Euler’s Method: Question 2-9.5 & Solution

Question 5

A normalized population satisfies $p'=2p(1-p),\qquad 0\le p(0)\le 1.$ The exact solution preserves the physical interval $[0,1]$ . For explicit Euler with step $h>0$ , write $P_{n+1}=F_h(P_n)$ . For the local discrete comparison, a small perturbation near $P=1$ is multiplied to first order by $F_h'(1)$ ; call the linearized update contracting when $|F_h'(1)|<1$ .

Tasks

Derive $F_h$ and find exactly which positive steps map all of $[0,1]$ back into $[0,1]$ .
Prove the condition is necessary, not merely sufficient, by producing an overshooting initial value whenever it fails.
Compute the first two Euler updates for $h=1$ , $P_0=3/4$ , and explain why they misrepresent the exact solution’s qualitative behavior.
Find the step range for linearized contraction at $1$ . Compare it with interval preservation, and sketch the update maps for $h=1/4$ and $h=1$ .

Original worksheet page 1: question and worked solution for 2-9-005

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Question 5 – Solution

Strategy. Check the whole physical interval, not only the behavior of tiny perturbations near its upper endpoint.

Step 1: Factor the interval constraint. Euler gives $F_h(p)=p+2hp(1-p)$ . On $[0,1]$ it is nonnegative. Also $1-F_h(p)=(1-p)(1-2hp).$ If $0<h\le 1/2$ , both factors are nonnegative for every $p\in[0,1]$ , so $F_h(p)\le 1$ . Therefore $\boxed{0<h\le 1/2}$ preserves the interval at every step by induction.

Step 2: Prove necessity. If $h>1/2$ , choose $p$ strictly between $1/(2h)$ and $1$ . Then $1-p>0$ but $1-2hp<0$ , so $F_h(p)>1$ . Such a value exists for every $h>1/2$ , proving the stated condition is exact.

Step 3: Exhibit overshoot. For $h=1$ , $F_1(p)=3p-2p^2$ , giving $\boxed{P_1=9/8=1.125,\qquad P_2=27/32=0.84375}.$ The numerical solution crosses above the capacity and then drops. The exact solution starting at $3/4$ increases toward $1$ from below and cannot cross that equilibrium, by uniqueness.

See the diagram in the original worksheet below.

Step 4: Distinguish two step restrictions. Since $F_h'(p)=1+2h-4hp$ , $F_h'(1)=1-2h$ . Linearized contraction therefore holds for $\boxed{0<h<1}$ , a larger range than interval preservation. At $h=1$ the linearized magnitude equals one, so the strict contraction test makes no decision about nonlinear asymptotic behavior.

A locally contracting update can still move some valid states above the physical bound. The map criterion protects the entire interval; the derivative criterion only describes first-order behavior near one fixed point.

Original worksheet page 2: question and worked solution for 2-9-005

Euler’s Method — Question 5

Question 5

Question 5 – Solution