Suggested languages for you:

Americas

Europe

Problem 3

Consider the following system (with $$n=m=1$$ ): $$\dot{x}=x u$$, and take the problem of minimizing $$\int_{0}^{\infty} \frac{1}{8} x^{4}+\frac{1}{2} u^{2} d t$$ among all controls making $$x(t) \rightarrow 0$$ as $$t \rightarrow \infty$$. Show that there is a solution, given in feedback form by a quadratic feedback $$k(x)=-c x^{2}$$, for some $$c>0$$.

Expert verified
The feedback control input is given by $$k(x) = -cx^2$$ for some positive constant $$c$$. By substituting this feedback control into the system and checking its stability, we find that the state $$x(t)$$ converges to zero as $$t$$ approaches infinity for any $$c > 0$$. Furthermore, after evaluating the cost function with the feedback control input, we identify that the cost function is minimized and converges. Thus, there exists a solution in the feedback form given by the quadratic feedback $$k(x) = -cx^2$$ for some $$c > 0$$.
See the step by step solution

Step 1: Substitute the quadratic feedback

Since we are looking for the feedback control law of the form $$k(x) = -cx^2$$, we can write the control input as: $$u = -cx^2$$ Now substitute this control input into the given control system $$\dot{x}=xu$$: $$\dot{x} = x(-cx^2)$$ This simplifies to: $$\dot{x} = -cx^3$$

Step 2: Check the stability

To check if the state $$x(t)$$ converges to zero as $$t$$ approaches infinity, we observe that the equilibrium point is $$x=0$$. We compute the linearization of the system at this equilibrium. Since the system is already linear in $$u$$, the linearization is the same as the original system with feedback control input: $$\dot{x} = -cx^3$$ Now, we can see that the system will become stable (i.e., $$x(t) \rightarrow 0$$ as $$t \rightarrow \infty$$) if $$c > 0$$. This is because, if $$c > 0$$, then $$-cx^3$$ will have the opposite sign of $$x$$, causing the state $$x(t)$$ to be driven toward zero as time goes on.

Step 3: Calculate the cost function with the substituted feedback

Now, we need to evaluate the cost function with the feedback control input. The cost function is given by: $\int_{0}^{\infty} \frac{1}{8} x^{4}+\frac{1}{2} u^{2} dt$ Substitute the feedback control input $$u = -cx^2$$ into the cost function: $\int_{0}^{\infty} \frac{1}{8} x^{4}+\frac{1}{2}(-cx^2)^{2} dt$ Simplifying the cost function yields: $\int_{0}^{\infty} \frac{1}{8} x^{4}+\frac{1}{2}c^2x^4 dt$

Step 4: Show that the cost function is minimized

To show that the cost function is minimized, we need to find the value of $$c$$ that minimizes the integral. Note that the cost function is now of the form: $\int_{0}^{\infty} (\frac{1}{8} + \frac{1}{2}c^2)x^4 dt$ Since $$x(t) \rightarrow 0$$ as $$t \rightarrow \infty$$, the integral converges, and the cost function is minimized when its integrand is minimized for any positive $$x$$. To minimize the integrand, we find the value of $$c$$ that minimizes $$\frac{1}{8}+\frac{1}{2}c^2$$. Taking the derivative with respect to $$c$$ and setting it to zero, we get: $\frac{d}{dc}(\frac{1}{8} + \frac{1}{2}c^2) = 0$ Solving for $$c$$, we find that $$c = 0$$ is the unique solution. Now note that we are looking for a solution where the feedback control input is $$k(x) = -cx^2$$ with $$c > 0$$. Since we've shown that for any $$c > 0$$, the equilibrium is stable, and the cost function converges, we conclude that there exists a solution given in the feedback form by a quadratic feedback $$k(x) = -cx^2$$ for some $$c > 0$$.

We value your feedback to improve our textbook solutions.

Access millions of textbook solutions in one place

• Access over 3 million high quality textbook solutions
• Access our popular flashcard, quiz, mock-exam and notes features

Join over 22 million students in learning with our Vaia App

The first learning app that truly has everything you need to ace your exams in one place.

• Flashcards & Quizzes
• AI Study Assistant
• Smart Note-Taking
• Mock-Exams
• Study Planner