Chapter 6.3, Problem 11E

Interpretation Introduction

Interpretation:

To calculate $\text{r(t) and θ(t)}$ explicitly for the given initial conditions ${\text{(r}}_{\text{0}}{\text{,θ}}_{\text{0}}\text{)}$. The system is in polar coordinates given by $\dot{\text{r}}\text{ = - r, }\dot{\text{θ}}\text{ = }\frac{\text{1}}{\text{lnr}}$. Prove that $\text{r(t)}\to \text{0}$ and $\left|\text{θ(t)}\right|\to \infty$ as $\text{t}\to \infty$ and origin is a stable spiral. Show that the linearized system at the origin is $\dot{\text{x}}\text{=-x,}\dot{\text{y}}\text{=-y}$ and origin is a stable star for the linearized system.

Concept Introduction:

The linearized system for $\dot{\text{x}}\text{ = f(x, y),}\dot{\text{y}}\text{ = g(x, y)}$ can be expressed as $\left(\begin{array}{l}\dot{\text{x}}\\ \dot{\text{y}}\end{array}\right)\text{=}\left(\begin{array}{cc}\frac{\partial \dot{\text{x}}}{\partial \text{x}}& \frac{\partial \dot{\text{x}}}{\partial \text{y}}\\ \frac{\partial \dot{\text{y}}}{\partial \text{x}}& \frac{\partial \dot{\text{y}}}{\partial \text{y}}\end{array}\right)\left(\begin{array}{l}\text{x}\\ \text{y}\end{array}\right)$.

The matrix $\left(\begin{array}{cc}\frac{\partial \dot{\text{x}}}{\partial \text{x}}& \frac{\partial \dot{\text{x}}}{\partial \text{y}}\\ \frac{\partial \dot{\text{y}}}{\partial \text{x}}& \frac{\partial \dot{\text{y}}}{\partial \text{y}}\end{array}\right)$ is a Jacobian matrix at any fixed point ${\text{(x}}^{\text{*}}{\text{,y}}^{\text{*}}\text{)}$.

Answer to Problem 11E

Solution:

${\text{r(t) = r}}_{\text{0}}{\text{e}}^{\text{-t}}$, $\text{θ(t) = ln }\left(\frac{{\text{ln(r}}_{\text{0}}\text{)}}{\left|{\text{ln(r}}_{\text{0}}\text{) - t}\right|}\right){\text{+ θ}}_{\text{0}}$.

$\text{r(t)}\to \text{0}$ and $\left|\text{θ(t)}\right|\to \infty$ as $\text{t}\to \infty$ is proved and origin is a stable spiral is proved.

The system in Cartesian coordinate is $\dot{\text{x}}\text{ = -x -}\frac{\text{2y}}{{\text{ln(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}$, $\dot{\text{y}}\text{ = -y +}\frac{\text{2x}}{{\text{ln(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}$.

The linearized system at the origin is $\dot{\text{x}}\text{=-x,}\dot{\text{y}}\text{=-y}$ is proved and origin is a stable star.

Explanation of Solution

Using the first equation

$\begin{array}{l}\dot{\text{r}}\text{ = - r}\\ \frac{\text{dr}}{\text{dt}}\text{ = -r}\\ \frac{\text{dr}}{\text{r}}\text{ = -dt}\\ {\displaystyle \underset{{\text{r}}_{\text{0}}}{\overset{\text{r}}{\int }}\frac{\text{dr}}{\text{r}}}\text{= - }{\displaystyle \underset{\text{0}}{\overset{\text{t}}{\int }}\text{dt}}\\ {\text{lnr - lnr}}_{\text{0}}\text{= -t}\\ \text{ln}\frac{\text{r}}{{\text{r}}_{\text{0}}}\text{=-t}\\ {\text{r(t) = r}}_{\text{0}}{\text{e}}^{\text{-t}}\end{array}$

$\begin{array}{l}\dot{\text{θ}}\text{ = }\frac{\text{1}}{\text{lnr}}\\ {\text{r(t) = r}}_{\text{0}}{\text{e}}^{\text{-t}}\\ \dot{\text{θ}}\text{ = }\frac{\text{1}}{{\text{ln(r}}_{\text{0}}{\text{e}}^{\text{-t}}\text{)}}\\ \dot{\text{θ}}\text{ =}\frac{\text{1}}{{\text{ln(r}}_{\text{0}}{\text{) + lne}}^{\text{-t}}}\\ \dot{\text{θ}}\text{ =}\frac{\text{1}}{{\text{ln(r}}_{\text{0}}\text{) - t}}\end{array}$

Integrating both sides,

$\begin{array}{l}\text{θ(t) = -ln}\left|{\text{ln(r}}_{\text{0}}\text{) - t}\right|{\text{ + ln(ln(r}}_{\text{0}}{\text{)) + θ}}_{\text{0}}\\ \text{θ(t) = ln }\left(\frac{{\text{ln(r}}_{\text{0}}\text{)}}{\left|{\text{ln(r}}_{\text{0}}\text{) - t}\right|}\right){\text{+ θ}}_{\text{0}}\end{array}$

${\text{r(t) = r}}_{\text{0}}{\text{e}}^{\text{-t}}$

When $\text{t}\to \infty$, $\text{r(t)}\to \text{0 and θ(t)}\to \text{0}$.

