Theorems and proofs for
Partial differential equations
TMA372
Edvin Listo Zec
[email protected]
Friday 27th March, 2015
Introduction
This text is written as an aid for those that are taking the course TMA372 Partial
differential equations. It contains the recommended theorems and proofs for the
year of 2015, taken from the lecture notes of Mohammad Asadzadeh’s lectures.
Contents
1 L∞ error estimates for linear interpolation in an interval
1
2 BVP ⇔ VF ⇔ MP
2
3 Error estimates for BVP
5
4 Stability estimates for IVP
9
5 Lax-Milgram
11
6 Error estimates for the Poisson equation
13
7 Stability and energy estimates for the heat equation in
Rn
15
Edvin Listo Zec
1
Theorems&Proofs, PDE
L∞ error estimates for linear interpolation in
an interval
Theorem 1.1. If f 00 ∈ L∞ (a,b) and q = 1 (two interpolation nodes), then there are
interpolation constants ci independent of f and the length of [a,b] such that
(i) ||π1 f − f ||L∞ (a,b) ≤ ci (b − a)2 ||f 00 ||L∞ (a,b)
(ii) ||π1 f − f ||L∞ (a,b) ≤ ci (b − a)||f 0 ||L∞ (a,b)
(iii) ||(π1 f )0 − f 0 ||L∞ (a,b) ≤ ci (b − a)||f 00 ||L∞ (a,b) .
Proof. Every linear function p(x) on [a,b] can be written as a linear combination of
the basis functions
λa (x) =
b−x
b−a
and
λb (x) =
x−a
.
b−a
In other words p(x) = p(a)λa (x) + p(b)λb (x). Further, we note that
λa (x) + λb (x) = 1
and
aλa (x) + bλb (x) = x.
We have that
π1 f (x) = f (a)λa (x) + f (b)λb (x)
(1.1)
is the linear function connecting the points (a,f (a)) and (b,f (b)).
Figure 1: Linear lagrange basis functions.
We now Taylor expand f (a) and f (b) about x ∈ (a,b):



f (a) = f (x) + (a − x)f 0 (x) + 21 (a − x)2 f 00 (ηa ),


f (b) = f (x) + (b − x)f 0 (x) + 21 (b − x)2 f 00 (ηb ),
ηa ∈ [a,x]
(1.2)
1
ηb ∈ [x,b].
Edvin Listo Zec
Theorems&Proofs, PDE
Now we insert f (a) and f (b) from (1.2) into (1.1), and we get
π1 f (x) = f (x)[λa (x) + λb (x)] + f 0 (x)[(a − x)λa (x) + (b − x)λb (x)]+
1
1
+ (a − x)2 f 00 (ηa )λa (x) + (b − x)2 f 00 (ηb )λb (x) =
2
2
1
1
= f (x) + (a − x)2 f 00 (ηa )λa (x) + (b − x)2 f 00 (ηb )λb (x).
2
2
Therefore
|π1 f (x) − f (x)| =
1
(a − x)2 f 00 (ηa )λa (x) +
2
1
(b − x)2 f 00 (ηb )λb (x) .
2
(1.3)
Further, we note that in the interval x ∈ [a,b] it holds that (a − x)2 ≤ (a − b)2
and (b − x)2 ≤ (a − b)2 . Also, λa (x) ≤ 1 and λb (x) ≤ 1, ∀x ∈ (a,b). Moreover,
we have from the definition of the maximum norm that |f 00 (ηa )| ≤ |f 00 |L∞ (a,b) and
|f 00 (ηb )| ≤ |f 00 |L∞ (a,b) . This gives us that (1.3) may be estimated as
1
1
|π1 f (x) − f (x)| ≤ (a − b)2 · 1 · |f 00 |L∞ (a,b) + (a − b)2 · 1 · |f 00 |L∞ (a,b) ,
2
2
(1.4)
and thus
|π1 f (x) − f (x)| ≤ (a − b)2 |f 00 |L∞ (a,b) corresponding to ci = 1.
The other estimates (ii) and (iii) are proved similarly.
