DerivationODEsadvanced

Derivation: one-step Adams-Moulton (AM2)

Full construction of AM2: Lagrange interpolant including the new node, change of variable, computed 1/2-1/2 weights, connection with the trapezoidal rule and local error.

Step 1: integral form

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AM2 uses the whole new interval: it interpolates the slope between fkf_k and fk+1f_{k+1}, so it is the implicit trapezoidal rule.
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AM2 uses the whole new interval: it interpolates the slope between fkf_k and fk+1f_{k+1}, so it is the implicit trapezoidal rule.

  1. Start from the IVP y′(t)=f(t,y(t))y'(t)=f(t,y(t)). Integrate both sides between tkt_k and tk+1t_{k+1}:

    ∫tktk+1y′(τ) dτ=∫tktk+1f(τ,y(τ)) dτ\int_{t_k}^{t_{k+1}} y'(\tau)\,d\tau=\int_{t_k}^{t_{k+1}} f\bigl(\tau,y(\tau)\bigr)\,d\tau

  2. By the Fundamental Theorem of Calculus, the left-hand side is exact. The only part to approximate is the integral of ff:

    y(tk+1)=y(tk)+∫tktk+1f(τ,y(τ)) dτy(t_{k+1})=y(t_k)+\int_{t_k}^{t_{k+1}} f\bigl(\tau,y(\tau)\bigr)\,d\tau

  3. In the numerical formula write yk≈y(tk)y_k\approx y(t_k) and yk+1≈y(tk+1)y_{k+1}\approx y(t_{k+1}):

    yk+1=yk+∫tktk+1f(τ,y(τ)) dτy_{k+1}=y_k+\int_{t_k}^{t_{k+1}} f\bigl(\tau,y(\tau)\bigr)\,d\tau

Step 2: interpolate f on the new interval

  1. Unlike AB2, AM2 interpolates at tkt_k and tk+1t_{k+1}. Thus it uses fk=f(tk,yk)f_k=f(t_k,y_k) and fk+1=f(tk+1,yk+1)f_{k+1}=f(t_{k+1},y_{k+1}):

    p1(τ)=fkLk(τ)+fk+1Lk+1(τ)p_1(\tau)=f_k L_k(\tau)+f_{k+1}L_{k+1}(\tau)

  2. The Lagrange basis functions are:

    Lk(τ)=τ−tk+1tk−tk+1=tk+1−τh,Lk+1(τ)=τ−tktk+1−tk=τ−tkhL_k(\tau)=\frac{\tau-t_{k+1}}{t_k-t_{k+1}}=\frac{t_{k+1}-\tau}{h},\qquad L_{k+1}(\tau)=\frac{\tau-t_k}{t_{k+1}-t_k}=\frac{\tau-t_k}{h}

  3. Substitution gives the line that approximates the slope inside the subinterval:

    p1(τ)=fk tk+1−τh+fk+1 τ−tkh=fk+1τ−tkh−fkτ−tk+1hp_1(\tau)=f_k\,\frac{t_{k+1}-\tau}{h}+f_{k+1}\,\frac{\tau-t_k}{h}=f_{k+1}\frac{\tau-t_k}{h}-f_k\frac{\tau-t_{k+1}}{h}

Step 3: change of variable and integrals

  1. To integrate, use the same change as in your notes: s=τ−tks=\tau-t_k. Then τ=tk+s\tau=t_k+s, dτ=dsd\tau=ds, and the limits are s=0s=0 and s=hs=h.

    s=τ−tk,τ=tk+s,τ−tk+1=s−hs=\tau-t_k,\qquad \tau=t_k+s,\qquad \tau-t_{k+1}=s-h

  2. With this change, the interpolant becomes:

    p1(tk+s)=fk+1sh−fks−hh=fk+1sh+fkh−shp_1(t_k+s)=f_{k+1}\frac{s}{h}-f_k\frac{s-h}{h}=f_{k+1}\frac{s}{h}+f_k\frac{h-s}{h}

  3. Integrate the fk+1f_{k+1} term:

    ∫0hfk+1sh ds=fk+11h[s22]0h=h2fk+1\int_0^h f_{k+1}\frac{s}{h}\,ds=f_{k+1}\frac{1}{h}\left[\frac{s^2}{2}\right]_0^h=\frac{h}{2}f_{k+1}

  4. Integrate the fkf_k term:

    ∫0hfkh−sh ds=fk1h[hs−s22]0h=h2fk\int_0^h f_k\frac{h-s}{h}\,ds=f_k\frac{1}{h}\left[hs-\frac{s^2}{2}\right]_0^h=\frac{h}{2}f_k

  5. Therefore the slope integral is approximated by the sum of those two weights:

    ∫tktk+1f(τ,y(τ)) dτ≈∫tktk+1p1(τ) dτ=h2(fk+1+fk)\int_{t_k}^{t_{k+1}} f\bigl(\tau,y(\tau)\bigr)\,d\tau\approx \int_{t_k}^{t_{k+1}}p_1(\tau)\,d\tau=\frac{h}{2}\bigl(f_{k+1}+f_k\bigr)

  6. Substituting into the integral form gives AM2:

    yk+1=yk+h2(fk+1+fk)y_{k+1}=y_k+\frac{h}{2}\bigl(f_{k+1}+f_k\bigr)

Step 4: why it is implicit

  1. The key detail is that fk+1f_{k+1} has not been computed yet:

    fk+1=f(tk+1,yk+1)f_{k+1}=f(t_{k+1},y_{k+1})

  2. So the formula is really an equation for the unknown yk+1y_{k+1}:

    yk+1=yk+h2(f(tk+1,yk+1)+f(tk,yk))y_{k+1}=y_k+\frac{h}{2}\Bigl(f(t_{k+1},y_{k+1})+f(t_k,y_k)\Bigr)

  3. If solving with Newton, define a residual R(z)R(z), keeping ff only for the ODE function:

    R(z)=z−yk−h2(f(tk+1,z)+fk)R(z)=z-y_k-\frac{h}{2}\Bigl(f(t_{k+1},z)+f_k\Bigr)

  4. Newton updates the approximation until R(z)R(z) is essentially zero:

    z(m+1)=z(m)−R(z(m))R′(z(m)),R′(z)=1−h2 fy(tk+1,z)z^{(m+1)}=z^{(m)}-\frac{R(z^{(m)})}{R'(z^{(m)})},\qquad R'(z)=1-\frac{h}{2}\,f_y(t_{k+1},z)

  5. In a predictor-corrector pair, one usually predicts yk+1y_{k+1} with AB2 and inserts that prediction into AM2:

    yk+1(p)=yk+h2(3fk−fk−1),yk+1(c)=yk+h2(f(tk+1,yk+1(p))+fk)y_{k+1}^{(p)}=y_k+\frac{h}{2}(3f_k-f_{k-1}),\qquad y_{k+1}^{(c)}=y_k+\frac{h}{2}\Bigl(f(t_{k+1},y_{k+1}^{(p)})+f_k\Bigr)

Step 5: error and link with the trapezoid

The formula obtained is exactly the trapezoidal rule applied to the slope integral y′(t)y'(t). Since the trapezoid has local error proportional to the third derivative,

ek+1=−112h3y′′′(ξ)=O(h3)e_{k+1}=-\frac{1}{12}h^3y'''(\xi)=\mathcal{O}(h^3)
Local error of AM2; after accumulating steps, the global error is order 2.

Thus AM2 has the same global order as AB2, but a smaller error constant. The improvement is paid for by solving the implicit equation or using a corrector.