Love Affairs Dynamics with One Delay in Losing Memory or Gaining Affection

Abstract

A dynamic model of a love affair between two people is examined under different conditions. First the two-dimensional model is analyzed without time delays in the interaction of the lovers. Conditions are derived for the existence of a unique as well as for multiple steady states. The nonzero steady states are always stable and the stability of the zero steady state depends on model parameters. Then a delay is assumed in the mutual-reaction process called the Gaining-affection process. Similarly to the no-delay case, the nonzero steady states are always stable. The zero steady state is either always stable or always unstable or it is stable for small delays and at a certain threshold stability is lost in which case the steady state bifurcates to a limit cycle. When delay is introduced to the self-reaction process called the Losing-memory process, then the asymptotic behavior of the steady state becomes more complex. The stability of the nonzero steady state is lost at a certain value of the delay and bifurcates to a limit cycle, while the stability of the zero steady state depends on model parameters and there is the possibility of multiple stability switches with stability losses and regains. All stability conditions and stability switches are derived analytically, which are also verified and illustrated by using computer simulation.

The author highly appreciate the financial supports from the MEXT-Supported Programe for the Strategic Research Foundation at Private Universities 2013-2017, the Japan Society for the Promotion of Science (Grant-in-Aid for Scientific Research (C), 24530202, 25380238, 26380316) and Chuo University (Joinet Research Grant). The usual disclaimer apply.

Appendix

Proof of Theorem 1

Proof

The zero steady state, $x_{0}^{*}=0$ and $y_{0}^{*}=0$, is clearly a solution of (3) and (4). Thus the two isoclines intersect at least once at the origin. We investigate whether such an intersection happens only once or not. To this end, we differentiate u(x) and $v(x),\$

$$\begin{aligned} u^{\prime }(x)=\frac{\dfrac{\alpha _{x}}{\beta _{x}}}{1-\left( \dfrac{\alpha _{x}}{\beta _{x}}x\right) ^{2}},\ u^{\prime \prime }(x)=\frac{2x\left( \dfrac{\alpha _{x}}{\beta _{x}}\right) ^{3}}{\left[ 1-\left( \dfrac{\alpha _{x}}{\beta _{x}}x\right) ^{2}\right] ^{2}} \end{aligned}$$

and

$$\begin{aligned} v^{\prime }(x)=\frac{\beta _{y}}{\alpha _{y}}\left( \frac{2}{e^{x}+e^{-x}} \right) ^{2},\ v^{\prime \prime }(x)=-\frac{8\beta _{y}}{\alpha _{y}}\frac{ e^{x}-e^{-x}}{\left( e^{x}+e^{-x}\right) ^{3}}. \end{aligned}$$

Although $\alpha _{x}>0$ and $\alpha _{y}>0$ by assumption, the signs of $\beta _{x}$ and $\beta _{y}\$are undetermined. We consider three cases, depending on the signs of $\beta _{x}$ and $\beta _{y}$.

• (i) Assume first that $\beta _{x}$ and $\beta _{y}$ have different signs. Then $u^{\prime }(x)$ and $v^{\prime }(x)$ also have different signs, so one is strictly increasing and the other is strictly decreasing. So $x_{0}^{*}=0$ and $y_{0}^{*}=0\$are the only steady state if $\alpha _{x}\alpha _{y}>0>\beta _{x}\beta _{y}$.

