Chapter 7.6, Problem 25E

Interpretation Introduction

Interpretation:

Consider the weakly nonlinear oscillator $\ddot{\text{x}}\text{ + x + εh(x,}\dot{\text{x}}\text{,t) = 0}$. Let $\text{x(t) = r(t)cos }\left(\text{t+}\varphi \text{(t)}\right)$, $\dot{\text{x}}\text{ = - r(t)sin}\left(\text{t}+\varphi \text{(t)}\right)$. This change of variable should be regarded as a definition of $\text{r(t)}$ and $\varphi \text{(t)}$.

Show that $\dot{\text{r}}\text{ = ε h sin}\left(\text{t+}\varphi \right)$, $r\dot{\varphi }=\text{ε h cos}\left(\text{t+}\varphi \right)$.

Let $\langle \text{r}\rangle \left(\text{t}\right)\text{ = }\frac{\text{1}}{\text{2π}}{\displaystyle \underset{\text{t-π}}{\overset{\text{t+π}}{\int }}\text{r}\left(\text{τ}\right)}\text{dτ}$ denote the running average of r over one cycle of the sinusoidal oscillation. Show that $\frac{\text{d}\langle \text{r}\rangle }{\text{dt}}\text{ = }\langle \frac{\text{dr}}{\text{dt}}\rangle$.

Show that $\frac{\text{d}\langle \text{r}\rangle }{\text{dt}}\text{ = ε}\langle \text{h}\left[\text{rcos(t+}\varphi \text{),-rsin(t+}\varphi \text{),t}\right]\text{sin(t+θ)}\rangle$.

Show that $\text{r(t)= }\overline{\text{r}}\text{+O}(\in )$ and $\varphi \text{(t) = }\overline{\varphi }(t)\text{+O}(\in )$, and therefore $\begin{array}{l}\frac{\text{d}\overline{\text{r}}}{\text{dt}}\text{ = ε}\langle \left(\text{h}\left(\overline{\text{r}}\text{ cos}\left(\text{t + }\overline{\varphi }\right)\right)\text{, - }\overline{\text{r}}\text{ sin}\left(\text{t + }\overline{\varphi }\right)\text{, t}\right)\text{sin}\left(\text{t + }\overline{\varphi }\right)\text{ + O}\left({\text{ε}}^{\text{2}}\right)\rangle \\ \overline{\text{r}}\frac{\text{d}\overline{\varphi }}{\text{dt}}\text{= ε}\langle \left(\text{h}\left(\overline{\text{r}}\text{ cos}\left(\text{t + }\overline{\varphi }\right)\right)\text{, - }\overline{\text{r}}\text{ sin}\left(\text{t + }\overline{\varphi }\right)\text{, t}\right)\text{cos}\left(\text{t + }\overline{\varphi }\right)\text{ + O}\left({\text{ε}}^{\text{2}}\right)\rangle \end{array}$.

Concept Introduction:

The expression for the averaging equation for magnitude is $\text{x}\left(\text{t,ε}\right){\text{ = x}}_{\text{0}}\text{+O}\left(\text{ε}\right)$.

The expression of initial displacement ${\text{x}}_{\text{0}}$ is ${\text{x}}_{\text{0}}\text{ = r}\left(\text{T}\right)\text{cos}\left(\varphi \left(\text{T}\right)\text{+τ}\right)$.

The expression of the amplitude of any limit cycle for the original system is ${\text{r=r}}_{\text{0}}\text{+O}\left(\text{ε}\right)$.

The expression of the frequency of any limit cycle for the original system is $\text{ω = 1+ε}\varphi \text{'}$.

Taylor series expansion of $\text{x}\left(\text{t,ε}\right)$ is $\text{x}\left(\text{t,ε}\right)\sim \text{x}\left(\text{t,T}\right)\text{+O}\left(\text{ε}\right)$. Here, $\text{O}\left(\text{ε}\right)$ are the higher order terms of the Taylor series expansion.

Answer to Problem 25E

Solution:

The result $\dot{\text{r}}\text{ = ε h sin}\left(\text{t+}\varphi \right)$, $r\dot{\varphi }=\text{ε h cos}\left(\text{t+}\varphi \right)$ is proved.

The result $\frac{\text{d}\langle \text{r}\rangle }{\text{dt}}\text{ = }\langle \frac{\text{dr}}{\text{dt}}\rangle$ is proved.

