An estimate of accuracy for interpolant numerical solutions of a PDE problem

0.14 MB·English

by  Gianluca Argentini

An estimate of accuracy for
interpolant numerical solutions of a
5
0
0 PDE problem
2

n
a Gianluca Argentini
J

2 [email protected]

Advanced Computing Laboratory
]
A
Information & Communication Technology Department
N
. Riello Group, 37045 Legnago (Verona), Italy
s
c
[
December 2004

2
v
0
2 Abstract
1
In this paper we present an estimate of accuracy for a piecewise
2
1 polynomial approximation of a classical numerical solution to a non
4 linear differential problem. We suppose the numerical solution u is
0
computed using a grid with a small linear step and interval time T ,
/ u
s whilethepolynomialapproximation v is an interpolation ofthevalues
c
: of a numerical solution on a less fine grid and interval time T <<T .
v v u
i The estimate shows that the interpolant solution v can be, under
X
suitable hypotheses, a good approximation and in general its compu-
r
a tational cost is much lower of the cost of the fine numerical solution.
We present two possible applications to linear case and periodic case.
1 The problem
Let [a,b], a < b, an interval on the real line R, and let C1([0,+∞)×[a,b]) the
space of continuously differentiable real functions defined on [0,+∞)×[a,b].
Let F : C1([0,+∞)×[a,b]) → R a continuously differentiable real functional.
Then, if u ∈ C1([0,+∞) × [a,b]), the partial differential equation, usually
associated to hyperbolic conservation laws,
1
∂u ∂
+ F(u) = 0 (1)
∂t ∂x
is in general non linear on the unknown u. Let u ∈ C1([a,b]) and u , u two
0 a b
real numbers. We consider the boundary problem
∂u(x)+ ∂F(u)(x) = 0 ∀x ∈ (a,b)
∂t ∂x
u(t,a) = u ,u(t,b) = u ∀t ∈ [0,+∞) (2)
 a b
u(0,x) = u (x) ∀x ∈ (a,b)
 0
and its numerical resolution. In this paper we consider only the Euler for-

ward in time - centered in space (EFC) schema (see [4]) for Finite Differences
method, but for other numerical schemas one can apply the same consider-
ations with analogous, but more complicated, demonstrations. Let h > 0 a
linear step for a partition (x ) , 0 ≤ j ≤ P, P = b−a, of the interval [a,b]
j j h
such that x = a, x = b, and let ∆t the time step of the numerical schema.
0 P
We denote by un = (un) the numerical solution of problem (2) computed at
j j
time step n-th, hence at instant t = n∆t.
In EFC schema the values at time step n+1 are so computed (see [1],[4]):
∆t
un+1 = un −F′(un) (un −un ), 0 < j < P −1 (3)
j j j 2h j+1 j−1
where F′(un) is the value of dF(u) obtained by substitution of u with the
j du
discretized value un. We suppose that the schema (3) satisfies the Courant-
j
Friedrichs-Lewy condition (CFL) (see [4]):
∆t
F′(un) ≤ 1 ∀j ∈ Z,∀n ∈ N,∀h ∈ R (4)
j h +
This condition is(cid:12)usually(cid:12)admitted in the literature about numerical resolu-
(cid:12) (cid:12)
tion of PDEs and is related to the question of stability of solutions.
Let N ∈ N, N > 1, r ∈ N , k = h, dt = ∆t, uN the numerical solution of
+ r r
(2) at time step N with linear step h, uM the solution at time step M = Nr
with linear step k. Note that CFL condition is satisfied in the second case
too, because ∆t = dt. The solution uM is computed on a grid finer than that
h k
of uN, being h > k, and hence it can be considered a better approximation of
the exact solution, if exists, of the problem (2). But uM has the disadvantage
of a much greater computational cost, which can be very high for r >> 1
2
(see [2]).
We construct a piecewise polynomial interpolation v of uN on the nodes
of the partition with grid step h. Function v can be considered as a global
continuous and differentiable representation of the discrete values uN on the
interval [a,b]. Then one can compute, for every couple (x ,x ) of nodes,
j j+1
the interpolating polynomial v = v (t) on a set of r values of t. In the
j j
following, we present an estimate of the error |v (t) −uM|, where t = x−xj
j x h
(see Section 3), with the aim of establishing if v can be considered a good
approximation for the solution of the problem (2).
2 Construction of interpolating polynomials
Let (x ,x ) a couple of nodes internal to [a,b], hence 1 ≤ j ≤ P − 2.
j j+1
For simplicity we don’t consider the boundary nodes x and x , but with a
0 P
suitable choice of a discretization formula for the spatial derivative ∂u (see
∂x
[1], [5]) we could apply analogous considerations. We write, for more simple
notation, p = x , p = x , d = uN, d = uN . We construct a cubic
1 j 2 j+1 1 j 2 j+1
polynomial q = q(t) = at3 +bt2 +ct+d, 0 ≤ t ≤ 1, imposing the following
four conditions:
1. q(0) = p ;
1
2. q(1) = p ;
2
3. q′(0) = d ;
1
4. q′(1) = d ;
2
where q′(t) = 3at2 + 2bt + c. Then, solving the following linear system for
the unknowns a, b, c, d
d = p
1
a+b+c+d = p
 2 (5)
c = d
 1

