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Let $F:\mathbb{R} \to \mathbb R$ be a bounded Lipschitz function and $G(x,y) = (0,\chi_{\{x \le F(y)\}})$.

Consider the ODE $$ \begin{cases} \partial_t \Phi(t,x) = G(\Phi), & t \in [0,T]\\ \Phi(0,x) = x & x \in \mathbb R^2 \end{cases} $$

  • How can we write the solution $\Phi$ explicitly?
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    $\begingroup$ BTW, this is a PDE, not an ODE. Please fix it in the question.. $\endgroup$
    – user64494
    Apr 25, 2019 at 4:06
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    $\begingroup$ Also $G$ has two arguments, but you wrote $G(\Phi)$. $\endgroup$
    – user64494
    Apr 25, 2019 at 4:20
  • $\begingroup$ @user64494 $\Phi:\mathbb R^2 \to \mathbb R^2$. We use $\Phi_1$ as $x$ and $\Phi_2$ as $y$. $\endgroup$
    – Riku
    Apr 25, 2019 at 11:48
  • $\begingroup$ this still makes no sense; if $\Phi=(\Phi_1,\Phi_2)\in\mathbb{R}^2$, then $\Phi_2\in\mathbb{R}$; but in the definition of $G(x,y)$ you use $F(y)$ and if as you say $y$ should be replaced by $\Phi_2$, then you wish to evaluate the function $F$ at a point in $\mathbb{R}$, however, you have defined $F:\mathbb{R}^2\rightarrow\mathbb{R}$. $\endgroup$
    – Carlo Beenakker
    Apr 25, 2019 at 11:54
  • $\begingroup$ @CarloBeenakker That is a typo. Thank you. I'll edit it. $\endgroup$
    – Riku
    Apr 25, 2019 at 11:55

2 Answers 2

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OK, from the comments I understand that the problem boils down to

$$\frac{\partial}{\partial t}\Phi_2(t,x_1,x_2)=\theta\bigl(F[\Phi_2(t,x_1,x_2)]-x_1\bigr),\;\;\Phi_2(0,x_1,x_2)=x_2,$$

where $\theta$ is the unit step function. Let me assume that $F$ is a nondecreasing function. Then the solution is

$$\Phi_2(t,x_1,x_2)=\begin{cases} t+x_2&\text{if}\;\;F(x_2)>x_1\\ x_2&\text{if}\;\;F(x_2)\leq x_1. \end{cases} $$

A more general choice of $F$ can be readily accommodated, by piecing together increasing and decreasing segments. For any $F$, the function $\Phi_2(t,x_1,x_2)=x_2$ whenever $F(x_2)\leq x_1$, so we only need to consider regions in which $F(x_2)>x_1$ and $\Phi_2$ increases linearly with $t$ until $F(\Phi_2)$ becomes smaller than $x_1$.

as requested, the Mathematica code

Manipulate[Module[{sol = NDSolve[{phi2'[t] == UnitStep[Sin[phi2[t]] - x1], phi2[0] == x2}, phi2, {t, 0, tfinal}]}, Plot[Evaluate[phi2[t] /. sol], {t, 0, tfinal}]], {x1, 0, 2}, {x2, 0, 2}, {tfinal, 1, 10}]

will produce a plot of $\Phi_2(t)$ (for $F=\sin$) where you can vary $t_{\rm final}$, $x_1$, and $x_2$.

alternatively, the Mathematica code

sol = NDSolve[{D[phi2[t, x1, x2], t] == UnitStep[Sin[phi2[t, x1, x2]] - x1], phi2[0, x1, x2] == x2}, phi2, {t, 0, 5}, {x1, 0, 2}, {x2, 0, 2}]; Manipulate[VectorPlot[{x1, Evaluate[phi2[t, x1, x2] /. sol]}, {x1, 0, 2}, {x2, 0, 2}], {t, 0, 5}] Manipulate[StreamPlot[{x1, Evaluate[phi2[t, x1, x2] /. sol]}, {x1, 0, 2}, {x2, 0, 2}], {t, 0, 5}]

will produce a vector plot or a stream plot of $(\Phi_1,\Phi_2)$ in the $x$-$y$ plane, with $t$ as a parameter that you can vary.

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A remark, too long for a comment. Let use the notation $H=\mathbf 1_{\mathbb R_+}$. You have to deal with an ODE whose flux is $$ a(x,y)=H(F(y)-x), $$ where $F$ is a Lipschitz-continuous function, $x$ is a parameter. We have also formally $$ \frac{\partial a}{\partial y}=\delta_0(F(y)-x)F'(y), $$ and assuming as in the previous answer that $F$ is monotone (say increasing), the rhs qualifies as the line-measure on the manifold with equation $F(y)=x$. As a result the flux is $BV$. Although there is a good theory for flow of $BV$ vector fields (say with bounded divergence), the second picture in the previous answer seems to indicate a non-uniqueness phenomenon, maybe related to the fact that the divergence of the vector field is unbounded.

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