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> [!proposition] Proposition. ([[SIR model on random 4-regular graph]])
> ~
Consider the SIR model on a [[configuration model]] [[network]] where all nodes have [[degree]] four (also known as a random 4-[[regular graph]]).
**a. What is the critical value $\phi_{c}$ of the transmission probability at the epidemic threshold?**
$\frac{\langle k \rangle}{\langle k^{2} \rangle - \langle k \rangle}=\frac{1}{3}$.
**b. When the transmission probability is $\phi=\frac{1}{2}$ , show that if an epidemic happens, the fraction $S$ of nodes infected is $S=\frac{3\sqrt{ 5 }-5}{2}$.**
Let $\phi=\frac{1}{2}$. To say 'an epidemic happens' is to say that a [[giant cluster]] forms following the bond percolation process. In this case $S=1-g_{0}(u)$. $g_{0}(u)$ is just $u^{4}$. So $S=1-u^{4}$, where $u$ is the probability that a given node is not connected to the [[giant cluster]] via a particular one of its edges and is seen in the equation $u=1-\phi + \phi g_{1}(u)$. Now, $g_{1}(u)=\frac{1}{4}u^{3}$. (see [[configuration model]] properties). Hence $u=1-\phi + \frac{\phi}{4}u^{3}$. After factoring out $u-1$ via [[synthetic division]], we get the equivalent equation $(u-1)(\phi u ^{2} + \phi u + \phi - 1)=0.$
Divide both sides by $u-1$,, then quadratic formula yields $u=-\frac{1}{2} \pm \frac{\sqrt{ 4 \phi - 3 \phi^{2} }}{2\phi}.$
Discarding the $u<0$ option we are left with $S= 1- \frac{1}{16}(\frac{\sqrt{ 4\phi - 3 \phi^{2} }}{\phi}-1)^{4}.$
Now let $\phi=\frac{1}{2}$ and the desired result follows.
> [!proof]- Proof. ([[SIR model on random 4-regular graph]])
> ~
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#### References
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