Math 5520 Homework 3 Solutions March, 2016

Math 5520 Homework 3 Solutions
March, 2016
1. Consider the boundary value problem
−∇ · A∇u = f (x), in Ω ⊂ Rd ,
(1)
u = gD on ΓD ,
(2)
γu + A∇u · ~n = gR on ΓR ,
(3)
where ∂Ω = ΓD ∪ ΓR , |ΓD | > 0, 0 ≤ γ(x) ≤ γ1 , and A(x) is a d × d, symmetric
matrix satisfying
α0 χT χ ≤ χT A(x)χ ≤ α1 χT χ, ∀χ ∈ Rd , ∀x ∈ Ω.
Here, 0 < α0 ≤ α1 . You may assume that the domain boundary and all
problem data are smooth and bounded as needed.
a) Derive the weak formulation of (1)-(3) .
b) Prove that there exists a unique weak solution.
c) Prove that if u ∈ H 2 (Ω), then it is a “strong” solution (in the L2 -sense)
to (1)-(3).
d) Give the energy minimization formulation of the problem and show
equivalence with the weak formulation.
Solutions:
(a) Multiply through (1) by v ∈ C ∞ (Ω), then integrate over the domain and
apply the Divergence Theorem:
Z
Z
Z
A∇u · ∇v dx −
(A∇u · n̂) v dσ =
f v dx.
Ω
∂Ω
Ω
Due to the Dirichlet boundary condition (2), we now restrict v = 0 on ΓD .
Furthermore, we apply (3) on ΓR and thereby eliminate the normal derivative
of the solution on the boundary. The result is:
Z
Z
Z
A∇u · ∇v dx +
(γu − gR ) v dσ =
f v dx.
Ω
ΓR
Ω
Now move the gR -term to the right-hand side. For notational convenience, we
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now present the weak problem as follows: Find u ∈ VD (Ω) such that
= L(v), for all v ∈ V0 , with
v ∈ H 1 (Ω) | v|ΓD = gD ,
V0 =
v ∈ H 1 (Ω) | v|ΓD = 0 ,
Z
Z
a(u, v) ≡
A∇u · ∇v dx +
γuv dσ,
Ω
ΓR
Z
Z
and L(v) ≡
f v dx +
gR v dσ.
a(u, v)
VD
=
Ω
(4)
(5)
(6)
(7)
(8)
ΓR
(b) One may apply the Riesz Representation Theorem, but I will show the
result using Lax-Milgram, since it requires less work to show that the
assumptions hold. First, note that V0 is a Hilbert space. We reformulate the
weak problem by choosing any known G ∈ VD , which we may assume to exist
given a smooth boundary and |ΓD | > 0, then decompose u = w + G so that
w ∈ V0 is sought to satisfy
a(w, v) = L(v) − a(G, v) ≡ L(v), for all v ∈ V0 .
(9)
Claim 1: a(·, ·) is coercive. Indeed, since we may use the Poincaré inequality
for |ΓD | > 0, given any v ∈ V0 we have
Z
Z
Z
1/2
2
2
a(v, v) =
|A ∇v| dx +
γv dσ ≥ α0
|∇v|2 dx
Ω
ΓR
Ω
Z
Z
α0
α0
2
|∇v| dx +
|∇v|2 dx
=
2 Ω
2 Ω
Z
Z
α0
α0
|∇v|2 dx +
v 2 dx ≥ Ckvk2H 1 ,
≥
2 Ω
2CP2 Ω
where CP is the appropriate Poincaré constant and
α0 α0
C ≡ min
,
> 0.
2 2CP2
Claim 2: a(·, ·) is continuous. We apply the standard series of inequalities to
get the desired bounds:
a(w, v) ≤ α1 |w|1 |v|1 + γ1 kwkL2 (ΓR ) kvkL2 (ΓR )
≤ α1 |w|1 |v|1 + γ1 kwkL2 (∂Ω) kvkL2 (∂Ω)
≤ α1 |w|1 |v|1 + C1 kwkH 1 (Ω) kvkH 1 (Ω)
≤ C2 kwkH 1 (Ω) kvkH 1 (Ω) ,
using the Trace Inequality and bounding the H 1 -semi-norm above by the full
H 1 -norm. Similary, we show
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Claim 3: L(·) is continuous.
|L(v)| ≤ kf kL2 (Ω) kvkL2 (Ω) + α1 |G|1 |v|1 + γ1 kGkL2 (ΓR ) + kgR kL2 (ΓR ) kvkL2 (ΓR )
≤ C3 kf kL2 (Ω) + α1 |G|1 + γ1 kGkL2 (ΓR ) + kgR kL2 (ΓR ) kvkH 1 (Ω) ,
so that there is some C4 > 0 such that
|L(v)| ≤ C4 kvkH 1 ⇒ kLk∗ ≡
sup
v6=0, v∈V0
|L(v)|
≤ C4 < ∞.
