Chapter 9.6, Problem 1E

Interpretation Introduction

Interpretation:

• a) To prove $\text{V=}\frac{\text{1}}{\text{2}}{\text{e}}_{\text{2}}^{\text{2}}{\text{+2}}_{\text{3}}^{\text{2}}$ decays exponentially fast by using $\dot{\text{V}}\le -\text{kV}$.

• b) To show from part (a), ${\text{e}}_{\text{2}}\text{(t),e(t)}\to 0$.

• c) To show ${\text{e}}_{\text{1}}\text{(t)}\to 0$ exponentially fast.

Concept Introduction:

• ➢ Lorenz equations

  $\begin{array}{l}\dot{\text{x}}=\sigma (\text{y}-\text{x})\\ \dot{\text{y}}=\text{rx}-\text{y}-\text{xz}\\ \dot{\text{z}}=\text{xy}-\text{bz}\\ \text{Here σ, r, b > 0}\end{array}$

  The solution of Lorenz equations oscillates irregularly for a wide range of parameters, never exactly repeating but always remains in a bounded region of phase space.

• ➢ The equations governing error dynamics are

  $\begin{array}{l}{\dot{\text{e}}}_{\text{1}}=\sigma \left({\text{e}}_{\text{2}}-{\text{e}}_{\text{1}}\right)\\ {\dot{\text{e}}}_{\text{2}}=-{\text{e}}_{\text{2}}-{\text{20u(t)e}}_{\text{3}}\\ {\dot{\text{e}}}_{\text{3}}={\text{5u(t)e}}_{\text{2}}-{\text{be}}_{\text{3}}\end{array}$

• ➢ The Liapunov function has the form

  $\frac{\text{1}}{\text{2}}\frac{\text{d}}{\text{dt}}\left({\text{e}}_{\text{2}}^{\text{2}}{\text{ + 4e}}_{\text{3}}^{\text{2}}\right)$

Answer to Problem 1E

Solution:

• a) It is proved that the $\text{V=}\frac{\text{1}}{\text{2}}{\text{e}}_{\text{2}}^{\text{2}}{\text{+2}}_{\text{3}}^{\text{2}}$ decaying exponentially fast.

• b) It is shown that ${\text{e}}_{\text{2}}\text{(t),e(t)}\to 0$(decaying exponentially fast).

• c) It is shown that ${\text{e}}_{\text{1}}\text{(t)}\to 0$ exponentially fast.

Explanation of Solution

• a)

  $\dot{\text{V}}=-{\text{e}}_{\text{2}}^{\text{2}}-{\text{4be}}_{\text{3}}^{\text{2}}$

  ${\text{e}}_{\text{2}}{\text{ and e}}_{\text{3}}$ are error dynamic terms.

  The given condition is

  $\dot{\text{V}}\le -\text{kV}$

  Putting $\dot{\text{V}}=-{\text{e}}_{\text{2}}^{\text{2}}-{\text{4be}}_{\text{3}}^{\text{2}}$ and $\text{V =}\frac{\text{1}}{\text{2}}{\text{e}}_{\text{2}}^{\text{2}}{\text{ + 2e}}_{\text{3}}^{\text{2}}$ in the above equation

  $\begin{array}{l}-{\text{e}}_{\text{2}}^{\text{2}}-{\text{4be}}_{\text{3}}^{\text{2}}\le -\text{k}\left(\frac{\text{1}}{\text{2}}{\text{e}}_{\text{2}}^{\text{2}}{\text{ + 2e}}_{\text{3}}^{\text{2}}\right)\\ -{\text{e}}_{\text{2}}^{\text{2}}-{\text{4be}}_{\text{3}}^{\text{2}}\le -\text{k}\frac{\text{1}}{\text{2}}{\text{e}}_{\text{2}}^{\text{2}}-{\text{k2e}}_{\text{3}}^{\text{2}}\end{array}$

  Comparing the coefficients of L.H.S and R.H.S

  $-1<-\text{k}\frac{\text{1}}{\text{2}}$

  $\therefore \text{k}<2$

  And

  $\text{4b < 2k}$

  $\therefore \text{k < 2b}$

  Thus the inequality holds until $\text{k < min}\left\{\text{ 2 , 2b }\right\}$

  Now from the given condition

  $\dot{\text{V}}\le -\text{kV}$

  Replacing $\dot{\text{V}}\text{ =}\frac{\text{dV}}{\text{dt}}$ in the above equation