The result $\frac{\text{d}\langle \text{r}\rangle }{\text{dt}}\text{ = ε}\langle \text{h}\left[\text{rcos(t+}\varphi \text{),-rsin(t+}\varphi \text{),t}\right]\text{sin(t+θ)}\rangle$ is proved.

The results $\text{r(t)= }\overline{\text{r}}\text{+O}(\in )$, $\varphi \text{(t) = }\overline{\varphi }(t)\text{+O}(\in )$ and

$\begin{array}{l}\frac{\text{d}\overline{\text{r}}}{\text{dt}}\text{ = ε}\langle \left(\text{h}\left(\overline{\text{r}}\text{ cos}\left(\text{t + }\overline{\varphi }\right)\right)\text{, - }\overline{\text{r}}\text{ sin}\left(\text{t + }\overline{\varphi }\right)\text{, t}\right)\text{sin}\left(\text{t + }\overline{\varphi }\right)\text{ + O}\left({\text{ε}}^{\text{2}}\right)\rangle \\ \overline{\text{r}}\frac{\text{d}\overline{\varphi }}{\text{dt}}\text{= ε}\langle \left(\text{h}\left(\overline{\text{r}}\text{ cos}\left(\text{t + }\overline{\varphi }\right)\right)\text{, - }\overline{\text{r}}\text{ sin}\left(\text{t + }\overline{\varphi }\right)\text{, t}\right)\text{cos}\left(\text{t + }\overline{\varphi }\right)\text{ + O}\left({\text{ε}}^{\text{2}}\right)\rangle \end{array}$ are proved.

Explanation of Solution

The expression for the averaging equation for magnitude is:

$\text{x}\left(\text{t,ε}\right){\text{ = x}}_{\text{0}}\text{+O}\left(\text{ε}\right)$

Here, x is the position and ${\text{x}}_{\text{0}}$ is the initial position of the system.

The expression of initial displacement ${\text{x}}_{\text{0}}$ is:

${\text{x}}_{\text{0}}\text{ = r}\left(\text{T}\right)\text{cos}\left(\varphi \left(\text{T}\right)\text{+τ}\right)$

Here, $\text{r}\left(\text{T}\right)$ is the radius of the polar coordinate system, and $\varphi$ is the angle formed by the radius.

The expression of the amplitude of any limit cycle for the original system is ${\text{r=r}}_{\text{0}}\text{+O}\left(\text{ε}\right)$.

The expression of the frequency of any limit cycle for the original system is:

$\text{ω = 1+ε}\varphi \text{'}$

Here, $\varphi \text{'}$ is the Taylor series expansion of the function $\varphi \left(\text{T}\right)$.

Taylor series expansion of $\text{x}\left(\text{t,ε}\right)$ is :

$\text{x}\left(\text{t,ε}\right)\sim \text{x}\left(\text{t,T}\right)\text{+O}\left(\text{ε}\right)$

Here, $\text{O}\left(\text{ε}\right)$ are the higher order terms of the Taylor series expansion.

(a)

The first derivative of $\text{r}\left(\text{t}\right)$ and $\varphi \left(\text{t}\right)$ is expressed as follows:

The system equation is:

$\ddot{\text{x}}\text{+x+εh}\left(\text{x,}\dot{\text{x}}\text{,t}\right)\text{ = 0}$

And,

$\begin{array}{l}\text{x}\left(\text{t}\right)\text{ = r}\left(\text{t}\right)\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)\\ \dot{\text{x}}\left(\text{t}\right)\text{ = -r}\left(\text{t}\right)\text{sin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\end{array}$

Differentiate function $\text{x}\left(\text{t}\right)$ with respect to time,

$\begin{array}{l}\frac{\text{d}}{\text{dt}}\text{x}\left(\text{t}\right)\text{ = }\frac{\text{d}}{\text{dt}}\text{r}\left(\text{t}\right)\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)\\ \dot{\text{x}}\left(\text{t}\right)\text{ = }\dot{\text{r}}\text{cos}\left(t+\varphi \right)\text{-r}\left(\text{1+}\dot{\varphi }\right)\text{sin}\left(\text{t+}\varphi \right)\end{array}$

Substitute $\text{-rsin}\left(\text{t+}\varphi \right)$ for $\dot{\text{x}}\left(\text{t}\right)$ in the above,

