Let's carefully solve the problem step by step. The given 1D dynamical system is:

$\dot{x} = r + \frac{1}{2}x - \frac{x}{1+x}$

We are tasked with finding the bifurcation points, sketching the bifurcation diagram, and drawing the phase portraits as the parameter $r$ is varied.

Step 1: Find the fixed points

Fixed points occur when $\dot{x} = 0$. Therefore, set the right-hand side equal to zero and solve for $x$:

$r + \frac{1}{2}x - \frac{x}{1+x} = 0$

To make this easier to solve, first find a common denominator for the $\frac{x}{1+x}$ term. Multiply both sides of the equation by $1 + x$ to eliminate the denominator:

$(1 + x)\left( r + \frac{1}{2}x - \frac{x}{1 + x} \right) = 0$

$r(1 + x) + \frac{1}{2}x(1 + x) - x = 0$

Now expand each term:

$r + rx + \frac{1}{2}x + \frac{1}{2}x^2 - x = 0$

Simplify the terms involving $x$:

$r + rx + \left( \frac{1}{2} - 1 \right)x + \frac{1}{2}x^2 = 0$

This simplifies further to:

$r + rx - \frac{1}{2}x + \frac{1}{2}x^2 = 0$

Rearranging:

$r + x\left( r - \frac{1}{2} \right) + \frac{1}{2}x^2 = 0$

Step 2: Analyze the nature of fixed points and bifurcations

This is a quadratic equation in $x$. We can solve for $x$ using the quadratic formula:

$x = \frac{-\left( r - \frac{1}{2} \right) \pm \sqrt{\left( r - \frac{1}{2} \right)^2 - 2r}}{1}$

Simplifying the discriminant:

$x = -\left( r - \frac{1}{2} \right) \pm \sqrt{\left( r - \frac{1}{2} \right)^2 - 2r}$

We now analyze the discriminant $\Delta = \left( r - \frac{1}{2} \right)^2 - 2r$.

Step 3: Determine bifurcation points

A bifurcation occurs when the discriminant $\Delta$ changes sign, i.e., when it becomes zero. Let's solve for when $\Delta = 0$:

$\left( r - \frac{1}{2} \right)^2 - 2r = 0$

Expanding the first term:

$r^2 - r + \frac{1}{4} - 2r = 0$

Simplifying:

$r^2 - 3r + \frac{1}{4} = 0$

Now apply the quadratic formula to solve for $r$:

$r = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(\frac{1}{4})}}{2(1)}$

$r = \frac{3 \pm \sqrt{9 - 1}}{2} = \frac{3 \pm \sqrt{8}}{2} = \frac{3 \pm 2\sqrt{2}}{2}$

Thus, the bifurcation points occur at:

$r = \frac{3 + 2\sqrt{2}}{2}, \quad r = \frac{3 - 2\sqrt{2}}{2}$

Step 4: Sketch the bifurcation diagram

The bifurcation diagram will plot the fixed points $x^*$ versus the parameter $r$. At the bifurcation points, we expect changes in the number or stability of fixed points.

Since the discriminant goes from positive to negative, a saddle-node bifurcation is expected at the critical values of $r$, where two fixed points merge and annihilate.

Step 5: Sketch the phase portraits

For different values of $r$:

• For $r$ less than the first bifurcation point: Two fixed points exist, one stable and one unstable.
• At the bifurcation points: The two fixed points merge (saddle-node bifurcation).
• For $r$ greater than the second bifurcation point: No fixed points exist, leading to divergent behavior of the system.

Final Steps

Now that we have the bifurcation points and the general structure of the bifurcation diagram, the next step is to sketch the phase portraits and bifurcation diagram based on this information.