Let $f:\mathbb R_+ \to \mathbb R_+$ be a function twice continuously differentiable (with derivative $f'$ and second derivative $f''$), and $a$, and $b$ be parameters in $\mathbb R_+$. Consider the system

\begin{align} \dfrac{dx(t)}{dt}=&\; f\left(x(t)\right)-y(t), \\[2ex] \dfrac{dy(t)}{dt}=&\; ay(t)\left(f'(x(t))-b\right), \end{align}

with $x(t)\ge 0$ and $y(t)\ge 0$ for all $t$, and boundary conditions

\begin{equation} x(0)= x_0, \qquad\text{and}\qquad\lim_{t\to\infty}e^{-bt}x(t)y(t)^{-a}=0. \end{equation}

By choosing $f$ appropriately it is possible to show that this system, without the initial condition, can have multiple stationary points. I would like to show, that even if that is the case, the following conjecture it true:

Conjecture $\;$ Given an $x_0$, there is a unique solution to the system that converges to one of the stationary points.

Why I think the conjecture is true

If $f'(0)>0$ and $f''(x)<0$ for all $x\in\mathbb R_+$, then there exists a unique stationary point and there is a simple proof of uniqueness of the solution given an arbitrary $x_0$ that involves drawing a phase diagram in the space $(x,y)$ and showing that there is a unique saddle path that the solution must be at all times (otherwise the second boundary condition would be violated) and that converges to the stationary point.

The reason why I think the conjecture is true is because, given an $f$ in $\mathcal C^2$, a similar phase diagram can be drawn. Here is a sketch of an example of a phase diagram with multiple stationary points:

enter image description here

Notice that the arrows flip between regions in a way that, given an $x_0$, there is only one saddle path (the lines in blue) that the solution could follow. My guess is that a proof for the conjecture would involve showing this is always the case.


If $x$ denotes the capital stock, $y$ the consumption level, $f$ is the production function, $a$ the inverse of the intertemporal elasticity of substitution, and $b$ is the discount rate, then this system is describes the equilibrium allocation of a simple economic growth model.

  • $\begingroup$ Please presenf $f'(x(t))$ via differentials $\endgroup$ – Yuri Negometyanov Apr 18 '18 at 22:35
  • $\begingroup$ $f'(x(t)) = \frac{df}{dx}$ or $\frac{df}{dt}?$ $\endgroup$ – Yuri Negometyanov Apr 18 '18 at 22:40
  • $\begingroup$ For those interested in constructing a counter-example, the Jacobian matrix of the mapping $ (x,y)^\top \mapsto (f(x) - y, a y (f'(x) - b))^\top $ has eigenvalues $$ \frac{(1+a)f'(x) - ab}{2} \pm \sqrt{ \left(\frac{(1-a)f'(x) + ab}{2}\right)^2 - a y f''(x) }$$ $\endgroup$ – Harry49 Apr 29 '18 at 8:36
  • $\begingroup$ @Harry49, would you mind explaining how you would go about constructing a counter-example using this information? I would be happy to give you the bounty for a broad explanation or just some references, since then at least I would have a way to move forward with this. $\endgroup$ – mzp May 3 '18 at 15:35
  • $\begingroup$ @YuriNegometyanov I edited the question to clarify that, the notation is $f'(x) = \frac{df(x)}{dx}$ and $f'(x(t)) = \frac{df(x(t))}{dx}$. $\endgroup$ – mzp Jun 11 '18 at 14:57