$\begin{array}{l}\text{-rsin}\left(\text{t+}\varphi \right)=\dot{\text{r}}\text{cos}\left(t+\varphi \right)\text{-r}\left(\text{1+}\dot{\varphi }\right)\text{sin}\left(\text{t+}\varphi \right)\\ \text{-rsin}\left(\text{t+}\varphi \right)\text{ = }\dot{\text{r}}\text{cos}\left(\text{t+}\varphi \right)\text{-rsin}\left(\text{t+}\varphi \right)\text{-r}\dot{\varphi }\text{sin}\left(\text{t+}\varphi \right)\\ \dot{\text{r}}\text{cos}\left(\text{t+}\varphi \right)-\text{r}\dot{\varphi }\text{sin}\left(\text{t+}\varphi \right)\text{ = 0}\end{array}$

Further differentiate the function $\dot{\text{x}}\left(\text{t}\right)$ from the above expression,

$\begin{array}{l}\frac{\text{d}\dot{\text{x}}\left(\text{t}\right)}{\text{dt}}\text{ = }\frac{\text{d}\left(\text{-r}\left(\text{t}\right)\text{sin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\right)}{\text{dt}}\\ \ddot{\text{x}}\left(\text{t}\right)\text{ = -}\dot{\text{r}}\text{sin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{-r}\left(\text{1+}\dot{\varphi }\right)\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)\end{array}$

Substitute $\text{-}\dot{\text{r}}\text{sin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{-r}\left(\text{1+}\dot{\varphi }\right)\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)$ for $\ddot{\text{x}}\left(\text{t}\right)$ and $\text{r}\left(\text{t}\right)\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)$ for $\text{x}\left(\text{t}\right)$ in the above expression of $\ddot{\text{x}}\text{+x+εh}\left(\text{x,}\dot{\text{x}}\text{,t}\right)$

$\begin{array}{l}\text{-}\dot{\text{r}}\text{sin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{-r}\left(\text{1+}\dot{\varphi }\right)\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)+\text{r}\left(\text{t}\right)\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{+εh = 0}\\ \text{-}\dot{\text{r}}\text{sin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{-r}\dot{\varphi }\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)+\text{εh = 0}\\ \dot{\text{r}}\text{sin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{+r}\dot{\varphi }\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{ = εh}\end{array}$

From the above calculation of $\text{r}\left(\text{t}\right)$ and $\varphi \left(\text{t}\right)$, there are two expressions obtained,

$\begin{array}{l}\dot{\text{r}}\text{cos}\left(\text{t+}\varphi \right)\text{-r}\dot{\varphi }\text{sin}\left(\text{t+}\varphi \right)\text{ = 0}\\ \dot{\text{r}}\text{sin}\left(\text{t+}\varphi \left(t\right)\right)\text{+r}\dot{\varphi }\text{cos}\left(\text{t+}\varphi \left(t\right)\right)\text{ = εh}\end{array}$

Multiply $\text{cos}\left(\text{t+}\varphi \right)$ in equation $\dot{\text{r}}\text{cos}\left(\text{t+}\varphi \right)\text{-r}\dot{\varphi }\text{sin}\left(\text{t+}\varphi \right)\text{ = 0}$ and $\text{sin}\left(\text{t+}\varphi \right)$ in equation $\dot{\text{r}}\text{sin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{+r}\dot{\varphi }\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{ = εh}$

Therefore, the equation is:

$\begin{array}{l}{\dot{\text{r}}}^{\text{cos}}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{-r}\dot{\varphi }\text{sin}\left(\text{t+}\varphi \left(t\right)\right)\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{ = 0}\\ {\dot{\text{r}}}^{\text{sin}}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{+r}\dot{\varphi }\text{cos}\left(\text{t+}\varphi \left(t\right)\right)\text{sin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{ = εhsin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\end{array}$

${\text{sin}}^{\text{2}}{\text{θ+cos}}^{\text{2}}\text{θ=1}$

Add the above two expression,

${\dot{\text{r}}\text{cos}}^{\text{2}}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{ - r}\dot{\varphi }\text{sin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right){\text{+}\dot{\text{r}}\text{sin}}^{\text{2}}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{+r}\dot{\varphi }\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)\sin \left(\text{t+}\varphi \left(\text{t}\right)\right)=\text{εhsin}\left(\text{t + }\varphi \left(\text{t}\right)\right)$ From the trigonometric expression,

