EXERCISES PDE 31.10.12-02.11.12
1. Exercise
N
Let U ∈ R be a bounded open set. We say that v ∈ C 2 (Ū ) is subharmonic iff −∆v ≤ 0 in U .
(a) Prove that subharmonic functions enjoy the following form of the mean-value property: for
every x ∈ U , for every r > 0 such that B(x, r) ⊂ U
v(x) ≤
v(y) dy
B(x,r)
(b) Use the previous property to prove the maximum principle for subharmonic functions:
max v = max v .
∂U
Ū
Try also to give an elementary direct proof of this result.
Hint: first consider the case −∆v < 0 . Then prove the general case by approximating v with
the functions vε (x) := v(x) + ε|x|2 .
(c) Consider φ ∈ C ∞ (R) convex and let u be harmonic in U . Prove that the composition φ(u)
is subharmonic.
(d) Let u be harmonic in U . Prove that |∇u|2 is subharmonic.
Solution: (a) Fix x ∈ U and r > 0 such that B(x, r) ⊂ U . Define the function
ψ(ρ) :=
v(ξ) dS(ξ) =
∂B(x,ρ)
v(x + ρζ) dS(ζ)
∂B(0,1)
for ρ ∈ (0, r) . Observe that
lim φ(ρ) = v(x) .
(1.1)
ρ→0+
Now using the divergence theorem one has
ψ 0 (ρ) =
∇v(x + ρζ) · ζ dS(ζ) =
∂B(0,1)
=
∂B(x,ρ)
∇v(ξ) ·
∂B(x,ρ)
∂v
r
(ξ) dS(ξ) =
∂ν
N
ξ−x
dS(ξ) =
ρ
∆v(y) dy ≥ 0
B(x,ρ)
so that the function ψ(ρ) is increasing. Taking into account (1.1) this gives
ψ(ρ) ≥ ψ(0+) = v(x)
whence, also using Fubini’s theorem, and denoting by αN the voulume of the unit sphere
ˆ
ˆ r ˆ
1
1
v(y) dy =
v(y)
dy
=
v(ξ)
dS(ξ)
dρ =
αˆN rN B(x,r)
αN ˆrN 0
B(x,r)
∂B(x,ρ)
r
r
v(x)
1
= N
N ρN −1 ψ(ρ) dρ ≥ N
N ρN −1 dρ = v(x) ,
r
r
0
0
as required.
(b) Assume that v takes its maximum at an interior point x0 . But for every r > 0 such that
B(x0 , r) ⊂ U one has, by part (a)
v(y) − v(x0 ) dy ≥ 0 ;
B(x0 ,r)
1
2
EX1
since the integrand must be nonpositive this implies v(y) = v(x0 ) for every y ∈ B(x0 , r) . It follows
that v is constant in the connected component of U containing x0 (strong maximum principle) which
implies the weaker conclusion. Alternative proof will appear here next week.
(c) By the usual rules of differentiation, one has
∆[φ(u)] = φ0 (u)∆u + φ00 (u)|∇u|2 .
Since ∆u = 0 and φ00 ≥ 0 by convexity, one gets ∆[φ(u)] ≥ 0 as required.
∂u
are still harmonic. By the previous step with
(d) For every i = 1, . . . , N the derivatives ∂x
i
2
∂u
φ(t) = t2 , the functions ∂xi
are subharmonic. Then
|∇u|2 =
N
X
∂u 2
∂xi
i=1
must be subharmonic as well.
2. Exercise
Let U ∈ RN be a bounded open set. Let v ∈ C 2 (Ū ) satisfy
(
−∆v = f in U
v = g on ∂U .
Prove that there exist a constant CU only depending on U such that
max |v| ≤ CU (max |g| + max |f |) .
∂U
Ū
Ū
Solution: Let L := maxŪ |f | , and define v̂(x) := v(x) +
L
2
2N |x| .
One has
−∆v̂ = f − L = f − max |f | ≤ 0
Ū
and therefore by the maximum principle for subharmonic functions and trivial estimates
1
|y|2 .
max v ≤ max v̂ = max v̂ ≤ max g + L max
∂U
∂U
y∈∂U 2N
Ū
Ū
Therefore, setting MU := maxy∈∂U
1
2
2N |y|
and recalling the expression of L, we get
max v ≤ max g + MU max |f | ≤ max |g| + MU max |f |
∂U
Ū
∂U
Ū
(2.1)
Ū
where the last estimate is obvious. The same argument with −v in place of v gives
max(−v) ≤ max | − g| + MU max | − f | = max |g| + MU max |f | .
Ū
∂U
Ū
∂U
(2.2)
Ū
Combining (2.1) and (2.2) gives
max |v| ≤ max |g| + MU max |f |
Ū
∂U
Ū
and eventually the conclusion, with CU the larger between 1 and MU .
EXERCISES PDE WS 2012-2013
1. Exercises PDE 07.11.12-09.11.12
N
Exercise 1. Let U ⊂ R be a bounded open set. Consider a sequence un of harmonic functions
such that un ⇒ u uniformly on U . What can we say about u?
Solution: u is a harmonic function, too. Actually, u is continuous since it is the limit of continuous
functions; moreover, x ∈ U and r > 0 such that B(x, r) ⊂ U one has
u(x) = lim un (x) = lim
n→+∞
n→+∞
un (ξ) dS(ξ) =
∂B(x,r)
u(ξ) dS(ξ) ,
∂B(x,r)
that is, it holds the mean-value property. This one implies first that u ∈ C ∞ (U ) (remember the
argument with the mollifiers!), and eventually that u is harmonic, since such is a C ∞ function with
the mean value property.
Exercise 2. Fix R > 0 and consider a nonnegative function u ≥ 0 which is harmonic in the ball
B(0, R) . Prove the following Harnack’s inequality: for every x such that |x| < R one has
RN −2
R + |x|
R − |x|
u(0) ≤ u(x) ≤ RN −2
u(0) .
(R + |x|)N −1
(R − |x|)N −1
(1.1)
Deduce that
sup u ≤ 3N
B(0, R
2 )
inf u .
B(0, R
2 )
Solution: Denote by αN the volume of the unit sphere. By Poisson’s formula, for every x such that
|x| < R one has
ˆ
u(x) =
KR (x, ξ) u(ξ) dS(ξ)
∂B(0,R)
with KR the Poisson’s kernel
R2 − |x|2
.
N αN R |ξ − x|N
Since |ξ| = R > |x| by means of elementary triangle inequalities we get
KR (x, ξ) :=
R + |x|
R − |x|
≤ KR (x, ξ) ≤
N
−1
N αN R (R + |x|)
N αN R (R − |x|)N −1
and since u is nonnegative, substituting into the Poisson’s formula we arrive at
ˆ
ˆ
R − |x|
R + |x|
u(ξ) dS(ξ) ≤ u(x) ≤
u(ξ) dS(ξ) .
N αN R (R + |x|)N −1 ∂B(0,R)
N αN R (R − |x|)N −1 ∂B(0,R)
Since by the mean-value formula
ˆ
u(ξ) dS(ξ) = N αN RN −1 u(0) ,
∂B(0,R)
we eventually get (1.1).
Now, for |x| = R2 , Harnack’s inequality reads, after simple computations,
2N −2
u(0) ≤ u(x) ≤ 3 · 2N −2 u(0) ,
3N −1
1
2
EXERCISES PDE
which gives
max u ≤ 3 · 2N −2 u(0)
∂B(0, R
2 )
min u ≥
and
∂B(0, R
2 )
2N −2
u(0) .
3N −1
Combining the two we arrive at
max u ≤ 3N
∂B(0, R
2 )
min u
∂B(0, R
2 )
which entails the conclusion, by the maximum and minimum principles.
Exercise 3. Find an explicit solution for the following Dirichlet problem on the square QL := (0, L)2
associated to the Laplace equation


∆u(x, y) = 0




u(0, y) = f (y)

u(L, y) = 0



u(x, 0) = 0



u(x, L) = 0
in the following cases:
(a) f (y) = sin( nπ
L y) , with n ∈ N;
(b) f is a generic C 1 function satisfying f (0) = f (L) = 0 and thus can be developed in a Fourier
series
+∞
X
nπ
f (y) =
an sin( y)
L
n=1
with
+∞
X
|an | < +∞ .
(1.2)
n=1
Hint: first, try to see which functions of the form u(x, y) = v(x)w(y) satisfy the Laplace equation
with 0 boundary conditions on three sides of the square.
Solution: We first observe that due to the boundary conditions, v and w must satisfy
w(0) = w(L) = 0
(1.3)
v(L) = 0 .
(1.4)
and
Moreover, the Laplace equation gives
v 00 (x)
w00 (y)
=−
v(x)
w(y)
for every x and y ; but this is possible only if both sides are equal to a constant, and thus there must
exist λ ∈ R such that
v 00 (x) − λv(x) = w00 (y) + λw(y) = 0
(1.5)
for every x and y . Taking into account the boundary conditions (1.3), the function w has to solve a
boundary-value problem of the form
(
w00 (y) + λw(y) = 0
(1.6)
w(0) = w(L) = 0 .
Without entering the details of the related (beautiful) theory, we only observe that by solving explicitly
and imposing the boundary conditions, one sees that the problem (1.6) has solution only for a discrete
set of values of λ (no wonder: this is not a Cauchy problem)! Precisely, it must be
nπ 2
, n∈N
(1.7)
λ = λn :=
L
EXERCISES PDE
3
and for given n the solutions of (1.6) are all multiples of the functions
wn (y) := sin
nπ y ,
L
n ∈ N.
For fixed n , now, and thus for fixed λn , by (1.5) and (1.4) we can recover the functions vn (x) as
solutions of the problem
(
vn00 (y) − λn vn (y) = 0
vn (L) = 0 .
The solutions of this problem are all multiples of the functions
vn (x) := sinh
nπ
(x − L) .
L
We recall that the function sinh(t) := 12 (et − e−t ) is an odd increasing function, sinh(t) ≥ 0 ⇐⇒
t ≥ 0.
We can now take a small breath by observing that we have answered the question posed by the
hint: the harmonic functions we searched are of the form
nπ nπ
(x − L) sin
y , µ ∈ R, n ∈ N .
(1.8)
un (x, y) = µ sinh
L
L
Now for f (y) = sin( nπ
L y) we immediately get that the solution can be recovered from (1.8) by
imposing the boundary condition (namely, recover µ by imposing equality for x = 0 ) and we get the
solutions
1
nπ
nπ un (x, y) = −
sinh
(x − L) sin
y .
sinh(nπ)
L
L
In the general case, we first proceed formally. Observe that finite linear combinations of the un
given by (1.8) are still harmonic and 0 on the three sides of the square; but since f is given by an
infinite sum, we look for a candidate solution in the form of an infinite linear combination
u(x, y) =
+∞
X
n=1
bn sinh
nπ
nπ (x − L) sin
y .
L
L
By imposing equality for x = 0 with the Fourier development of f we get that our candidate solution
is
+∞
X
an
nπ
nπ −
u(x, y) =
sinh
(x − L) sin
y .
(1.9)
sinh(nπ)
L
L
n=1
But since the solution is given by an infinite series, we are not done. We must first show that the series
converges, to have a well defined function! And actually, we want to show that it converges uniformly
on QL : thus, the limit will be a harmonic function by Exercise 1 (by construction the partial sums
are harmonic functions), while the boundary conditions are attained by the choice of the coefficients.
By the properties of sinh , we now observe that for every x ∈ [0, L] we have
sinh nπ
L (x − L)
0≤−
≤1
sinh(nπ)
and thus, for every (x, y) ∈ QL
an
nπ
nπ sinh
(x − L) sin
y ≤ |an | ;
−
sinh(nπ)
L
L
the required uniform convergence follows now by (1.2).
4
EXERCISES PDE
2. Exercises PDE 14.11.12-16.11.12
Exercise 1. Harnack’s Theorem: consider an increasing sequence un : B(0, R) → R of harmonic
functions, that is un (x) ≤ un+1 (x) for every x ∈ B(0, R) . Assume that un (0) is a Cauchy sequence.
Then there exists a harmonic function u such that un ⇒ u uniformly in B(0, r) for every r < R .
Solution: It suffices to show that un (x) is a Cauchy sequence uniformly for x ∈ B(0, r) . To this aim,
fix h ∈ N , and observe that, by the hypothesis, for every k ≤ h we have that uk − uh is a positive
harmonic function. Thus, by Harnack’s inequality there exists a constant C(N, r, R) such that
|uk (x) − uh (x)| =
sup
x∈B(0,r)
≤ C(N, r, R)
inf
sup (uk (x) − uh (x))
x∈B(0,r)
(uk (x) − uh (x)) ≤ C(N, r, R)(uk (0) − uh (0)) .
x∈B(0,r)
Since the last term of this chain of inequalities is by the hypothesis a Cauchy sequence, we get the
conclusion.
N
Exercise
ˆ 2. Let u : R → R be a harmonic function. Prove that:
(a)
u2 (x) dx < +∞ ⇒ u ≡ 0 ;
ˆRn
(b)
|∇u|2 (x) dx < +∞ ⇒ u ≡ const .
Rn
Solution: For part (a), observe that u2 is subharmonic (Ex.1 (c), 31.10.12). Then by the mean-value
property one has, for every x ∈ Rn and every R > 0
ˆ
ˆ
1
C
1
2
u
(x)
dx
≤
u2 (x) dx =
.
u2 (x) ≤
αN RN B(x,R)
αN RN Rn
αN RN
Letting R to +∞ , we get u2 (x) ≤ 0 for every x , which implies the conclusion.
∂
u with i = 1, . . . , N to obtain ∇u ≡ 0
For part (b), apply part (a) to the harmonic functions ∂x
i
and thus the conclusion.
Exercise 3.
• Consider a holomorphic function f : C → C and its real and imaginary part u : R2 → R and
v : R2 → R, that is
f (z) = u(x, y) + i v(x, y)
(2.1)
with z = x + iy . Prove that u and v are harmonic.
• Prove the following converse. Let u : B(0, R) ⊂ R2 → R be a harmonic function. Then,
there exists v : B(0, R) ⊂ R2 → R harmonic such that f (z) defined by (2.1) is a holomorphic
function. Is v unique?
Solution: We need to recall the following preliminaries about holomorphic functions. By the CauchyRiemann theorem, f is holomorphic in region of the complex plane D if and only if the following
conditions are satisfied