 3a+2b+c = d
2
one obtains the expressionfor the derivative of each cubic:


q′(t) = (6p −6p +3d +3d )t2 +(−6p +6p −4d −2d )t+d (6)
1 2 1 2 1 2 1 2 1
3
3 Auxiliary technical results
Let w = uN the numerical solution of problem (2) obtained with a EFC
schemawithgridsteph(h-grid)andinitialconditionu sampledusingsteph.
0
Weconsider acoupleofinternalnodes(x ,x )onh-grid,andasubpartition
j j+1
(x ) , where r ∈ N , mod(r,2) = 0, x = x ,...,x = x . Let
j,i 0≤i≤r + j j,0 j+1 j,r
v = v (t) = q′(t), where q is the cubic computed solving the system (5). We
j j j
call v an interpolant solution of the problem (2). We define the set t = m,
m r
where m ∈ N, 0 ≤ m ≤ r. If s = xj+1−xj, we have x = x = ms + x ,
r m j,m j
therefore
m mx −x x −x 1 ms x −x
t = = j+1 j = m j+1 j = = m j (7)
m
r r h r h h h
Let k = h, and let u = uM the solution obtained with grid step k and
r
u sampled using k. For simplicity we denote u the value of u on the node
0 m
x = ms+x , where j, as before, is index of the h-grid. In the following we
m j
demonstrate some preliminary results for estimate the error |v −u |, where
m m
v = v(t ).
m m
Proposition 1 If t ∈ [0,1], exists a value s ∈ [0,1] such that
s
t2 = t− (8)
4
Proof. If y = t2−t, then y ≤ 0 and y′ = 2t−1, therefore t = 1 is the so-
0 2
lution of y′(t) = 0. At t the function y has a minimum, because y′′ = 2 > 0.
0
The value is y(1) = −1. Therefore we have −1 ≤ t2−t ≤ 0, from which (8)
2 4 4
follows with s ∈ [0,1]. (cid:3)
Proposition 2 For every m such that 0 ≤ m ≤ r:
3 3
|v −v | ≤ h+2|d −d |+ |d +d | (9)
m r 1 2 1 2
2 4
Proof. From (6) we have v = v(1) = d , therefore
r 2
|v −v | ≤ |(6p −6p +3d +3d )t2 +(−6p +6p −4d −2d )t+d −d |
m r 1 2 1 2 1 2 1 2 1 2
≤ |(6p −6p )(t2 −t)|+|(3d +3d )t2 −(4d +2d )t|+|d −d |
1 2 1 2 1 2 1 2
4
Then, using (8), we obtain
3 3
|v −v | ≤ |p −p |+2|d −d |+ |d +d | (10)
m r 1 2 1 2 1 2
2 4
and hence (9), because |p −p | = h. (cid:3)
1 2
In the same manner, using the fact that v = v(0) = d , one proofs the
0 1
following
Proposition 3 For every m such that 0 ≤ m ≤ r:
3 3
|v −v | ≤ h+2|d −d |+ |d +d | (11)
m 0 1 2 1 2
2 4
In the following propositions ǫ is a fixed positive real number, w0 is the
initial condition u of (2) sampled on the nodes of h-grid and u0 the initial
0
condition sampled on k-grid nodes.
Proposition 4 If |u0 − u0 | ≤ ǫ ∀i, then |un − un | ≤ ǫ ∀i,
i i+1 3M i i+1 3(M−n)
∀n ≤ M,n ∈ N .
+
Proof. By induction. Let n = 1. We have
|u1 −u1 | ≤ |u1 −u0|+|u0 −u0 |+|u0 −u1 | (12)
i i+1 i i i i+1 i+1 i+1
Using the EFC schema (3), the CFL condition (4) and the hypothesis, we
have
1 1
|u1 −u1 | ≤ |u0 −u0 |+|u0 −u0 |+ |u0 −u0|
i i+1 2 i+1 i−1 i i+1 2 i+2 i
1 1 1 1
≤ |u0 −u0|+ |u0 −u0 |+|u0−u0 |+ |u0 −u0 |+ |u0 −u0|
2 i+1 i 2 i i−1 i i+1 2 i+2 i+1 2 i+1 i
ǫ ǫ
≤ 3 =
3M 3(M−1)
that is the first inductive step. The intermediate inductive step is |un −
i
un | ≤ ǫ . Then
i+1 3(M−n)
|un+1 −un+1| ≤ |un+1 −un|+|un −un |+|un −un+1|
i i+1 i i i i+1 i+1 i+1
5
1 1
≤ |un −un |+|un −un |+ |un −un|
2 i+1 i−1 i i+1 2 i+2 i
1 1 1 1
