Stability Analysis of Explicit and Implicit Euler Methods for a First Order ODE, Assignments of Mathematics

An analysis of the stability of explicit and implicit euler methods for solving the initial value problem of a first order ordinary differential equation (ode) y′ = −y + 1. How to find the region of absolute stability for each method, and discusses the behavior of the solutions when the step size h is equal to 0.5, 1.5, and 2.1.

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Pre 2010

Uploaded on 08/31/2009

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Math 151B Homework #3 due Friday 1/30, in class
1. stability
Consider the intial value problem
y0=y+1
t,1t100, y(1) = 1
a. Find the region of absolute stability for explicit Euler’s method.
Answer:
Explicit Euler’s method uses the discrete difference equation
yi+1 =yi+hf(ti, yi)
=yi+h(yi+ 1/ti)
= (1 h)yi+h/ti
If we introduce an error of δiinto the computation of yi, we get
yi+1 +δi+1 = (1 h)(yi+δi) + h/ti
Subract the original equation from this and get
δi+1 = (1 h)δi
As long as h2, Euler’s method is absolutely stable.
b. What do you expect to happen when h= 0.5?
Answer:
If h= 0.5, then (1 h) is positive and less than 1. So the solution should be stable and
free from overshoot.
c. What do you expect to happen when h= 1.5?
Answer:
If h= 1.5, then (1 h) is negative and greater than 1. So the solution will be absolutely
stable, but with overshoot problems.
d. What do you expect to happen when h= 2.1?
Answer:
If h > 2 then the solution will be unstable, and numerical errors will grow exponentially.
e. Confirm the above numerically/graphically.
Answer:
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Math 151B Homework #3 – due Friday 1/30, in class

  1. stability

Consider the intial value problem

y′^ = −y +

t

, 1 ≤ t ≤ 100 , y(1) = 1

a. Find the region of absolute stability for explicit Euler’s method. Answer: Explicit Euler’s method uses the discrete difference equation

yi+1 = yi + hf (ti, yi) = yi + h(−yi + 1/ti) = (1 − h)yi + h/ti

If we introduce an error of δi into the computation of yi, we get

yi+1 + δi+1 = (1 − h)(yi + δi) + h/ti

Subract the original equation from this and get

δi+1 = (1 − h)δi

As long as h ≤ 2, Euler’s method is absolutely stable. b. What do you expect to happen when h = 0.5? Answer: If h = 0.5, then (1 − h) is positive and less than 1. So the solution should be stable and free from overshoot.

c. What do you expect to happen when h = 1.5? Answer: If h = 1.5, then (1 − h) is negative and greater than −1. So the solution will be absolutely stable, but with overshoot problems.

d. What do you expect to happen when h = 2.1? Answer: If h > 2 then the solution will be unstable, and numerical errors will grow exponentially. e. Confirm the above numerically/graphically. Answer:

0

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Exact Euler 0. Euler 1.

When h = 0.5, we get a very good approximation. When h = 1.5, we have some overshoot problems, but the absolute magnitude decays in time. Finally, when h = 2.1, the errors grow without bound.

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    • Implicit Euler 1. Exact
    • Implicit Euler 2.