Ecole doctorale de Physique de la région Parisienne (ED107)

178 Spectral methods and vectorial equations
at our choice of spectral basis. Then, we shall discuss the further difficulties that arise in
vectorial PDE and explain the way we overcome them.
B.1.1 The scalar heat equation
Consider the heat equation :
∂F
∂t
= α∆F + S (B.1)
where ∆F = ∂2F ∂r2 + 2 ∂F 1 + r ∂r r2 ( ∂2F ∂ϑ2 cos ϑ ∂F 1
+ + sin ϑ ∂ϑ sin ϑ2 ∂2F ∂ϕ2 ) is the Laplacian in spherical
coordinates, α the heat conductivity supposed to be constant, and S a source term (that
may include nonlinear terms if they exist). The idea of spectral methods is to look for the
solution of the Equation (B.1) on the form
F [r, ϑ, ϕ, t] =
∞
n,l,m=0
Fnlm[t] H l n[r] K m l [ϑ] Lm[ϕ] (B.2)
with (0 ≤ r ≤ 1; 0 ≤ ϑ ≤ π; 0 ≤ ϕ ≤ 2π) and where H l n[r], K m l [ϑ], Lm[ϕ] are a well
chosen complete set of functions. The problem is then to find the time evolution of the
coefficients Fnlm[t].
In a spherical geometry, it is quite natural to choose Lm[ϕ] = expimϕ and Km l [ϑ] =
P m
l [ϑ], where P m
l [ϑ] are the Legendre functions. With these choices, the Equation (B.1)
can be written
∂Flm[r,t]
∂t
∂2Flm[r,t] − α ∂r2 + 2 ∂Flm[r,t]
r ∂r
lp(l+1)
−
r 2
Flm[r, t] = Slm[r, t] (B.3)
where Slm[r, t] are the Fourier-Legendre coefficients of the function S at the radius r and
the instant t.
In order to handle the singularity at r = 0, we shall consider separately the cases l = 0
, l = 1 and l > 1 :
- For l = 0, we use the fact that a C 1 function symmetric with respect to the inversion
r → −r has its first derivative that vanishes at least as r at r = 0. Therefore, for
such a function, the term ∂2F ∂r2 + 2 ∂F is regular. Even Chebyshev polynomials T2n[r]
r ∂r
have this property, and the choice H 0 n[r] = T2n[r] (n ≥ 0) then satisfies the regularity
conditions.
- For l = 1, it is almost the same, but the final choice is H 1 n[r] = T2n+1[r].
- The case l > 1 is more delicate to handle. Indeed, it can be shown [See Bonazzola
& Marck (1990)] that for F [r, ϑ, ϕ, t] to be a C ∞ function, the coefficients Flm must
vanish as rl at r = 0. It means that the Flm[r, t] are symmetric with respect to the
inversion r → −r if l is even, and anti-symmetric in the opposite case. We shall