4. Use Liapunov's function V(x) = x² + x² + x² to prove that the origin isan asymptotically stable equilibrium of the system.18||−x₂ − x₁x² + x² – x³x₁ + x²-x²-X1X3 — X3x² − x2x² – x²]

Noted that for given system of non linear differential equations and given Liapunov function…

Answered: 4. Use Liapunov's function V(x) = x² +…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

4. Use Liapunov's function V(x) = x² + x² + x² to prove that the origin is
an asymptotically stable equilibrium of the system
.18
||
−x₂ − x₁x² + x² – x³
x₁ + x²-x²
-X1X3 — X3x² − x2x² – x²]

Expert Solution

Step 1: Asymptotically stability check

Noted that for given system of non linear differential equations

$x with. on top equals open square brackets table row cell negative x subscript 2 minus x subscript 1 x subscript 2 squared plus x subscript 3 squared minus x subscript 1 cubed end cell row cell x subscript 1 plus x subscript 3 cubed minus x subscript 2 cubed end cell row cell negative x subscript 1 x subscript 3 minus x subscript 3 x subscript 1 squared minus x subscript 2 x subscript 3 squared minus x subscript 3 to the power of 5 end cell end table close square brackets$ and given Liapunov function $V left parenthesis x right parenthesis equals x subscript 1 squared plus x subscript 2 squared plus x subscript 3 squared$

equilibrium point $open parentheses 0 comma 0 close parentheses$ is said to be asymptotically stable if the following expression

$fraction numerator partial differential V over denominator partial differential x subscript 1 end fraction. space stack x subscript 1 with. on top plus fraction numerator partial differential V over denominator partial differential x subscript 2 end fraction. space stack x subscript 2 with. on top plus fraction numerator partial differential V over denominator partial differential x subscript 3 end fraction. space stack x subscript 3 with. on top$ is negative definite,

In other words, $fraction numerator partial differential V over denominator partial differential x subscript 1 end fraction. space stack x subscript 1 with. on top plus fraction numerator partial differential V over denominator partial differential x subscript 2 end fraction. space stack x subscript 2 with. on top plus fraction numerator partial differential V over denominator partial differential x subscript 3 end fraction. space stack x subscript 3 with. on top less than 0 comma f o r space a l l space open parentheses x subscript 1 comma x subscript 2 comma x subscript 3 close parentheses not equal to open parentheses 0 comma 0 comma 0 close parentheses$

Now noted that $fraction numerator partial differential V over denominator partial differential x subscript 1 end fraction. space stack x subscript 1 with. on top plus fraction numerator partial differential V over denominator partial differential x subscript 2 end fraction. space stack x subscript 2 with. on top plus fraction numerator partial differential V over denominator partial differential x subscript 3 end fraction. space stack x subscript 3 with. on top equals$ $negative open square brackets 2 open parentheses x subscript 1 x subscript 2 close parentheses squared plus 2 open parentheses x subscript 1 x subscript 3 close parentheses squared plus 2 open parentheses x subscript 1 close parentheses to the power of 4 plus 2 open parentheses x subscript 2 close parentheses to the power of 4 plus 2 open parentheses x subscript 3 close parentheses to the power of 6 close square brackets$

Obviously ,R.H.S term of above $less than 0 comma$ $f o r space a l l space open parentheses x subscript 1 comma x subscript 2 comma x subscript 3 close parentheses not equal to open parentheses 0 comma 0 comma 0 close parentheses$ ,because inside the braket ,this is sum of square terms ,equal to zero implies each variavle indicate zero.