Consider a biological system described by the nonlinear multiple switch model

$S ≔\mathrm{Vector}\left(\left[\stackrel{\.}{x}\left(t\right)\=-x\left(t\right)\+\frac{s}{1\+y{\left(t\right)}^2}\,\stackrel{\.}{y}\left(t\right)\=-y\left(t\right)\+\frac{s}{1\+x{\left(t\right)}^2}\right]\right)$

where the unknowns $x\,y$ represent two protein concentrations and $s$ represents the strength of unrepressed protein expression. This latter quantity is regarded as a time-constant parameter.

The equilibria of $S$ correspond to $\stackrel{\.}{x}\=\stackrel{\.}{y}\=0$, or equivalently, $F\=0$, where

$f≔\genfrac{}{}{0ex}{}{\mathrm{rhs}~\left(S\right)}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{\left[x\left(t\right)\=x\,y\left(t\right)\=y\right]}$

The following two Hurwitz determinants determine the stability of the hyperbolic equilibria of $S$:

$d_1≔-\left(\frac{\partial }{\partial x}{f}_1\+\frac{\partial }{\partial y}{f}_2\right)$

$d_2≔\left(\frac{\partial }{\partial x}{f}_1\right)\cdot \left(\frac{\partial }{\partial y}{f}_2\right)-\left(\frac{\partial }{\partial y}{f}_1\right)\cdot \left(\frac{\partial }{\partial x}{f}_2\right)$

The semi-algebraic system below encodes the asymptotically stable hyperbolic equilibria:

$$\begin{gathered} P≔\left[\mathrm{numer}\left(f_1\right)\=0\,\mathrm{numer}\left(f_2\right)\=0\,x\>0\,y\>0\,s\>0\,\mathrm{numer}\left(d_2\right)\>0\right]\: \\ \mathrm{print}~\left(P\right)\: \end{gathered}$$

We solve this problem by first computing a real comprehensive triangular decomposition of $P$ with respect to the parameter $s$, using the new command RealComprehensiveTriangularize:

$R≔\mathrm{PolynomialRing}\left(\left[y\,x\,s\right]\right)\:$

$C≔ \mathrm{RealComprehensiveTriangularize}\left(P\,1\,R\right)$

This is a decomposition of the original system into several simpler triangular systems and some additional conditions on the parameter $s$:

$\mathrm{Display}\left(C_1\,R\right)$

$\mathrm{Display}\left(C_2\,R\right)$

Suppose we are interested in those values of $s$ for which the biological system is bistable (that is, there are at least two stable equilibria). This means that we are looking for values of $s$ such that $P$ has at least 2 positive real solutions. We use the RealComprehensiveTriangularize command again and apply it to our previous result $C$:

$\mathrm{result}≔\mathrm{RealComprehensiveTriangularize}\left(C\, R\, 2\right)$

The condition on $s$ that we are looking for is given by the second entry:

$\mathrm{Display}\left({\mathrm{result}}_2\,R\right)$

The locations of the stable equilibria are described by the following triangular system from the first entry:

$\mathrm{Display}\left({\mathrm{result}}_1\,R\right)$

We now illustrate the result by discussing the special case $s\=3$. From the equations above, there are $2$ stable equilibria given by:

$\mathrm{equilibria}≔\left[\mathrm{allvalues}\left(\mathrm{solve}\left(\genfrac{}{}{0ex}{}{\left[x y-1\=0\, x^2-3 x\+1\=0\right]}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{s\=3}\right)\right)\right]$

$\mathrm{evalf}\left(\mathrm{equilibria}\right)$

We verify that the last inequality is satisfied:

$\mathrm{condition}≔\genfrac{}{}{0ex}{}{8 x s^3-6 x s^5-4 s^2\+5 s^4-s^6\+x s^7\>0}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{s\=3}$

$\mathrm{evalf}\left(\genfrac{}{}{0ex}{}{\mathrm{condition}}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{{\mathrm{equilibria}}_1}\right)$

$\mathrm{evalf}\left(\genfrac{}{}{0ex}{}{\mathrm{condition}}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{{\mathrm{equilibria}}_2}\right)$

Below, we graphically display trajectories of the dynamical system in the special case. The first plot is a 2D animation, and the second one is a 3D static plot with time as the z-axis. We can see the two stable equilibria well in both plots, but there is also a third, unstable, equilibrium that can be seen best in the 3D plot.

