Sine oscillations F = 0.01

In the study of dynamical systems, the Biryukov equation (or Biryukov oscillator), named after Vadim Biryukov (1946), is a non-linear second-order differential equation used to model damped oscillators.^[1]

The equation is given by

$${\displaystyle {\frac {d^{2}y}{dt^{2}}}+f(y){\frac {dy}{dt}}+y=0,\qquad \qquad (1)}$$

where ƒ(y) is a piecewise constant function which is positive, except for small y as

$${\displaystyle {\begin{aligned}&f(y)={\begin{cases}-F,&|y|\leq Y_{0};\\[4pt]F,&|y|>Y_{0}.\end{cases}}\\[6pt]&F={\text{const.}}>0,\quad Y_{0}={\text{const.}}>0.\end{aligned}}}$$

Eq. (1) is a special case of the Lienard equation; it describes the auto-oscillations.

Solution (1) at a separate time intervals when f(y) is constant is given by^[2]

$${\displaystyle y_{k}(t)=A_{1,k}\exp(s_{1,k}t)+A_{2,k}\exp(s_{2,k}t)\qquad \qquad (2)}$$

where exp denotes the exponential function. Here

$${\displaystyle s_{k}={\begin{cases}\displaystyle {\frac {F}{2}}\mp {\sqrt {\left({\frac {F}{2}}\right)^{2}-1}},&|y|<Y_{0};\\[2pt]\displaystyle -{\frac {F}{2}}\mp {\sqrt {\left({\frac {F}{2}}\right)^{2}-1}}&{\text{otherwise.}}\end{cases}}}$$

Expression (2) can be used for real and complex values of s_k.

The first half-period’s solution at ${\displaystyle y(0)=\pm Y_{0}}$ is

Relaxation oscillations F = 4

$${\displaystyle {\begin{aligned}y(t)&={\begin{cases}y_{1}(t),&0\leq t<T_{0};\\[4pt]y_{2}(t),&\displaystyle T_{0}\leq t<{\frac {T}{2}}.\end{cases}}\\[4pt]y_{1}(t)&=A_{1,k}\cdot \exp(s_{1,k}t)+A_{2,k}\cdot \exp(s_{2,k}t),\\[2pt]y_{2}(t)&=A_{3,k}\cdot \exp(s_{3,k}t)+A_{4,k}\cdot \exp(s_{4,k}t).\end{aligned}}}$$

The second half-period’s solution is

$${\displaystyle y(t)={\begin{cases}\displaystyle -y_{1}\left(t-{\frac {T}{2}}\right),&\displaystyle {\frac {T}{2}}\leq t<{\frac {T}{2}}+T_{0};\\[4pt]\displaystyle -y_{2}\left(t-{\frac {T}{2}}\right),&\displaystyle {\frac {T}{2}}+T_{0}\leq t<T.\end{cases}}}$$

The solution contains four constants of integration A₁, A₂, A₃, A₄, the period T and the boundary T₀ between y₁(t) and y₂(t) needs to be found. A boundary condition is derived from continuity of y(t) and dy/dt.^[3]

Solution of (1) in the stationary mode thus is obtained by solving a system of algebraic equations as

$${\displaystyle {\begin{array}{ll}&y_{1}(0)=-Y_{0}&y_{1}(T_{0})=Y_{0}\\[6pt]&y_{2}(T_{0})=Y_{0}&y_{2}\!\left({\tfrac {T}{2}}\right)=Y_{0}\\[6pt]&\displaystyle \left.{\frac {dy_{1}}{dt}}\right|_{T_{0}}=\left.{\frac {dy_{2}}{dt}}\right|_{T_{0}}\qquad &\displaystyle \left.{\frac {dy_{1}}{dt}}\right|_{0}=-\left.{\frac {dy_{2}}{dt}}\right|_{\frac {T}{2}}\end{array}}}$$

The integration constants are obtained by the Levenberg–Marquardt algorithm. With ${\displaystyle f(y)=\mu (-1+y^{2})}$, ${\displaystyle \mu ={\text{const.}}>0,}$ Eq. (1) named Van der Pol oscillator. Its solution cannot be expressed by elementary functions in closed form.

References