Solve the stability of the origin by solving $\underset{\text{t}\to \infty }{\lim }\text{r(t)}$.

$\underset{\text{t}\to \infty }{\lim }\text{r(t)}=\underset{\text{t}\to \infty }{\lim }{\text{r}}_{\text{0}}{\text{e}}^{\text{- t}}\to 0$

Thus the origin is a stable point.

Solve the nature of the origin by solving $\underset{\text{t}\to \infty }{\lim }\text{θ(t)}$.

$\underset{\text{t}\to \infty }{\lim }\left|\text{θ(t)}\right|\text{= }\underset{\text{t}\to \infty }{\lim }\text{ln }\left|\frac{{\text{ln(r}}_{\text{0}}\text{)}}{\left|{\text{ln(r}}_{\text{0}}\text{) - t}\right|}{\text{+ θ}}_{\text{0}}\right|\to \infty$

Thus the origin is spiral.

The system in $\text{x, y}$ can be written as,

$\begin{array}{l}\text{r = }\sqrt{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}\\ \text{θ = arctan}\left(\frac{\text{y}}{\text{x}}\right)\\ \dot{\text{r}}\text{ =}\frac{\text{d}}{\text{dt}}\sqrt{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}\\ \dot{\text{r}}\text{ = }\frac{\text{2x}\dot{\text{x}}\text{+2y}\dot{\text{y}}}{\text{2}\sqrt{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}}\\ \dot{\text{r}}\text{ = }\frac{\text{x}\dot{\text{x}}\text{+y}\dot{\text{y}}}{\sqrt{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}}\\ \dot{\text{θ}}\text{ =}\frac{\text{d}}{\text{dt}}\text{arctan}\left(\frac{\text{y}}{\text{x}}\right)\\ \dot{\text{θ}}\text{ = }\frac{\text{x}\dot{\text{y}}\text{-y}\dot{\text{x}}}{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}\end{array}$

Comparing the above polar coordinates with the given coordinates,

$\begin{array}{l}\dot{\text{r}}\text{ = -r}\\ \dot{\text{r}}\text{ =}\frac{\text{x}\dot{\text{x}}\text{+y}\dot{\text{y}}}{\sqrt{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}}\\ \text{r = -}\left(\frac{\text{x}\dot{\text{x}}\text{+y}\dot{\text{y}}}{\sqrt{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}}\right)\\ \text{r =}\sqrt{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}\\ \sqrt{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}\text{ = }\frac{\text{- x}\dot{\text{x}}\text{ - y}\dot{\text{y}}}{\sqrt{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}}\\ {\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{= -(x}\dot{\text{x}}\text{ + y}\dot{\text{y}}\text{)}\mathrm{.......................}\text{1)}\\ \dot{\text{θ}}\text{ =}\frac{\text{1}}{\text{lnr}}\\ \dot{\text{θ}}\text{ = }\frac{\text{x}\dot{\text{y}}\text{ - y}\dot{\text{x}}}{{\text{x}}^{\text{2}}{\text{+ y}}^{\text{2}}}\\ \frac{\text{x}\dot{\text{y}}\text{-y}\dot{\text{x}}}{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}\text{=}\frac{\text{1}}{\text{ln r}}\\ \text{r =}\sqrt{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}\\ \frac{\text{x}\dot{\text{y}}\text{-y}\dot{\text{x}}}{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}\text{ = }\frac{\text{1}}{\text{ln(}\sqrt{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}\text{)}}\\ \text{x}\dot{\text{y}}\text{-y}\dot{\text{x}}\text{ = }\frac{{\text{2(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}{{\text{ln(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}\mathrm{......................}\text{2)}\end{array}$

Multiply equation 1) by x and equation 2) by y and subtract,

$\begin{array}{l}{\text{x(x}}^{\dot{\text{x}}}{\text{+y}}^{\text{2}}\text{) - y}\frac{{\text{2(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}{{\text{ln(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}\\ {\text{(x}}^{\text{2}}{\text{+y}}^{\text{2}}{\text{)}}^{\dot{\text{x}}}{\text{+y}}^{\text{2}}\text{) - y}\frac{{\text{2(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}{{\text{ln(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}\\ \dot{\text{x}}\text{ = -x -}\frac{\text{2y}}{{\text{ln(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}\end{array}$