∂
∂

 u=
v
∂x
∂y
(2.2)
∂
∂

 u=− v
∂y
∂x
and if this happens one has
∂
∂
f 0 (z0 ) =
u(x0 , y0 ) + i v(x0 , y0 )
(2.3)
∂x
∂x
for every z0 ∈ D , with z0 = x0 + iy0 .
It also follows that if Γ is a Jordan curve surrounding a domain D and f is holomorphic in D ,
then the line integral of f over Γ vanishes: in formulas
ˆ
f (z) dz = 0
(2.4)
Γ
EXERCISES PDE
5
Indeed, by (2.1), dz = dx + i dy we get
ˆ
ˆ
ˆ
f (z) dz =
P (x, y) · ds + i Q(x, y) · ds ,
Γ
Γ
Γ
where the vector fields P and Q are given by P = (u, −v) and Q = (v, u) respectively; by Stokes’
Theorem and (2.2) we get
ˆ
ˆ
P (x, y) · ds =
curlP (x, y) dxdy = 0
ˆΓ
ˆD
Q(x, y) · ds =
curlQ(x, y) dxdy = 0 ,
Γ
D
proving (2.4). With these preliminaries we can solve the exercise.
First, implication (a) is quite easy. Indeed, if f is holomorphic, by (2.2) we get
∆u =
∂2
∂2
∂2
∂2
u + 2u =
v−
v=0
2
∂x
∂y
∂x∂y
∂y∂x
since we can exchange the order or derivations by smoothness of u. A similar proof shows that v is
harmonic, too.
For part (b), we start by observing that if f holomorphic having u as its real part exists, then by
(2.3) and (2.2) it must be
∂
∂
f 0 (z) =
u(x, y) − i u .
(2.5)
∂x
∂y
Then, defining g(z) as the right-hand side, which depends only on u, it only suffices to show that
there exists a holomorphic primitive f of g satisfying f (0) = u(0) . If it exists, its imaginary part v
will be harmonic by part (a), and uniquely determined up to a constant, thus concluding the exercise.
By analogy with integral calculus in one variable, we set
ˆ
f (z) = u(0) +
g(w) dw
[0,z]
where [0, z] is the line segment connecting 0 and z , oriented from 0 to z . We observe that g satisfies
the Cauchy-Riemann conditions; indeed, since u is smooth and harmonic

∂
∂2
∂2
∂


 Reg =
u
=
−
u=
Img
2
2
∂x
∂x
∂y
∂y
2
2
∂
∂
∂
∂


u=−
u = − Reg .
 Img = −
∂x
∂x∂y
∂y∂x
∂x
From this, by (2.4) with Γ the union of the three segments [0, z0 ] , [z0 , z] and [z, 0] all oriented from
the first to the second endpoint (observe that this curve is the boundary of a domain interely contained
in the ball, thus g is holomorphic inside!) we easily get
ˆ
ˆ
ˆ
g(w) dw −
g(w) dw =
g(w) dw ,
[0,z]
[0,z0 ]
[z0 ,z]
´
g(w) dw
f (z) − f (z0 )
[z ,z]
= 0
z − z0
z − z0
for every z and z0 in B(0, R) ; and now, only continuity of g suffices to say that the righ-hand side
has a limit when z goes to z0 , which is equal to g(z0 ) , as we wanted.
so that
Remark 2.1. In Exercise 3 part (a) it was used the fact that when f is holomorphic, then u and
v are C ∞ . Actually, this is a consequence of the fact that the derivative of a holomorphic function
is still holomorphic, usually recovered as a consequence of Cauchy’s formula for the derivatives. It is
interesting to show that this fact can be also derived only using the Cauchy-Riemann conditions and
the theory of harmonic functions presented so far. I give a sketch of the proof.
6
EXERCISES PDE
Let f , u and v as in (2.1) and consider fn = ρn ? u + i(ρn ? v) , where ρn are the usual mollifiers.
It is easy to see that ρn ? u and ρn ? v (which are now C ∞ ) still satisfy (2.2) (why?) so that fn are
holomorphic and ρn ? u and ρn ? v harmonic. As ρn ? u and ρn ? v converge locally uniformly to u
and v with n going to +∞ , we get that u and v are harmonic, and in particular C ∞ . Now, (2.5)
holds, and since u is harmonic we get (how?) that f 0 is holomorphic, as required.
3. Exercises PDE 21.11.12-23.11.12
Exercise 1. Assume N = 1 and u(x, t) = v( √xt ) .
(a) Show that ut = uxx if and only if v solves the differential equation
z
v”(z) + v 0 (z) = 0
2
and compute the solutions of it in dependance of two arbitrary constants c and d.
(b) Differentiate with respect to x and select the constant c properly, to obtain the fundamental
solution for N = 1 . Why this procedure gives the fundamental solution? (Hint: What is the
initial condition for u?)
Solution: (a) It holds, denoting with primes the derivatives of v with respect to its argument z :
−x
zv 0 (z)
∂
u(x, t) = √ v 0 (z) = −
∂t
2t
2t t
2
∂
1
u(x, t) = v 00 (z)
∂x2
2t
so that ut = uxx if and only if v solves the differential equation
z
v”(z) + v 0 (z) = 0 .
2
(b) Setting w = v 0 one has that it must be
w0 (z) = ce
for some arbitrary constant c ∈ R, whence
ˆ
v(z) = c
z
e
−z 2
4
−s2
4
ds + d
0
with d ∈ R again arbitrary. It is easy to check that
∂
c −x2
U (x, t) :=
u(x, t) = √ e 4t
∂x
t
solves again the heat equation. Moreover we can impose to this function the “initial condition” for
the fundamental solution, that is it must approximate the Dirac delta when t tends to 0 . This results
in the coupling of the two conditions
lim U (x, t) = 0 if |x| =
6 0
t→0
and
ˆ
U (x, t) dx = 1
(3.1)
R
for t in a neighborhood of 0 . The first is automatically satisfied, while the second gives the required
constant c = 2√1 π .
Also, observe that (3.1) asserts conservation of the total heat, a property that follows from the
equation for a solution decaying at infinity with its derivative. It is actually, for t 6= 0
ˆ
ˆ
d
−
U (x, t) dx =
Uxx (x, t) dx = Ux (+∞, t) − Ux (−∞, t) = 0
dt R
R
which implies (3.1) possibly with a constant different from 1 , provided that the total heat keeps
bounded for t going to 0 (if not we have identically +∞ , as it happens for u(x, t) !)
EXERCISES PDE
7
Exercise 2. Write down an explicit formula for a solution of
(
ut − ∆u + cu = f in Rn × (0, +∞)
u = g in Rn × {t = 0}
(3.2)
where c ∈ R.
Solution: Set v(t, x) = ect u(t, x) where u solves (3.2). By a direct computation one gets
vt − ∆v = ect (ut − ∆u + cu)
in Rn × (0, +∞) therefore v solves
(
vt − ∆v = ect f (x, t) in Rn × (0, +∞)
v = g in Rn × {t = 0} .
A solution of the previous problem is given via Duhamel’s principle by
ˆ tˆ
ˆ
−|x−y|2
−|x−y|2
1
ecs
4(t−s) f (y, s) dsdy
4t
v(x, t) =
g(y)
dy
+
e
N
N e
(4πt) 2 Rn
0
Rn (4π(t − s)) 2
Now, setting u(x, t) := e−ct v(t, x) , with v as above, gives a solution to (3.2).
1
Exercise 3. Tychonov’s counterexample: Consider the holomorphic function g(z) = e− z2 for z ∈
C\{0} and denoting by g (k) its k -th derivative, define the function
+∞
X g (k) (t)


x2k if t > 0 , x ∈ R
u(x, t) = k=0 (2k)!


