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# 数学代写|非线性动力系统代写Nonlinear Dynamics代考|MATH326 General N = 2 Systems

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## 数学代写|非线性动力系统代写Nonlinear Dynamics代考|The damped driven pendulum

Another example is that of the damped and driven harmonic oscillator,
$$\frac{d^2 \phi}{d s^2}+\gamma \frac{d \phi}{d s}+\sin \phi=j .$$
This is equivalent to a model of a resistively and capacitively shunted Josephson junction, depicted in fig. 3.3. If $\phi$ is the superconducting phase difference across the junction, the current through the junction is given by $I_J=I_{\mathrm{c}} \sin \phi$, where $I_{\mathrm{c}}$ is the critical current. The current carried by the resistor is $I_R=V / R$ from Ohm’s law, and the current from the capacitor is $I_C=\dot{Q}$. Finally, the Josephson relation relates the voltage $V$ across the junction to the superconducting phase difference $\phi: V=(\hbar / 2 e) \dot{\phi}$. Summing up the parallel currents, we have that the total current $I$ is given by
$$I=\frac{\hbar C}{2 e} \ddot{\phi}+\frac{\hbar}{2 e R} \dot{\phi}+I_{\mathrm{c}} \sin \phi,$$
which, again, is equivalent to a damped, driven pendulum.
This system also has a mechanical analog. Define the ‘potential’
$$U(\phi)=-I_{\mathrm{c}} \cos \phi-I \phi .$$
The equation of motion is then
$$\frac{\hbar C}{2 e} \ddot{\phi}+\frac{\hbar}{2 e R} \dot{\phi}=-\frac{\partial U}{\partial \phi} .$$

## 数学代写|非线性动力系统代写Nonlinear Dynamics代考|Classification of N = 2 fixed points

Suppose we have solved the fixed point equations $V_x\left(x^, y^\right)=0$ and $V_y\left(x^, y^\right)=0$. Let us now expand about the fixed point, writing
\begin{aligned} & \dot{x}=\left.\frac{\partial V_x}{\partial x}\right|{\left(x^, y^\right)}\left(x-x^\right)+\left.\frac{\partial V_x}{\partial y}\right|{\left(x^, y^\right)}\left(y-y^\right)+\ldots \ & \dot{y}=\left.\frac{\partial V_y}{\partial x}\right|{\left(x^, y^\right)} ^{\left(x-x^\right)}+\left.\frac{\partial V_y}{\partial y}\right|{\left(x^, y^\right)} ^{\left(y-y^\right)}+\ldots . \ & \end{aligned}
We define
$$u_1=x-x^* \quad, \quad u_2=y-y^*,$$
which, to linear order, satisfy
The formal solution to $\dot{\boldsymbol{u}}=\boldsymbol{M} \boldsymbol{u}$ is
$$\boldsymbol{u}(t)=\exp (M t) \boldsymbol{u}(0),$$

where $\exp (M t)=\sum_{n=0}^{\infty} \frac{1}{n !}(M t)^n$ is the exponential of the matrix $M t$.
The behavior of the system is determined by the eigenvalues of $M$, which are roots of the characteristic equation $P(\lambda)=0$, where
\begin{aligned} P(\lambda) & =\operatorname{det}(\lambda \mathbb{I}-M) \ & =\lambda^2-T \lambda+D, \end{aligned}
with $T=a+d=\operatorname{Tr}(M)$ and $D=a d-b c=\operatorname{det}(M)$. The two eigenvalues are therefore
$$\lambda_{\pm}=\frac{1}{2}\left(T \pm \sqrt{T^2-4 D}\right) .$$

## 数学代写|非线性动力系统代写Nonlinear Dynamics代考|The damped driven pendulum

$$\frac{d^2 \phi}{d s^2}+\gamma \frac{d \phi}{d s}+\sin \phi=j .$$

$$I=\frac{\hbar C}{2 e} \ddot{\phi}+\frac{\hbar}{2 e R} \dot{\phi}+I_{\mathrm{c}} \sin \phi,$$

$$U(\phi)=-I_{\mathrm{c}} \cos \phi-I \phi .$$

$$\frac{\hbar C}{2 e} \ddot{\phi}+\frac{\hbar}{2 e R} \dot{\phi}=-\frac{\partial U}{\partial \phi} .$$

## 数学代写|非线性动力系统代写Nonlinear Dynamics代考|Classification of $\mathrm{N}=2$ fixed points

$$u_1=x-x^* \quad, \quad u_2=y-y^*,$$

$$\boldsymbol{u}(t)=\exp (M t) \boldsymbol{u}(0),$$

$$P(\lambda)=\operatorname{det}(\lambda \mathbb{I}-M) \quad=\lambda^2-T \lambda+D,$$

$$\lambda_{\pm}=\frac{1}{2}\left(T \pm \sqrt{T^2-4 D}\right) .$$

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