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Time-Dependent Problems 229
We now write Pn(x,t) = n−1 −x2
m=0 cm(t)Hm(x)e , where the cm(t)’s are the
Fourier coefficients of Pn at t ∈]0, ˆ T]. Substituting Pn in (10.2.22) and recalling
(1.7.1), we obtain
(10.2.24)
d
dt cm(t) = −2ζ(m + 1)cm(t), 0 ≤ m ≤ n − 1, ∀t ∈]0, ˆ T].
Now, we can find the explicit expression of the coefficients. We note that, from (10.2.23),
the initial datum is Pn(x,0) = (I∗ v,nU0)( √ 4ζx), v(x) := ex2, x ∈ R (see (6.7.6)).
Another scheme is obtained with the initial condition Pn(x,0) = (Π ∗ v,nU0)( √ 4ζx),
x ∈ R (see (6.7.3)). This amounts to assigning the values cm(0), 0 ≤ m ≤ n −1, as the
Fourier coefficients of U0. As the reader can easily check, after this modification, the
solution to (10.2.22) turns out to be the approximation of V by the Galerkin method.
Indeed, Pn also satisfies
(10.2.25)
R
∂Pn
∂(Pnv) ∂φ
φv dx = −ζ
∂t R ∂x ∂x dx, ∀φ ∈ Sn−1, ∀t ∈]0, ˆ T].
10.3 Approximation of linear first-order problems
We now focus the attention to another time-dependent partial differential equation. The
unknown U : [−1,1] × [0,T] → R, satisfies
(10.3.1)
∂U ∂U
(x,t) = −ζ (x,t), ∀x ∈] − 1,1], ∀t ∈]0,T],
∂t ∂x
(10.3.2) U(x,0) = U0(x), ∀x ∈ [−1,1],
(10.3.3) U(−1,t) = σ(t), ∀t ∈]0,T],