Solving a recursive equation system with max operator

Question

I have this set of equations:

$y=1+0.5x$

$x=1+max(0.5y,0.5c)$

Where $c$ is some constant.

I am wondering what way is there to solve this.

Answer 1

As

$$\max(a,b)=\frac12(a+b+|a-b|)$$

your issue can be transformed into

$$x=\frac12(y+c+|y-c|) \iff 2x=0.5x+c+1+|1-c+0.5x|$$

$$\iff \frac32x-(c+1)=|\frac12x+1-c|$$

The solution $x$ of equation (1) can be represented in a graphical way as the abscissa of the intersection point of the line and broken line with resp. equations:

$$y=\frac32x-(c+1) \ \ \& \ \ y=|\frac12x+1-c|$$

Observing the different slopes $\frac32$ and $\pm \frac12$, they will always share a unique point of intersection, with the particularity that

• for $c \le 2$, the abscissa of this intersection point is always equal to $2$, whereas

• for $c>2$ it is, by suppression of the absolute value, $x=\frac12c+1$.

See the following Geogebra picture:

Answer 2

This is equivalent to $x=1+\max(0.5(1+0.5x),0.5c)$.

If $0.5(1+0.5x) \geq 0.5c$ then you have $x = 1+0.5(1+0.5x) \implies x = 2$ and $c\leq 2$.

If $0.5(1+0.5x) \leq 0.5c$ then you have $x = 1+0.5c$ and $c\geq 2$.

Thus for $c\leq 2$ the unique solution is $x=2$ and for $c\geq 2$ the unique solution is $1+0.5c$.