0 if t = 0 , x ∈ R
Rigorously justify that this is a solution of the heat equation with 0 Cauchy datum by showing
uniform convergence of the series (and of the series of its time and space derivatives involved in the
equation) on any semi-strip of the type
(x, t) ∈ [−a, a] × [δ, +∞)
with a, δ > 0 .
Hint: apply Cauchy’s formula for the derivatives of holomorphic functions in this form
ˆ
g(z)
k!
g (k) (t) =
dz
2π i ∂B(t, 2t ) (z − t)k+1
(3.3)
to estimate g (k) (t) . Obviously, if you find a method not using complex analysis, it is also valid!
Solution: For every z ∈ Ct one has that there exists ω ∈ [0, 2π) such that
1
z = t(1 + eiω )
2
whence
1
1
1
1 2(2 + e−iω )
=
=
.
z
t 1 + 21 eiω
t 5 + 4 cos ω
Therefore
4 (2 + e−iω )2
1
= 2
.
2
z
t (5 + 4 cos ω)2
so that
1
4 4 + cos(2ω) − 4 cos(ω)
4
Re 2 = 2
≥
,
2
z
t
(5 + 4 cos ω)
81t2
where the last estimate is simply obtained by observing that 0 is a minimizer of the function at the
numerator, plus the trivial fact that (5 + 4 cos ω)2 ≤ 81 .
With this, we get
1
1
4
|e− z2 | = e−Re z2 ≤ e− 81t2
8
EXERCISES PDE
for every z ∈ Ct , therefore (3.3) gives
ˆ
ˆ 2π
−Re z12
2 k
4
k!
|g(z)|
k!
t
e
|g (k) (t)| ≤
|dz| ≤
2k+1
e− 81t2 .
dθ ≤ k!
k+1
k+1
2π ∂B(t, 2t ) |z − t|
2π 0
t
2
t
By recalling the easy estimate
1
2k k!
≤
(2k)!
k!
for every k ∈ N, we conclude that
g (k) (t)
x2 k 1
4
x2k ≤ e− 81t2
(2k)!
t
k!
for every k ∈ N and every (x, t) . Therefore the series defining u(x, t) is totally and thus uniformly
convergent in [−a, a] × [δ, +∞) ; moreover, we have the estimate
4
|u(x, t)| ≤ e− 81t2 +
a2
t
for every x ∈ [−a, a] and t > 0 . This implies
lim |u(x, t)| = 0
t→0+
uniformly in x ∈ [−a, a] , which means that u ∈ C(R × [0, +∞)) .
A similar proof shows that the series
+∞ (k+1)
X
g
(t)
k=0
(2k)!
x2k ,
(3.4)
formally giving ut (x, t) , as well as
+∞
X
g (k) (t) 2k−1
x
(2k − 1)!
k=1
and
+∞
X
k=1
g (k) (t)
x2(k−1)
(2(k − 1))!
(3.5)
formally giving ux (x, t) , and uxx (x, t) , respectively, uniformly converge in [−a, a]×[δ, +∞) . Therefore
u(x, t) is twice differentiable in x and once t ; a trivial algebraic computation using (3.4) and (3.5)
shows that it solves the heat equation, as required.
4. Exercises PDE 28.11.12-30.11.12
Exercise 1.
(a) Show that the general solution of the PDE uxy = 0 is
u(x, y) = F (x) + G(y)
(4.1)
for arbitrary functions F and G.
(b) Using the change of variables ξ = x + t , η = x − t, show that utt − uxx = 0 if and only if
uξη = 0 .
(c) Rederive D’Alembert’s formula.
(d) Under which conditions on the initial data g, h is the solution a right-moving wave? A leftmoving wave?
Solution: (a) Clearly any function of the form (4.1) solves the equation, and it must be only shown
the converse. Take a solution u of the PDE uxy = 0 and, for fixed x , set fx (y) := ux (x, y) . One has
that
d
fx (y) = 0
dy
EXERCISES PDE
9
therefore
fx (y) ≡ Cx ,
(4.2)
a constant depending only on x . We now set f (x) = Cx for every x , we call F (x) a primitive of f ,
and we observe that by definition of fx we can rewrite (4.2) as follows: for every x and y , it holds
ux (x, y) =
d
F (x) .
dx
We now set, for fixed y , gy (x) := u(x, y) . We have
d
d
gy (x) = ux (x, y) =
F (x)
dx
dx
for every x and y . This gives
gy (x) − F (x) ≡ Gy ,
a constant depending only on y . Setting G(y) := Gy , the previous equality gives the conclusion.
(b) Setting ξ = x + t , η = x − t is equivalent to t = 21 (ξ − η) and x = 12 (ξ + η) . Consequently, let
v(ξ, η) := u( 12 (ξ − η), 12 (ξ + η)) . denoting with ut differentiation with respect to the first argument
and with ux differentiation with respect to the second argument, we have
1 1
∂
1 1
1
1
v(ξ, η) = − ut (ξ − η), (ξ + η) + ux (ξ − η), (ξ + η)
∂η
2
2
2
2
2
2
and therefore
1 1
∂2
1 1
1
1
v(ξ, η) = − utt (ξ − η), (ξ + η) − uxt (ξ − η), (ξ + η) +
∂ξ∂η
4
2
2
2
2 14
1 1
1
1
1
+ utx (ξ − η), (ξ + η) + uxx (ξ − η), (ξ + η)
4
2
2
4
2
2
so that utt − uxx = 0 if and only if vξη = 0 as we wanted.
(c) By part (a) and (b) we have, denoting with u a solution of the wave equation, and for ξ = x + t ,
η = x − t:
1
1
u (ξ − η), (ξ + η) = F (ξ) + G(η)
2
2
for arbitrary functions F and G, that is
u(t, x) = F (x + t) + G(x − t) ,
the sum of the left-travelling wave F (x + t) and of the right-travelling wave G(x − t) .
(d) A pure left-travelling wave corresponds to G = const., and similarly a pure right-travelling
wave corresponds to F = const. Imposing the initial conditions g(x) = u(0, x) and h(x) = ut (0, x)
we get that it must be
(
F (x) + G(x) = g(x)
F 0 (x) − G0 (x) = h(x)
for every x , that is

F (x) + G(x) = g(x)
ˆ x
F (x) − G(x) =
h(s) ds + const. .
0
´x
We then have G(x) ≡ const. if and only if g(x) − 0 h(s) ds ≡ const., which is to say that we have
a left-traveling wave if and only if g 0 = h . Similarly we have a right-traveling wave if and only if
g 0 = −h.
10
EXERCISES PDE
Exercise 2.
(a) Let E:= (E1 , E2 , E3 ) and B:= (B1 , B2 , B3 ) be solutions of the Maxwell equations
(
Et = curlB, Bt = −curlE
div E = div B = 0 .
Show
Ett − ∆E = 0,
Btt − ∆B = 0 .
(b) Assume that u := (u1 , u2 .u3 ) solves the evolution equation of linear elasticity
in R3 × (0, +∞) .
utt − µ∆u − (λ + µ)∇(div u) = 0
(4.3)
Show that w := div u and w := curl u each solve wave equation, but with different speed of
propagation.
Solution: (a) We use the following vector calculus identity:
curl(curlF) = ∇(div F) − ∆F
(4.4)
for every C 2 vector field F . Taking into account that div E = 0 , (4.4) yields
curl(curlE) = −∆E .
Observing that by Maxwell’s equations Ett = curlBt = −curl(curlE) we get that Ett − ∆E = 0 . The
proof of the other case is similar.
(b) Using (4.3), one has
wtt = curl utt = µ curl∆u + (λ + µ) curl(∇(div u)) .
The curl of a gradient vector field being zero, we arrive at
wtt = µ curl ∆u = µ ∆w
by simply exchanging the order of differentiation. This is a wave equation with speed of propagation
µ.
Analogously
wtt = div (utt ) = µ div (∆u) + (λ + µ) div (∇(div u)) = µ ∆w + (λ + µ) div (∇w)
again exchanging the order of differentiation and recalling that w = div u . Since div (∇w) = ∆w we
get
wtt = (λ + 2µ) ∆w ,
that is a wave equation with speed of propagation λ + 2µ .
Exercise 3. Let V be a (complex) pre-Hilbertian space and A a linear mapping from V to V .
Suppose preliminarly that the dimension of V is finite and equal to n ∈ N and suppose that there
exists an orthonormal basis {v1 , . . . , vn } of V consisting of eigenvectors of A relative to the eigenvalues
(λ1 , . . . , λn ) .
(a) Consider the following Cauchy problem associated to a linear ODE in V :
d
w = Aw,
dt
and show that
w(t) :=
n
X
w(0) = g
(4.5)
hg, vi ieλi t vi
i=1
is its only solution.
(b) Give an analogous statement for this other initial value problem
d2
w = Aw,
dt2
w(0) = g,
d
w(0) = h .
dt
(4.6)
EXERCISES PDE
11
We now want to generalise this method to an infinite-dimensional situation where we look to the
solution of the wave equation with periodic boundary conditions, that is


utt − uxx = 0 (x, t) ∈ (0, 2π) × (0, +∞)
(4.7)
u(0, x) = g(x), ut (0, x) = h(x)