2
BVP ⇔ VF ⇔ MP
(BVP)1
(VF)1
Z 1
 0

 − a(x)u0 (x)

 u(0)
= f (x),
0 < x < 1,
(2.1)
= u(1) = 0.
a(x)u0 (x)v 0 (x) dx =
0
Z 1
0
f (x)v(x) dx, ∀v(x) ∈ H01
(2.2)
(MP)1 Find u ∈ H01 : F (u) ≤ F (w), ∀w ∈ H01 , where F (w) is the total potential
energy of the displacement w(x):
Z 1
1Z 1
0 2
F (w) =
a(w ) dx −
f w dx
|2 0 {z
}
| 0 {z }
Internal (elastic) energy
2
Load potential
(2.3)
Edvin Listo Zec
Theorems&Proofs, PDE
Theorem 2.1. The following two properties are equivalent:
(i) u satisfies (BVP)1
(ii) u is twice differentiable and satisfies (VF)1
Proof. [⇒] We assume that a(x) is piecewise continuous in (0,1), bounded and
strictly positive for 0 ≤ x ≤ 1. Let v(x) and v 0 (x) be square integrable functions,
i.e. v, v 0 ∈ L2 (0,1). We define the following space:
H01 = v(x) :
Z 1
(v(x)2 + v 0 (x)2 ) dx < ∞,
v(0) = v(1) = 0 .
0
Now multiply (BVP)1 with the test function v(x) ∈ H01 (0,1) and integrate over
(0,1). This yields
−
Z 1
0
a(x)u0 (x) v(x) dx =
Z 1
f (x)v(x) dx.
0
0
Integration by parts gives us
h
− a(x)u0 (x)v(x)
i1
0
+
Z 1
a(x)u0 (x)v 0 (x) dx =
Z 1
f (x)v(x) dx.
0
0
However, since v(0) = v(1) = 0 we get (VF)1 :
Z 1
a(x)u0 (x)v(x) dx =
Z 1
∀v(x) ∈ H01 .
f (x)v(x) dx,
0
0
[⇐] We start by integrating the left hand side of (VF)1 by parts:
−
Z 1
0
a(x)u0 (x) v(x) dx =
Z 1
f (x)v(x) dx,
0
0
∀v(x) ∈ H01 .
This can be rewritten as:
Z 1 0
0
− a(x)u (x) − f (x) v(x) dx = 0,
0
∀v(x) ∈ H01 .
To show that u satisfies (BVP)1 is equivalent to claim that
0
(2.4) ⇒ − a(x)u0 (x) − f (x) ≡ 0,
∀x ∈ (0,1).
Suppose that the above claim does not hold. Then
0
∃ξ ∈ (0,1) : − a(ξ)u0 (ξ) − f (ξ) 6= 0,
3
(2.4)
Edvin Listo Zec
Theorems&Proofs, PDE
where we may assume WLOG that
0
− a(ξ)u0 (ξ) − f (ξ) > 0 (or < 0).
Thus, with the assumptions that f ∈ C(0,1) and a ∈ C 1 (0,1), we have by continuity
that ∃δ > 0, such that in a δ-neighbourhood of ξ,
0
g(x) := − a(x)u0 (x) − f (x) > 0,
∀x ∈ (ξ − δ,ξ + δ).
Now choose v(x) from (2.4) as the hat function v ∗ (x) > 0 with v ∗ (ξ) = 1 and the
support Iδ := (ξ − δ,ξ + δ). Then v ∗ (x) ∈ H01 and
Z 1 0
0
∗
− a(x)u (x) − f (x) v (x) dx =
Z
Iδ
0
g(x) v ∗ (x) dx > 0.
| {z } | {z }
>0
>0
This is a contradiction of (2.4), and thus our claim holds and the proof is complete.
Theorem 2.2. The following two properties are equivalent:
(i) u satisfies (VF)1
(ii) u is the solution for the minimisation problem (MP)1 .
In other words
Z 1
0
0 0
au v dx =
Z 1
0
f v dx, ∀v ∈ H01 ⇐⇒ F (u) ≤ F (w), ∀w ∈ H01
Proof. [⇒] For w ∈ H01 , let v = w − u. Then v ∈ H01 since H01 is a vector space and
u ∈ H01 . Also
Z 1
2
1Z 1 a (u + v)0 dx −
f (u + v) dx =
2 0
0
1Z 1
1Z 1
1Z 1
=
2au0 v 0 dx +
a(u0 )2 dx +
a(v 0 )2 dx−
2
2
2
0
| 0 {z
} | 0 {z
}
F (w) = F (u + v) =
(I)
−
Z 1
f u dx −
| 0 {z
(III)
}
(II)
Z 1
f v dx .
| 0 {z
(IV )
}
Using (VF)1 we get that (I) − (IV ) = 0. Further, by the definition of the functional
F, (II) − (III) = F (u). Thus
1Z 1
F (w) = F (u) +
a(x)(v 0 (x))2 dx,
2 0
4
Edvin Listo Zec
Theorems&Proofs, PDE
and since a(x) > 0 we get that F (w) ≥ F (u) and the first implication is proven.
