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For $s>0$, the SIR model equations have the first integral of an ordinary differential equationHere $\ln$ is the natural logarithm. A short proof outline is to use the chain rule and substitute the SIR model equations:The SIR first integral gives $i$ directly from $s$, so it describes the curve in a phase portrait even before its time parametrization is known.

If $i_0>0$ and $s_0>\gamma /\beta$, the infectious compartment initially grows. The SIR model equations give its unique later maximum whenBefore this SIR infection peak, $\beta s-\gamma >0$; afterward it is negative because the susceptible compartment keeps shrinking. Substituting $s_{\ast }$ into the SIR first integral givesFor the stipulated initial value problem,Thus about $30.38\%$ of the modelled population belongs to the infectious compartment simultaneously at the SIR infection peak. This is a peak of the infectious compartment, not of incidence, whose flow is $\beta si$. If $s_0\le \gamma /\beta$, the maximum is already $i_0$ at the start and the interior SIR infection peak formula does not apply.

The numerical solution of a differential equation puts the SIR infection peak at about day $26.63258$ with $i\approx 0.303812682386$, agreeing with the SIR first integral to within $10^{-11}$. The accompanying generator checks nonnegative fractions and conservation of population without clipping or rescaling. For steps $0.4$, $0.2$ and $0.1$ day, the largest differences between successive numerical trajectories at shared times are approximately $1.086\times 10^{-7}$ and $6.985\times 10^{-9}$. Their ratio is $15.55$, close to the factor $16$ expected when a fourth-order Runge-Kutta method's step is halved.