sign}(X_1)}\right)\left(\frac{[\dot{f}-X_1+X_1(0)+\beta\text{sign}(X_1(0))](|X_1|+\delta)\text{sign}(X_1)}{\dot{g}-\varepsilon\text{sign}(X_2)}\right) - \dot{f}\\ &= -X_1+X_1(0)+\beta\text{sign}(X_1(0)) \end{aligned} $$ Note that $X_1(t) = \text{sign}(X_1(0))(|X_1(0)|+\beta(1-e^{-t}))$ is the unique solution to $\dot{X}_1(t) = -X_1(t)+X_1(0)+\beta\text{sign}(X_1(0))$. Moreover, since we defined $\text{sign}(0) = 1$, thus, $|X_1(t)|$ is bounded from above by $|X_1(0)| + \beta$ and is bounded from bellow for any $t>0$ even when $X_1(0)=0$. Thus, dividing by $|X_1|+\delta$ in the expression for $\dot{U}_1$ is not problematic (no singularity) and $\dot{U}_1(t)$ will remain bounded.

Now, lets analyze $\dot{X}_2$. First, note that there are no $t\geq 0$ such that $\dot{g}(t)-\varepsilon\text{sign}(X_2(t))= 0$. Otherwise, there would exist $t\geq 0$ such that $ \dot{g}(t) = \varepsilon\text{sign}(X_2(t)) $ which implies $|\dot{g}(t)| = \varepsilon$ for some $t$, which is not possible since $\varepsilon>|\dot{g}|$. Hence, dividing by $\dot{g}-\varepsilon\text{sign}(X_2)$ is not problematic in the expression for $U_2$ (no singularity). Now, to obtain boundedness of $X_2$: $$ \begin{aligned} \dot{X}_2 &= \dot{U_1}X_1-\dot{g} \\&= \left(\frac{\dot{g}-\varepsilon\text{sign}(X_2)}{(|X_1|+\delta)\text{sign}(X_1)}\right)X_1 - \dot{g} \\&= -\varepsilon\text{sign}(X_2)\sigma(X_1) + \left(\sigma(X_1)-1\right)\dot{g} \end{aligned} $$ where we defined $$\sigma(X_1):=\frac{X_1}{(|X_1|+\delta)\text{sign}(X_1)} = \frac{|X_1|\text{sign}(X_1)}{(|X_1|+\delta)\text{sign}(X_1)}=\frac{|X_1|}{|X_1|+\delta}$$ and used the fact that $X_1=|X_1|\text{sign}(X_1)$.

Moreover, note that $|X_1(t)|$ basically goes from $|X_1(0)|$ towards $|X_1(0)| + \beta$ asymptotically and without touching zero (except perhaps at $t=0$). Hence, there exist some $t^*$ at which $|X_1(t^*)|\geq |X_1(0)|+\beta/2$. Hence, we can bound $\sigma(X_1)$ from bellow by: $$ \frac{|X_1(t)|}{|X_1(t)|+\delta}\geq \frac{|X_1(0)|+\beta/2}{|X_1(0)|+\beta + \delta}=\underline{\sigma} $$ for $t\geq t^*$. Similarly, we obtain $$ \frac{|X_1(t)|}{|X_1(t)|+\delta}\leq\frac{|X_1(0)|+\beta}{|X_1(0)|+\beta/2 + \delta}=\bar{\sigma} $$

We check the boundedness of $X_2$ by a Lyapunov function argument for $t\geq t^*$ and verify boundedness for $0\leq t\leq t^*$ in a later step. Let $V = X_2^2/2$ and compute $\dot{V}$ for $t\geq t^*$: $$ \begin{aligned} \dot{V} &= X_2\dot{X}_2 \\&= X_2(-\varepsilon\text{sign}(X_2)\sigma(X_1) + \left(\sigma(X_1)-1\right)\dot{g}) \\&= -\sigma(X_1)\varepsilon |X_2| + (\sigma(X_1)-1)\dot{g}X_2\\&\leq - \underline{\sigma}\varepsilon |X_2| + |{\sigma(X_1)}-1||\dot{g}||X_2| \\ &\leq - \underline{\sigma}\varepsilon |X_2| +({\bar{\sigma}}+1)|\dot{g}||X_2| \\ &= -|X_2|(\underline{\sigma}\varepsilon- (\bar{\sigma}+1)|\dot{g}|)<0 \end{aligned} $$ since $\varepsilon>|\dot{g}|(1+\bar{\sigma})/\underline{\sigma}$. Therefore, $V=X_2^2/2$ always decreases, and thus stability (and boundedness) for $X_2$ follows for $t\geq t^*$.

Now, since the controller $\dot{U_1}$ and $X_1,\dot{g}$ remain bounded, then $X_2$ will remain bounded for $0\leq t \leq t^*$ (no escapes in finite time before $t=t^*$). Thus we have showed that the controllers have no singularities (thus are bounded) and they effectively make $X_1,X_2$ remain bounded.