Computer Algebra Recipes

2.3. NUMERICAL SOLUTION OF ODES 103
The semi-implicit algorithm is now implemented in the following recipe,
taking a total of 5000 time steps of size h =0:004.
> restart: Digits:=10: total:=5000: h:=0.004:
The parameter values are entered, with g at the lower end of its range.
> a:=1: b:=0.5: beta:=79.67: g:=0.1: alpha:=73: sigma:=13:
Initially, we take x0 =4,y0 =0:4, and z0 =0:1.
> t[0]:=0: x[0]:=4: y[0]:=0.4: z[0]:=0.1:
The initial plotting point is formed, and the starting time of the do loop that
iterates the algorithm is recorded.
> pnt[0]:=[t[0],x[0],y[0]]: begin:=time():
> for n from 0 to total do
The various coe±cients are evaluated on the nth step.
> N[n]:=x[n]+y[n]+z[n];
> A1[n]:=1+h*(b-a)+h*g*(N[n]+x[n])+h*beta*z[n];
> B1[n]:=h*g*x[n];
> C1[n]:=h*(beta+g)*x[n];
> D1[n]:=h*g*N[n]*x[n]+h*beta*x[n]*z[n];
> A2[n]:=h*g*y[n]-h*beta*z[n];
> B2[n]:=1+h*(sigma+b)+h*g*(N[n]+y[n]);
> C2[n]:=h*g*y[n]-h*beta*x[n];
> D2[n]:=h*g*N[n]*y[n]-h*beta*x[n]*z[n];
> A3[n]:=h*g*z[n];
> B3[n]:=h*g*z[n]-h*sigma;
> C3[n]:=1+h*(alpha+b)+h*g*(N[n]+z[n]);
> D3[n]:=h*g*N[n]*z[n];
The linear equations for xn+1, yn+1, zn+1 are entered,
> E1:=A1[n]*x[n+1]+B1[n]*y[n+1]+C1[n]*z[n+1]=x[n]+D1[n];
> E2:=A2[n]*x[n+1]+B2[n]*y[n+1]+C2[n]*z[n+1]=y[n]+D2[n];
> E3:=A3[n]*x[n+1]+B3[n]*y[n+1]+C3[n]*z[n+1]=z[n]+D3[n];
and numerically solved.
> sol[n+1]:=(fsolve(fE1,E2,E3g,fx[n+1],y[n+1],z[n+1]g));
The solution is assigned and the values of xn+1, yn+1, andzn+1 are recorded
at time tn+1 = tn + h.
> assign(sol[n+1]);
> x[n+1],y[n+1],z[n+1];
> t[n+1]:=t[n]+h;
Finally, a plotting point is formed and the loop ended.
> pnt[n+1]:=[t[n+1],x[n+1],y[n+1]];
> end do: