Show body Body Write
h = Δ x ,
k = Δ t and
δ xx u i = ( u i + 1 − 2 u i + u i − 1 ) / h 2 . Insert
a smooth exact
heat equation solution into the
Crank-Nicolson method . Expanding about the
midpoint ( x i , t n + k /2 ) gives
k u ( t + k /2 ) − u ( t − k /2 ) = u t + 24 k 2 u ttt + O ( k 4 ) ,
2 D [ δ xx u ( t − k /2 ) + δ xx u ( t + k /2 )] = D [ u xx + 12 h 2 u xxxx + 8 k 2 u xx tt ] + O ( h 4 + k 4 + h 2 k 2 ) .
Thus the residual, defined
as left side minus right side, is
T = 24 k 2 u ttt − 8 D k 2 u xx tt − 12 D h 2 u xxxx + O ( h 4 + k 4 + h 2 k 2 ) .
Since
u t = D u xx implies
u ttt = D u xx tt ,
T = − 12 k 2 u ttt − 12 D h 2 u xxxx + O ( h 4 + k 4 + h 2 k 2 ) .
This proves
second -order
numerical consistency in both
space and time . About the old
time level, uncombined
Taylor remainders can include
O ( k 3 ) ; the
midpoint expression displays the symmetric leading errors most cleanly. Multiplying by
k gives the one-step defect
O ( k 3 + k h 2 ) . Stable
evolution with suitable initial and boundary
data then gives global error
O ( k 2 + h 2 ) ;
consistency alone does not prove global
convergence .
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