Multiplying of the system equations leads to the equation in the form of $$\left(f(x(t))-y(t)\right)\dfrac{dy}{dt} = ay(t)\left(f'(x(t))-b\right)\dfrac{dx(t)}{dt}.\tag1$$ Presentation of the function $y(t)$ as the superposition $y(x(t)),$ $$\dfrac{dy}{dx} = \dfrac{y'_t}{x'_t}$$ allows to eliminate variable $t$ from $(1):$ $$(f(x)-y)\dfrac{dy}{dx} = ay\dfrac{d}{dx}(f(x)-bx).\tag2$$ Then $$(y-f(x))\dfrac{d}{dx}(y-f(x))+(y-f(x))f'(x)+ay(f'(x)-b)=0,$$ $$(y-f(x))\dfrac{d}{dx}(y-f(x))+((a+1)f'(x)-ab)y -f(x)f'(x)=0,$$ $$(y-f(x))\dfrac{d}{dx}(y-f(x))+((a+1)f'(x)-ab)(y-f(x))+a(f'(x)-b)f(x)=0.\tag3$$ Equation $(3)$ allows the substitution $$z(x)= y-f(x)\tag4,$$ $$\dfrac{dx}{dt}= -z(x(t)),$$ which changes the issue system to the form of \begin{cases} \dfrac{dt}{dx}= -\dfrac1z\\[4pt] z\dfrac{dz}{dx}+((a+1)f'(x)-ab)z+a(f'(x)-b)f(x)=0, \end{cases} \begin{cases} t= -\int\limits_{x_0}^x\dfrac{\,\mathrm d\xi}{z(\xi)}\\[4pt] \left(\dfrac{dz}{dx}+((a+1)f'(x)-ab)\right)z+a(f'(x)-b)f(x)=0.\tag5 \end{cases} Easy to see that the variable $t$ has the gaps in the points $x,$ where $z(x)= 0,$ or $y(x)=f(x).$

On the other hand, such points presents as the stationary points by the coordinate $t.$

Let us consider behavior of the equation $(5.2)$ at the simple case $$f(x)= cx+d,\quad s=h(cx+d).\tag6$$ Then $$zz'+(a(c-b)+c)z+a(c-b)(cx+d)=0,$$ $$\dfrac{dz}{dx}=\dfrac{dz}{ds}\dfrac{ds}{dx}= ch\dfrac{dz}{ds},$$ or $$z\dfrac{dz}{ds}+gz+s=0,\tag7$$ where $$\quad h = \sqrt{a\left(1-\dfrac bc\right)},\quad a(c-b)=ch^2,\quad g=h+\dfrac1h.\tag8$$ Equation $(7)$ has the common solution $$\begin{cases} \ln|z^2+gsz+s^2|-\dfrac{2g}{\sqrt{4-g^2}}\arctan \dfrac{2z+gs}{s\sqrt{4-g^2}}=c,\text{ if }|g|<2,\\ (s\pm z)\exp\dfrac{s}{s\pm z}=C,\text{ if }g=\pm2,\\ C_1|z-gs|^\alpha+C_2|z-gs|^\beta,\text{ if }|g|>2,\tag9 \end{cases}$$ wherein $$|C_1|+|C_2|>0,\quad \alpha,\ \beta = -\dfrac12\pm\dfrac12\sqrt{g^2-4}.\tag{10}$$ Exact solution $(9)-(10)$ has two branches. The reason is that the equation $(5.2)$ can be written in the form of $$\dfrac12\dfrac{dz^2}{dx}+((a+1)f'(x)-ab)\sqrt{z^2}+a(f'(x)-b)f(x)=0,$$ which allows existance of multiple solutions.

  • 1
    $\begingroup$ Thank you for your answer! There are some parts that I am having a hard time understanding: (i) how did $\frac{dy(t)}{dt}$ become $\frac{dy}{dx}$ in equation $(1)$; (ii) where does $\frac{dt}{dx}=-\frac{1}{z}$ come from; and (iii) how does the system and argument you obtain at the end help decide whether or not the solution is unique. Would you mind adding a bit more detail to clarify those? $\endgroup$ – mzp Jul 4 '18 at 12:50
  • $\begingroup$ @mzp You are welcome! About first and the second questions - I'm waiting comments to updated text. The third answer requres detalized answer, and I'm preparing it now. $\endgroup$ – Yuri Negometyanov Jul 4 '18 at 14:33
  • $\begingroup$ @mzp The third answer is ready too. Waiting to your comments. $\endgroup$ – Yuri Negometyanov Jul 4 '18 at 17:01
  • $\begingroup$ Thank again. Your edits do clarify the issues I was having trouble with. However, with respect to the last point, don't the boundary conditions select one of the solutions? $\endgroup$ – mzp Jul 5 '18 at 11:11
  • $\begingroup$ @mzp I think, this depends from the choosen method of solution. If we deal with the theoretical method, then the boundary conditions can be satisfied or not. If we deal with the numerical method that can depend of the calculation method (either the boundary conditions or details of algorithm.) $\endgroup$ – Yuri Negometyanov Jul 6 '18 at 0:42

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