u(t, 0) − u(t, 2π) = ux (t, 0) − ux (t, 2π) = uxx (t, 0) − uxx (t, 2π) = 0 for every t .
To this end, let
V := {v ∈ C 2 ([0, 2π], C) : v(0) − v(2π) = vx (0) − vx (2π) = vxx (0) − vxx (2π) = 0} .
(4.8)
(c) Interpret (4.7) as a linear ODE of the type (4.6) in the pre-Hilbertian space V , endowed with
the usual L2 scalar product
ˆ 2π
1
hu, vi :=
u(x)v(x) dx ,
2π 0
where the operator A is given by
d2
: V ⊂ L2 ((0, 2π)) → L2 ((0, 2π)) .
(4.9)
dx2
What are the initial conditions? What happens of the boundary value conditions?
(d) Show that (eikx )k∈Z is an orthonormal basis of V consisting of eigenvectors of A and compute
the relative eigenvalues (hint: Fourier series).
(e) Consider now (4.7) with g := x2 (x − 2π)2 and h ≡ 0 . Verify that g and h belong to V ,
then find a formal solution to (4.7) assuming that the statement you found in (b) can be
generalised to an infinite-dimensional situation, and using the basis (eikx )k∈Z . Then show
that you actually found a solution, in this way!
A :=
Solution: (a) Uniqueness is a well-known fact. By a direct calculation, since λi vi = Avi we get
n
n
n
X
X
X
d
hg, vi ieλi t vi ,
w(t) :=
hg, vi ieλi t λi vi =
hg, vi ieλi t Avi = A
dt
i=1
i=1
i=1
Pn
d
so that dt w = Aw as we wanted. On the other hand w(0) =
i=1 hg, vi ivi = g since vi is a
orthonormal basis. In view of point (b) it is important to remark that if vi is a basis of eigenvectors,
but no orthonormal basis, then, denoting with gi the components of g with respect to vi , the
solution to (4.5) is given by
n
X
w(t) :=
gi eλi t vi ;
(4.10)
i=1
only, gi 6= hg, vi i, in general!
d
(b) Setting v(t) := dt
w and W := (v, w) ∈ V × V the problem is equivalent to the following one
d
W = BW, W (0) = (g, h)
dt
where the linear operator B acts on V × V as follows:
B(w, v) := (v, Aw) .
Observe that the only hypothesis that A is diagonalisable in V does not assure that B is diagonalisable
in V × V ! We have actually to distinguish two cases.
(b1) ker A = {0}: this is the case where B possesses a basis of eigenvectors.
We make the following
√
convention: for every λ ∈ C with λ 6= 0 we denote with λ the complex square root of λ
with argument in [0, π) . Then, a basis in V × V made of eigenvectors of B is easily given by
the set of vectors
{(W1 )+ , (W1 )− , . . . , (Wn )+ , (Wn )− }
12
EXERCISES PDE
where for every i = 1, . . . , n we have set
(Wi )+ = p
1
1 + |λi |
and
(Wi )− = p
√
1
1 + |λi |
√
(vi , λi vi )
√
(vi , − λi vi ) .
√
The corresponding eigenvalues are λi and − λi , respectively. But since this basis is no
orthonormal one, we can apply case (a) only in the form (4.10), to get
W (t) =
n
X
√
(wi )+ e
λi t
(Wi )+ +
i=1
n
√
X
(wi )− e− λi t (Wi )− .
i=1
where (wi )+ and (wi )− are the coefficients of W (0) with respect to the basis
{(Wi )+ , (Wi )− }i=1,...,n
. Projecting on the first factor we obtain the expression of w in this form
n
n
√
√
X
X
bi e− λi t vi
w(t) =
ai e λi t vi +
with ai = √(wi )+
1+|λi |
(4.11)
i=1
i=1
and bi = √(wi )+ . But the key point is that now we can easily determine
1+|λi |
ai and bi by imposing the initial conditions, since vi is an orthonormal basis! Precisely, it
must be
(
ai + bi = hg, vi i
√
√
λi ai − λi bi = hh, vi i
for every i. Solving the system gives that the coefficients ai and bi in (4.11) are given by
1
hh, vi i 1
hh, vi i ai =
hg, vi i + √
, bi =
hg, vi i − √
.
(4.12)
2
2
λi
λi
(b2) dim ker A = m 6= 0 . Denoting with {vi }n−m
the eigenvectors of A relative to the nonzero
i=1
eigenvalues {λi }n−m
we
see
that
the
analogous
of (4.11), that is the function
i=1
w1 (t) =
n−m
X
√
ai e
i=1
λi t
vi +
n−m
X
bi e −
√
λi t
vi ,
(4.13)
i=1
wtih ai and bi given by (4.12) still solves the equation in (4.6), but with a different initial
d
datum: precisely w1 (0) = g − Pker A g and dt
w1 (0) = h − Pker A h where Pker A denotes the
ker
orthogonal projection onto ker A. But now, given an orthonormal basis {v1ker , . . . , vm
} of
ker A it is easy to check that the unique solution of (4.6) is w(t) = w1 (t) + z(t) where w1 (t)
is given by (4.13) and (4.12) and z(t) is given by:
m
X
z(t) :=
[hg, vjker i + hh, vjker it]vjker .
(4.14)
j=1
2
d
Indeed z(t) ∈ ker A for every t , dt
2 z(t) ≡ 0 = Az(t) , and z(0) = Pker A g as well as
d
z(0)
=
P
h
so
that
we
conclude
by linearity.
ker A
dt
(c) Simply interpret (4.7) as a generalised Cauchy problem
d
d2
w = Aw, w(0) = g,
w(0) = h .
dt2
dt
with w : R → V , where this one is given by (4.8), and A is given by (4.9). Setting u(t, ·) = w(t) the
boundary conditions ar already incorporated in the request that w(t) must belong to V . Obviously,
for the problem to have sense, it must be g and h in V , too!
EXERCISES PDE
13
(d) It is a very-well known fact that (eikx )k∈Z is an orthonormal basis of L2 by the theory of Fourier
series; since (eikx )k∈Z ⊂ V for, it constitutes an orthonormal basis of V , too. A direct calculation
shows that for every k ∈ Z
A(eikx ) = −k 2 eikx
so that they are eigenvectors of A with eigenvalues −k 2 .
(e) Checking
√ that g belongs to V is straightforward. Now, for k 6= 0 , we have according to our
convention λk = i|k|, so that, setting
αk := hg, eikx i
we get that the coefficients ak and bk in (4.12) satisfy ak = bk = 21 αk . Therefore the function w1 (t)
defined in (4.13) reduces in our case to
X 1
X
w1 (t) =
αk [ei|k|t + e−i|k|t ]eikx =
αk cos(|k|t)eikx .
2
k∈Z, k6=0
By a direct computation, αk =
k∈Z, k6=0
− k244
for every k 6= 0 , so that
∞
w1 (t) = −
X
k∈Z, k6=0
X 1
24
ikx
cos(|k|t)e
=
−48
cos(kt) cos(kx) .
k4
k4
k=1
The eigenvector corresponding to the eigenvalue 0 is the constant 1 , so that in our case
z(t) = hg, 1i =
8π 4
15
therefore we get to our candidate solution
∞
w(t) =
X 1
8π 4
− 48
cos(kt) cos(kx) .
15
k4
(4.15)
k=1
The verification that (4.15) is actually the solution of (4.7) is an easy exercise in differentiation of
series whose proof I omit.
5. Exercises PDE 5.12.12-7.12.12
Exercise 1. Let u(t, x) : R × R → R be a solution of the wave equation utt − uxx = 0 . Show that
ˆ +∞ − s2
e 4t
√ u(s, x) ds
v(t, x) :=
t
−∞
satisfies the heat equation ut − uxx = 0 for every t > 0 and x ∈ R .
Solution: By D’Alembert’s formula, u(s, x) = F (x + s) + G(x − s) for two functions of a real variable
F and G. With the change of variable z = x + s one has
ˆ +∞ − s2
ˆ +∞ − (x−z)2
4t
e
e 4t
√ F (x + s) ds =
√
F (z) dz
t
t
−∞
−∞
and similarly, with z = x − s we get
ˆ +∞ − s2
ˆ +∞ − (x−z)2
4t
e 4t
e
√ G(x − s) ds =
√
G(z) dz
t
t
−∞
−∞
so that
ˆ +∞ − (x−z)2
4t
e
√
v(t, x) =
(F (z) + G(z)) dz
t
−∞
√
solving the heat equation with initial datum 2 πu(0, x) .
14
EXERCISES PDE
Exercise 2. We say that a function u : [0, 1] → R is absolutely continuous if for every ε > 0 there
exists δ > 0 such that if 0 ≤ t0 < s0 < t1 < s1 < . . . tn < sn ≤ 1 are points in [0, 1] satisfyng
n
X
(si − ti ) ≤ δ
i=1
then
n
X
|u(ti ) − u(si )| ≤ ε
i=1
An absolutely continuous function u is continuous and u̇(t) (the pointwise derivative) exists for almost
every t ∈ [0, 1] .
Show that if u ∈ W 1,p (0, 1) for 1 ≤ p then there exists v absolutely continuous such that u = v
almost everywhere; moreover the weak derivative of u is given exactly by the pointwise derivative v̇
and v̇ ∈ Lp (0, 1) . Moreover, if p > 1 :
ˆ 1
1
1
|u0 (t)|p dt) p
(5.1)
|u(x) − u(y)| ≤ |x − y|1− p (
0
for every x and y in [0, 1] .
Solution: The key of the exercise, important in itself, is that u satisfies the fundamental Theorem
of calculus, that is
ˆ x
u(x) = u(0) +
u0 (t) dt
(5.2)
0
for every x ∈ [0, 1] , with u0 the weak derivative. Observe that the right-hand side is well defined,
if we assume that the weak derivative belongs to Lp . To prove (5.2), no knowledge of absolutely
continuous function is needed, but only Fubini’s
theorem and exercise 4 below. Let us see how. Given
´x
a function v ∈ L1 ([0, 1]) , define V (x) := 0 v(y) dy and we claim that
V0 = v,
(5.3)
Cc∞ ([0, 1])
that is, the weak derivative of V is v . To this aim, take ϕ ∈
Theorem
ˆ 1
ˆ 1ˆ x
ˆ 1ˆ
−
V (x)ϕ0 (x) dx = −
v(y) dy ϕ0 (x) dx = −
0
0
0
1
and observe that by Fubini’s
ˆ
ϕ0 (x) dx v(y) dy =
0
y
´x 0
u (y) dy ,
0
1
v(y)ϕ(y) dy .
0
This proves (5.3). Now, if you define U (x) := u(0) +
(5.3) gives that (U − u)0 = 0 , so
that by Exercise 4 below it must be U (x) − u(x) ≡ const. for every x ∈ [0, 1] . But since U (0) = u(0) ,
the constant is 0 and (5.2) is proved.
Let us also observe that (5.1) is an immediate consequence of (5.2) and the Holder’s (or Jensen’s
inequality), when p > 1 ; it is indeed for every 0 ≤ x < y ≤ 1
ˆ 1
p p1
ˆ y
1 ˆ y
1
1
|u(y) − u(x)| = u0 (t) dt = |y − x| u0 (t) dt
≤ |y − x|1− p (
|u0 (t)|p dt) p .
|y
−
x|
x
x
0
1
Absolute
´ x continuity of u follows from (5.2) with the only observation that for every v ∈ L ([0, 1]) ,
V (x) := 0 v(y) dy is absolutely continuous. To show this, we can suppose without loss of generality
that v ≥ 0 . For fixed ε > 0 , there exist N ∈ N, depending only on ε , such that defining vN (x) :=
min{v(x), N } , one has
ˆ 1
ε
|v(x) − vN (x)| dx ≤ .
(5.4)
2
0
ε
This is because the sequence vk (x) := min{v(x), k} is L1 convergent to v . Now, fix δ = 2N
; for
every finite sequence of points 0 ≤ t0 < s0 < t1 < s1 < . . . tn < sn ≤ 1 satisfyng
n
X
(si − ti ) ≤ δ
i=1
EXERCISES PDE
one has
n
X
15
ˆ
|V (ti ) − V (si )| =
v(y) dy
Bn
i=1
with Bn := ∪ni=0 [ti , si ] . Since now, using (5.4) and our choice of δ
ˆ
ˆ
ˆ
ε
ε
v(y) dy ≤
|v(y) − vN (y)| dy +
vN (y) dy ≤ + N |Bn | ≤ + N δ = ε ,
2
2
Bn
Bn
Bn
our claim is proved.
Finally, define
1
(u(t + h) − u(t)) .
h
By (5.2) and the Lebesgue differentiation Theorem we have that
ˆ
1 t+h 0
Uh (t) =
u (y) dy → u0 (t)
h t
Uh (t) :=
in L1 ([0, 1]) when h → 0 , and therefore almost everywhere. Being Uh (t) exactly the difference
quotients of u at t , we have shown that u is pointwise almost everywhere differentiable, and that its
pointwise derivative coincides a.e. with u0 .
Exercise 3. Let U := (−1, 1) × (−1, 1) . Define u


1 − x1 in T1

1 + x in T
1
2
u(x1 , x2 ) :=
1 − x2 in T3



1 + x2 in T4
as follows
:= {x1
:= {x1
:= {x2
:= {x2
> 0,
< 0,
> 0,
< 0,
|x2 | < x1 }
|x2 | < −x1 }
|x1 | < x2 }
|x1 | < −x2 }
For which 1 ≤ p ≤ ∞ does u belong to W 1,p (U ) ?
Solution: Let us take ϕ := (ϕ1 , ϕ2 ) ∈ Cc∞ (U ; R2 ) and define the vectors ν+ := (− √12 , √12 ) (unit
normal to the line {x1 = x2 } ) and ν− := ( √12 , √12 ) (unit normal to the line {x1 = −x2 } ). First, we
compute
ˆ
x1 div ϕ(x1 , x2 ) dx1 dx2 .
T1
Since div (x1 ϕ) − (1, 0) · ϕ = x1 div ϕ, by the divergence theorem (taking into account the orientation
of the exterior normal!) we get
ˆ
ˆ
x1 div ϕ(x1 , x2 ) dx1 dx2 = −
(1, 0) · ϕ(x1 , x2 ) dx1 dx2 +
ˆ T1
ˆ T1
(5.5)
ξ1 ϕ(ξ) · ν+ ds(ξ) −
ξ1 ϕ(ξ) · ν− ds(ξ) .
{x1 =x2 ; x1 >0}
{x1 =−x2 ; x1 >0}
Here we took into account that ϕ(1, x2 ) ≡ (0, 0) . In a similar way we get
ˆ
ˆ
x2 div ϕ(x1 , x2 ) dx1 dx2 = −
(0, 1) · ϕ(x1 , x2 ) dx1 dx2 +
ˆ T4
ˆT4
ξ2 ϕ(ξ) · ν− ds(ξ) −
ξ2 ϕ(ξ) · ν+ ds(ξ) .
{x1 =−x2 ; x2 <0}
{x1 =x2 ; x2 <0}
But this is clearly equivalent to
ˆ
ˆ
x2 div ϕ(x1 , x2 ) dx1 dx2 = −
(0, 1) · ϕ(x1 , x2 ) dx1 dx2 +
T
ˆ T4
ˆ4
−
ξ1 ϕ(ξ) · ν− ds(ξ) +
ξ1 ϕ(ξ) · ν+ ds(ξ) .
{x1 =−x2 ; x1 >0}
{x1 =x2 ; x1 <0}
(5.6)
16
EXERCISES PDE
The same reasonings lead to
ˆ
ˆ
x1 div ϕ(x1 , x2 ) dx1 dx2 = −
(1, 0) · ϕ(x1 , x2 ) dx1 dx2 +
ˆ T2
ˆT2
ξ1 ϕ(ξ) · ν− ds(ξ) −
ξ1 ϕ(ξ) · ν+ ds(ξ)
{x1 =−x2 ; x1 <0}
{x1 =x2 ; x1 <0}
ˆ
and
ˆ
(5.7)
ˆ
x2 div ϕ(x1 , x2 ) dx1 dx2 = −
T3
ˆT3
(0, 1) · ϕ(x1 , x2 ) dx1 dx2 +
(5.8)
ξ1 ϕ(ξ) · ν− ds(ξ) −
{x1 =−x2 ; x1 <0}
ξ1 ϕ(ξ) · ν+ ds(ξ)
{x1 =x2 ; x1 >0}
ˆ
Since
div ϕ(x1 , x2 ) dx1 dx2 = 0
U
again by the divergence Theorem, we have that
ˆ
ˆ
ˆ
−
u div ϕ(x1 , x2 ) dx1 dx2 =
x1 div ϕ(x1 , x2 ) dx1 dx2 −
x1 div ϕ(x1 , x2 ) dx1 dx2 +
U
T1
T2
ˆ
ˆ
x2 div ϕ(x1 , x2 ) dx1 dx2 −
x2 div ϕ(x1 , x2 ) dx1 dx2 .
T3
T4
Using (5.5), (5.7), (5.8), and (5.6) we conclude that
ˆ
ˆ
−
u div ϕ(x1 , x2 ) dx1 dx2 =
v · ϕ(x1 , x2 ) dx1 dx2
U
U
where v ∈ L∞ (U ; R2 ) is defined as