$\mathrm{special}≔\genfrac{}{}{0ex}{}{\mathrm{convert}\left(S\,\mathrm{list}\right)}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{s\=3}$

$\mathrm{with}\left(\mathrm{DEtools}\right)\:$

$$\begin{gathered} {\mathrm{IC}}_2 ≔ \left[\left[x\left(0\right)\=0.5\,y\left(0\right)\=0\right]\,\mathrm{seq}\left(\left[x\left(0\right)\=i\,y\left(0\right)\=0\right]\,i\=1..5\right)\, \\ \left[x\left(0\right)\=0\,y\left(0\right)\=0.5\right]\,\mathrm{seq}\left(\left[x\left(0\right)\=0\,y\left(0\right)\=i\right]\,i\=1..5\right)\, \\ \left[x\left(0\right)\=4.5\,y\left(0\right)\=5\right]\,\mathrm{seq}\left(\left[x\left(0\right)\=i\,y\left(0\right)\=5\right]\,i\=1..4\right)\, \\ \left[x\left(0\right)\=5\,y\left(0\right)\=4.5\right]\,\mathrm{seq}\left(\left[x\left(0\right)\=5\,y\left(0\right)\=i\right]\,i\=1..4\right)\, \\ \mathrm{seq}\left(\left[x\left(0\right)\=i\,y\left(0\right)\=i\right]\,i\=0..5\right)\right]\: \end{gathered}$$

$\mathrm{DEplot}\left(\mathrm{special}\, \left[x\left(t\right)\, y\left(t\right)\right]\, t \= 0.. 20\, x \= 0.. 5\, y \= 0.. 5\, {\mathrm{IC}}_2\, \mathrm{arrows}\=\mathrm{none}\, \mathrm{animate}\=\mathrm{true}\, \mathrm{numframes}\=40\, \mathrm{linecolor}\=\sqrt{t}\right)$

$$\begin{gathered} {\mathrm{IC}}_3 ≔ \left[ \\ \mathrm{seq}\left(\left[x\left(0\right)\=0.1\, y\left(0\right)\=0.04 j\right]\, j\=0..25\right)\, \\ \mathrm{seq}\left(\left[x\left(0\right)\=0.1\, y\left(0\right)\=1\+0.04 j\right]\, j\=0..25\right)\, \\ \mathrm{seq}\left(\left[x\left(0\right)\=0.1\, y\left(0\right)\=4\+0.1 j\right]\, j\=0..10\right)\, \\ \mathrm{seq}\left(\left[x\left(0\right)\=0.1\, y\left(0\right)\=3\+0.05 j\right]\, j\=0..20\right)\, \\ \mathrm{seq}\left(\left[x\left(0\right)\=0.1\, y\left(0\right)\=2\+0.05 j\right]\, j\=0..20\right)\, \\ \mathrm{seq}\left(\left[y\left(0\right)\=0.1\, x\left(0\right)\=0.04 j\right]\, j\=0..25\right)\, \\ \mathrm{seq}\left(\left[y\left(0\right)\=0.1\, x\left(0\right)\=1\+0.04 j\right]\, j\=0..25\right)\, \\ \mathrm{seq}\left(\left[y\left(0\right)\=0.1\, x\left(0\right)\=4\+0.1 j\right]\, j\=0..10\right)\, \\ \mathrm{seq}\left(\left[y\left(0\right)\=0.1\, x\left(0\right)\=3\+0.05 j\right]\, j\=0..20\right)\, \\ \mathrm{seq}\left(\left[y\left(0\right)\=0.1\, x\left(0\right)\=2\+0.05 j\right]\, j\=0..20\right)\, \\ \mathrm{seq}\left(\left[x\left(0\right)\=0.2 j\, y\left(0\right)\=0.2 j\right]\, j\=0..25\right) \\ \right]\: \end{gathered}$$

    $\mathrm{DEplot3d}\left(\mathrm{special}\, \left\{x\left(t\right)\,y\left(t\right)\right\}\, t \= 0.. 50\, x\=0.. 5\, y\=0.. 5\, {\mathrm{IC}}_3\, \mathrm{scene} \= \left[x\left(t\right)\,y\left(t\right)\,t\right]\, \mathrm{thickness}\=1\, \mathrm{linecolor}\= \sqrt{t}\, \mathrm{axes} \= \mathrm{frame}\right)$

Let $z$ be a complex variable. Which of the following two identities holds for all $z \in \mathrm{\ℂ}$?

