Computer Algebra Recipes

102 CHAPTER 2. PHASE-PLANE ANALYSIS
Using the BDA for the time derivative to connect time step n to the previous
step n ¡ 1 and approximating the rhs with its value at step n, thebackward
Euler approximation to equations (2.19) is
xn ¡ xn¡1
h
= Xn; yn ¡ yn¡1
h
= Yn;
zn ¡ zn¡1
h
= Zn;
or letting n ! n + 1 and substituting the forms of X, Y ,andZ,
xn+1 = xn + h (a ¡ b) xn+1 ¡ hgNn+1 xn+1 ¡ h¯xn+1 zn+1
yn+1 = yn + h¯xn+1 zn+1 ¡ h (¾ + b) yn+1 ¡ hgNn+1 yn+1
zn+1 = zn + h¾yn+1 ¡ h (® + b) zn+1 ¡ hgNn+1 zn+1
(2.20)
with Nn+1 = xn+1 + yn+1 + zn+1.
This algorithm is nonlinear in the unknowns xn+1, yn+1, andzn+1, andis
an example of an implicit algorithm because one cannot explicitly express the
unknowns in terms of the known values xn, yn, andzn. The standard approach
is to make the scheme semi-implicit by linearizing it as follows. For a nonlinear
function f(xn+1; zn+1), we Taylor expand thus:
μ
μ
@f
@f
f(xn+1;zn+1) =f(xn;zn)+(xn+1¡xn)
+(zn+1¡zn)
+¢¢¢
@x
@z
xn;zn
xn;zn
For our fox-rabies example, this gives us, for example,
f ´ (xz) n+1 = xn zn + zn(xn+1 ¡ xn)+xn(zn+1 ¡ zn): (2.21)
Applying this procedure to all of the quadratic terms on the rhs of (2.20) and
gathering all the \new" values xn+1, etc.,onthelhs,weobtain
with
A1n xn+1 + B1n yn+1 + C1n zn+1 = xn + D1n;
A2n xn+1 + B2n yn+1 + C2n zn+1 = yn + D2n;
A3n xn+1 + B3n yn+1 + C3n zn+1 = zn + D3n;
(2.22)
A1n =1+h (b ¡ a)+hg(Nn + xn)+h¯zn; B1n =hgxn; C1n =h (¯ + g) xn;
A2n =hgyn ¡ h¯zn; B2n =1+h (¾ + b)+hg(Nn + yn); C2n =hgyn ¡ h¯xn;
A3n =hgzn; B3n =hgzn ¡ h¾; C3n =1+h (® + b)+hg(Nn + zn);
D1n =hgNn xn + h¯xn zn; D2n =hgNn yn ¡ h¯xn zn; D3n =hgNn zn:
We now have three linear equations (2.22), with known numerical coe±cients
determined from the previous step, for the three unknowns xn+1, yn+1, zn+1.