Stability of a difference sche1ne 91
so that
With these entries, we look for a solution of problem (6) in the form
n
u(t) = 2= ocs(t) es.
s=l
Substitution of this expression into equation (6) yields
which implies that
providing the same grounds for the series
n
(7) u(t) = L Ock(O) exp {->.kt} ek.
k=l
Under the initial condition
n
u(O) = Uo = L ock(O) ek'
k=l
we arrive at the relations
n
II Uo 112 = L oc~(O).
k=l
From (7) it follows that
n n
II u(t) 112 = ~ ~ ock(O) ocs(O) exp {->.kt} exp {->.st} (ek 'es)
k=l s=l
n n
k=l k=l