≤ |un −un|+ |un −un |+|un −un |+ |un −un |+ |un −un|
2 i+1 i 2 i i−1 i i+1 2 i+2 i+1 2 i+1 i
ǫ ǫ
≤ 3 =
3(M−n) 3(M−(n+1))
(cid:3)
and this is the final inductive step.
Proposition 5 If |u0 − u0 | ≤ ǫ ∀i, then |un+1 − un| ≤ ǫ ∀i,
i i+1 3M i i 3(M−n)
∀n ≤ M.
Proof. Let n ≥ 0. Then
1
|un+1 −un| ≤ |un −un |
i i 2 i+1 i−1
1
≤ (|un −un|+|un −un |)
2 i+1 i i i−1
and hence, from Proposition (4)
ǫ
|un+1−un| ≤ (13)
i i 3(M−n)
(cid:3)
that is the thesis.
In the next proposition, the index i is associated to the node x in the
i
k-grid, while the index j is associated to the node x in the h-grid.
j
Proposition 6 If |u0 −u0 | ≤ ǫ ∀i, then |u0 −u0 | ≤ ǫ ∀j.
i i+1 3M j j+1 3N
Proof. Using the fact that r < 3r and M = Nr ≥ (N +r) if r,N > 1, we
have
r−1 r−1 r−1
ǫ ǫ ǫ
|u0 −u0 | ≤ |u0 −u0 | ≤ ≤ =
j j+1 j+mr j+mr+1 3Nr r3N 3N
m=0 m=0 m=0
X X X
(cid:3)
therefore the proposition is verified.
Note that by definition w0 = u0, hence from Proposition (6) follows that
j j
6
ǫ
|w0−w0 | ≤ (14)
j j+1 3N
Then the Propositions (4) and (5) hold for the solutions (wk) too, using
N instead of M:
Corollary 1 If |u0−u0 | ≤ ǫ ∀i, then |wk+1−wk| ≤ ǫ , |wk−wk | ≤
i i+1 3M j j 3(N−k) j j+1
ǫ ∀j, ∀k ≤ N.
3(N−k)
In particular, for k = N, writing wN = d and wN = d , follows that
j 1 j+1 2
Corollary 2 If |u0 −u0 | ≤ ǫ ∀i, then |d −d | ≤ ǫ.
i i+1 3M 1 2
In the next proposition the simplified symbol u = uM is the value of
0 j
solution u at node x of the k-grid, i.e. the node x of the h-grid.
jr j
Proposition 7 If |u0 −u0 | ≤ ǫ ∀i, then |v −u | ≤ ǫ.
i i+1 3M 0 0
Proof. By definition, v = d = w = wN, where wN is the solution at
0 1 j j
time step N of (2) with initial condition w . Then, using a telescopic sum
0
like (a −a ) = N (a −a ):
N 0 k=1 k k−1
P
N M
|v −u | = |w −u | ≤ |wk −wk−1|+|w0 −u0|+ |uh−1−uh| (15)
0 0 j 0 j j j j j j
k=1 h=1
X X
By definition, w0 = u0, hence the second term of the second member is null.
j j
For the first sum, from hypothesis and from Corollary (1) follows
N N
ǫ
|wk −wk−1| ≤
j j 3(N−k+1)
k=1 k=1
X X
ǫ N ǫ 3(N+1) −1 ǫ
= 3k = ≤
3(N+1) 3(N+1) 2 2
k=1
X
For the second sum the computation, using the Proposition (5), is the same.
(cid:3)
Therefore the proposition is verified.
7
Using the simplified symbol u = uM for the value of u at the node
r (j+1)r
x at time step M, in the same manner we can proof the following
(j+1)r
Proposition 8 If |u0 −u0 | ≤ ǫ ∀i, then |v −u | ≤ ǫ.
i i+1 3M r r
In the next Proposition we remember that u is the value of u = uM on
m
the node x = ms+x .
m j
Proposition 9 Let 1 ≤ m ≤ r. If |u0−u0 | ≤ ǫ ∀i, then |u −u | ≤ mǫ.
2 i i+1 3M 0 m
Proof. Using Proposition (4) with n = M, we have
m
|u −u | ≤ |uM −uM| ≤ mǫ (cid:3)
0 m k−1 k
k=1
X
Proposition 10 Let r ≤ m ≤ r. If |u0 −u0 | ≤ ǫ ∀i, then |u −u | ≤
2 i i+1 3M m r
(r −m)ǫ.
Proof. Using Proposition (4) with n = M, we have
r−m
|u −u | ≤ |uM −uM | ≤ (r −m)ǫ (cid:3)
m r m+k−1 m+k
k=1
X
4 The main results
Now we can stated the main results on the estimate about the accuracy of
the interpolant solution v.
Theorem 1 Let ǫ > 0, m ∈ N , 1 ≤ m ≤ r. If |u0 −u0 | ≤ ǫ ∀i, then