  $\frac{\text{dV}}{\text{dt}}\le -\text{kV}$

  $\text{dV}\le -\text{kV}\text{.dt}$

  $\frac{\text{dV}}{\text{V}}\le -\text{k}\text{.dt}$

  Integrating the above equation

  ${\displaystyle \int \frac{\text{dV}}{\text{V}}}\le -{\displaystyle \int \text{k}\text{.dt}}$

  $\log \text{V}\le -\text{kT}+\mathrm{l}{\text{ogV}}_{\text{0}}$

  Rearranging the above equation

  $\log \left(\frac{\text{V}}{{\text{V}}_{\text{0}}}\right)\le -\text{kT}$

  $\left(\frac{\text{V}}{{\text{V}}_{\text{0}}}\right)\le e^{-\text{kT}}$

  $\therefore \text{V}\le {\text{V}}_{\text{0}}{e}^{-\text{kT}}$

  From the above expression, it is proved that the function decays exponentially.

• b)

  From the part (a)

  $\frac{1}{2}{\text{e}}_2^2{\text{< V <V}}_{\text{0}}{e}^{-\text{kT}}$

  Rearranging above equation,

  ${\text{e}}_2^2{\text{< 2V < 2V}}_{\text{0}}{e}^{-\text{kT}}$

  Taking the square root of the above equation

  ${\text{e}}_2\text{< }\sqrt{\text{2V}}\text{ < }\sqrt{{\text{2V}}_{\text{0}}{e}^{-\text{kT}}}$

  $\therefore {\text{e}}_2\text{< }\sqrt{{\text{2V}}_0}{e}^{\frac{-\text{kT}}{2}}$

  From the above expression, ${\text{e}}_2$ tends to zero exponentially fast.

  Again from the part (a)

  $2{\text{e}}_3^2{\text{< V <V}}_{\text{0}}{e}^{-\text{kT}}$

  ${\text{e}}_3\text{ < }\sqrt{\frac{{\text{V}}_{\text{0}}}{2}}{e}^{\frac{-\text{kT}}{2}}$

  Hence ${\text{e}}_3$ tend to zero exponentially fast.

• c)

  From the system

  ${\dot{e}}_1=\sigma \left(e_2-e_1\right)$

  Putting ${\text{e}}_2=\sqrt{{\text{2V}}_0}{e}^{\frac{-\text{kT}}{2}}$ in above equation

  ${\dot{e}}_1+\sigma e_1=\sigma \sqrt{{\text{2V}}_0}{e}^{\frac{-\text{kT}}{2}}$

  $\frac{{\text{de}}_{\text{1}}}{\text{dt}}+\sigma e_1=\sigma \sqrt{{\text{2V}}_0}{e}^{\frac{-\text{kT}}{2}}$

  This is the first-order differential equation,

  Put $\frac{\text{d}}{\text{dt}}\text{=D}$ in the above equation

  $\left(\text{D+σ}\right){\text{e}}_{\text{1}}=\text{σ }\sqrt{{\text{2V}}_0}{e}^{\frac{-\text{kT}}{2}}$

  To find the solution of the above differential equation, first, find the solution of complementary function.

  $\begin{array}{l}\left(\text{D+σ}\right){\text{e}}_{\text{1}}=0\\ \left(\text{D+σ}\right)=0\\ \text{D}=-\sigma \end{array}$

  Therefore the solution of complementary function is

  ${\text{CF = c}}_{\text{1}}{\text{e}}^{{\text{m}}_{\text{1}}\text{t}}$

  $\therefore {\text{CF = c}}_{\text{1}}{\text{e}}^{\text{- }\sigma \text{t}}$

  The particular integral of the function is

  $\text{P}\text{.I=}\frac{\text{σ }\sqrt{{\text{2V}}_{\text{0}}}{\text{e}}^{\frac{\text{-kT}}{\text{2}}}}{\left(\text{D+σ}\right)}$

  Put $D=-\frac{k}{2}$

  $\text{P}\text{.I= - }\frac{\text{2σ }\sqrt{{\text{2V}}_{\text{0}}}{\text{e}}^{\frac{\text{-kT}}{\text{2}}}}{\left(\text{k - 2σ}\right)}$

  The solution of the differential equation is

  ${\text{e}}_{\text{1}}\text{=P}\text{.I + C}\text{.F}$

  $\therefore {\text{e}}_{\text{1}}\text{= -}\frac{\text{2σ }\sqrt{{\text{2V}}_{\text{0}}}{\text{e}}^{\frac{\text{-kT}}{\text{2}}}}{\left(\text{k - 2σ}\right)}{\text{ + c}}_{\text{1}}{\text{e}}^{\text{- }\sigma \text{t}}$

  Since in the above equation, both terms contain exponentially decaying terms, ${\text{e}}_{\text{1}}$ decays exponentially fast.