[⇐] We assume that F (u) ≤ F (w) ∀w ∈ H01 . Let gv () = F (u + v) for an arus that g (as a function of ) has a minimum at
bitrary v ∈ H01 . Then (MP)1 gives
d
= 0. We have that
= 0. In other words, d gv ()
=0
Z 1
1Z 1 0 2
gv () = F (u + v) =
f (u + v) dx =
a (u + v) dx −
2 0
0
Z 1
Z 1
Z 1
1
0 2
2 0 2
0 0
=
{a(u ) + a (v ) + 2au v } dx −
f u dx − f v dx,
2 0
0
0
and that the derivative of gv () is
Z 1
dgv ()
1Z 1
0 2
0 0
f v dx.
{2a(v ) + 2au v } dx −
=
d
2 0
0
But,
dgv () d
=0
= 0, which gives
Z 1
0 0
au v dx −
Z 1
f v dx = 0,
0
0
which is our desired variational formulation (VF)1 . In conclusion, we can state that
F (u) ≤ F (w) ∀w ∈ H01 ⇒ (VF)1 , which completes the proof.
3
Error estimates for BVP
Definition
(0)
A finite element formulation for the Dirichlet BVP (2.1) is given by: find uh ∈ Vh
such that:
(FEM)
Z 1
0
a(x)u0h (x)v 0 (x) dx =
Z 1
f (x)v(x) dx,
0
(0)
∀v ∈ Vh ,
(3.1)
where
(0)
Vh
= {v : v ∈ C(I, P1 (Ik )), v(0) = v(1) = 0},
Ik = [xk−1 , xk ] a subinterval of I = [0,1]. So, C(I, P1 (Ik )) denotes the set of all
continuous piecewise linear functions on the partition τh = {0 = x0 < x1 < · · · <
xM < xM +1 = 1} of I.
(0)
Note that Vh
is a finite dimensional subspace of H01 .
5
Edvin Listo Zec
Theorems&Proofs, PDE
Theorem 3.1. If u(x) is a solution to the Dirichlet BVP (2.1) and uh (x) its finite
element approximation defined by (3.1), then
(0)
||u − uh ||E ≤ ||u − v||E ,
∀v(x) ∈ Vh .
(3.2)
(0)
In other words, the finite element solution uh ∈ Vh is the best approximation in
(0)
the energy norm of the solution u, by functions in Vh .
(0)
Proof. Take v ∈ Vh
||u − uh ||2E =
=
Z 1
0
Z 1
0
=
Z 1
0
+
Z 1
arbitrarily. We have that
a(x)(u0 (x) − u0h (x))2 dx =
a(x) u0 (x) − u0h (x) u0 (x) − v 0 (x) + v 0 (x) − u0h (x) dx =
a(x) u0 (x) − u0h (x) u0 (x) − v 0 (x) dx+
a(x) u0 (x) − u0h (x) v 0 (x) − u0h (x) dx =
|0
{z
}
(0)
=0 by Galerkin orthogonality since v−uh ∈Vh ⊂H01
=
Z 1
0
≤
a(x)1/2 u0 (x) − u0h (x) a(x)1/2 u0 (x) − v 0 (x) dx ≤
Z 1
0
a(x) u (x) −
0
2
u0h (x)
= ||u − uh ||E · ||u − v||E ,
1/2 Z 1
dx
0
0
2
a(x) u (x) − v (x)
1/2
dx
=
0
(0)
∀v ∈ Vh ,
where we used Cauchy-Schwartz’ inequality in the last estimate. The result is thus
||u − uh ||E ≤ ||u − v||,
(0)
∀v ∈ Vh .
Theorem 3.2 (A priori error estimate). If u and uh are the solutions to the (BVP)1
and and (FEM) respectively, then there exists an interpolation constant Ci depending only on a(x) such that
||u − uh ||E ≤ Ci ||hu00 ||a
(0)
Proof. Since πh u(x) ∈ Vh , we may choose v = πh u(x) in (3.2) and use an estimate
from an interpolation theorem (generalised to the weighted norm || · ||a ) to get
||u−uh ||E ≤ ||u−πh u||E = ||u0 −(πh u)0 ||a ≤ Ci ||hu00 ||a = Ci
Z 1
which is the desired result and thus the proof is complete.