${\text{sin}}^{\text{2}}{\text{θ+cos}}^{\text{2}}\text{θ=1}$

$\dot{\text{r}}\text{ = εhsin}\left(\text{t + }\varphi \left(\text{t}\right)\right)$

Multiply $\text{sin}\left(\text{t+}\varphi \right)$ in equation $\dot{\text{r}}\text{cos}\left(\text{t+}\varphi \right)\text{- r}\dot{\varphi }\text{sin}\left(\text{t+}\varphi \right)\text{ = 0}$ and $\text{cos}\left(\text{t+}\varphi \right)$ in equation $\dot{\text{r}}\text{sin}\left(\text{t+}\varphi \left(t\right)\right)\text{+r}\dot{\varphi }\text{cos}\left(t+\varphi \left(\text{t}\right)\right)\text{=εh}$

$\begin{array}{l}\dot{\text{r}}\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{sin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{-r}\dot{\varphi }{\text{sin}}^{\text{2}}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{=0}\\ \dot{\text{r}}\text{sin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{+r}\dot{\varphi }{\text{cos}}^{\text{2}}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{sin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{ = εhcos}\left(\text{t+}\varphi \left(\text{t}\right)\right)\end{array}$

Adding the above expressions,

$\text{r}\dot{\varphi }\text{ = εhcos}\left(\text{t+}\varphi \left(\text{t}\right)\right)$

Hence, expression of $\dot{\text{r}}$ and $\text{r}\dot{\varphi }$ is $\begin{array}{l}\text{r}\dot{\varphi }=\text{εh cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)\\ \dot{\text{r}}=\text{εh sin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\end{array}$.

(b)

The expression for $\langle \text{r}\rangle \left(\text{t}\right)$ is given in the question is:

$\langle \text{r}\rangle \left(\text{t}\right)\text{ = }\frac{\text{1}}{\text{2π}}{\displaystyle \underset{\text{t-π}}{\overset{\text{t+π}}{\int }}\text{r}\left(\text{τ}\right)}\text{dτ}$

From Leibniz’s integral rule, considered the function $\dot{\text{r}}\left(\text{t}\right)$ is continuous for $\text{τ}\in \left[\text{t-π, t+π}\right]$.

Differentiate the above expression,

$\begin{array}{l}\frac{\text{d}\langle \text{r}\rangle }{\text{dt}}\text{ = }\frac{\text{d}}{\text{dt}}\left(\frac{\text{1}}{\text{2π}}{\displaystyle \underset{\text{t-π}}{\overset{\text{t+π}}{\int }}\text{r}\left(\text{t}\right)\text{dτ}}\right)\\ \frac{\text{d}\langle \text{r}\rangle }{\text{dt}}\text{ = }\frac{\text{1}}{\text{2π}}\left(\text{r}\left(\text{t+π}\right)\left(\frac{\text{d}}{\text{dt}}\left(\text{t+π}\right)\right)\text{-r}\left(\text{t-π}\right)\left(\frac{\text{d}}{\text{dt}}\left(\text{t-π}\right)\right){\displaystyle {\int }_{\text{t-π}}^{\text{t+π}}\frac{\partial }{\partial \text{t}}\text{r}\left(\text{τ}\right)}\text{dτ}\right)\\ \frac{\text{d}\langle \text{r}\rangle }{\text{dt}}\text{ = }\frac{\text{1}}{\text{2π}}\left(\text{r}\left(\text{t+π}\right)\left(\text{1}\right)\text{-r}\left(\text{t-π}\right)\left(\text{1}\right)\text{+}{\displaystyle {\int }_{\text{t-π}}^{\text{t+π}}\text{0 }}\text{dτ}\right)\\ \frac{\text{d}\langle \text{r}\rangle }{\text{dt}}\text{ = }\frac{\text{1}}{\text{2π}}\left(\text{r}\left(\text{t+π}\right)\text{-r}\left(\text{t-π}\right)\right)\end{array}$

$\begin{array}{l}\frac{\text{d}\langle \text{r}\rangle }{\text{dt}}\text{ = }\frac{\text{1}}{\text{2π}}{\displaystyle {\int }_{\text{t-π}}^{\text{t+π}}\frac{\text{d}}{\text{dτ}}}\text{r}\left(\text{τ}\right)\text{dτ}\\ \frac{\text{d}\langle \text{r}\rangle }{\text{dt}}\text{ = }\langle \frac{\text{dr}}{\text{dt}}\rangle \end{array}$

Hence, expression $\frac{\text{d}\langle \text{r}\rangle }{\text{dt}}\text{ = }\langle \frac{\text{dr}}{\text{dt}}\rangle$ is proved.