(−1, 0) in T1



(1, 0) in T
2
v(x1 , x2 ) :=

(0, −1) in T3



(0, 1) in T4 .
Therefore u ∈ W 1,p (U ) for every p.
Exercise 4. Suppose U is connected and u ∈ W 1,p (U ) satisfies Du = 0 almost everywhere in U ,
with Du the Sobolev gradient. Prove that u is (almost everywhere) constant in U .
Solution: Fix a relatively compact open subset U 0 of U ; it suffices to show u constant in U 0 . For ε
sufficiently small, now, the convolutions with the simmetric mollifiers ρε are well defined; furthermore,
for every ϕ ∈ C ∞ (U 0 ; Rn ) , one has, by simmetry of mollifiers and since ρε ? ϕ ∈ C ∞ (U ; Rn ) for ε
small enough, that
ˆ
ˆ
ˆ
div ϕ(x)(ρε ? u)(x) dx =
u(x)(ρε ? div ϕ)(x) dx =
u(x)div (ρε ? ϕ)(x) dx .
U0
U
U
By the hypothesis, the right-hand side is 0 , so that D(ρε ? u) = 0 on U 0 . By smoothness of ρε ? u
this implies that ρε ? u is constant in U 0 ; by Lp convergence, when ε is going to 0 we get that u is
almost everywhere equal to a constant, as required.
6. Exercises PDE 12.12.12-14.12.12
Exercise 1. (a) Let Ω be an open bounded subset of Rn . For every u ∈ Cc∞ (Ω) , prove that the
following interpolation inequality holds:
ˆ
2
ˆ
ˆ
|Du(x)|2 dx ≤ C
u(x)2 dx
|D2 u(x)|2 dx ,
(6.1)
Ω
2
Ω
Ω
with D u the Hessian matrix.
(b) Assume ∂Ω ∈ C 1 , let u ∈ W 2,2 (Ω) ∩ W01,2 (Ω) . Prove that u satisfies (6.1).
EXERCISES PDE
17
Solution: (a) For every v ∈ C ∞ (Ω) , it holds
v∆v + |Dv|2 = div (v · Dv) .
(6.2)
For v = u, integrating on Ω , by the divergence theorem we get
ˆ
ˆ
|Du(x)|2 dx = −
u(x)∆u(x) dx
Ω
Ω
since u = 0 on ∂Ω . By the Cauchy-Schwarz inequality we then have
ˆ
2 ˆ
ˆ
ˆ
ˆ
2
2
2
2
|Du(x)| dx ≤
u(x) dx
|∆u(x)| dx ≤ C
u(x) dx
|D2 u(x)|2 dx ,
Ω
Ω
Ω
Ω
where the last estimate is obvious. This proves (6.1).
(b) By (6.2) and the divergence Theorem, we get
ˆ
ˆ
ˆ
|Dvn (x)|2 dx = −
vn (x)∆vn (x) dx +
Ω
Ω
vn (ξ)
∂Ω
Ω
∂
vn (ξ) dS(ξ) .
∂ν
(6.3)
Now, considering wn as in the hint, we obviously have
ˆ
ˆ
∂
∂
vn (ξ) vn (ξ) dS(ξ) =
(vn − wn )(ξ) vn (ξ) dS(ξ)
∂ν
∂ν
∂Ω
∂Ω
since wn (ξ) ≡ 0 . By the divergence Theorem, we then get
ˆ
ˆ
∂
vn (ξ) vn (ξ) dS(ξ) =
div [(vn (x) − wn (x))Dvn (x)] dx =
∂ν
ˆ ∂Ω
Ω ˆ
(vn (x) − wn (x))∆vn (x) dx +
D(vn (x) − wn (x)) · Dvn (x) dx .
Ω
Ω
2,2
Now, by W
convergence of vn to u, ∆vn and Dvn are bounded in L2 , while kvn − wn kW 1,2 (Ω) →
ku − ukW 1,2 (Ω) = 0 as n goes to +∞ , therefore
ˆ
∂
vn (ξ) vn (ξ) dS(ξ) = 0 .
(6.4)
lim
n→+∞ ∂Ω
∂ν
By (6.4) and the W 2,2 convergence of vn to u, taking the limit in (6.3) we get
ˆ
ˆ
2
|Du(x)| dx = −
u(x)∆u(x) dx
Ω
Ω
and we conclude as in part (a).
Exercise 2. Poincaré’s inequality:
• Let u ∈ W01,2 ((0, d)) . Prove that
ˆ d
ˆ
u(x)2 dx ≤ d2
0
d
u0 (x)2 dx .
• Let u ∈ W01,2 ((0, d) × RN −1 ) . Prove that
ˆ
ˆ
u(x)2 dx ≤ d2
[0,d]×RN −1
(6.5)
0
|Du(x)|2 dx .
[0,d]×RN −1
Hint: Use density of Cc∞ functions.
• Does the inequality hold for u ∈ W 1,2 ((0, d) × RN −1 ) ?
Solution: Let x and y ∈ [0, d] . We have already seen (where?) that
ˆ d
1
1
|u0 (t)|p dt) 2
|u(x) − u(y)| ≤ |x − y| 2 (
0
(6.6)
18
EXERCISES PDE
and that u is absolutely continuous (in particular, continuous). Therefore u ∈ W01,2 ([0, d]) implies
u(0) = 0 and then for y = 0 we have
ˆ d
ˆ d
1
1
1
1
|u(x)| ≤ |x| 2 (
|u0 (t)|p dt) 2 ≤ d 2 (
|u0 (t)|p dt) 2 ,
0
0
where the last estimate is trivial. Taking squares and integrating in [0, d] with respect to x , we get
(6.5).
Now, let u ∈ Cc∞ ((0, d) × RN −1 ) . Writing x = (x1 , x̂) for every x ∈ [0, d] × RN −1 , with x1 ∈ [0, d]
and x̂ ∈ RN −1 , we have that for fixed x̂ the function u(·, x̂) belongs to Cc∞ ((0, d)) . We therefore
have, by (6.5), that
ˆ d
ˆ d
ˆ d
2
2
d
∂
2
2
2
u(x1 , x̂) dx1 ≤ d
u(x1 , x̂) dx1 = d
u(x) dx1
dx1
∂x1
0
0
0
for every x̂ ∈ RN −1 . Now, by Fubini’s Theorem
ˆ
ˆ
ˆ d
u(x)2 dx =
u(x1 , x̂)2 dx1 dx̂ ≤
N −1
0
RN −1
ˆ
ˆ [0,d]×R
2
ˆ d ∂
2
|Du(x)|2 dx .
u(x) dx1 dx̂ ≤ d2
d
∂x1
[0,d]×RN −1
RN −1
0
This proves (6.6) when u ∈ Cc∞ ((0, d) × RN −1 ) . In the general case, there exists un ∈ Cc∞ (Ω) such
that un → u in W 1,2 (Ω) . In particular
ˆ
ˆ
2
lim
un (x) dx =
u(x)2 dx
n→∞
[0,d]×RN −1
[0,d]×RN −1
ˆ
and
ˆ
|Dun (x)|2 dx =
lim
n→∞
[0,d]×RN −1
ˆ
Since
|Du(x)|2 dx .
[0,d]×RN −1
ˆ
un (x)2 dx ≤ d2
[0,d]×RN −1
|Dun (x)|2 dx ,
[0,d]×RN −1
(6.6) follows by simply taking the limit.
Such an inequality cannot hold in general in W 1,2 (Ω) . Constant functions in dimension N = 1
provide an easy counterexample.
W 1,∞ (Ω)
Exercise 3. Let Ω be an open subset of Rn . We denote by C ∞ (Ω)
the set of functions v
in W 1,∞ (Ω) such that there exists a sequence vn of C ∞ (Ω) -functions converging to v in the W 1,∞
norm on compact subsets of Ω .
• C ∞ (Ω)
W 1,∞ (Ω)
=?
• In case that C ∞ (Ω)
to C ∞ (Ω)
W 1,∞ (Ω)
W 1,∞ (Ω)
6= W 1,∞ (Ω) , find a function v ∈ W 1,∞ (Ω) that does not belong
.
1,∞
Solution: W
convergence corresponds to uniform convergence of the sequence of functions and
of the sequence of gradients. Therefore, since the limit of continuous functions must be continuous,
one easily has that
W 1,∞ (Ω)
C ∞ (Ω)
⊆ C 1 (Ω) .
But also the converse inclusion holds! To prove this, it simply suffices to follow exactly the steps of
the Meyers-Serrin Theorem (Theorem 2, Section 5.3.2 in the book of Evans); observe actually that in
step 2, since now the functions ζi u are Cc1 (U ) , by mollification one can find ui ∈ Cc∞ (Ω) such that
supp ui ⊂ Wi and kui − ζi ukW 1,∞ (Ω) ≤ 2δi (why?). Therefore
C ∞ (Ω)
W 1,∞ (Ω)
= C 1 (Ω) .
EXERCISES PDE
19
To find the required example, given for instance Ω = B(0, 1) , it suffices to find v ∈ W 1,∞ (B(0, 1))
that does not belong to C 1 (B(0, 1)) . Consider for instance v(x) := |x| . This is clearly no C 1
function in B(0, 1) , since it is not differntiable at 0 . On the other hand, for every ε > 0 one has
v ∈ C 1 (B(0, 1) \ B(0, ε)) , so that, given ϕ ∈ Cc∞ (B(0, 1); Rn ) one has
ˆ
ˆ
ˆ
x
v(x) div ϕ(x) dx = −
· ϕ(x) dx −
|ξ| ϕ(ξ) · ν(ξ) dS(ξ) .
B(0,1)\B(0,ε) |x|
∂B(0,ε)
B(0,1)\B(0,ε)
ˆ
Since
∂B(0,ε)
and
x
|x|
|ξ| ϕ(ξ) · ν(ξ) dS(ξ) ≤ N α(N )εN kϕkL∞ (B(0,1);Rn )
∈ L∞ (B(0, 1); Rn ) , by taking the limit as ε → 0 we get by dominated convergence
ˆ
ˆ
x
· ϕ(x) dx ,
v(x) div ϕ(x) dx = −
|x|
B(0,1)
B(0,1)
which proves that v ∈ W 1,∞ (B(0, 1)) .
7. Exercises PDE 19.12.12
Exercise 1. Let Ω be the open subset of RN defined by Ω := B(0, 1) \ {xN = 0}. Show that the
function
(
1 if xN > 0
u(x1 , x2 , . . . , xN ) :=
0 if xN < 0
belongs to W 1,p (Ω) for every 1 ≤ p ≤ ∞ , but there exists no sequence vn of C ∞ (Ω) functions such
that vn → u in W 1,p (Ω) .
Solution: Being u a C ∞ function on each connected component of Ω , it is certainly in W 1,p (Ω)
for every 1 ≤ p ≤ ∞. Also observe that u is defined almost everywhere in B(0, 1) , and actually
belongs to L∞ (B(0, 1)) , but does not belong to W 1,1 (B(0, 1)) ! Indeed, suppose by contradiction that
W 1,1 (B(0, 1)) . Then, it exists g ∈ L1 (B(0, 1); Rn ) such that
ˆ
ˆ
u(x)div ϕ(x) dx = −
g(x) · ϕ(x) dx
B(0,1)
B(0,1)
Cc∞ (B(0, 1); Rn ) .
Now, let us define B(0, 1)+ := B(0, 1) ∩ {xN > 0} and similarly
for every ϕ ∈
−
B(0, 1) := B(0, 1)∩{xN < 0}. Since Du = 0 in B(0, 1)+ , and obviously every ϕ ∈ Cc∞ (B(0, 1)+ ; Rn )
also belongs to Cc∞ (B(0, 1); Rn ) , it must be
ˆ
g(x) · ϕ(x) dx = 0
B(0,1)+
Cc∞ (B(0, 1)+ ; Rn ) .
for every ϕ ∈
Thus, g = 0 a.e. in B(0, 1)+ . Arguing similarly in B(0, 1)− , we
get g = 0 a.e. in B(0, 1) . We should therefore have
ˆ
u(x)div ϕ(x) dx = 0
(7.1)
B(0,1)
for every ϕ ∈
we have
ˆ
Cc∞ (B(0, 1); Rn ) .
But on the other hand, by definition of u and the divergence theorem
ˆ
B(0,1)
ˆ
div ϕ(x) dx = −
u(x)div ϕ(x) dx =
B(0,1)+
ϕ(ξ) · eN (ξ) dS(ξ)
B(0,1)∩{xN =0}
for every ϕ ∈ Cc∞ (B(0, 1); Rn ) , and this has clearly no reason of being 0 ! So, u ∈
/ W 1,1 (B(0, 1)) .
∞
1,p
Now, if a sequence vn of C (Ω) functions such that vn → u in W (Ω) exists, we have that vn
and Dvn are Cauchy sequences in Lp (Ω) and Lp (Ω; Rn ) respectively. Since B(0, 1) and Ω differ by
a set of Lebesgue measure 0 , we also have that vn and Dvn are Cauchy sequences in Lp (B(0, 1)) and
Lp (B(0, 1); Rn ) , respectively. Thus, vn has a limit ũ in W 1,p (B(0, 1)) , and it must be ũ = u almost
20
EXERCISES PDE
everywhere in B(0, 1) . But since the definition of weak derivative does not depend on the Lebesgue
representative, this would imply u ∈ W 1,p (B(0, 1)) , a contradiction.
Exercise 2. Let Ω be a bounded open subset of Rn .
(a) Let f ∈ W 1,p (Ω) and g ∈ W 1,q (Ω) with 1 ≤ p, q ≤ ∞ . Find a sufficient condition on p and q
such that f g ∈ W 1,1 (Ω) and compute the weak gradient D(f g) .
(b) Show that when N = 1 it suffices to take p = q = 1 .