6
0
a(x)h2 (x)u00 (x)2 dx
1/2
,
Edvin Listo Zec
Theorems&Proofs, PDE
Remark 1. The interpolation theorem is not in the weighted norm. The a(x)
dependence of the interpolation constant Ci can be shown as follows:
0
0
||u − (πh u) ||a =
≤
Z 1
0
0
2
0
a(x) u (x) − (πh u) (x)
1/2
dx
≤
max a(x)1/2 · ||u0 − (πh u)0 ||L2 ≤ ci max a(x)1/2 ||hu00 ||L2 =
x∈[0,1]
x∈[0,1]
= ci max a(x)1/2
Z 1
x∈[0,1]
≤ ci
h(x)2 u00 (x)2 dx
1/2
≤
0
max a(x)1/2 Z
1
x∈[0,1]
min a(x)1/2
a(x)h(x)2 u00 (x)2 dx
1/2
.
0
x∈[0,1]
So, we have that
max a(x)1/2
C i = ci
x∈[0,1]
min a(x)1/2
,
x∈[0,1]
where ci is the interpolation constant from a theorem (second estimate in theorem
3.3 from the lecture notes).
Theorem 3.3 (A posteriori error estimate). There is an interpolation constant ci
depending only on a(x) such that the error in the finite element approximation of
the (BVP) satisfies
||e(x)||E ≤ ci
Z 1
0
1
h(x)2 R(uh (x))2 dx
a(x)
!1/2
,
where R(uh (x)) = f + (a(x)u0h (x))0 is the residual and e(x) := u(x) − uh (x) ∈ H01 .
Proof. We start with the definition of the energy norm:
||e(x)||2E =
=
Z 1
a(x)(e0 (x))2 dx =
Z 1
0
0
Z 1
a(x)u0 (x)e0 (x) dx −
0
a(x) u0 (x) − u0h (x) e0 (x) dx =
Z 1
0
a(x)u0h (x)e0 (x) dx.
Further, we know that e ∈ H01 , so the variational formulation (V F )1 gives
Z 1
0
0
a(x)u (x)e (x) dx =
0
Z 1
f (x)e(x) dx.
0
This gives
||e(x)||2E
=
Z 1
f (x)e(x) dx −
0
Z 1
0
7
a(x)u0h (x)e0 (x) dx.
Edvin Listo Zec
Theorems&Proofs, PDE
We now add and subtract the interpolant πh e(x) and its derivative (πh e)0 (x) to e
and e0 in the integrands, which yields
||e(x)||2E =
Z 1
0
f (x) e(x) − πh e(x) dx +
Z 1
0
f (x)πh e(x) dx −
|
−
Z 1
a(x)u0h (x)
0
0
{z
0
}
(i)
e (x) − (πh e) (x) dx −
Z 1
0
a(x)u0h (x)(πh e)0 (x) dx .
|
{z
}
(ii)
(0)
We now have that (i)−(ii) = 0, since uh (x) is a solution of (FEM) and πh e(x) ∈ Vh .
Thus we have
||e(x)||2E
=
Z 1
0
=
Z 1
0
Z 1
0
M
+1 Z xk
X
f (x) e(x) − πh e(x) dx −
f (x) e(x) − πh e(x) dx −
a(x)u0h (x) e0 (x) − (πh e)0 (x) dx =
xk−1
k=1
a(x)u0h (x) e0 (x) − (πh e)0 (x) dx.
Further, we integrate by parts in the integrals in the sum:
−
Z xk
xk−1
=−
a(x)u0h (x) e0 (x) − (πh e)0 (x) dx =
a(x)u0h (x)
xk
e(x) − πh e(x)
+
Z xk xk−1
xk 1
0 a(x)u0h (x)
e(x) − πh e(x) dx.
Now we use the fact that e(xk ) = πh e(xk ) for k = 1,2, . . . ,M + 1, where xk are the
interpolation nodes. This makes the boundary terms vanish and we have
−
Z xk
xk−1
a(x)u0h (x)
0
0
e (x) − (πh e) (x) dx =
Z xk xk−1
0 a(x)u0h (x)
e(x) − πh e(x) dx.
Now we sum over k to get
−
Z 1
0
a(x)u0h (x)
0
0
e (x) − (πh e) (x) dx =
Z 1
0
0 a(x)u0h (x)
e(x) − πh e(x) dx,
where we interpret a(x)u0h (x) on each local subinterval [xk−1 , xk ], since u0h in
general is discontinuous which implies that u00h does not exist globally on [0,1]. Thus
||e(x)||2E
=
Z 1
0
=
f (x) e(x) − πh e(x) dx +
Z 1
0
0 f (x) + a(x)u0h (x)
Z 1
0
a(x)u0h (x)
0 e(x) − πh e(x) dx.