(c)

$\begin{array}{l}\frac{\text{d}\langle \text{r}\rangle }{\text{dt}}\text{ = }\langle \frac{\text{dr}}{\text{dt}}\rangle \\ \frac{\text{d}\langle \text{r}\rangle }{\text{dt}}\text{ = }\langle \dot{\text{r}}\left(\text{t}\right)\rangle \end{array}$

Substitute $\in \text{h}\left(\text{x,}\dot{\text{x}}\text{,t}\right)\text{sin}\left(\text{t+}\varphi \right)$ for $\dot{\text{r}}\left(\text{t}\right)$ in above,

$\langle \dot{\text{r}}\left(\text{t}\right)\rangle \text{ = }\langle \in \text{h}\left(\text{x,}\dot{\text{x}}\text{,t}\right)\text{sin}\left(\text{t+}\varphi \right)\rangle$

Substitute $\text{-r}\left(\text{t}\right)\text{sin}\left(\text{t+}\varphi \left(\text{t}\right)\right)$ for $\dot{\text{x}}\left(\text{t}\right)$ and $\text{r}\left(\text{t}\right)\text{cos}\left(\text{t+}\varphi \left(\text{t}\right)\right)$ for $\text{x}\left(\text{t}\right)$ in the above expression,

$\langle \dot{\text{r}}\left(\text{t}\right)\rangle \text{ = }\langle \in \text{h}\left(\text{r}\left(\text{t}\right)\cos \left(\text{t+}\varphi \left(\text{t}\right)\right)\text{, - rsin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{,t}\right)\text{sin}\left(\text{t+}\varphi \right)\rangle$

Therefore the expression of $\langle \dot{\text{r}}\left(\text{t}\right)\rangle$ is

$\langle \dot{\text{r}}\left(\text{t}\right)\rangle \text{ = }\langle \in \text{h}\left(\text{r}\left(\text{t}\right)\cos \left(\text{t+}\varphi \left(\text{t}\right)\right)\text{, - rsin}\left(\text{t+}\varphi \left(\text{t}\right)\right)\text{,t}\right)\text{sin}\left(\text{t+}\varphi \right)\rangle$

(d)

The magnitude and phase expression for the system equation is calculated as:

Apply Taylor series expansion of $\text{r}\left(\text{t}\right)$ and $\varphi \left(\text{t}\right)$,

$\begin{array}{l}\text{r}\left(\text{t}\right)\text{ = }\overline{\text{r}}\text{ + O}\left(\text{ε}\right)\\ \varphi \left(\text{t}\right)\text{ = }\overline{\varphi }\text{+ O}\left(\text{ε}\right)\end{array}$

The above expression is true only because the average value of one oscillation and the rate of change over that cycle,

$\begin{array}{l}\text{r = }\overline{\text{r}}\text{+}\langle \dot{\text{r}}\rangle \\ \text{r = }\overline{\text{r}}\text{+ε}\langle \text{hsin}\left(\text{t+}\varphi \right)\rangle \\ \text{r(t) = }\overline{\text{r}}\text{+O}(\in )\end{array}$

$\begin{array}{l}\varphi \text{= }\overline{\varphi }\text{+}\langle \dot{\varphi }\rangle \\ \varphi \text{= }\overline{\varphi }+\frac{\in }{\overline{\text{r}}}\langle \text{hcos}\left(\text{t+}\varphi \right)\rangle \\ \varphi \text{(t) = }\overline{\varphi }(t)\text{+O}(\in )\end{array}$

Therefore, from above expression for $\langle \dot{\text{r}}\left(\text{t}\right)\rangle$ and $\dot{\varphi }\left(\text{t}\right)$,