(c) Kettenregel: assume that F : R → R ∈ C 1 ∩ W 1,∞ (R) . Show that for every u ∈ W 1,p (Ω) the
composition F (u) ∈ W 1,p (Ω) and compute the weak gradient.
Solution: (a) Assume that p and q are conjugate exponents, that is p1 + 1q = 1 . By Meyers-Serrin’s
Theorem, there exist two sequences fn and gn of C ∞ (Ω) functions such that fn → f in W 1,p (Ω)
and gn → g in W 1,q (Ω) . Now, by Holder’s and Minkowsky’s inequalities
kfn gn − fm gm k1 = kfn (gn − gm ) + (fn − fm )gm k1 ≤ kfn kp kgn − gm kq + kgm kq kfn − fm kp
for every n and m in N. Thus, fn gn is a Cauchy sequence in L1 (Ω) , and similarly we can prove
that D(fn gn ) = fn Dgn + gn Dfn is a Cauchy sequence in L1 (Ω; Rn ) . So, fn gn is Cauchy in W 1,1 (Ω)
and is converging to f g in L1 . It follows that f g is in W 1,1 (Ω) and that fn gn is converging to f g
in W 1,1 (Ω) . In particular,
D(f g) = f Dg + gDf .
(b) It suffices to consider the case of an interval, say (0, 1) . So, let f and g ∈ W 1,1 ((0, 1)) . We
begin with the case where f and g ∈ W01,1 ((0, 1)) . Take fn and gn sequences of Cc∞ functions
converging to f , and g , respectively, in W 1,1 ((0, 1)) . Since for every x ∈ [0, 1] , one has
ˆ 1
|fn (x) − f (x)| ≤
|fn0 (y) − f 0 (y)| dy
0
we get that actually fn converges to f uniformly in [0, 1] , and clearly, also gn converges to g uniformly
in [0, 1] ! Thus, fn gn converges to f g uniformly in [0, 1] . On the other hand
(fn gn )0 = fn gn0 + gn fn0 ;
by the uniform convergences of fn and gn , and the L1 convergences of fn0 and gn0 , we easily get that
(fn gn )0 → f g 0 + gf 0
in L1 ((0, 1)) . This proves that f g belongs to W01,1 ((0, 1)) .
In the general case, we only can take fn and gn sequences of C ∞ ([0, 1]) functions converging to
f , and g , respectively, in W 1,1 ((0, 1)) . By the fundamental Theorem of calculus, we easily get, for
every m and n ∈ N
ˆ 1
0
|fn (0) − fm (0)| ≤ |fn (x) − fm (x)| +
|fn0 (y) − fm
(y)| dy
0
so that integrating from 0 to 1 , it is
ˆ
|fn (0) − fm (0)| ≤
ˆ
1
0
1
0
|fn0 (y) − fm
(y)| dy
|fn (x) − fm (x)| dx +
0
which yields that fn (0) is a Cauchy sequence. Now, again by the fundamental theorem of calculus,
we have
ˆ 1
0
|fn (x) − fm (x)| ≤ |fn (0) − fm (0)| +
|fn0 (y) − fm
(y)| dy
0
for every x ∈ (0, 1) , so that fn is uniromly Cauchy. It follows that fn converges to f uniformly
in [0, 1] , and similarly gn converges to g uniformly in [0, 1] . Arguing as in the previous step, the
conclusion follows.
EXERCISES PDE
21
(c) Since F : R → R ∈ C 1 ∩ W 1,∞ (R) , for every u ∈ W 1,p (Ω) the functions F (u) , F 0 (u) , and
F (u)Du are a.e. well defined and belong to Lp (Ω) , Lp (Ω) , and Lp (Ω; Rn ) , respectively. Now, taking
a sequence un of C ∞ (Ω) functions such that un → u in W 1,p (Ω) , we first have
0
kF (un ) − F (u)kLp (Ω) ≤ kF 0 kL∞ (R) kun − ukLp (Ω)
which gives F (un ) → F (u) in Lp (Ω) . By boundedness of F 0 , the convergence
F 0 (un )Du → F 0 (u)Du in Lp (Ω; Rn )
can be proved by dominated convergence. Finally, one has
lim kF 0 (un )Dun − F 0 (u)DukLp (Ω;Rn ) = lim kF 0 (un )Dun − F 0 (un )DukLp (Ω;Rn )
n
n
≤ kF 0 kL∞ (R) lim kDun − DukLp (Ω;Rn ) = 0 ,
n
whence it follows that F (u) ∈ W 1,p (Ω) , and that the weak gradient is given by F 0 (u)Du.
Exercise 3. Let Ω be a bounded open subset of Rn , and u ∈ W 1,p (Ω) . Define as usual u+ (x) :=
max{u(x), 0} , and u− (x) := max{−u(x), 0} . Show that u+ , u− , and |u| ∈ W 1,p (Ω) , too, and
compute the corresponding weak gradients.
Hint: u+ = limε→0 Fε (u) , with
(
1
(z 2 + ε2 ) 2 − ε if z ≥ 0
Fε (z) :=
0 if z < 0 .
Solution: Take Fε as in the hint, and observe that Fε ∈ C 1 ∩ W 1,∞ (R) . Therefore, the previous
exercise gives that Fε (u) ∈ W 1,p (Ω) and that
1
D(Fε (u)) = Fε0 (u)Du = u+ (ε2 + u2 )− 2 Du ,
1
where the last equality follows by a direct computation. Since u+ (ε2 + u2 )− 2 → χ{u>0} a.e. in Ω ,
by dominated convergence we get that
D(Fε (u)) → χ{u>0} Du
in Lp (Ω; Rn ) . It follows that u+ belongs to W 1,p (Ω) , and that
D(u+ ) = χ{u>0} Du .
Since u− = u+ − u and |u| = u+ − u− we get that u− , and |u| ∈ W 1,p (Ω) , too, and the weak
gradients are given by
D(u− ) = −χ{u≤0} Du and D(|u|) = χ{u>0} Du − χ{u≤0} Du .
8. Exercises PDE 9-11.01.13
Exercise 1. Prove the following multiplicative form of the trace inequality: if u ∈ W 1,2 (Rn+ ) , denoting with T u the trace of u on the hyperplane RN −1 one has
kT uk2L2 (RN −1 ) ≤ CkukL2 (RN
kDukL2 (RN
.
+)
+)
Solution: Being Rn+ an extension domain, for every u ∈ W 1,2 (Rn+ ) there exists a sequence un of
functions in Cc∞ (Rn ) converging to u in W 1,2 (Rn+ ) . Now, for every x̂ ∈ RN −1 by the fundamental
Theorem of calculus one has
ˆ +∞
ˆ +∞
∂ 2
∂
2
un (x̂, 0) = −
un (x̂, xN ) dxN = −2
un (x̂, xN )
un (x̂, xN ) dxN
∂xN
∂xN
0
0
for every n ∈ N . This implies by Fubini’s Theorem
ˆ
ˆ
∂
2
un (x̂, 0) dx̂ = −2
un (x)
un (x) dx .
n
∂x
N
−1
N
R
R+
22
EXERCISES PDE
Since for every n and every x̂ the trace T un (x̂) of un on the hyperplane RN −1 is simply equal to
un (x̂, 0) , the previous equality and the Cauchy-Schwarz inequality yield
kT un k2L2 (RN −1 ) ≤ 2 kun kL2 (Rn+ ) kDun kL2 (Rn+ ) .
Since un is converging to u in W 1,2 (Rn+ ) , by continuity of the trace operator T un is converging
to T u in L2 (RN −1 ) , so that we get
kT uk2L2 (RN −1 ) ≤ 2 kukL2 (Rn+ ) kDukL2 (Rn+ ) .
by letting n going to +∞ , as required.
Exercise 2. Let Ω be an open set of Rn , and Γ a C 1 hypersurface such that Ω = Ω1 ∪ Ω2 ∪ Γ with
Ω1 and Ω2 open connected disjoint.
(a) Let f1 ∈ C 1 (Ω1 ) and f2 ∈ C 1 (Ω2 ) . Under which conditions the function f ∈ L∞ (Ω) defined
by f = f1 in Ω1 and f = f2 in Ω2 belongs to W 1,∞ (Ω) ?
(b) Let p ≥ 1 , f1 ∈ W 1,p (Ω1 ) and f2 ∈ W 1,p (Ω2 ) . Under which conditions the function f ∈ Lp (Ω)
defined by f = f1 in Ω1 and f = f2 in Ω2 belongs to W 1,p (Ω) ?
Solution: (a) Denote with νΓ the normal vector to the hypersurface Γ pointing from Ω1 to Ω2 . By
the Divergence Theorem we have, for every ϕ ∈ Cc∞ (Ω) ,
ˆ
ˆ
ˆ
f2 (x)div ϕ(x) dx =
f1 (x)div ϕ(x) dx +
f (x)div ϕ(x) dx =
Ω2
Ω
Ω1
ˆ
ˆ
ˆ
−
Df1 (x) · ϕ(x) dx −
Df2 (x) · ϕ(x) dx + (f1 (ξ) − f2 (ξ)) ϕ(ξ) · νΓ (ξ) dS(ξ)
Ω1
therefore f ∈ W
Ω2
1,∞
Γ
(Ω) if and only if f1 = f2 on Γ . In that case one also has
(
Df1 in Ω1
Df =
Df2 in Ω2 .
(b) It suffices to repeat the previous reasonings with suitable using of the notion of trace to deduce
that the required condition is T f1 = T f2 a.e. in Γ , where T denotes the trace operator from W 1,p (Ω1 )
to Lp (∂Ω1 ) , and from W 1,p (Ω2 ) to Lp (∂Ω2 ) , respectively.
Exercise 3. Give an example of a connected open set Ω ∈ Rn and of a function u ∈ W 1,∞ (Ω) such
that u is not Lipschitz continuous on Ω .
Solution: It simply suffices to modify the example in Exercise 1, 19.12.12. Namely, set Ω : =
B(0, 2) \ {xN = 0, |x| ≤ 21 } , which is connected, and let for instance
(
(|x|2 − 41 )2 if xN > 0, |x| ≤ 21
f (x) :=
0 elsewhere .
+
Denoting with B 0, 12
the upper hemisphere B 0, 12 ∩ {xN > 0}, one has by the Divergence
theorem that for any ϕ ∈ Cc∞ (Ω; Rn )
ˆ
ˆ
ˆ
2
f (x)div ϕ(x) dx =
+ f (x)div ϕ(x) dx = −4
+ (|x| − 14 ) x · ϕ(x) dx ;
Ω
1
B 0, 2
1
B 0, 2
indeed, no boundary term appears since either f or ϕ are 0 on ∂B 0, 12
W 1,∞ (Ω) . On the other hand, for every 0 < ε < 21 , one has
+
. It follows that f ∈
(ε2 − 41 )2
|f (0, . . . , 0, ε) − f (0, . . . , 0, −ε)|
=
|(0, . . . , 0, ε) − (0, . . . , 0, −ε)|
2ε
which is unbounded when ε is close to 0 . Therefore f cannot be Lipschitz continuous.
EXERCISES PDE
23
9. Exercises PDE 16-18.01.13
Exercise 1. Let Ω be a bounded domain with smooth boundary and u ∈ W 1,1 (Ω) . Assume that
Du ∈ Lp (Ω) for some 1 < p < ∞ . Prove that u ∈ W 1,p (Ω) .
Hint: Prove initially that u ∈ Lploc (Ω) , to understand the exercise. Taking a closer look to the
proof of Theorem 3, Section 5.3.3 of Evans’ book can help near to the boundary.
Solution: First step: let us prove that u ∈ Lploc (Ω) , which requires no regularity of the boundary.
Define un = %n ? u ∈ C ∞ (Rn ) , and fix a smooth open subset Ω0 ⊂⊂ Ω . For n large, depending on
Ω0 , it holds Dun = %n ? Du in Ω0 (the formula Dun = D%n ? u is instead true on the whole Rn : why
this difference?). We therefore deduce by Jensen’s inequality (how?) that
kDun kLp (Ω0 ) ≤ kDukLp (Ω) .
(9.1)
0
By Poincaré-Wirtinger and (9.1), if ūn,Ω0 denotes the integral mean of un on Ω we get that there
exists a constant CΩ0 depending on Ω0 such that
ˆ
|un (x) − ūn,Ω0 |p dx ≤ CΩ0 kDukpLp (Ω) .
Ω0
1
Since un is converging to u in L (Ω0 ) , by Fatou’s Lemma, denoting with ūΩ0 the integral mean of u
on Ω0 , we arrive at
ˆ
|u(x) − ūΩ0 |p dx ≤ CΩ0 kDukpLp (Ω)
