DerivationODEsintermediate

Derivation: implicit Euler

Approximating the derivative at the new node with a backward difference produces the implicit Euler method and the nonlinear equation to solve at each step.

Backward difference at the new node

tₖtₖ₊₁h f(tₖ₊₁, yₖ₊₁)right rectanglef(τ, y(τ))
The integral derivation of implicit Euler uses the right endpoint: the rectangle height contains the unknown yk+1y_{k+1}.
Expand diagram

The integral derivation of implicit Euler uses the right endpoint: the rectangle height contains the unknown yk+1y_{k+1}.

  1. Instead of approximating the derivative at tkt_k looking forward, approximate it at tk+1t_{k+1} looking backward, with the backward difference:

    y′(tk+1)≈y(tk+1)−y(tk)hy'(t_{k+1})\approx\frac{y(t_{k+1})-y(t_k)}{h}

  2. Substitute into the differential equation evaluated at the new node, y′(tk+1)=f(tk+1,y(tk+1))y'(t_{k+1})=f\bigl(t_{k+1},y(t_{k+1})\bigr), and solve:

    yk+1=yk+h f(tk+1,yk+1)y_{k+1}=y_k+h\,f(t_{k+1},y_{k+1})

  3. The same scheme follows from the integral form of the IVP, approximating the integrand by its value at the right endpoint (right-hand rectangle), in exact parallel to Route 3 of the explicit Euler derivation.

  4. Since yk+1y_{k+1} appears inside ff, each step requires solving an (in general nonlinear) equation in the unknown yk+1y_{k+1}, for instance with the Newton-Raphson method:

    g(yk+1)=yk+1−yk−h f(tk+1,yk+1)=0g(y_{k+1})=y_{k+1}-y_k-h\,f(t_{k+1},y_{k+1})=0