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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 $$c > 0$$.

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