Let me elaborate on what I mean by the title of this question.

In studying (specifically non-linear) PDEs, one finds that each term in the equations contributes a certain amount to the behavior of the solution and the resulting behavior of the said solution depends on the tension between these effects. Such effects are for example non-linearity, transport, diffusion, dispersion, and dissipation. I am interested in example where these phenomena are manifested in simpler ODEs, and these examples would then be considered toy models.

One example would be a tension between linear dissipation (the first term) and non-linear growth (the second term) manifested in this ODE: $$ \dot{y}(t)=-y+y^2, y(0)=y_0\neq 0, 1. $$ Let $y_0>0$. This is well manifested in the solution, which takes the form $$y(t)=\left(1+e^t\left(\frac{1-y_0}{y_0}\right)\right)^{-1}$$ is globally defined for $t\ge 0$ if $y_0<1$ (the linear term dominates), and in fact $y(t)\to 0$ as $t\to\infty$. However if $y_0>1$, then the solution blows up at $t=-\log(1-1/y_0)$, originating from the fact that the non-linear term is dominating.

I am looking for similar examples where the solution to an ODE qualitatively represents the tension between effects seen in more complicated PDEs.


2 Answers 2


Consider the sine-Gordon partial differential equation $$\tag{1} \partial_t^2 u -\partial_x^2u +\sin (u)=0, \quad t\in\mathbb R, x\in\mathbb R. $$ The solutions $u=u(t)$ that are constant in space satisfy the ordinary differential equation of the mathematical pendulum; letting $y(t)=u(t)$, $$\tag{2} \ddot y + \sin(y)=0.$$

I am very far from being an expert of these models, but this Wolfram Mathematica notebook nicely demonstrates how (1) can actually be seen as an infinite system of coupled pendulums (2). The constant-in-space situation that we just described corresponds to all pendulums swinging at unison.

Thus, the mathematical pendulum can be seen as a "building block" for the sine-Gordon equation. I submit that this does count as a "toy model". Even without going too deep into the theory, you can check in the Wikipedia page that solutions are often identified up to equivalence mod $2\pi$. The variable $u$ ($\varphi$ in Wikipedia) is often referred to as a twist, that is, an angle. The pendulum identification gives at least an intuition for these mental models.


The method of characteristics allows the transformation of hyperbolic PDEs into ODEs. Consequently, influences of diffusion vs. transport are excluded, but you can study the effect of nonlinear terms.

A classic example for this related to Burgers' Equation (which itself is a model-problem for more complex PDEs) which reads

$$ \partial_t u + \frac12 \partial_x \big(u^2\big) = \partial_t u + u \partial_x u= 0. \tag{1} \label{1}$$

Considering the Riemann Problem

$$ u(t = t_0, x) = u_0(x) = \begin{cases} u_l & x < 0 \\ u_r & x > 0 \end{cases} \tag{2} \label{2} $$ it is well-known that the solution $u(t,x) = u\big(t, x(t) \big)$ of \eqref{1} stays constant along the characteristic curves $$ x(t) = u_0(x_0)t + x_0. \tag{3} \label{3}$$

Define now $v := \partial_x u$ and differentiate \eqref{1} with respect to $x$ (assuming existence of partial derivatives): \begin{align} 0&=\partial_t u_x + u_x u_x + u u_{xx} \tag{4} \\ \Leftrightarrow 0&=\partial_t v + v^2 + u v_x \tag{5} \label{5} \end{align} Along characteristics \eqref{3} $u$ and thus $v$ are only functions of time $t$ and thus $$v_x\big(t, x(t) \big) = v_x\big(t, u_0(x_0)t + x_0 \big) = 0. \tag{6}$$ Consequently, \eqref{5} simplifies to $$v'(t) = - v^2(t) \tag{7} $$ with solution $$v(t) = \frac{1}{t+c} \tag{8}$$ which can, depending on the initial condition, blow up in finite time. For instance, for $v(t_0 = 0) = v_0 = -1$, the solution is given by $$v(t) = \frac{1}{t-1}. \tag{9}$$

As a consequence, the spatial derivative $\partial_x u(t) = v(t)$ can blow up in finite time, even if the initial spatial derivative is small. This is related to the fact that Burgers equation develops a discontinuity (i.e., infinite spatial derivatives) at time $ t_\text{Blow up} $ if there exists $x$ such that $u_0'(x) < 0$. This time can be computed as $$t_\text{Blow up} = \frac{-1}{\min_x u_0'(x) }. \tag{10}$$


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