Backward Euler method: Difference between revisions

Revision as of 10:33, 29 November 2012

Revision as of 23:05, 28 January 2013

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== Description ==

Consider the [[ordinary differential equation]]

Consider the [[ordinary differential equation]] <math> y' = f(t,y) </math> with initial value <math> y(t_0) = y_0. </math> A numerical method produces a sequence <math> y_0, y_1, y_2, \ldots </math> such that <math> y_k </math> approximates <math> y(t_0+kh) </math>, where <math> h </math> is called the step size.

:<math> \frac{\mathrm{d} y}{\mathrm{d} t} = f(t,y) </math>

Considerwith theinitial [[ordinary differential equation]]value <math> y'(t_0) = f(t,y)y_0. </math> withHere initialthe valuefunction <math>f</math> y(and the initial data <math>t_0)</math> =and <math>y_0.</math> are known; the function <math>y</math> depends on the real variable <math>t</math> and is unknown. A numerical method produces a sequence <math> y_0, y_1, y_2, \ldots </math> such that <math> y_k </math> approximates <math> y(t_0+kh) </math>, where <math> h </math> is called the step size.

The backward Euler method computes the approximations using

:<math> y_{k+1} = y_k + h f(t_{k+1}, y_{k+1}). </math> <ref>{{harvnb|Butcher|2003|p=57}}</ref>

This differs from the (forward) Euler method in that the latter uses <math> f(t_k, y_k) </math> in place of <math>f(t_{k+1}, y_{k+1})</math>.

The backward Euler method is an implicit method: the new approximation <math> y_{k+1} </math> appears on both sides of the equation, and thus the method needs to solve an algebraic equation for the unknown <math> y_{k+1} </math>. Sometimes, this can be done by [[fixed-point iteration]]:

:<math> y_{k+1}^{[0]} = y_k, \quad y_{k+1}^{[i+1]} = y_k + h f(t_{k+1}, y_{k+1}^{[i]}). </math>

If this sequence converges (within a given tolerance), then the method takes its limit as the new approximation

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== Derivation ==

Integrating the differential equation <math> \frac{\mathrm{d} y'}{\mathrm{d} t} = f(t,y) </math> from <math> t_k </math> to <math> t_{k+1} = t_k + h </math> yields

: <math> y(t_{k+1}) - y(t_k) = \int_{t_k}^{t_{k+1}} f(\taut, y(\taut)) \,\mathrm{d}\taut. </math>

Now approximate the integral on the right by the right-hand [[rectangle method]] (with one rectangle):

: <math> y(t_{k+1}) - y(t_k) \approx h f(t_{k+1}, y(t_{k+1})) \,\mathrm{d}\tau. </math>

Finally, use that <math> y_k </math> is supposed to approximate <math> y(t_k) </math> and the formula for the backward Euler method follows.<ref>{{harvnb|Butcher|2003|p=57}}</ref>