A nonlinear evolutionary equation modelling a dockless bicycle-sharing system

Abstract

We model the operation of a dockless bicycle-sharing system by two groups of interacting components, bicycles on a sidewalk and users of the system. The model illustrates the bicycle-sharing system by a nonlinear evolutionary equation about the density of bikes on the sidewalk. Users’ behaviours, which are some straightforward and necessary actions, determine coefficients of the nonlinear evolutionary equation. Some nontrivial solutions to the equation show that even if every user has no malice, and the environment is stable, it is still possible that heaps of shared-bicycles appear somewhere along the road. Based on the data of heaps, parameters of users’ psychological models can be obtained. A numerical simulation shows how to calculate features of users and change the supply of bicycles into the system.

Appendices

Appendix 1 Proof of Lemma 1

Proof

For any u, the numerator f(u) is positive and it does not influence the sign of f(u). We consider the denominator $\phi (u)$ of f(u):

$$\begin{aligned} \phi (u)=\eta -\gamma \alpha ^{-2}(1+\alpha u)\exp (-\alpha u). \end{aligned}$$

(39)

As $0<\eta < \gamma \alpha ^{-2}$, it is obvious that $\phi (0)<0$. Moreover, $\phi (\infty )=\eta >0$. And for $u>0$, $\phi '(u)=\gamma u \exp (-\alpha u)>0$. So there is a unique $B>0$ such that $\phi (B)=0$ and for $u\in [0, B)$, $\phi (u)<0$. Then the result of 1) follows.

We write $g(u)=f(u)\exp (\alpha u)$. It follows from

$$\begin{aligned} f'(u)=-\alpha f(u)+g'(u)\exp (-\alpha u)\end{aligned}$$

(40)

and (37) that

$$\begin{aligned} \frac{1}{g^2(u)}g'(u)=-\gamma u \exp (-\alpha u) . \end{aligned}$$

(41)

Therefore

$$\begin{aligned} \frac{1}{g(u)} =\eta -\gamma \alpha ^{-2}(1+\alpha u) \exp (-\alpha u). \end{aligned}$$

(42)

Notice that it follows from 1) that for $u\in [0,B)$, the above function is not zeros at u and hence (36) follows. That is, f is a solution to (37) on [0, B).

As f(u) is continuous at any $u\in [0,B)$, (38) is a proper definition to F(u). It follows from $f(u)<0$ that F(u) is decreasing. From its definition, we have that $F(0)=0$. Finally, to prove that $F(B-0)=-\infty$, we only need to prove that for some $\delta >0$,

$$\begin{aligned} F(B-0) - F(B-\delta )=\int _{B-\delta }^{B} f(u) du=-\infty . \end{aligned}$$

(43)

Now we study

$$\begin{aligned} \frac{1}{f(u)}=\exp (\alpha u) \phi (u), \end{aligned}$$

(44)

where $\phi (u)$ is given in (39). As for $u>0$, $\phi '(u)=\gamma u \exp (-\alpha u)>0$,

$$\begin{aligned} \left. \frac{d((f(u))^{-1})}{d u}\right| _{u=B}=\exp (\alpha B) \phi '(B)\ne 0. \end{aligned}$$

(45)

Moreover, it follows from $f(B-0)=-\infty$ that $1/f(B-0)=0$. Therefore, there is a constant C and $\delta >0$ such that for $u\in [B-\delta , B]$

$$\begin{aligned} \frac{1}{f(u)} \ge C(u-B), \end{aligned}$$

(46)

and hence $f(u) \le C^{-1}(u-B)^{-1}$. So we can obtain

$$\begin{aligned} \int _{B-\delta }^{B} f(u) du\le \int _{B-\delta }^{B} \frac{1}{C(u-B)} du =-\infty .\end{aligned}$$

(47)

Appendix 2 Proof of Theorem 3

.