Ω0
and this implies the claim.
Second step: suitably changing coordinates like in Theorem 3, Section 5.3.3 of Evans’ book, for
every x0 ∈ ∂Ω one can find a neighborhood V := B(x0 , r) ∩ Ω and a positive number λ > 0 such
that the ball B(x + nλ eN , n1 ) ⊂ Ω for all x ∈ V . Therefore, if one defines un (x) = u(x + nλ eN ) one
has un ∈ W 1,1 (V ) and since we only did a translation of the argument
kDun kLp (V ) ≤ kDukLp (Ω) .
(9.2)
Now, defining vn = %n ? un ∈ C ∞ (Rn ) , by condition B(x + nλ eN , n1 ) ⊂ Ω it holds Dvn = %n ? Dun .
Notice also that un and thus vn converge to u in L1 (V ) . Therefore, the same argument used in the
first step plus (9.2) proves that u in Lp (V ) .
Third step: Ω can be written as a finite union Ω = ∪m
i=1 Vi where V0 ⊂⊂ Ω and each Vi , i ≥ 1 is
such that the second step can be applied. Since
ˆ
m ˆ
X
p
|u(x)| dx ≤
|u(x)|p dx
Ω
i=1
p
Vi
the finiteness of the right-hand side gives u ∈ L (Ω) . Since Du ∈ Lp (Ω) by the hypothesis, the
exercise is concluded.
Exercise 2. The geometric counterpart of the Sobolev inequality.
(a) (optional, if not simply assume the result). Let f ∈ L1 (Rn ; Rm ) . Prove that
ˆ
nˆ
o
|f (x)| dx = sup
f (x) · ϕ(x) dx : ϕ ∈ Cc∞ (Rn ; Rm ), kϕk∞ ≤ 1 .
Rn
(9.3)
Rn
(b) Let C1,N the Sobolev constant in Rn , i.e. the least possible constant such that the Sobolev
inequality
kuk N ≤ CkDuk1
N −1
holds for every u ∈ W 1,1 (Rn ) . Prove the following isoperimetric inequality: for every smooth bounded
open set Ω ⊂ Rn one has
N −1
|Ω| N ≤ C1,N Hn−1 (∂Ω) ,
(9.4)
n−1
(∂Ω) is the (Hausdorff, or surface) measure of the
where |Ω| is the Lebesgue measure of Ω and H
boundary. Hint: approximate the characteristic function χΩ by convolution and use (9.3) and the
divergence theorem to estimate the L1 norm of the gradients.
24
EXERCISES PDE
Solution: (a) Define
(
f1 (x) :=
f (x)
|f (x)|
0
if f (x) 6= 0
otherwise
Let %n be the usual mollifiers, and, given a sequence ϕn of Cc∞ (Rn ) functions with 0 ≤ ϕn ≤ 1 with
ϕn (x) → 1 for every x , set gn := ϕn (%n ? f1 ) . It is clear that gn ∈ Cc∞ (Rn ; Rm ) and that kgn k∞ ≤ 1 .
Furthermore gn converges to f1 a.e. Observing that f (x) · f1 (x) = |f (x)| for every x , by dominated
convergence one has
ˆ
ˆ
nˆ
o
|f (x)| dx = lim
f (x) · gn (x) dx ≤ sup
f (x) · ϕ(x) dx : ϕ ∈ Cc∞ (Rn ; Rm ), kϕk∞ ≤ 1 .
Rn
n
Rn
Rn
The converse inequality is trivial so that we get (9.3).
(b) By (9.3) and integrating by parts we get that for every u ∈ W 1,1 (Rn ) one has
nˆ
o
kDuk1 = sup
u div ϕ : ϕ ∈ Cc∞ (Rn ; Rn ), kϕk∞ ≤ 1 .
(9.5)
Rn
Now, if %n are the usual mollifiers, we have for every ϕ ∈ Cc∞ (Rn ; Rn ) with kϕk∞ ≤ 1 that also
k%n ? ϕk∞ ≤ 1 . Therefore, by symmetry of the mollifiers, standard properties of the convolution and
also using the divergence theorem, we have
ˆ
ˆ
ˆ
(%n ? χΩ )(x) div ϕ(x) dx =
χΩ (x) div (%n ? ϕ)(x) dx =
div (%n ? ϕ)(x) dx =
Rn
ˆ
Rn
Ω
(%n ? ϕ)(ξ) · ν(ξ) dHn−1 (ξ) ≤ Hn−1 (∂Ω)
∂Ω
for every smooth bounded open set Ω ⊂ Rn . It follows from this and (9.5) that for every n ∈ N one
has
kD(%n ? χΩ )k1 ≤ Hn−1 (∂Ω)
(9.6)
for every smooth bounded open set Ω ⊂ Rn .
Now, combining (9.6) with the Sobolev inequality gives
k%n ? χΩ k
N
N −1
≤ C1,N Hn−1 (∂Ω)
for every smooth bounded open set Ω ⊂ Rn and every n ∈ N. Letting n to +∞ we get
kχΩ k
N
N −1
≤ C1,N Hn−1 (∂Ω)
which is exactly (9.4).
Exercise 3. Let p > 2 . Show with a counterexample that the Morrey’s Imbedding Theorem is not
verified in the nonsmooth two-dimensional domains
Dα := {(x, y) : 0 < x < 1,
0 < y < xα }
for α sufficiently large.
Hint: prove that there are unbounded funtions in W 1,p (Dα ) .
Solution: Let α > p − 1 . We can choose γ > 0 such that
α+1
γ+1
α − (γ + 1)p > −1 .
> p, which is equivalent to
(9.7)
Now, consider the function
uγ (x, y) := x−γ .
The function uγ is unbounded in Dα so that it cannot belong to a space of Hölder functions. On the
other hand by Fubini’s theorem one has
ˆ
ˆ 1
|uγ (x, y)|p dxdy =
xα−γp dx < +∞
Dα
0
EXERCISES PDE
25
since by (9.7) one gets α − γp > −1 . Similarly, being Du(x, y) = (−γx−(1+γ) , 0) one gets
ˆ
ˆ 1
xα−(γ+1)p dx < +∞
|Duγ (x, y)|p dxdy = γ p
0
Dα
again using Fubini’s theorem and (9.7). Therefore uγ ∈ W 1,p (Dα ) but it cannot satisfy Morrey’s
imbedding Theorem.
10. Exercises PDE 23-25.01.13
Exercise 1. Fourier analysis in Sobolev spaces. Let f ∈ L2 ((0, π)) . Define for every N ∈ N
N
2X
an sin(nt)
π n=1
fN (t) :=
´π
where an = 0 f (t) sin(nt) dt. It is known that fN converges to f in L2 ((0, π)) (if you need to
convince yourself that no cosine is needed to have L2 convergence, simply apply the general theorem
to the odd extension of f to [−π, π] ).
(a) Assume that f ∈ H01 ((0, π)) . Show that fN converges to f uniformly in [0, π] . Hint: prove
fN to be a Cauchy sequence in some Sobolev space.
(b) If f ∈ H01 ((0, π)) ∩ H 2 ((0, π)) prove that there exists a constant C such that
kfN − f kH 1 ≤
C
kf kH 2 .
N
(10.1)
Solution: (a) Since f (0) = f (π) = 0 integrating by parts we obtain
ˆ π
nan = −
f 0 (t) cos(nt) dt
(10.2)
0
for every n ∈ N. Since {
gives
q
2
π
cos(nt) : n ∈ N} is an orthonormal system of L2 (0, π) , Bessel’s inequality
N
X
N ˆ
X
π
2
π
f 0 (t) cos(nt) dt ≤ kf 0 k22
2
0
n=1
n=1
PN
for every N ∈ N . Being a convergent
series, n=1 n2 a2n is a Cauchy sequence.
q
n2 a2n
=
On the other hand, since { π2 cos(nt) : n ∈ N} is an orthonormal system of L2 (0, π) , for every
N ∈ N and every M ∈ N with M ≥ N , one has
0
0 2
kfM
− fN
k2 = k
M
X
nan cos(n·)k22 =
n=N
M
X
n2 a2n k cos(n·)k22 =
n=N
M
X
n2 a2n .
(10.3)
n=N
It follows that fN is a Cauchy sequence in L2 ((0, π)) . Since fN was converging to f in L2 ((0, π)) ,
combining the two gives that fN is converging to f in H01 ((0, π)) , and therefore uniformly by Morrey’s
imbedding Theorem.
(b) If f ∈ H 2 ((0, π)) we can once more integrate by parts in (10.2), obtaining
ˆ π
ˆ
1 π 00
0
f (t) cos(nt) dt = −
f (t) sin(nt) dt
n 0
0
ˆ
so that
π
f 00 (t) sin(nt) dt
n2 an =
0
and, again using Bessel’s inequality,
N
X
n=1
n4 a2n ≤
π 00 2
kf k2
2
26
EXERCISES PDE
for every N ∈ N . Combining this with (10.3) for every N ∈ N and every M ∈ N with M ≥ N , one
has
M
M
X
π
1 X 4 2
0
0 2
kf 00 k22 .
kfM
− fN
k2 =
n an ≤
n2 a2n ≤ 2
N
2N 2
n=N
n=N
When M goes to +∞ we therefore obtain
0
kf 0 − fN
k2 ≤
r
π 1 00
kf k2
2N
Since by Poincaré’s inequality
0
kf − fN k2 ≤ πkf 0 − fN
k2
we get that
r
kf − fN kH01 ((0,π)) ≤
1
π
(1 + π) kf 00 k2
2
N
as required.
Exercise 2. Céa’s Lemma: Consider a symmetric, bounded, coercive bilinear form B : H × H → R
on a Hilbert space H , that is
B(u, v) = B(v, u) ,
B(u, v) ≤ C1 kukH kvkH ,
B(u, u) ≥ c0 kuk2H .
(a) Fix a linear continuous functional F : H → R. Justify that there exists a unique u ∈ H such
that
B(u, v) = F (v)
for every v ∈ H .
(b) Fix additionally an N -dimensional subspace HN of H . Justify the following fact: there exist
a unique approximate solution uN ∈ HN , that is a unique uN ∈ HN such that
B(uN , vN ) = F (vN )
for every vN ∈ HN .
(c) Prove the following equality
B(u − uN , u − uN ) + B(vN − uN , vN − uN ) = B(u − vN , u − vN )
for every vN ∈ HN . Deduce Céa’s estimate:
q
ku − uN kH ≤ Cc01 inf{ku − vN k : vN ∈ HN } .
(10.4)
(10.5)
(d) Consider now H = H01 ((0, π)) and HN := span{t 7→ sin(nt) : n = 1, . . . , N } . For a(x) ∈
C 1 [0, π] with 1 ≤ a(x) ≤ 2 for every x and f ∈ L2 ((0, π)) set
ˆ π
B(u, v) :=
a(t) u0 (t) v 0 (t) dt
(10.6)
´π
0
and F (v) := 0 f (t) v(t) dt. Check that u and uN as in (a) and (b), respectively, exist, and show
that uN → u in H01 ((0, π)) as N goes to +∞ . If additionally u ∈ H 2 ((0, π)) show that there exists
a constant C such that
C
ku − uN kH 1 ≤ kukH 2 .
N
Solution: (a) follows from the Lax-Milgram theorem. The same holds for (b) since the restriction of
B to the Hilbert subspace HN × HN is clearly bilinear, coercive and continuous.
(c) By bilinearity and simmetry of the form B we have
B(u − uN , u − uN ) = B(u, u) − 2B(u, uN ) + B(uN , uN )
B(vN − uN , vN − uN ) = B(vN , vN ) − 2B(uN , vN ) + B(uN , uN )
B(u − vN , u − vN ) = B(u, u) − 2B(u, vN ) + B(vN , vN )
EXERCISES PDE
27
therefore (10.4) is equivalent to prove that
2(B(uN , uN ) − B(u, uN ) + B(u, vN ) − B(uN , vN )) = 0 .
This can be easily proved to be true by observing that by definition of u and uN , for every vN ∈ HN
one has
B(u, vN ) = F (vN ) = B(uN , vN )
and in particular, for vN = uN ,
B(u, uN ) = B(uN , uN ) .
Now, by (10.4), and the coercivity and continuity of B we get for every vN ∈ HN
c0 ku − uN k2H ≤ B(u − uN , u − uN ) ≤ B(u − uN , u − uN ) + B(vN − uN , vN − uN ) =
B(u − vN , u − vN ) ≤ C1 ku − vN k2H