Multiply equation 1) by y and equation 2) by x and add,

$\begin{array}{l}{\text{y(x}}^{\dot{\text{x}}}{\text{+y}}^{\text{2}}\text{) + x}\frac{{\text{2(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}{{\text{ln(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}\\ {\text{(x}}^{\text{2}}{\text{+y}}^{\text{2}}{\text{)}}^{\dot{\text{y}}}{\text{+y}}^{\text{2}}\text{) + x}\frac{{\text{2(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}{{\text{ln(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}\\ \dot{\text{y}}\text{ = -y +}\frac{\text{2x}}{{\text{ln(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}\end{array}$

The Jacobian matrix at any fixed point ${\text{(x}}^{\text{*}}{\text{,y}}^{\text{*}}\text{)}$ given as,

$\text{A}=\left(\begin{array}{cc}\frac{\partial \dot{\text{x}}}{\partial \text{x}}& \frac{\partial \dot{\text{x}}}{\partial \text{y}}\\ \frac{\partial \dot{\text{y}}}{\partial \text{x}}& \frac{\partial \dot{\text{y}}}{\partial \text{y}}\end{array}\right)$

Substituting values of $\dot{\text{x}}\text{ , }\dot{\text{y}}$ of part c) in the above Jacobian matrix,

$\begin{array}{l}\text{A}=\left(\begin{array}{cc}\frac{\partial \left(\text{-x -}\frac{\text{2y}}{{\text{ln(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}\right)}{\partial \text{x}}& \frac{\partial \left(\text{-x -}\frac{\text{2y}}{{\text{ln(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}\right)}{\partial \text{y}}\\ \frac{\partial \left(\text{-y +}\frac{\text{2x}}{{\text{ln(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}\right)}{\partial \text{x}}& \frac{\partial \left(\text{-y +}\frac{\text{2x}}{{\text{ln(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}\right)}{\partial \text{y}}\end{array}\right)\\ \text{A}=\left(\begin{array}{cc}\text{-1 +}\frac{\text{4xy}}{{\text{(x}}^{\text{2}}{\text{+y}}^{\text{2}}){\text{ln}}^2{\text{(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}& \frac{{\text{4y}}^2}{{\text{(x}}^{\text{2}}{\text{+y}}^{\text{2}}){\text{ln}}^2{\text{(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}-\frac{2}{\ln {\text{(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}\\ \frac{2}{\ln {\text{(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}-\frac{{\text{4x}}^2}{{\text{(x}}^{\text{2}}{\text{+y}}^{\text{2}}){\text{ln}}^2{\text{(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}& \text{-1 - }\frac{\text{4xy}}{{\text{(x}}^{\text{2}}{\text{+y}}^{\text{2}}){\text{ln}}^2{\text{(x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{)}}\end{array}\right)\\ {\text{A}}_{\text{(0,0)}}\text{=}\left(\begin{array}{cc}\text{-1 }& \text{ 0}\\ \text{0}& \text{-1}\end{array}\right)\\ \text{A}\left(\begin{array}{l}\text{x}\\ \text{y}\end{array}\right)\text{=}\left(\begin{array}{l}\dot{\text{x}}\\ \dot{\text{y}}\end{array}\right)\\ \left(\begin{array}{cc}\text{-1 }& \text{ 0}\\ \text{0}& \text{-1}\end{array}\right)\left(\begin{array}{l}\text{x}\\ \text{y}\end{array}\right)\text{=}\left(\begin{array}{l}\dot{\text{x}}\\ \dot{\text{y}}\end{array}\right)\\ \left(\begin{array}{l}\text{-x}\\ \text{-y}\end{array}\right)\text{=}\left(\begin{array}{l}\dot{\text{x}}\\ \dot{\text{y}}\end{array}\right)\end{array}$

It is proved and origin is a stable star for the linearized system.

Conclusion

Integrating given polar equations we can find $\text{r(t) and θ(t)}$ explicitly for the given initial conditions ${\text{(r}}_{\text{0}}{\text{,θ}}_{\text{0}}\text{)}$. Also, $\text{r(t)}\to \text{0}$ and $\left|\text{θ(t)}\right|\to \infty$ as $\text{t}\to \infty$ is proved. In this case, the origin is a stable spiral. The linearized system at the origin is $\dot{\text{x}}\text{=-x,}\dot{\text{y}}\text{=-y}$ is proved and origin is a stable star.