8
e(x) − πh e(x) dx =
Edvin Listo Zec
Theorems&Proofs, PDE
0
Now we let the residual error be R(uh (x)) = f (x) + a(x)u0h (x) . This is a well0
defined function except in in the set {xk }, k = 1, . . . ,M − 1 since a(xk )u0h (xk )
not defined there. By now using Cauchy-Schwartz’ inequality we get
||e(x)||2E
=
Z 1
0
=
R(uh (x)) e(x) − πh e(x) dx =
Z 1
0
≤
is
e(x) − πh e(x)
q
h(x)R(uh (x)) a(x)
h(x)
a(x)
Z 1
0
1
q
1 2
h (x)R2 (uh (x)) dx
a(x)
!1/2 Z

0
1
!
dx ≤
e(x) − πh e(x)
a(x)
h(x)
!2
1/2
dx
.
Now, by the definition of the L2 -norm we get that
e(x) − π e(x) h
h(x)
=

Z 1

a(x)
0
a
e(x) − πh e(x)
h(x)
1/2
!2
dx
.
(3.3)
To estimate equation (3.3) we can use an interpolation estimate for e(x) in each
subinterval which gives us
e(x) − π e(x) h
h(x)
≤ Ci ||e0 (x)||a = Ci ||e(x)||E ,
a
for a constant Ci dependent on a(x). Thus,
||e(x)||2E
≤
Z 1
0
1 2
h (x)R2 (uh (x)) dx
a(x)
!1/2
Ci ||e(x)||E ,
which completes the proof.
4
Stability estimates for IVP
Here we shall consider the following problem.
(
u̇(t) + a(t)u(t) = f (t),
u(0) = u0 ,
0<t≤T
(4.1)
where f (t) is the source term and a(t) a bounded function. If a(t) ≥ 0, the problem
is called parabolic, while a(t) ≥ α > 0 yields a dissipative problem.
9
Edvin Listo Zec
Theorems&Proofs, PDE
Theorem 4.1 (Stability estimates). Using the solution formula
u(t) = u0 e
−A(t)
+
Z t
e−(A(t)−A(s)) f (s) ds,
(4.2)
0
where A(s) = 0t a(s) ds and eA(t) is the integrating factor, we can derive the following stability estimates:
R
(i) If a(t) ≥ α > 0, then
|u(t)| ≤ e−αt |u0 | +
1
(1 − e−αt ) max |f (s)|.
0≤s≤t
α
(ii) If a(t) ≥ 0, i.e. α = 0 (the parabolic case), then
|u(t)| ≤ |u0 | +
Z t
|f (s)| ds
or |u(t)| ≤ |u0 | + ||f ||L1 (0,t) .
0
Proof. We begin with (i). A(t) = 0t a(s) ds will be an increasing function of t, when
a(t) ≥ α > 0, A(t) ≥ αt. This gives us that
R
A(t) − A(s) =
Z t
a(r) dr −
Z s
a(r) dr =
a(r) dr ≥ α(t − s).
t
0
0
Z s
Thus we get e−A(t) ≤ e−αt and e−(A(t)−A(s)) ≤ e−α(t−s) . Now we use equation (4.2)
to get
Z
|u(t)| ≤ |u0 |e−αt +
t
e−α(t−s) |f (s)| ds.
(4.3)
0
By integrating we will get the following
|u(t)| ≤ e−αt |u0 | + max |f (s)|
0≤s≤t
|u(t)| ≤ e−αt |u0 | +
1 −α(t−s)
e
α
s=t
, i.e.
s=0
1
(1 − e−αt ) max |f (s)|.
0≤s≤t
α
This proves (i).
We go further to (ii). We let α = 0 in equation (4.3), so we get
|u(t)| ≤ |u0 | +
which completes the proof.
10
Z t
0
|f (s)| ds,
Edvin Listo Zec
5
Theorems&Proofs, PDE
Lax-Milgram
Firstly, we recall:
(u,v) =
Z 1
0
0
u (x)v (x) dx and `(v) =
Z 1
0
f (x)v(x) dx.
0
Secondly, in this section we will use the following expressions.
1. Bilinear: a(u,v) satisfies the same properties as scalar products, however it
does not need to be symmetric.
2. Bounded: |a(u,v)| ≤ β||u|| ||v||, β > 0 constant.
3. Elliptic: a(u,v) ≥ α||v||2 , α > 0 constant.
Theorem 5.1 (Lax-Milgram theorem). Assuming `(v) is a bounded, q
linear functional on V (a Hilbert space with scalar product (u,v) and norm ||u|| = (u,u)) and
a(u,v) is bilinear bounded and elliptic in V , then there is a unique u ∈ V such that
a(u,v) = `(v),
∀v ∈ V.
