Linear System of Differential Equations - 3

Calculus Level 3

{ x ˙ 1 = 3 x 1 − x 2 + u x ˙ 2 = 5 x 1 − x 2 \begin{cases} \dot{x}_1 = 3x_1 - x_2 + u \\ \dot{x}_2 = 5x_1 - x_2 \end{cases}

Let x 1 x_1 , x 2 x_2 and u u be functions of time t t satisfying the system of equations above which governs the behaviour of an arbitrary physical system. The energy of this physical system is:

E = x 1 2 + x 2 2 2 E = \frac {x_1^2 + x_2^2}2

The goal of this question is to choose u u which is dependent on the variables x 1 x_1 and x 2 x_2 such that the energy of the system always decreases (for any non-zero initial conditions) with time. Make the appropriate choice for u u among the given options.

Bonus: What happens to the energy when u = 0 u=0 ? What is the significant observation one can make through this problem?

u = − 4 x 1 − 4 x 2 u = -4x_1 - 4x_2 None of the above u = − 3 x 1 + 5 x 2 u = -3x_1 + 5x_2 u = − 4 x 1 + 6 x 2 u = -4x_1 + 6x_2

Karan Chatrath by Karan Chatrath , 27, Netherlands

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2 solutions

Steven Chase
Sep 14, 2019

Time-differentiating the energy and substituting in the differential equations, we get the following requirement:

3 x 1 2 + 4 x 1 x 2 + u x 1 − x 2 2 < 0 3 x_1^2 + 4 x_1 x_2 + u x_1 - x_2^2 < 0

What remains is to plug in the candidate u u expressions and see whether the results satisfy the condition. What we want is for the resulting expression to consist purely of − x 1 2 -x_1^2 and − x 2 2 -x_2^2 terms.

For example, u = − 4 x 1 − 4 x 2 u = -4 x_1 - 4 x_2 yields:

3 x 1 2 + 4 x 1 x 2 + ( − 4 x 1 − 4 x 2 ) x 1 − x 2 2 = 3 x 1 2 + 4 x 1 x 2 + − 4 x 1 2 − 4 x 1 x 2 − x 2 2 = 3 x 1 2 − 4 x 1 2 − x 2 2 = − x 1 2 − x 2 2 3 x_1^2 + 4 x_1 x_2 + (-4 x_1 - 4 x_2) x_1 - x_2^2 \\ = 3 x_1^2 + 4 x_1 x_2 + -4 x_1^2 - 4 x_1 x_2 - x_2^2 \\ = 3 x_1^2 - 4 x_1^2 - x_2^2 \\ = - x_1^2 - x_2^2

So this works. Recall that this system is inherently unstable due to its eigenvalues having positive real parts. For u = 0 u = 0 , the system diverges. The lesson is that we can stabilize an inherently unstable system by applying a properly-chosen forcing function.

2 Helpful 2 Interesting 2 Brilliant 0 Confused

Thanks for the solution. Your explanation of the significance of this problem really sums up why I enjoy control theory. The idea of 'making nature dance to our tunes' using mathematics is one I find really cool. There will be more variants coming up.

Karan Chatrath - 1 year, 9 months ago
Chew-Seong Cheong
Sep 15, 2019

Given that E = x 1 2 + x 2 2 2 E = \dfrac {x_1^2+x_2^2}2 , ⟹ E ˙ = x 1 x ˙ 1 + x 2 x ˙ 2 \implies \dot E = x_1\dot x_1 + x_2 \dot x_2 . For the energy of the system to be always decreasing then E ˙ < 0 \dot E < 0 . That is:

x 1 x ˙ 1 + x 2 x ˙ 2 < 0 ⟹ x 1 ( 3 x 1 − x 2 + u ) + x 2 ( 5 x 1 − x 2 ) < 0 Let u = a x 1 + b x 2 3 x 1 2 + 4 x 1 x 2 + a x 1 2 + b x 1 x 2 − x 2 2 < 0 Note that LHS = − x 2 2 , when a = − 3 , b = − 4 − x 1 2 − x 2 2 < 0 when u = − 4 x 1 − 4 x 2 \begin{aligned} x_1\dot x_1 + x_2 \dot x_2 & < 0 \\ \implies x_1 (3x_1-x_2 + u) + x_2(5x_1 - x_2) & < 0 & \small \color{#3D99F6} \text{Let }u = ax_1 + bx_2 \\ 3x_1^2 + 4x_1x_2 + ax_1^2 + bx_1x_2 - x_2^2 & < 0 & \small \color{#3D99F6} \text{Note that LHS } = - x_2^2 \text{, when }a=-3, b = -4 \\ -x_1^2 - x_2^2 & < 0 & \color{#3D99F6} \text{when }u = \boxed{-4x_1-4x_2} \end{aligned}

1 Helpful 1 Interesting 1 Brilliant 0 Confused

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