$\begin{array}{l}\frac{\text{d}\overline{\text{r}}}{\text{dt}}\text{= }\langle \dot{\text{r}}\rangle \\ \frac{\text{d}\overline{\text{r}}}{\text{dt}}\text{= ε}\langle \left(\text{h}\left(\overline{\text{r}}\text{ cos}\left(\text{t + }\overline{\varphi }\right)\right)\text{, - }\overline{\text{r}}\text{ sin}\left(\text{t + }\overline{\varphi }\right)\text{, t}\right)\text{sin}\left(\text{t + }\varphi \right)\rangle \\ \frac{\text{d}\overline{\text{r}}}{\text{dt}}\text{= ε}\langle \left(\text{h}\left(\overline{\text{r}}\text{ cos}\left(\text{t + }\overline{\varphi }\right)\right)\text{, - }\overline{\text{r}}\text{ sin}\left(\text{t + }\overline{\varphi }\right)\text{, t}\right)\text{sin}\left(\text{t + }\varphi \right)\text{ + O}\left(\text{ε}\right)\rangle \\ \frac{\text{d}\overline{\text{r}}}{\text{dt}}\text{= ε}\langle \left(\text{h}\left(\overline{\text{r}}\text{ cos}\left(\text{t + }\overline{\varphi }\right)\right)\text{, - }\overline{\text{r}}\text{ sin}\left(\text{t + }\overline{\varphi }\right)\text{, t}\right)\text{sin}\left(\text{t + }\varphi \right)\text{ + O}\left({\text{ε}}^{\text{2}}\right)\rangle \end{array}$

$\begin{array}{l}\overline{\text{r}}\frac{\text{d}\overline{\varphi }}{\text{dt}}\text{ = }\langle \text{r}\dot{\varphi }\rangle \\ \overline{\text{r}}\frac{\text{d}\overline{\varphi }}{\text{dt}}\text{= ε}\langle \left(\text{h}\left(\text{r(t) cos}\left(\text{t + }\overline{\varphi }\right)\right)\text{, - r(t) sin}\left(\text{t + }\overline{\varphi }\right)\text{, t}\right)\text{cos}\left(\text{t + }\overline{\varphi }\right)\rangle \\ \overline{\text{r}}\frac{\text{d}\overline{\varphi }}{\text{dt}}\text{= ε}\langle \left(\text{h}\left(\overline{\text{r}}\text{ cos}\left(\text{t + }\overline{\varphi }\right)\right)\text{, - }\overline{\text{r}}\text{ sin}\left(\text{t + }\overline{\varphi }\right)\text{, t}\right)\text{cos}\left(\text{t + }\overline{\varphi }\right)\text{ + O}\left(\text{ε}\right)\rangle \\ \overline{\text{r}}\frac{\text{d}\overline{\varphi }}{\text{dt}}\text{= ε}\langle \left(\text{h}\left(\overline{\text{r}}\text{ cos}\left(\text{t + }\overline{\varphi }\right)\right)\text{, - }\overline{\text{r}}\text{ sin}\left(\text{t + }\overline{\varphi }\right)\text{, t}\right)\text{cos}\left(\text{t + }\overline{\varphi }\right)\text{ + O}\left({\text{ε}}^{\text{2}}\right)\rangle \end{array}$

Therefore, the expression of $\frac{\text{d}\overline{\text{r}}}{\text{dt}}$ and $\overline{\text{r}}\frac{\text{d}\overline{\varphi }}{\text{dt}}$ is

$\begin{array}{l}\frac{\text{d}\overline{\text{r}}}{\text{dt}}\text{ = ε}\langle \left(\text{h}\left(\overline{\text{r}}\text{ cos}\left(\text{t + }\overline{\varphi }\right)\right)\text{, - }\overline{\text{r}}\text{ sin}\left(\text{t + }\overline{\varphi }\right)\text{, t}\right)\text{sin}\left(\text{t + }\overline{\varphi }\right)\text{ + O}\left({\text{ε}}^{\text{2}}\right)\rangle \\ \overline{\text{r}}\frac{\text{d}\overline{\varphi }}{\text{dt}}\text{= ε}\langle \left(\text{h}\left(\overline{\text{r}}\text{ cos}\left(\text{t + }\overline{\varphi }\right)\right)\text{, - }\overline{\text{r}}\text{ sin}\left(\text{t + }\overline{\varphi }\right)\text{, t}\right)\text{cos}\left(\text{t + }\overline{\varphi }\right)\text{ + O}\left({\text{ε}}^{\text{2}}\right)\rangle \end{array}$.