• (ii) Assume next that $\beta _{x}$ and $\beta _{y}$ are both positive. Then

  $$\begin{aligned} u(0)=0,\ u\left( \frac{\beta _{x}}{\alpha _{x}}\right) =\infty ,\ u\left( - \frac{\beta _{x}}{\alpha _{x}}\right) =-\infty ,\ u^{\prime }(x)>0\text {, } u^{\prime \prime }(x)\left\{ \begin{array}{l}>0\text { if }x>0, \\ \\<0\text { if }x<0 \end{array} \right. \end{aligned}$$

  and

  $$\begin{aligned} v(0)=0,\ v\left( \infty \right) =\frac{\beta _{y}}{\alpha _{y}},\ v\left( -\infty \right) =-\frac{\beta _{y}}{\alpha _{y}},\ v^{\prime }(x)>0\text {, } v^{\prime \prime }(x)\left\{ \begin{array}{l}<0\text { if }x>0, \\ \\ >0\text { if }x<0. \end{array} \right. \end{aligned}$$

  Furthermore

  $$\begin{aligned} u^{\prime }(0)=\frac{\alpha _{x}}{\beta _{x}}\text { and }\ v^{\prime }(0)= \frac{\beta _{y}}{\alpha _{y}}. \end{aligned}$$

  Only zero solution is possible if $u^{\prime }(0)\ge v^{\prime }(0)$, that is, if

  $$\begin{aligned} \frac{\alpha _{x}}{\beta _{x}}\ge \frac{\beta _{y}}{\alpha _{y}}\ \text {or } \alpha _{x}\alpha _{y}\ge \beta _{x}\beta _{y}. \end{aligned}$$

  If $\alpha _{x}\alpha _{y}<\beta _{x}\beta _{y}$, then there are two nonzero solutions in addition to the zero steady state: one in the positive region $(x_{1}^{*},y_{1}^{*})>0$ due to the convexity of u(x) and the concavity of v(x) for positive x and the other in the negative region $(x_{2}^{*},y_{2}^{*})<0$ due to the concavity of u(x) and the convexity of v(x) for negative x.

• (iii) Assume finally that $\beta _{x}<0$ and $\beta _{y}<0$. Equation (3) remains same if $\beta _{x}$ and $\beta _{y}$ are replaced by $-\beta _{x}$ and $-\beta _{y}$, so previous case may apply for existence of nonzero solutions.

   $\blacksquare$

Proof of Theorem 2

Proof

We omit to prove the first four cases, (1), (2), (3) and (4). For the last case in which $\beta _{x}\beta _{y}>\alpha _{x}\alpha _{y},\$we consider two cases depending of the signs of $\beta _{x}$ and $\beta _{y}$.

(i) We first assume $\beta _{x}>0$ and $\beta _{y}>0$. At a non-zero solution $v^{\prime }(x_{k}^{*})<u^{\prime }(x_{k}^{*})$, that is,

$$\begin{aligned} \frac{\beta _{y}}{\alpha _{y}}d_{x}<\frac{\dfrac{\alpha _{x}}{\beta _{x}}}{ 1-\left( \dfrac{\alpha _{x}}{\beta _{x}}x\right) ^{2}}. \end{aligned}$$

(23)

Since from the first equation in (2),

$$\begin{aligned} \frac{\alpha _{x}}{\beta _{x}}x=\tanh (y), \end{aligned}$$

the right hand side of (23) is

$$\begin{aligned} \frac{\dfrac{\alpha _{x}}{\beta _{x}}}{1-\left( \dfrac{e^{y}-e^{-y}}{ e^{y}+e^{-y}}\right) ^{2}}=\frac{\dfrac{\alpha _{x}}{\beta _{x}}}{\left( \dfrac{2}{e^{y}+e^{-y}}\right) ^{2}}=\frac{\dfrac{\alpha _{x}}{\beta _{x}}}{ d_{y}}. \end{aligned}$$

So we have

$$\begin{aligned} \frac{\beta _{y}}{\alpha _{y}}d_{x}<\frac{\dfrac{\alpha _{x}}{\beta _{x}}}{ d_{y}} \end{aligned}$$

(24)

or

$$\begin{aligned} \alpha _{x}\alpha _{y}>\beta _{x}\beta _{y}d_{x}d_{y}. \end{aligned}$$

(25)