  $\sqrt{z^2-1}\=\sqrt{z-1}\cdot \sqrt{z\+1}$

$\sqrt{1-z^2}\=\sqrt{1-z}\cdot \sqrt{1\+z}$

In Maple, the branch cut for the square root function $\sqrt{z}$ is the negative real axis $z<0$. If we rewrite $z\&equals;x\&plus;\mathrm{i} y$ in Cartesian coordinates, we can express that condition by the simple semi-algebraic system $\mathrm{\&Im;}\left(z\right)\&equals; y\&equals;0\&comma;\mathrm{\&real;}\left(z\right)\&equals; x<0$. For example (2.1) above, the branch cuts of the three square root functions are, from left to right:

$\mathrm{assume}\left(x\&comma;\mathrm{real}\right)$

$\mathrm{assume}\left(y\&comma;\mathrm{real}\right)$

$\left[\mathrm{\&Im;}\left({\left(x\&plus;\mathrm{i} y\right)}^2-1\right)\&equals;0\&comma;\mathrm{\&real;}\left({\left(x\&plus;\mathrm{i} y\right)}^2-1\right)<0\right]$

$\left[\mathrm{\&Im;}\left(x\&plus;\mathrm{i} y-1\right)\&equals;0\&comma; \mathrm{\&real;}\left(x\&plus;\mathrm{i} y-1\right)<0\right]$

$\left[\mathrm{\&Im;}\left(x\&plus;\mathrm{i} y\&plus;1\right)\&equals;0\&comma; \mathrm{\&real;}\left(x\&plus;\mathrm{i} y\&plus;1\right)<0\right]$

The following plot depicts these three branch cuts.

$\mathrm{with}\left(\mathrm{plottools}\right)\&colon;$

$\mathrm{with}\left(\mathrm{plots}\right)\&colon;$

$$\begin{gathered} \mathrm{branchcuts}≔\mathrm{display}\left(\mathrm{line}\left(\left[0\&comma;-10\right]\&comma;\left[0\&comma;10\right]\&comma;\mathrm{color}\&equals;\mathrm{red}\right)\&comma; \mathrm{line}\left(\left[-1\&comma;-0.12\right]\&comma;\left[1 \&comma;-0.12\right]\&comma;\mathrm{color}\&equals;\mathrm{red}\right)\&comma; \\ \mathrm{line}\left(\left[-10\&comma;0.05\right]\&comma;\left[1\&comma;0.05\right]\&comma;\mathrm{color}\&equals;\mathrm{green}\right)\&comma; \\ \mathrm{line}\left(\left[-10\&comma;-0.12\right]\&comma;\left[-1\&comma;-0.12\right]\&comma;\mathrm{color}\&equals;\mathrm{blue}\right)\&comma; \\ \mathrm{thickness}\&equals;4\&comma; \mathrm{view}\&equals;\left[-5..5\&comma;-5..5\right]\right)\&colon; \end{gathered}$$

$\mathrm{branchcuts}$

By continuity arguments, it is now sufficient to check the proposed identity (2.1) at finitely many points covering all possible combinations of branch cuts. Such a set of points can be computed using the command CylindricalAlgebraicDecompose:

$b_1≔\left[\&comma;\&comma;\right]$

$R≔\mathrm{PolynomialRing}\left(\left[y\&comma;x\right]\right)\&colon;$

$S≔\mathrm{CylindricalAlgebraicDecompose}\left(b_1\&comma;R\&comma;\mathrm{output}\&equals;\mathrm{rootof}\right)$

The union of all branch cuts was decomposed into the 7 disjoint intervals above. We pick one sample point for each interval:

$S≔\left[\left[x\&equals;-2\&comma;y\&equals;0\right]\&comma;S_2\&comma;\left[x\&equals;-\frac{1}{2}\&comma;y\&equals;0\right]\&comma;\left[x\&equals;0\&comma;y\&equals;-1\right]\&comma;S_5\&comma;\left[x\&equals;0\&comma;y\&equals;1\right]\&comma;\left[x\&equals;\frac{1}{2}\&comma;y\&equals;0\right]\right]$

We also add one sample point $\left(1\&comma;1\right)$ not contained on any branch cut:

$S≔\left[S\left[\right]\&comma;\left[x\&equals;1\&comma;y\&equals;1\right]\right]$

The following graph displays the sample points in relation to the branch cuts.

