UNIVERSITY OF OSLO
Faculty of Mathematics and Natural
Sciences
Examination in: MAT-INF4300 — Partial Differential Equations
and Sobolev Spaces I
Solution, exam 13.12.2010
Problem 1
1a
We use Stokes’ theorem and get
Z
Z
c dxdy =
−∆u dxdy
U
U
Z
∂u
dS.
=−
∂U ∂ν
R
R
Since U c dxdy = cπ and ∂U dS = 2π, we get cπ = −2π, which yields
c = −2.
1b
We have
1
− (rv 0 )0 = −2,
r
which gives rv 0 = r2 +C, for some constant C. Using the boundary condition
∂u
0
∂ν = v = 1, for r = 1, we get C = 0. We integrate one more time and obtain
v(r) =
r2
+C
2
for some constant C.
1c
We multiply the equation (1a) with a test function v and integrate by part.
Using Stokes’ theorem, we obtain
Z
Z
Z
∂u
v dS +
Du · Dv dxdy =
v dxdy
−
∂U ∂ν
U
U
which gives, by the boundary condition (1b),
Z
Z
B(u, v) = c
v dxdy +
U
= f (v).
∂U
v dS
Examination in MAT-INF4300 (Monday 13. December 2010)
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Solution Page 2
We have
Z
|f (v)| ≤ |c| kvkL2 (
1
2
Z
dxdy) + (
U
∂U
1
dS) 2 kvkL2 (∂U ) .
(1)
By the trace theorem, we have
kvkL2 (∂U ) ≤ C kvkH 1 (U )
for some constant C which is independent on v. Thus, (1) yields
√
√
|f (v)| ≤ (|c| π + C 2π) kvkH 1
and the mapping f is continuous
1d
Let u0 (x, y) =
r2
2.
The function u0 is a weak solution of (1). We have
B(u0 , v) = f (v) and B(u, v) = f (v)
for all v ∈ H 1 . After substracting these two identities, we obtain
B(u − u0 , v) = 0
and, taking v = u − u0 , it yields
kD(u − u0 )k2L2 = 0
so that D(u − u0 ) = 0. Since the domain U is connected, u − u0 is constant,
that is,
u = u0 + C,
for some constant C and
u(x, y) =
x2 + y 2
+ C.
2
Problem 2
2a
The Banach space X is compactly embedded in Y if, given any bounded
sequence in X, there exists a subsequence which converges in Y . The Banach
space Y is continuously embedded in Z if the injection u 7→ u from Y to Z
is continuous, that is, there exists a constant C such that
kukZ ≤ C kukY
for all u ∈ Y .
Examination in MAT-INF4300 (Monday 13. December 2010)
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Solution Page 3
2b
Let us assume that there exists ε0 > 0 such that, for all C ≥ 0, there exists
u ∈ X such that
kukY > ε0 kukX + C kukZ .
Taking C = k, we construct the sequence uk such that
kuk kY ≥ ε0 kuk kX + k kuk kZ .
We set vk =
uk
kuk kX .
We have
kvk kY ≥ ε0 + k kvk kZ .
(2)
Since kvk kX = 1, there exists a subsequence vkj which converges to some v
in Y . Since Y is continuously embedded in Z, vkj converges to v also in Z.
From (2), we get
vk ≤ 1 vk j Y
j Z
kj
and, therefore, limj→∞ vkj = 0 so that v = 0. However, (2) also yields
vk ≥ ε0 .
j Y
Letting j tend to ∞, we get kvk ≥ ε0 which leads to a contradiction.
2c
Let u ∈ W 1,p (0, 1). We set
Y = L∞ (0, 1),
X = W 1,p (0, 1),
Z = L1 (0, 1).
We have X ⊂⊂ Y and L∞ (0, 1) is continuously embedded in L1 (0, 1).
Indeed,
Z
1
kukL1 (0,1) =
0
|u| dx ≤ kukL∞ (0,1) .
Hence, we can use the result of the previous question and get that for any
ε̄ ≥ 0 there exists C̄ such that
kukL∞ ≤ ε̄ kukW 1,p + C̄ kukL1
or
kukL∞ ≤ ε̄C1 (kukLp + kDukLp ) + C̄ kukL1
1
(Here we use that (|a|p + |b|p ) p ≤ C1 (|a| + |b|) for some constant C1 which is
R1
1
independent on a and b). Since kukLp = ( 0 |u|p ) p ≤ kukL∞ , it follows that
kukL∞ ≤ ε̄C1 kukL∞ + ε̄C1 kDukLp + C̄ kukL1
or
kukL∞ ≤
ε̄C1
C̄
kDukLp +
kukL1 .
1 − ε̄C1
1 − ε̄C1
For any given ε > 0, we choose ε̄ such that
ε̄C1
1−ε̄C1
W 1,p (0, 1)
(3)
≤ ε. For such ε̄, there
exists C̄ for which (3) holds for all u ∈
and, denoting C =
we get
kukL∞ ≤ ε kukLp + C kukL1 .
C̄
1−ε̄C1 ,
Examination in MAT-INF4300 (Monday 13. December 2010)
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Solution Page 4
Problem 3
3a
We multiply (2a) by a test function and integrate by part. We get
Z
Z
(Lu)v dxdy = (−(5 + x)uxx + (x2 + y 2 − 5)uyy )v dxdy
U
ZU
= (ux ((5 + x)v) − uy ((x2 + y 2 − 5)v)y dxdy
ZU =
(5 + x)ux vx + ux v + (5 − x2 − y 2 )uy vy − 2yuy v dxdy
U
= B(u, v).
3b
We have
2
X
Z
B(u, v) =
U
ai,j (x, y)uxi uxj +
i,j=1
2
X
bi (x, y)ux v dxdy
i=1
where a1,1 = 5+x, a2,2 = 5−x2 −y 2 and a1,2 = a2,1 = 0. Since a1,1 (x, y) ≥ 4
and a2,2 (x, y) ≥ 5 − x2 − y 2 for all (x, y) ∈ U , we get
2
X
ai,j (x, y)ξi ξj ≥ 4(ξ12 + ξ22 ) = 4 |ξ|2
i,j=1
and B is uniformly elliptic.
3c
We have
Z
4 kDukL2 ≤ B(u, u) −
Z
(ux u) dxdy + 2
U
(yuy u) dxdy
U
1
≤ B(u, u) + (kux k2L2 + kukL2 ) + (kuy k2L2 + kukL2 )
2
3
3
≤ B(u, u) + kDuk2L2 + kuk2L2 .
2
2
Hence,
B(u, u) ≥
5
3
kDuk2L2 − kuk2L2 .
2
2
3d
The bilinear mapping B is continuous (it follows from Cauchy-Schwarz). Let
us prove that B is coercive. We have
5
3
kDuk2L2 − kDuk2L2
2
8
17
2
≥
kDukL2
8
B(u, u) ≥
Examination in MAT-INF4300 (Monday 13. December 2010)
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Solution Page 5
Since
kuk2H 1 = kuk2L2 + kDuk2L2
1
≤ kDuk2L2 + kDuk2L2
4
5
≤ kDuk2L2 ,
4
it yields
17
kukH 1
10
and B is coercive. By Lax-MilgramR theorem, it follows that there exists a
unique u ∈ H01 such that B(u, v) = U f v dxdy for all v ∈ H01 .
B(u, u) ≥