kvkH 1
The above claims justify the use of Lax-Milgram to conclude that there exists
a unique function w = w(G) for our choice of G such that (9) is satisfied. It
follows that for u = w + G,
a(u, v) = L(v), for all v ∈ V0 ,
with u ∈ VD by construction. This u must be unique, meaning that it is in
fact independent of G. Indeed, given any two functions u1 ∈ VD and u2 ∈ VD
satisfying the weak problem, it follows that for e ≡ u1 − u2 ∈ V0 , we have
a(u1 , v) − a(u2 , v) = a(e, v) = L(v) − L(v) = 0,
for all v ∈ V0 . We may choose v = e, so that by coercivity of the bilinear form
we have
0 ≤ kek2H 1 ≤ Ca(e, e) = 0 ⇒ kekH 1 = 0,
(for some fixed C > 0) thus u1 = u2 and the solution is unique. We have thus
proved the well-posedness of (4)-(8).
(c) Now let u ∈ H 2 solve the weak problem; we show it is then an L2 -strong
solution. We are allowed to assume enough smoothness of problem data and
the domain boundary, etc., so that from the weak problem we apply the
Divergence Theorem again to see that if a(u, v) = L(v), then
Z
Z
Z
Z
(γu + A∇u · n̂) v dσ =
f v dx +
gR v dσ,
(−∇ · A∇u) v dx +
Ω
Ω
ΓR
for all v ∈ V0 . Now define
F ≡ −∇ · A∇u − f and G ≡ γu + A∇u · n̂ − gR ,
with G defined on ΓR . It follows that
Z
Z
Fv dx +
Gv dσ = 0, for all v ∈ V0 .
Ω
ΓR
Note that F ∈ L (Ω). Given any v ∈ C0∞ (Ω) ⊂ V0 , we see that
Z
Fv dx = 0,
2
Ω
3
ΓR
implying that F = 0 by density of C0∞ in L2 . This shows that (1) holds in an
L2 -sense. We need not prove anything for (2). Now, since F = 0, it follows
that
Z
Gv dσ = 0, for all v ∈ V0 .
(10)
ΓR
Again, we are allowed to assume what we want in this problem for the domain
smoothness and data smoothness, so since u ∈ H 2 (Ω) we may assume
γu + A∇u · n̂ ∈ H 1/2 (∂Ω).
Then with appropriate assumptions on gR , (10) implies that G = 0 and
thus (3) holds in a weak sense.
(d) Define the functional
J (v) ≡
1
a(v, v) − L(v),
2
for all v ∈ V0 . The energy minimization (EM) formulation is to find w ∈ V0
such that J (w) ≤ J (v) holds for all v ∈ V0 .
(Weak implies EM). Given w ∈ V0 solving (9), we note that
1
s2
a(w, w) + s a(w, v) + a(v, v) − L(w) − sL(v)
2
2
s2
1
= a(w, w) + a(v, v) − L(w)
2
2
s2
= J (w) + a(v, v) ≥ J (w),
2
J (w + sv) =
proving that w solves the EM problem.
(EM implies weak). We note that
d
J (w + sv)|s=0 = 0,
ds
for any v ∈ V0 with v 6= 0. Then from the above expansion, we see that
a(w, v) − L(v) = 0, proving the weak problem holds.
2. Recall the operator A : V → V in the proof of the Lax-Milgram theorem
(see Lecture 11, slides 3-4): for φ ∈ V ,
hAφ, vi = a(φ, v), ∀v ∈ V.
Finish the proof of the Lax-Milgram theorem. Follow the approach outlined in
class:
a) Prove that A is linear,
b) prove that Range(A) is a closed subspace of V ,
c) prove that A is onto V (Projection Theorem),
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d) conclude that Range(A) = V , and then that the final result holds by way of
the Riesz Representation Theorem.
Solutions:
(a) Given scalars ci and functions φi ∈ V , i = 1, 2, we see that
hA (c1 φ1 + c2 φ2 ) , vi = a (c1 φ1 + c2 φ2 , v) = c1 a (φ1 , v) + c2 a (φ2 , v)
= c1 hAφ1 , vi + c2 hAφ2 , vi = hc1 Aφ1 + c2 Aφ2 , vi , for all v ∈ V.
It follows that A (c1 φ1 + c2 φ2 ) = c1 Aφ1 + c2 Aφ2 and thus A is linear.
(b) Given scalars ci and functions ψi = Aφi ∈ V , i = 1, 2, we see from (a) that
c1 ψ1 + c2 ψ2 = c1 Aφ1 + c2 Aφ2 = A (c1 φ1 + c2 φ2 ) ,
so that c1 ψ1 + c2 ψ2 ∈ Range(A) and thus Range(A) is a linear subspace. Let
ψn = Aφn be a sequence with
lim kψn − ψkV = 0,
n→∞
for some ψ ∈ V . Note then that by coercivity of a(·, ·), which is assumed in
the statement of Lax-Milgram,
kφn − φm k2V ≤ Ca(φn − φm , φn − φm ) = C hAφn − Aφm , φn − φm i
= C hψn − ψm , φn − φm i ≤ Ckψn − ψm kV kφn − φm kV
⇒ kφn − φm kV ≤ Ckψn − ψm kV .