We write $\alpha =\frac{2v\beta ^2}{\sigma ^2\lambda _0^2}>0$, $\gamma =\frac{2\beta }{\sigma ^2}>0$ and let $\eta$ be a positive number such that $\eta < \gamma \alpha ^{-2}$. It follows from 3) of Lemma 1 that F has an inverse $F^{-1}: (-\infty , 0] \rightarrow [0, B)$ such that $F^{-1}(-\infty )= B$ and $F^{-1}(0)= 0$. We write

$$\begin{aligned} h(y)=\left\{ \begin{array}{ll} F^{-1}(y), &{} y\le 0,\\ 0, &{} y\ge 0. \end{array} \right. \end{aligned}$$

(48)

We will show that u(t, y) given by (34) satisfies (10) at any (t, y) such that $\lambda _0 y+v\beta t\ne 0$.

Let $k=v\beta /\lambda _0$. It is clear that $u_t(t,y)=ku_y(t,y)$ for $\lambda _0 y+v\beta t\ne 0$. Substituting $u_t(t,y)=ku_y(t,y)$ into (10), we have

$$\begin{aligned}&\frac{\sigma ^2k}{2}u_{yyy}-k^2u_{yy}\nonumber \\&\quad -k(bu_y)_{y} -\lambda _0ku_y+v\,b_{y}=0. \end{aligned}$$

(49)

And hence

$$\begin{aligned} \frac{\sigma ^2k}{2}u_{yy}-k^2u_{y}-k(bu_y)-\lambda _0ku+v\,b=C, \end{aligned}$$

(50)

where C is a constant.

We will show that for $C=0$, (50) has two solutions: $u(t,y)=0$ and $u(t,y)=F^{-1}\left( y+\frac{v\beta }{\lambda _0}t\right)$ and hence u(t, y) given by (34) is a weak solution to (50). It is clear that $u(t,y)=0$ is a trivial solution. Substituting $b(u,u_y)$ given in (33) into (50) and notice $C=0$, we have that $\bar{u}(y)=u(0,y)$ satisfies

$$\begin{aligned}&\frac{\sigma ^2k}{2}\bar{u}_{yy}-k^2\bar{u}_{y}-k\beta \bar{u} \bar{u}_y -k\frac{v\beta ^2}{\lambda _0^2}(\bar{u}_y)^2\nonumber \\&-\lambda _0k\bar{u}+v\beta \bar{u}+ \frac{v^2\beta ^2}{\lambda _0^2}\bar{u}_y=0. \end{aligned}$$

(51)

The coefficient of $\bar{u}$ in the above (51) is $-\lambda _0k+v\beta =0$, that follows from $k=v\beta /\lambda _0$, and the coefficient of $\bar{u}_y$ is $-k^2+v^2\beta ^2\lambda _0^{-2}=0$. Hence (51) yields

$$\begin{aligned} \frac{\sigma ^2}{2}\bar{u}_{yy}-\beta \bar{u} \bar{u}_y -\frac{v\beta ^2}{\lambda _0^2}(\bar{u}_y)^2=0. \end{aligned}$$

(52)

Consider the inverse $y=y(\bar{u})$ of $\bar{u}=\bar{u}(y)$. Substituting

$$\begin{aligned} \bar{u}_y=\frac{1}{y_{\bar{u}}}, \quad \bar{u}_{yy}=-\frac{y_{\bar{u}\bar{u}}}{(y_{\bar{u}})^3} \end{aligned}$$

(53)

into (52), we have that

$$\begin{aligned} \frac{\sigma ^2}{2}y_{\bar{u}\bar{u}}+\beta \bar{u} y_{\bar{u}}^2 +\frac{v\beta ^2}{\lambda _0^2}y_{\bar{u}}=0. \end{aligned}$$

(54)

Let $f(\bar{u})=y_{\bar{u}}$, then we have

$$\begin{aligned} f'+\frac{2\beta }{\sigma ^2}\bar{u} f^2 +\frac{2v\beta ^2}{\sigma ^2\lambda _0^2}f=0. \end{aligned}$$

(55)

As $\alpha =\frac{2v\beta ^2}{\sigma ^2\lambda _0^2}$ and $\gamma =\frac{2\beta }{\sigma ^2}$, the above (55) and (37) is the same one. It follows from Lemma 1 that $y(\bar{u})=F(\bar{u})$. Therefore $\bar{u}=F^{-1}(y)$ is a solution to (51), and hence $u(t,y)=F^{-1}\left( y+\frac{v\beta }{\lambda _0}t\right)$ is a solution to (50).