+ i i+1 3M
3 3
|v −u | ≤ h+ |d +d |+(min[m,r −m]+3)ǫ (16)
m m 1 2
2 4
8
Proof. If m ≤ r, let p = 0, else p = r. Using Propositions (2) or (3), (7)
2
or (8), (9) or (10) and Corollary (1), we have
|v −u | ≤ |v −v |+|v −u |+|u −u |
m m m p p p p m
3 3
≤ h+2ǫ+ |d +d |+(min[m,r −m]+1)ǫ
1 2
2 4
because if m ≤ r then min[m,r −m] = m, else min[m,r −m] = r −m. (cid:3)
2
Corollary 3 Let ǫ > 0, m ∈ N , 1 ≤ m ≤ r. There exists δ > 0 such that
+
if |u0 −u0 | ≤ δ ∀i, then
i i+1
3 3
|v −u | ≤ h+ |d +d |+ǫ (17)
m m 1 2
2 4
Proof. Let µ = ǫ . Then µ > 0, and from Theorem (1), defined
min[m,r−m]+3
δ = µ , follows that if |u0 −u0 | ≤ δ, then
3M i i+1
3 3
|v −u | ≤ h+ |d +d |+(min[m,r −m]+3)µ (18)
m m 1 2
2 4
But (min[m,r −m]+3)µ = ǫ, so the thesis is verified. (cid:3)
The estimate stated in Corollary (3) is of the same kind of the esti-
mates established, in the case of finite element method, by Kuznetsov and
Tadmor-Tang for approximated numerical solutions to scalar conservation
laws equations (see [6]).
In particular, inequality (17) shows that the accuracy, respect to the fine u
solution computed using a grid step k, of the interpolant solution v, com-
puted using a grid step h = rk, is of the first order on h. In (17) the role
of the time steps N and M is hidden in the third addendum ǫ, as stated by
Corollary (2).
The next theorem gives a sufficient condition on the initial function u for
0
the validity of the hypothesis in Theorem (1).
Theorem 2 Let ǫ > 0, h > 0, N ∈ N, N > 1. If u ∈ C0([a,b]), then exists
0
r ∈ R such that, if 0 ≤ m ≤ r,
+
3 3
|v −u | ≤ h+ |d +d |+ǫ (19)
m m 1 2
2 4
9
Proof. The function u , continuous on a closed and limited subset of R,
0
is uniformly continuous, therefore exists δ > 0 such that if x ,x ∈ [a,b],
1 2
|x − x | ≤ δ, then |u (x ) − u (x )| ≤ ǫ. We can choose the parameter
1 2 0 1 0 2
r ∈ N , and hence the time step M = rN, such that ǫ ≤ δ. Therefore,
+ 3M
with this choice of r, it can be applied the theorem (1) and, with the right
choice of δ, the Corollary (3). (cid:3)
In the next section we discuss the role of the term |d + d | in some
1 2
particular cases.
5 Applications to particular cases
As first application of the previous estimate, let F(u) = au, with a ∈ R,
a 6= 0. Hence we suppose that F is linear. In this case the numerical
1
schema EFC is stable in the norm |u| = h |u |2 2 until time step N,
2 j j
i.e. |u|2 ≤ CN|u0|2 with CN = C(N) ∈ R+,(cid:16)if ∆Pt ≤ ha(cid:17)2; the constant CN is
equal to eN∆t (see [4]), while j is the index on h-grid. If we choose the grid
2 (cid:0) (cid:1)
step h such that h ≤ |a|, than the CFL condition is satisfied because
∆t h
|a| ≤ ≤ 1 (20)
h |a|
Lemma 1 Let m ∈ N, m ≥ 1, and (C ) a set of positive real numbers.
j 0≤j≤m
If A, B ∈ (C ) , then
j j
1
m 2
A+B ≤ 2 C2 (21)
j
!
j=0
X
Proof. We have A2 ≤ m C2 and B2 ≤ m C2, therefore
j=0 j j=0 j
P 1 P 1
m 2 m 2
A ≤ C2 , B ≤ C2 (22)
j j
! !
j=0 j=0
X X
(cid:3)
hence the thesis is verified.
If u ∈ C0([a,b]), let |u| = sup |u(x)|.
∞ x∈[a,b]
10