Proof. We begin by consider the variational formulation (VF) and the minimisation
problem (MP) in abstract forms:
(V)
Find u ∈ H01 such that (u,v) = `(v) ∀v ∈ H01
(M)
1
Find u ∈ H01 such that F (u) = min1 F (v), with F (v) = ||v||2 − `(v).
2
v∈H0
We have before proven that (V) and (M) are equivalent. We will now show that
(M) has a unique solution and thus we have proven the same for (V) since they are
equivalent, which in turn results in that we’ve proven the Lax-Milgram theorem.
We begin by noting that ∃ a real number σ such that F (v) > σ, ∀v ∈ H01 (otherwise
it would not be possible to minimise F ). In other words, we can write
1
1
F (v) = ||v||2 − `(v) ≥ ||v||2 − γ||v||,
2
2
where γ is the constant bounding `, i.e. |`(v)| ≤ γ||v||. However,
1
1
1
0 ≤ (||v|| − γ)2 = ||v||2 − γ||v|| + γ 2 ,
2
2
2
thus we have
1
1
F (v) ≥ ||v||2 − γ||v|| ≥ − γ 2 .
2
2
11
Edvin Listo Zec
Theorems&Proofs, PDE
Let now σ ∗ be the largest real number σ such that
F (v) > σ, ∀v ∈ H01 .
(5.1)
Take now a sequence of functions {uk }∞
k=0 such that
F (uk ) → σ ∗ .
(5.2)
To show that there exists a unique solution for (V) and (M) we will use the following:
∗
(i) It is always possible to find a sequence {uk }∞
k=0 such that F (uk ) → σ (because
is complete).
R
(ii) The parallelogram law:
||a + b||2 + ||a − b||2 = 2||a||2 + 2||b||2 .
Using (ii) and the linearity of ` we get
||uk − uj ||2 = 2||uk ||2 + 2||uj ||2 − ||uk + uj ||2 − 4`(uk ) − 4`(uj ) + 4`(uk + uj ) =
= 2||uk ||2 − 4`(uk ) + 2||uj ||2 − 4`(uj ) − ||uk + uj ||2 + 4`(uk + uj ) =
uk + uj
= 4F (uk ) + 4F (uj ) − 8F
,
2
where we used the definition F (v) = 12 ||v||2 − `(v), with v = uk , v = uj and
v = (uk + uj )/2. Also, by the linearity of ` we got:
uk
−||uk + uj ||2 + 4`(uk + uj ) = −4 uk + uj
+ uj 2
+ 8`
2
2
= −8F
uk + uj
.
2
Now, since F (uk ) → σ ∗ and F (uj ) → σ ∗ we have that
||uk + uj ||2 ≤ 4F (uk ) + 4F (uj ) − 8σ ∗ −−−−→ 0.
k,j→∞
1
1
We have thus shown that {uk }∞
k=0 is a Cauchy sequence. Since {uk } ⊂ H0 and H0 is
complete, {uk } is a convergent sequence. In other words, ∃u ∈ H01 : u = limk→∞ uk .
So by the continuity of F we have
lim F (uk ) = F (u).
k→∞
(5.3)
Now (5.2) and (5.3) gives us that F (u) = σ ∗ and by (5.1) and the definition of σ ∗
we have
F (u) < F (v), ∀v ∈ H01 .
Thus we have shown that there is a unique solution for (M), which means there
is a unique solution for (V) since (M) ⇔(V). Also, we didn’t use symmetry of the
scalar products, which means that we’ve proven the Lax-Milgram theorem with the
bilinear, bounded and elliptic a(u,v).
12
Edvin Listo Zec
6
Theorems&Proofs, PDE
Error estimates for the Poisson equation
In this section we will study the Poisson equation in higher dimensions:
(
−∆u = f, in Ω ⊂
u = 0, on ∂Ω.
Here Ω is a bounded domain in
Rd, d = 2,3
(6.1)
Rd, with polygonal boundary ∂Ω.
Further notation we are going to use is
τh = {K : ∪K = Ω},
where τh is a triangulation of the domain Ω by the element K with the maximum
diagonal h = max diag(K). Also, we shall consider continous, piecewise linear
approximations for the solution u ∈ H01 (Ω) in a finite dimensional subspace defined
as:
(0)
Vh = {v(x) : v continous, piecewise linear on τh , v = 0 on ∂Ω}.
