Memristor

Equation

We have an equation $$ \muU \cdot \frac{R+}{D} \cdot U \left( t \right) \cdot \theta\left( \omega \right) \cdot \theta\left( D - \omega \right) = \frac{d\omega}{dt} \cdot \left[ R_+ \cdot \frac{\omega\left( t \right)}{D}

         + R_- \cdot \left( 1 - \frac{\omega\left( t \right)}{D} \right) \right]

$$ Heaviside functions product can be folded to indicator $$

\frac{d\omega}{dt} \cdot \left[ R_+ \cdot \frac{\omega\left( t \right)}{D}

         + R_- \cdot \left( 1 - \frac{\omega\left( t \right)}{D} \right) \right]

= \muU \cdot \frac{R+}{D} \cdot U \left( t \right) \cdot \mathbb{1}\left( 0 < \omega\left( t \right) < D \right) $$

This means that either $$ \dot{\omega} = 0,\qquad \omega \notin \left( 0; D \right) \Longrightarrow \begin{cases} \omega\left( t \right) = 0,\qquad \omega\left( t' \right) \le 0,\qquad t' \le t, \\ \omega\left( t \right) = D,\qquad \omega\left( t' \right) \ge D,\qquad t' \le t \end{cases} $$ or $$ \omega\left( t \right) = - \frac{D \cdot R_-}{R_+ - R_-}. $$

The second equation is possible if $D=0$, which is not possible due to equation $0 < \omega\left( t \right) < D$, or $R_- = 0$, which is physically impossible in this problem: impedance should be greater than zero.

This means that when $\omega$ reaches $0$ or $D$, it cannot move further and we have a stable state there. We can solve the differential equation without indicator, keeping in mind spikes of derivative.

In order to solve this differential equation we need to integrate it $$ \int\limits{w\left( 0 \right)}^{w\left( t \right)} \left[ \omega\left( t \right) \cdot \frac{R+ - R_-}{D}

         + R_- \right] d\omega

= \muU \cdot \frac{R+}{D} \cdot \int\limits_0^t U \left( t \right) dt $$ Let us denote $$ \Phi\left( t \right) = \int\limits_0^t U \left( t \right) dt $$

Result looks like an equation $$ \omega^2 \cdot \frac{R+ - R-}{2 \cdot D} + \omega \cdot R_- - \omega_c

• \muU \frac{R+}{D} \cdot \Phi\left( t \right) = 0, \qquad \omegac = \omega\left( 0 \right)^2 \cdot \frac{R+ - R-}{2 \cdot D} + \omega\left( 0 \right) \cdot R- $$

Solutions are $$ \omega_{1, 2}\left( t \right) = \frac{

    - R_- \pm \sqrt{R_-^2
    + 4 \cdot \frac{R_+ - R_-}{2 \cdot D} \cdot \left( \omega_c + \mu_U \cdot \frac{R_+}{D} \cdot \Phi\left( t \right)
    \right)
}}
{\frac{R_+ - R_-}{D}}

$$ Can be simplified to $$

\omega_{1, 2}\left( t \right) = D \cdot \frac{

    - R_- \pm \sqrt{R_-^2
    + 2 \cdot \frac{R_+ - R_-}{D} \cdot \left( \omega_c + \mu_U \cdot \frac{R_+}{D} \cdot \Phi\left( t \right)
    \right)
}}
{R_+ - R_-}

$$

Electric current

$$ I\left( t \right) = \frac{D \cdot U\left( t \right)}{R_+ \cdot w\left( t \right) + R_- \cdot \left(D - w\left( t \right) \right)} $$

Constant voltage

Assume $U = U_0$ $$ \omega_{1, 2}\left( t \right) = D \cdot \frac{

    - R_- \pm \sqrt{R_-^2
    + 2 \cdot \frac{R_+ - R_-}{D} \left( \omega_c + \mu_U \cdot \frac{R_+}{D} \cdot U_0 \cdot t
    \right)
}}
{R_+ - R_-}

$$ and $$

\Phi\left( t \right) = t \cdot U_0 $$

Sine voltage

Assume $U = U_0 \cdot \sin{\left( \nu \cdot t \right)}$ $$ \omega_{1, 2}\left( t \right) = D \cdot \frac{

