The Computable Differential Equation Lecture ... - Bruce E. Shapiro

84 CHAPTER 4. IMPROVING ON EULER’S METHOD
The iteration formula is
y (i+1)
n
= y (i)
n
= y (i)
n
− g(y(i) n )
g ′ (y n (i) )
− y(i) n
− y n−1 − hf(t, y (i)
n )
1 − hf y (t, y (i)
n )
For the vector initial value problem, the corresponding iteration formula is
y (i+1)
n
= y (i)
n
(
− I − h ∂f ) −1 ( )
y n
(i) − y n−1 − hf(t n , y n (i) )
∂y
(4.109)
(4.110)
(4.111)
Again, it is not generally efficient to calculate a matrix inverse; it is better to solve
a linear system of equations. We can reformulate the last equation as
(
y (i+1)
n
) (
− y n
(i) I − h ∂f ) (
)
= − y n
(i) − y n−1 − hf(t n , y n (i) )
∂y
(4.112)
We can solve the linear system for δ = y n
(i+1)
δ − y n (i) .
− y (i)
n
and then calculate y (i+1)
n =
We then have the following Mathematica Implementation.
=⇒
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=⇒
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=⇒
BackwardEulerNewtonsMethod[f , {t0 , y0 }, h , tmax , tol :0.003,
nmax :5] :=
Module[{ t, y, n, time, yval, yvaln, yvalp, delta, r, J, JV,
fv},
r = {{t0, y0}};
yval = y0;
J = D[f[t, y], {y, 1}];
For[time = t0, time < tmax, time += h,
(***** calculate first guess at next grid point *****)
fv = f[time+h, yval];
JV = J /. {t -> time+h, y -> yval};
yvaln =(-yval + h (-fv + JV *yval))/(-1+h*JV);
(***** iterate using Newton’s method *****)
For[i = 1, i ≤ nmax, i++,
yvalp = yvaln;
(***** update derivative *****)
JV = J /. {t -> time+h, y -> yvalp};
(***** update function value *****)
fv = f[time+h, yvalp];
(***** Newton Iteration Formula *****)
yvaln=(yval + h(fv - JV* yvalp))/(1-h*JV);
(***** Within desired tolerance? *****)
Math 582B, Spring 2007
California State University Northridge
c○2007, B.E.Shapiro
Last revised: May 23, 2007