(ii) If $\beta _{x}<0$ and $\beta _{y}<0$, then $v^{\prime }(x_{k}^{*})>u^{\prime }(x_{k}^{*})$ for $k=1,2\$at any nonzero solution, so inequality (23) has opposite direction, as well as inequality (24) has opposite direction and by multiplying it by $\alpha _{y}\beta _{x}d_{y}<0$, Eq. (25) remains valid.   $\blacksquare$

Proof of Theorem 3

Proof

If any eigenvalue is multiple, then it also solves the following equation obtained by differentiating the left hand side of Eq. (7),

$$\begin{aligned} 2\lambda +\left( \alpha _{x}+\alpha _{y}\right) +\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}e^{-\lambda \tau _{x}}\tau _{x}=0. \end{aligned}$$

(26)

From Eq. (7),

$$\begin{aligned} \beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}e^{-\lambda \tau _{x}}=\lambda ^{2}+\left( \alpha _{x}+\alpha _{y}\right) \lambda +\alpha _{x}\alpha _{y} \end{aligned}$$

that is substituted into Eq. (26),

$$\begin{aligned} 2\lambda +\left( \alpha _{x}+\alpha _{y}\right) +\lambda ^{2}\tau _{x}+\left( \alpha _{x}+\alpha _{y}\right) \lambda \tau _{x}+\alpha _{x}\alpha _{y}\tau _{x}=0 \end{aligned}$$

or

$$\begin{aligned} \lambda ^{2}\tau _{x}+\left( 2+\alpha _{x}\tau _{x}+\alpha _{y}\tau _{x}\right) \lambda +(\alpha _{x}+\alpha _{y}+\alpha _{x}\alpha _{y}\tau _{x})=0. \end{aligned}$$

This equation cannot have pure complex root since multiplier of $\lambda$ is positive.   $\blacksquare$

Proof of Theorem 6

Proof

The characteristic equation for $\alpha _{x}=\alpha _{y}=\alpha$ is simplified as

$$\begin{aligned} \lambda ^{2}+\alpha \lambda -\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}+\alpha (\lambda +\alpha )e^{-\lambda \tau _{x}}=0. \end{aligned}$$

If $\lambda$ is a multiple root, then it also satisfies equation,

$$\begin{aligned} 2\lambda +\alpha +\alpha e^{-\lambda \tau _{x}}-\tau _{x}\alpha (\lambda +\alpha )e^{-\lambda \tau _{x}}=0. \end{aligned}$$

From the first equation

$$\begin{aligned} e^{-\lambda \tau _{x}}=-\lambda +\frac{\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k} }{\lambda +\alpha } \end{aligned}$$

and by substituting it into the second equation, we have

$$\begin{aligned} 2\lambda +\alpha +\left( -\lambda +\frac{\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}}{\lambda +\alpha }\right) -\tau _{x}\left( -\lambda (\lambda +\alpha )+\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}\right) =0 \end{aligned}$$

which can be written as

$$\begin{aligned} \lambda ^{3}\tau _{x}+\lambda ^{2}\left( 1+2\alpha \tau _{x}\right) +\lambda \left( 2\alpha +\alpha ^{2}\tau _{x}-\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}\tau _{x}\right) +\left( \alpha ^{2}+\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}(1-\alpha \tau _{x})\right) =0. \end{aligned}$$

If $\lambda =i\omega$, then

$$\begin{aligned} \omega ^{2}=\frac{2\alpha +\alpha ^{2}\tau _{x}-\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}\tau _{x}}{\tau _{x}}=\frac{\alpha ^{2}+\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}(1-\alpha \tau _{x})}{1+2\alpha \tau _{x}} \end{aligned}$$

This equation can be simplified as follows:

$$\begin{aligned} 2\alpha +2\tau _{x}(2\alpha ^{2}-\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k})+\alpha \tau _{x}^{2}\left( 2\alpha ^{2}-\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}\right) =0. \end{aligned}$$

If $\beta _{x}\beta _{y}\le 0$, then the left hand side is positive, so no solution exists. If $\beta _{x}\beta _{y}>0$, then $\omega _{+}^{2}>0$ if and only if $\alpha ^{2}>\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}$. In this case the left hand side is positive again showing that no solution exists.   $\blacksquare$

Proof of Theorem 7

Proof

Select $\tau _{x}$ as the bifurcation parameter and consider $\lambda$ as the function of $\tau _{x},\ \lambda =\lambda (\tau _{x}).\$Implicitly differentiating the characteristic equation with respect to $\tau _{x}$ gives

$$\begin{aligned} \left[ 2\lambda +\alpha +\alpha e^{-\lambda \tau _{x}}-\alpha \tau _{x}(\lambda +\alpha )e^{-\lambda \tau _{x}}\right] \frac{d\lambda }{d\tau _{x}}-\alpha \lambda (\lambda +\alpha )e^{-\lambda \tau _{x}}=0 \end{aligned}$$

implying that

$$\begin{aligned} \frac{d\lambda }{d\tau _{x}}= & {} \frac{\alpha \lambda (\lambda +\alpha )e^{-\lambda \tau _{x}}}{2\lambda +\alpha +\alpha e^{-\lambda \tau _{x}}-\alpha \tau _{x}(\lambda +\alpha )e^{-\lambda \tau _{x}}} \\&\\= & {} \frac{-\lambda ^{4}-2\lambda ^{3}\alpha -\lambda ^{2}\alpha ^{2}+\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}\lambda (\lambda +\alpha )}{2\lambda ^{2}+\alpha \lambda +2\lambda \alpha +\alpha ^{2}+\left( 1-\tau _{x}\lambda -\tau _{x}\alpha \right) \left( -\lambda ^{2}-\alpha \lambda +\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}\right) }. \end{aligned}$$

Assume that $\lambda =i\omega$, then the numerator becomes

$$\begin{aligned} \left( -\omega ^{4}+\omega ^{2}\left( \alpha ^{2}-\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}\right) \right) +i\omega \left( 2\omega ^{2}\alpha +\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}\alpha \right) \end{aligned}$$

and the denominator is simplified as

$$\begin{aligned} -\omega ^{2}(1+2\alpha \tau _{x})+\left( \alpha ^{2}+\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}(1-\alpha \tau _{x})\right) +i\left( -\tau _{x}\omega ^{3}+\omega \left( 2\alpha +\alpha ^{2}\tau _{x}-\tau _{x}\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}\right) \right) . \end{aligned}$$

Multiplying the numerator and the denominator by the complex conjugate of the denominator shows that Re$\left[ d\lambda /d\tau _{x}\right]$ has the same sign as

$$\begin{aligned} \omega ^{4}+\omega ^{2}\left( 2\alpha ^{2}\right) +\left[ \alpha ^{4}+2\alpha ^{2}\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}-\left( \beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}\right) ^{2}\right] . \end{aligned}$$

At $\omega ^{2}=\omega _{+}^{2}=\alpha ^{2}-\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k},\$this expression becomes

$$\begin{aligned} 2\alpha ^{2}\left( 2\alpha ^{2}-\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}\right) >0 \end{aligned}$$

showing that at the stability switch stability is lost or instability is retained. At $\omega ^{2}=\omega _{-}^{2}=-\left( \alpha ^{2}+\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k}\right)$, Re$\left[ d\lambda /d\tau _{x} \right]$ has the same sign as

$$\begin{aligned} 2\alpha ^{2}\beta _{x}\beta _{y}d_{x}^{k}d_{y}^{k} \end{aligned}$$

which is positive if $\beta _{x}\beta _{y}>0$ and negative if $\beta _{x}\beta _{y}<0$. In the first case stability is lost or instability is retained and in the second case stability is regained or stability is retained.   $\blacksquare$