Since ψn is strongly-convergent in V , it is also Cauchy and thus so is φn by
the above inequality. Since V is Hilbert (hence complete), we see that φn → φ
strongly for some φ in V . Next, note that for all v ∈ V ,
kAvk2V = hAv, Avi = a(v, Av) ≤ C kvkV kAvkV ⇒ kAvkV ≤ C kvkV ,
by continuity of the bilinear form. It follows that
kAφ − ψkV ≤ kAφ − Aφn kV + kAφn − ψkV = kA(φ − φn )kV + kψn − ψkV
≤ Ckφ − φn kV + kψn − ψkV → 0
as n → ∞, so that kAφ − ψkV = 0 and thus Aφ = ψ. Hence the range of A is
closed.
(c) The Projection
L Theorem says that we may write V as the direct sum
V = Range(A) Range(A)⊥ . In other words, each v ∈ V may be written
uniquely as v = P v + (I − P )v with P v the orthogonal projection operator onto
Range(A). We apply coercivity of the bilinear form to see that for any v ∈ V ,
k(I − P )vk2V ≤ Ca((I − P )v, (I − P )v) = C hA(I − P )v, (I − P )vi = 0,
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since A(I − P )v ∈ Range(A) and I − P )v ∈ Range(A)⊥ . Then v = P v for all
v ∈ V and hence V = Range(A).
(d) Since L is assumed to be a continuous linear functional on V , there exists
ψ ∈ V such that L(v) =< ψ, v >, for all v ∈ V by the Riesz Representation
Theorem. Then since ψ = Au for some u ∈ V , it follows that a(u, v) = L(v)
holds for all v ∈ V . The uniqueness of u is shown via coercivity; if
a(u1 , v) = a(u2 , v) = L(v) for all v, then
a(u1 − u2 , v) = 0 ⇒ a(u1 − u2 , u1 − u2 ) = 0 ≥ Cku1 − u2 kV ,
hence u1 = u2 .
3. a) Derive the weak formulation of the BVP
−(a(x)u0 (x))0 + b(x)u0 (x) + c(x)u(x)
=
f (x), 0 < x < 1,
(11)
u(0) = u(1)
=
0,
(12)
where 0 < a0 ≤ a(x) ≤ a1 < ∞, 0 < a0 ≤ c(x) ≤ c1 < ∞ and |b(x)| ≤ a0 for all
x ∈ [0, 1].
b) Prove that the weak formulation has a unique solution, assuming all
problem data to be smooth and bounded as needed.
Solutions:
(a) It should be clear that the weak problem is: Find u ∈ H01 (0, 1) such that
1
Z
au0 v 0 + bu0 v + cuv dx =
1
Z
0
f v dx, ∀v ∈ H01 (0, 1).
0
(b) We associate the bilinear form and linear functional
Z
a(u, v) ≡
1
au0 v 0 + bu0 v + cuv dx, L(v) ≡
0
Z
1
f v dx.
0
The continuity of both a(·, ·) and L(·) should be clear, so I will just show
coercivity. Then the well-posedness follows by Lax-Milgram. Indeed,
Z
1
0 2
0
Z
2
a|v | + bv v + cv dx ≥ a0
a(v, v) =
0
1
0 2
2
Z
1
|v | + v dx − a0
0
|v 0 ||v| dx
0
≥ a0 kvk2H 1 − a0 kv 0 kL2 kvkL2 .
Now apply Young’s inequality in the form
kv 0 kL2 kvkL2 ≤
1
1
1 0 2
kv kL2 + kvk2L2 = kvk2H 1 .
2
2
2
Coercivity follows from inserting this in the previous result above.
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4. Modify your code to approximate
−(a(x)u0 (x))0 + u0 (x)
=
0, 0 < x < 1,
(13)
u(0) = 1, u(1)
=
2.
(14)
Choose a(x) as in Homework 1, problem 2(c). Run the code with
n = 10, 20, 40, 80, 160, 320 elements and compute the convergence rates for the
error in the L2 and H 1 norms. Use the Richardson extrapolation method to
do this; do not compute errors by deriving and using the true solution (see
Lecture 18, slide 10).
In terms of convergence rates, we observe in Table 1 precisely the
optimal-order results that one expects for our smooth solution. Results could
vary somewhat depending on how you coded your methods and calculated the
errors. But the convergence rates should be optimal.
n
10
20
40
80
160
320
L2 difference
—
8.03E − 4
2.01E − 4
5.04E − 5
1.26E − 5
3.15E − 6
Rate
—
—
1.996
1.999
2.000
2.000
H 1 difference
—
2.87E − 2
1.44E − 2
7.21E − 3
3.60E − 3
1.80E − 3
Rate
—
—
0.997
0.999
1.000
1.000
Table 1: The convergence rates are optimal in both norms.
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