$\mathrm{display}\left(\mathrm{branchcuts}\&comma;\mathrm{seq}\left(\mathrm{point}\left(\mathrm{rhs}~\left(P\right)\right)\&comma;P \in S\right)\&comma;\mathrm{symbol}\&equals;\mathrm{solidcircle}\&comma;\mathrm{symbolsize}\&equals;20\right)$

Now we check whether the identity (2.1) holds for all of these sample points:

$f_1≔\genfrac{}{}{0ex}{}{\left(\mathrm{lhs}-\mathrm{rhs}\right)\left(\right)}{\phantom{x\&equals;a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{z\&equals;x\&plus;\mathrm{i} y}$

$\mathrm{seq}\left(\mathrm{radnormal}\left(\mathrm{eval}\left(f_1\&comma;P\right)\right)\&comma;P \in S\right)$

We conclude that (2.1) is not an identity, and we can even say precisely when it fails: on the first interval $x<-1 \mathbf{and} y\&equals;0$, and on the fourth interval $x\&equals;0 \mathbf{and} y<0$.

Now we proceed in the same way for the second proposed identity (2.2). The branch cuts are:

$\left[\mathrm{\&Im;}\left(1-{\left(x\&plus;\mathrm{i} y\right)}^2\right)\&equals;0\&comma;\mathrm{\&real;}\left(1-{\left(x\&plus;\mathrm{i} y\right)}^2\right)<0\right]$

$\left[\mathrm{\&Im;}\left(1-\left( x\&plus;\mathrm{i} y\right)\right)\&equals;0\&comma; \mathrm{\&real;}\left(1-\left( x\&plus;\mathrm{i} y\right)\right)<0\right]$

$\left[\mathrm{\&Im;}\left(1\&plus; x\&plus;\mathrm{i} y\right)\&equals;0\&comma; \mathrm{\&real;}\left(1\&plus; x\&plus;\mathrm{i} y\right)<0\right]$

$$\begin{gathered} \mathrm{branchcuts}≔\mathrm{display}\left(\mathrm{line}\left(\left[-10\&comma;0\right]\&comma;\left[-1\&comma;0\right]\&comma;\mathrm{color}\&equals;\mathrm{red}\right)\&comma; \mathrm{line}\left(\left[1\&comma;0\right]\&comma;\left[10 \&comma;0\right]\&comma;\mathrm{color}\&equals;\mathrm{red}\right)\&comma; \\ \mathrm{line}\left(\left[1\&comma;0.12\right]\&comma;\left[10\&comma;0.12\right]\&comma;\mathrm{color}\&equals;\mathrm{green}\right)\&comma; \\ \mathrm{line}\left(\left[-10\&comma;-0.12\right]\&comma;\left[-1\&comma;-0.12\right]\&comma;\mathrm{color}\&equals;\mathrm{blue}\right)\&comma; \\ \mathrm{thickness}\&equals;4\&comma; \mathrm{view}\&equals;\left[-5..5\&comma;-5..5\right]\right)\&colon; \end{gathered}$$

$\mathrm{branchcuts}$

$b_2≔\left[\&comma;\&comma;\right]$

$S≔\mathrm{CylindricalAlgebraicDecompose}\left(b_2\&comma;R\&comma;\mathrm{output}\&equals;\mathrm{rootof}\right)$

We need only 2+1 sample points in this case:

$S ≔ \left[\left[x\&equals;-2\&comma;y\&equals;0\right]\&comma;\left[x\&equals;0\&comma;y\&equals;0\right]\&comma;\left[x\&equals;2\&comma;y\&equals;0\right]\right]$

$\mathrm{display}\left(\mathrm{branchcuts}\&comma;\mathrm{seq}\left(\mathrm{point}\left(\mathrm{rhs}~\left(P\right)\right)\&comma;P \in S\right)\&comma;\mathrm{symbol}\&equals;\mathrm{solidcircle}\&comma;\mathrm{symbolsize}\&equals;20\right)$

$f_2≔\genfrac{}{}{0ex}{}{\left(\mathrm{lhs}-\mathrm{rhs}\right)\left(\right)}{\phantom{x\&equals;a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{z\&equals;x\&plus;\mathrm{i} y}$

$\mathrm{seq}\left(\mathrm{radnormal}\left(\mathrm{eval}\left(f_2\&comma;P\right)\right)\&comma;P \in S\right)$

Thus we have proved that (2.2) is indeed an identity.