Theorem 6.1 (A priori error estimate for the Poisson equation). Let e = u − U
be the error in the continous, piecewise linear approximation U of the solution u of
the Poisson equation (6.1). Then
||∇e|| = ||∇(u − U )|| ≤ C||hD2 u||,
for some constant C, and where
D2 u = (u2xx + 2u2xy + u2yy )1/2
Proof. For e = u − U we have ∇e = ∇u − ∇U = ∇(u − U ). We consider the
variational formulation of the problem:
(VF)
Z
∇u · ∇v dx =
Ω
Z
f v dx,
Ω
∀v ∈ H01 (Ω),
(6.2)
and the same for the approximation U :
(0)
(Vh )
Z
∇U · ∇v dx =
Ω
Z
Ω
f v dx,
(0)
∀v ∈ Vh .
(6.3)
(0)
We now subtract equation (6.3) from (6.2) where we restrict v to Vh , and we get
the Galerkin orthogonality:
Z
Ω
(∇u − ∇U ) · ∇v dx =
Z
∇e · ∇v dx = 0,
Ω
13
(0)
∀v ∈ Vh .
(6.4)
Edvin Listo Zec
Theorems&Proofs, PDE
Further, we get that
||∇e||2 =
Z
Z
∇e · ∇e dx =
Ω
∇e · ∇(u − U ) dx =
Ω
Z
∇e · ∇u dx −
Ω
Z
∇e · ∇U dx.
Ω
(0)
By using the Galerkin orthogonality (6.4), which we can do since U ∈ Vh , we get
that
Z
∇e · ∇U dx = 0.
Ω
Thus, we can remove the ∇U term and insert
2
||∇e|| =
Z
∇e · ∇u dx −
Ω
Z
∇e · ∇v dx =
Z
Ω
R
Ω
(0)
∇e · ∇v dx = 0, ∀v ∈ Vh :
∇e · ∇(u − v) dx ≤ ||∇e|| ||∇(u − v)||.
Ω
This means that
||∇(u − U )|| ≤ ||∇(u − v)||,
(0)
∀v ∈ Vh .
(6.5)
Thus, ∇U is, in the L2 -norm, closer to the exact solution ∇u than any other ∇v,
(0)
v ∈ Vh . In other words, measuring in H01 -norm, the error u − U is orthogonal to
(0)
(0)
Vh . It is possible to show that there is an interpolant v ∈ Vh such that
||∇(u − v)|| ≤ C||hD2 u||,
(6.6)
where h = h(x) =diag(K), x ∈ K and C is a constant independent of h. If we now
combine equations (6.5) and (6.6) we get
||∇e|| = ||∇(u − U )|| ≤ C||hD2 u||,
which indicates that the error is small if h(x) is sufficiently small depending on D2 u.
And thus the proof is done.
Theorem 6.2 (A posteriori error estimate for the Poisson equation). Let u be the
solution of the Poisson equation (6.1) and U its continous, piecewise linear finite
element approximation. Then there is a constant C independent of u and h, such
that
||u − U || ≤ C||h2 r||,
where r = f + ∆h U is the residual with ∆h being the discrete Laplacian defined by
X
(∆h U, v) =
(∇U, ∇v)K .
K∈τh
Proof. We begin by considering the dual problem
(
−∆ϕ(x) = e(x),
ϕ(x) = 0, x ∈ ∂Ω,
x∈Ω
e(x) = u(x) − U (x).
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Edvin Listo Zec
Theorems&Proofs, PDE
Then we have e(x) = 0, ∀x ∈ ∂Ω, and by using Green’s formula we get that
||e||2 =
Z
ee dx =
Z
Ω
Z
e(−∆ϕ) dx =
Ω
∇e · ∇ϕ dx.
Ω
By the Galerkin orthogonality and the boundary condition ϕ(x) = 0, ∀x ∈ ∂Ω, we
get
2
||e|| =
=
Z
ZΩ
∇e · ∇ϕ dx −
Z
∇e · ∇v dx =
Z
Ω
∇e · ∇(ϕ − v) dx =
Ω
(−∆e)(ϕ − v) dx ≤ ||h2 r|| ||h−2 (ϕ − v)|| ≤
Ω
≤ C||h2 r|| ||∆ϕ|| ≤ C||h2 r|| ||e||,
where we use that −∆e = −∆u + ∆h U = f + ∆h U = r is the residual and choose
v as an interpolant of ϕ. Hence, the final result becomes
||u − U || ≤ C||h2 r||.
