M371 - Alexiades
Problem Set 9
(do by hand, on a calculator)
Consider the IVPs:
(A) y'+2y = 1, 0<t<1 , y(0)=2.
(B) y' = y(1-y), 0<t<1 , y(0)=1/2.
1. For each one, do the following:
a. Find the exact solution y(t) and evaluate it at t=1.
b. Apply Euler's method with Δt=1/4 to find Y4 ≈ y(1).
Make a table of tn, Yn for n=0,1,2,3,4.
c. Find the error at t=1.
2. Euler's method is obtained by approximating y'(tn) by a forward finite difference.
Use the backward difference approximation to y'(tn+1) to derive the
Backward Euler Method: Yn+1 = Yn + Δt f(tn+1, Yn+1) , n=0,1,2,...
Note that now the unknown Yn+1 appears inside f(.,.), so this equation needs to be
solved for Yn+1 at each time-step!!! whence it is also called Implicit Euler Method.
For the simple ODEs (A), (B) above, the updating equation can be solved by hand,
but in general a root-finder (like Newton-Raphson) is needed.
This scheme is also 1st order, but it has better stability properties than Explicit Euler.
For each IVP problem (A), (B), do the following:
a. Apply the Backward Euler Method with Δt=1/4 to find Y4 ≈ y(1).
b. Find the error at t=1 and compare with Explicit Euler.
3. Use the centered difference approximation to y'(tn) to derive the so called
Midpoint Method: Yn+1 = Yn-1 + 2 Δt f(tn, Yn) , n=1,2,...
Note that this requires both Yn-1 and Yn to produce Yn+1. It is a 2-step method,
hence not self-starting (need Y0 and Y1 before it can be applied),
so some single-step method (like Euler) must be used to start it off.
However, it has the advantage of being a 2nd order method, and explicit.
For each IVP problem (A), (B), do the following:
a. Apply the Midpoint Method with Δt=1/4 to find Y4 ≈ y(1).
b. Find the error at t=1 and compare with Explicit Euler and with Implicit Euler.
Which method seems to be doing better in this case ?