Lecture Notes in Differential Equations - Bruce E. Shapiro

364 LESSON 33. NUMERICAL METHODS
The Backwards Euler Method
Now consider the IVP
y ′ = −5ty 2 + 5 t − 1 , y(1) = 1 (33.35)
t2 The exact solution is y = 1/t. The numerical solution is plotted for three
different step sizes on the interval [1, 25] in the following figure. Clearly
something appears to be happening here around h = 0.2, but what is it?
For smaller step sizes, a relatively smooth solution is obtained, and for
larger values of h the solution becomes progressively more jagged (figure
??).
Figure 33.2: Solutions for equation 33.35 for various step sizes using Euler’s
method.
0.4
y
0.3
0.2
0.1
5 10 15 20 25
t
This example illustrates a problem that occurs in the solution of differential
equations, known as stiffness. Stiffness occurs when the numerical method
becomes unstable. One solution is to modify Euler’s method as illustrated
in figure 33.3 to give the Backward’s Euler Method:
y n = y n−1 + h n f(t n , y n ) (33.36)
The problem with the Backward’s Euler method is that we need to know the
answer to compute the solution: y n exists on both sides of the equation, and
in general, we can not solve explicitly for it. The Backwards Euler Method
is an example of an implicit method, because it contains y n implicitly.
In general it is not possible to solve for y n explicitly as a function of y n−1
in equation 33.36, even though it is sometimes possible to do so for specific
differential equations. Thus at each mesh point one needs to make some
first guess to the value of y n and then perform some additional refinement