7
Stability and energy estimates for the heat equation in Rn
We will consider the initial boundary value problem for the heat conduction in Rn :







u̇(x,t) − ∆u(x,t) = f (x,t),
u(x,t) = 0,
u(x,0) = u0 (x),
Ω⊂
Rd, d = 1,2,3
(DE)
0 < t < T, (BC)
x ∈ Ω (IC)
(7.1)
Theorem 7.1 (Energy estimates). Let f ≡ 0. Then the solution u of the heat
equation (7.1) satisfies the following stability estimates:
max ||u||(t), 2
Z t
0
2
||∇u|| (s) ds ≤ ||u0 ||2
1
||∇u||(t) ≤ √ ||u0 ||
2t
Z t
1/2
2
s||∆u|| (s) ds
0
1
≤ ||u0 ||
2
1
||∆u||(t) ≤ √ ||u0 ||
2t
Z t
(7.2)
(7.3)
(7.4)
(7.5)
s
1
t
||u̇||(s) ds ≤
ln ||u0 ||
2
15
(7.6)
Edvin Listo Zec
Theorems&Proofs, PDE
Proof. Let f ≡ 0. To derive the first two estimates, we multiply (7.1) by u and
integrate over Ω:
Z
Z
u̇u dx − (∆u)u dx = 0.
(7.7)
Ω
Ω
1 d 2
u,
2 dt
so by using Green’s formula with the Dirichlet boundary
We note that u̇u =
condition u = 0 on ∂Ω we get
−
Z
(∆u)u dx = −
Z
(∇u · n)u ds +
∂Ω
Ω
Z
∇u · ∇u dx =
Ω
Z
|∇u|2 dx.
Ω
We can thus write equation (7.7) as:
Z
1 d
1 d Z 2
u dx + |∇u|2 dx = 0 ⇔
||u||2 + ||∇u||2 = 0,
2 dt Ω
2 dt
Ω
(7.8)
where || · || denotes the L2 (Ω)-norm. Now we substitute t with s and integrate
equation (7.8) over s ∈ (0,t) to get
Z t
Z t
1
1
1Z t d
||∇u||2 (s) ds = ||u||2 (t) − ||u||2 (0) +
||∇u||2 ds = 0.
||u||2 (s) ds +
2 0 ds
2
2
0
0
Now by inserting u(0) = u0 we have
2
||u|| (t) + 2
Z t
0
||∇u||(s) ds = ||u0 ||2 .
Thus we have the first two estimates:
||u||(t) ≤ ||u0 ||,
and
1
||∇u||2 (s) ds ≤ ||u0 ||2 .
2
Z t
0
To derive estimates (7.3) and (7.4) we now instead multiply u̇ − ∆u = 0 with −t∆u
and integrate over Ω to get
−t
Z
u̇∆u dx + t
Z
Ω
(∆u)2 dx = 0.
Ω
Green’s formula (u = 0 on ∂Ω ⇒ u̇ = 0 on ∂Ω) gives us that
Z
u̇∆u dx = −
Ω
Z
∇u̇ · ∇u dx = −
Ω
1 d
||∇u||2 ,
2 dt
so equation (7.9) can be rewritten as
t
1 d
||∇u||2 + t||∆u||2 = 0.
2 dt
16
(7.9)
Edvin Listo Zec
Theorems&Proofs, PDE
d
||∇u||2 =
Further, the above equation can be rewritten using the relation t dt
2
||∇u|| :
d t||∇u||2 + 2t||∆u||2 = ||∇u||2 .
dt
Again, we substitute t with s and integrate over (0,t):
Z t
0
d
(t||∇u||2 )−
dt
Z t
Z t
d 2
2
s||∇u|| ds +
2s||∆u|| (s) ds =
||∇u||2 (s) ds.
ds
0
0
Now, from equation (7.2) we have that the last equality is estimated as
t||∇u||2 (t) + 2
Z t
0
1
s||∆u||2 (s) ds ≤ ||u0 ||2 .
2
From this we have
1
||∇u||(t) ≤ √ ||u0 ||
2t
Z t
and
2
1/2
s||∆u|| (s) ds
0
1
≤ ||u0 ||,
2
which are (7.3) and (7.4). The estimate (7.5) is proved analogously.
To prove the last estimate we use u̇ = ∆u and (7.5) to write
Z t
||u̇||(s) ds =
Z t
||∆u||(s) ds =
Z t
≤
Z t
1
ds
s
1/2 Z t
1 √
√ s||∆u||(s) ds ≤
s
s||∆u||2 (s) ds
1/2
s
1
t
≤
ln ||u0 ||.
2
In the last two inequalities we used Cauchy-Schwartz inequality and (7.4), respectively. Thus all estimates are proven, and the proof is done.
17