    - R_- \pm \sqrt{R_-^2
    + 2 \cdot \frac{R_+ - R_-}{D}
        \cdot \left( \omega_c
        + \mu_U \cdot \frac{R_+}{D} \cdot \frac{U_0}{\nu}
        \cdot \left( 1 - \cos{\left(\nu \cdot t \right)} \right)
    \right)
}}
{R_+ - R_-}

$$ and $$

\Phi\left( t \right) = \frac{U_0}{\nu} \cdot \left( 1 - \cos{\left( \nu \cdot t \right)} \right) $$

Sine squared voltage

Assume $U = U_0 \cdot \sin^2{\left( \nu \cdot t \right)}$ $$ \omega_{1, 2}\left( t \right) = D \cdot \frac{

    - R_- \pm \sqrt{R_-^2
    + 2 \cdot \frac{R_+ - R_-}{D}
        \cdot \left( \omega_c
        + \mu_U \cdot \frac{R_+}{D} \cdot \frac{U_0}{4 \cdot \nu}
        \cdot \left( 2 \cdot \nu \cdot t - \sin{\left(2 \cdot \nu \cdot t \right)} \right)
    \right)
}}
{R_+ - R_-}

$$ and $$

\Phi\left( t \right) = \frac{U_0}{4 \cdot \nu} \cdot \left( 2 \cdot \nu \cdot t - \sin{\left(2 \cdot \nu \cdot t \right)} \right) $$

Electric charge to magnetic flux

As long as $$ \begin{cases} I &= \dot{q}, \\ U &= \dot{\Phi} \end{cases} $$ We have $$ \begin{cases} q\left( t \right) &= \int\limits_{t_0}^t I\left( t \right) \\ \Phi\left( t \right) &= \int\limits_{t_0}^t U\left( t \right) \end{cases} $$ It's hard to integrate $I$ analytically, so numeric method can be used $$ \int\limits_{t_0}^t I\left( t \right) \approx \sum\limits_1^n I\left( t_i \right) \cdot \Delta t, \qquad \Delta t = \frac{t - t_0}{n}, \qquad t_i = t_0 + \Delta t \cdot \left( n - 1 \right) $$

Another constraint

We have an equation $$ \muU \cdot \frac{R+}{D} \cdot U \left( t \right) \cdot \frac{\omega \cdot \left( D - \omega \right)}{D^2} = \frac{d\omega}{dt} \cdot \left[ R_+ \cdot \frac{\omega\left( t \right)}{D}

         + R_- \cdot \left( 1 - \frac{\omega\left( t \right)}{D} \right) \right]

$$ Can be solved via integration $$

\int\limits{w\left( 0 \right)}^{w\left( t \right)} \frac{R+ \cdot \omega\left( t \right)

  + R_- \cdot \left( D - \omega\left( t \right) \right)}{\omega\left( t \right) \cdot \left( D - \omega\left( t \right) \right)}

d\omega = \int\limits_{t_0}^{t} \muU \cdot \frac{R+}{D^2} \cdot U \left( t \right) dt $$

$$ \int\limits_{w\left( 0 \right)}^{w\left( t \right)} \left( \frac{R_+}{D - \omega\left( t \right)} + \frac{R_-}{\omega\left( t \right)} \right) d\omega = \mu_U \cdot \frac{R_+}{D^2} \cdot \int\limits_{t_0}^{t} U \left( t \right) dt $$

Result is $$ \frac{R-}{R+} \cdot \ln{\left| \frac{\omega\left( t \right)}{\omega\left( t_0 \right)} \right|}

• \ln{\left| \frac{D - \omega\left( t \right)}{D - \omega\left( t_0 \right)} \right|} = \frac{\muU}{D^2} \cdot \Phi\left( t \right) $$ thus $$ \frac{\omega\left( t \right)^{\frac{R-}{R_+}}}{D - \omega\left( t \right)} = \frac{\omega\left( t0 \right)^{\frac{R-}{R_+}}}{D - \omega\left( t_0 \right)} \cdot \exp{\left{ \frac{\mu_U}{D^2} \cdot \Phi\left( t \right) \right}} $$