whence
ku − uN kH ≤
q
C1
c0 ku
− vN kH
for every vN ∈ HN . Taking the infimum in the rigth-hand side gives (10.5).
(d) To prove existence of u and uN when B is given by (10.6), it suffices to check that the LaxMilgram theorem can be applied. The only relevant point to this end is to prove coercivity of B ,
which follows from the Poincaré inequality. Namely, one has
1
B(u, u) ≥ ku0 k22 ≥
kuk2H 1 .
1 + π2
It follows now from the previous exercise that
inf{ku − vN kH01 ((0,π)) : vN ∈ HN } → 0
as N goes to +∞ . In particular, by (10.1), if u ∈ H 2 one has for every N ∈ N
inf{ku − vN kH01 ((0,π)) : vN ∈ HN } ≤
C
kukH 2 .
N
Combining these two facts with (10.5) gives the conclusion.
Exercise 3. Consider a bounded smooth open subset Ω ⊂ Rn , and ΓD ⊂ ∂Ω with Hn−1 (ΓD ) > 0 .
Let
HΓ1D (Ω) := {u ∈ H 1 (Ω), T u = 0 on ΓD } ,
with T the trace operator. For λ > 0 define
ˆ
ˆ
1
Bλ (u, v) =
∇u(x) · ∇v(x) dx −
T u(ξ) T v(ξ) dHn−1 (ξ)
λ ∂Ω
Ω
for u and v ∈ HΓ1D (Ω) . Prove that, when λ is sufficiently large, for every f ∈ L2 (Ω) there exists a
unique uf ∈ HΓ1D (Ω) such that
ˆ
Bλ (uf , v) =
f (x) v(x) dx
Ω
for every v ∈ HΓ1D (Ω) . Hint: does the Poincaré inequality apply in HΓ1D (Ω) ?
Solution: Following the hint, let us first prove that there exists a constant C such that
kukL2 (Ω) ≤ Ck∇ukL2 (Ω)
(10.7)
for every u ∈ HΓ1D (Ω) . To prove this, it suffices to argue as in the proof of the Poincaré-Wirtinger
inequality (Theorem 1, Section 5.8.1 in Evans’book): contradicting (10.7) would produce the existence
of a sequence un in HΓ1D (Ω) with kun k|L2 (Ω) = 1 for every n which is converging to a constant c in
H 1 (Ω) . By continuity of the trace operator, HΓ1D (Ω) is a Hilbert subspace of H 1 (Ω) , so it must be
c = 0 , since this one is the only constant in HΓ1D (Ω) . On the other hand, by strong L2 convergence
1
we must have kckL2 (Ω) = c|Ω| 2 = 1 , and this gives a contradiction.
28
EXERCISES PDE
Now, let C be given by (10.7), and let CT the continuity constant of the trace operator. Assume
that λ > CT (1 + C 2 ) . Then for every u ∈ HΓ1D (Ω) one has, by the continuity of the trace and by
(10.7), that
Bλ (u, u) = k∇uk2L2 (Ω) −
1
CT
1
kuk2H 1 (Ω) −
kT uk2L2 (∂Ω) ≥
kuk2H 1 (Ω) = c0 kuk2H 1 (Ω)
2
λ
1+C
λ
CT
1
> 0 due to the assumption on λ . Therefore for λ > CT (1 + C 2 ) the bilinear
with c0 := 1+C
2 − λ
form Bλ is coercive on HΓ1D (Ω) . For any λ > 0 one has also by the Cauchy-Schwarz inequality and
the continuity of the trace operator that
Bλ (u, v) ≤ (1 +
CT
)kukH 1 (Ω) kvkH 1 (Ω)
λ
so that Bλ is continuous. The conclusion follows by the Lax-Milgram theorem.
11. Exercises PDE 30.01.13-01.02.13
Exercise 1. Let H be an Hilbert space, a : H × H → R a bilinear, symmetric, coercive, and
continuous form, and L : H → R a linear continuous functional. Prove that u ∈ H solves
a(u, v) = L(v)
for every v ∈ H
(11.1)
if and only if
n1
o
1
a(u, u) − L(u) = min
a(v, v) − L(v) : v ∈ H .
2
2
(11.2)
Hint: if u is a minimizer, fix v ∈ H and consider the function F (t) := 21 a(u + tv, u + tv) − L(u + tv) ,
t ∈ R.
Solution: “If ” part: assume that u ∈ H satisfies (11.2). Fix v ∈ H and consider the function
F (t) := 21 a(u + tv, u + tv) − L(u + tv) , t ∈ R . The function F attains then its minimum value for
t = 0 . Since by a direct computation
F (t) =
1
t2
a(u, u) − L(u) + t(a(u, v) − L(v)) + a(v, v)
2
2
(11.3)
imposing F 0 (0) = 0 we get (11.1).
“Only if ” part: assume that u ∈ H satisfies (11.1). Fix v ∈ H and, again, consider the function
F (t) := 21 a(u + tv, u + tv) − L(u + tv) , t ∈ R . By (11.3) and (11.1) we obtain
F (t) =
1
t2
a(u, u) − L(u) + a(v, v) .
2
2
By coerciveness of a, a(v, v) ≥ 0 which implies that F attains its minimum at 0 . In particular
F (0) ≤ F (1) , therefore
1
1
a(u, u) − L(u) ≤ a(u + v, u + v) − L(u + v)
2
2
(11.4)
for every v ∈ H . Now, for an arbitrary w ∈ H , applying (11.4) with v = w − u one gets
1
1
a(u, u) − L(u) ≤ a(w, w) − L(w)
2
2
that is (11.2).
EXERCISES PDE
29
Exercise 2. Let Ω ⊂ Rn be a bounded open set with smooth boundary, and f ∈ L2 (Ω) . We say
that u ∈ H 1 (Ω) is a weak solution of the Neumann problem
(
−∆u = f in Ω
(11.5)
∂u
∂ν = 0 on ∂Ω
ˆ
if
ˆ
∇u(x) · ∇v(x) dx =
Ω
f (x)v(x) dx
(11.6)
Ω
for any v ∈ H 1 (Ω) .
(a) Justify that actually (11.6) is a weak formulation of the problem (11.5) by showing that if u
2
solves (11.6) and in addition u ∈ H 2 (Ω) then −∆u = f in L2 (Ω) and ∂u
∂ν = 0 in L (∂Ω) in the sense
of traces.
(b) Show that a necessary and sufficient condition for the existence of a solution of (11.6) is
ˆ
f (x) dx = 0 .
(11.7)
Ω
Hint: use Lax-Milgram’s theorem in the Hilbert space
ˆ
1
H := {u ∈ H (Ω) :
u(x) dx = 0} .
(11.8)
Ω
(c) Let g ∈ L2 (∂Ω) . Generalise part (a) and (b) to the nonhomogeneuous Neumann problem
(
−∆u = f in Ω
(11.9)
∂u
∂ν = g on ∂Ω
by finding its weak formulation and a necessary and sufficient condition for existence of a weak solution.
Solution: (a) If u solves (11.6) and in addition u ∈ H 2 (Ω) , for every v ∈ Cc∞ (Ω) integrating by
parts and using (11.6) we have
ˆ
ˆ
ˆ
−
∆u(x)v(x) dx =
∇u(x) · ∇v(x) dx =
f (x)v(x) dx
Ω
Ω
Ω
which implies −∆u = f in L2 (Ω) . Using this additional information, integrating by parts and using
again (11.6) we get for every v ∈ H 1 (Ω)
ˆ
f (x)v(x) dx =
Ω
ˆ
ˆ
ˆ
∂u
−
∆u(x)v(x) dx =
∇u(x) · ∇v(x) dx −
(ξ)v(ξ) dHn−1 (ξ) =
Ω
Ω
∂Ω ∂ν
ˆ
ˆ
∂u
f (x)v(x) dx −
(ξ)v(ξ) dHn−1 (ξ) ,
∂ν
Ω
∂Ω
2
which gives ∂u
∂ν = 0 in L (∂Ω) .
Observe in addition that via integration by parts also the converse holds: if u ∈ H 2 (Ω) solves
(11.5), then (11.6) holds.
(b) Taking v = 1 in (11.6) gives that (11.7) is necessary. To prove that it is sufficient, defining the
subspace H as in (11.8), we first prove that, if u ∈ H satisfies
ˆ
ˆ
∇u(x) · ∇v(x) dx =
f (x)v(x) dx
(11.10)
Ω
Ω
for every v ∈ H , then (11.6) holds. Indeed, when v ∈ H 1 (Ω) , then ṽ = v −
H ; furthermore, ∇ṽ = ∇v and, by (11.7),
ˆ
ˆ
f (x)v(x) dx =
f (x)ṽ(x) dx .
Ω
Ω
1
|Ω|
´
Ω
v(x) dx belongs to
30
EXERCISES PDE
Therefore (11.10) implies (11.6). Now, existence of a unique solution to (11.10) in H follows by the
Lax-Milgram theorem once it is checked that the bilinear form B : H × H → R defined by
ˆ
B(u, v) :=
∇u(x) · ∇v(x) dx
Ω
is continuous and coercive. Continuity is obvious, while coercivity follows by the Poincaré-Wirtinger
inequality
k∇uk2 ≥ CΩ kuk2
for every u ∈ H (add the details!).
(c) The solution is very similar to the previous case, so I only sketch it. Integrating by parts, one
easily gets that a smooth solution of (11.9) satisfies
ˆ
ˆ
ˆ
∇u(x) · ∇v(x) dx =
f (x)v(x) dx +
g(ξ)v(ξ) dHn−1 (ξ) .
(11.11)
Ω
Ω
∂Ω
This is therefore our candidate weak formulation. Testing (11.11) on Cc∞ functions we get that if u
solves it and in addition u ∈ H 2 (Ω) then −∆u = f in L2 (Ω) . Using this additional information,
integrating by parts and using again (11.11) we get for every v ∈ H 1 (Ω)
ˆ
f (x)v(x) dx =
Ω
ˆ
ˆ
ˆ
∂u
(ξ)v(ξ) dHn−1 (ξ) =
−
∆u(x)v(x) dx =
∇u(x) · ∇v(x) dx −
∂ν
Ω
Ω
∂Ω
ˆ
ˆ ∂u
f (x)v(x) dx −
(ξ) − g(ξ) v(ξ) dHn−1 (ξ) ,
Ω
∂Ω ∂ν
2
which gives ∂u
∂ν = g in L (∂Ω) .
We then show that (11.11) has a solution if and only if
ˆ
ˆ
f (x) dx +
g(ξ) dHn−1 (ξ) = 0 .
Ω
(11.12)
∂Ω
Indeed, taking v = 1 in (11.11) produces (11.12) as a necessary condition. On the other hand, if
(11.12) holds, exactly as in part (b) we can show that, defining the subspace H as in (11.8), if u ∈ H
satisfies
ˆ
ˆ
ˆ
∇u(x) · ∇v(x) dx =
Ω
g(ξ)v(ξ) dHn−1 (ξ) .
f (x)v(x) dx +
Ω
(11.13)
∂Ω
for every v ∈ H , then (11.11) holds. Existence of a solution to (11.13) in H follows again by the
Lax-Milgram Theorem; indeed, the bilinear form B : H × H → R defined by
ˆ
B(u, v) :=
∇u(x) · ∇v(x) dx
Ω
is continuous and coercive, while the linear functional
ˆ
ˆ
L(v) :=
f (x)v(x) dx +
Ω
g(ξ)v(ξ) dHn−1 (ξ)
∂Ω
is continuous from H to R by continuity of the trace operator.
Exercise 3. Let Ω ⊂ Rn be a bounded open set with smooth boundary, with ∂Ω = Γ1 ∪ Γ2 , and
assume that Hn−1 (Γ1 ) > 0 . Let f ∈ L2 (Ω) . Find the weak formulation of the mixed boundary-value
problem

−∆u = f in Ω

(11.14)
u = 0 on Γ1

 ∂u
∂ν = 0 on Γ2
and prove existence of a weak solution by using Lax-Milgram’s theorem in a suitable Hilbert subspace
of H 1 (Ω) .
EXERCISES PDE
31
Solution: The presence of the Dirichlet boundary condition suggests to operate in the Hilbert subspace
H := {u ∈ H 1 (Ω), u = 0 on Γ1 } ,
where we have already seen that the Poincaré inequality
k∇uk2 ≥ CΩ kuk2
holds (short reminder: the only constant function in H is 0 and ∂Ω is smooth).
The weak formulation is: finding u ∈ H such that
ˆ
ˆ
∇u(x) · ∇v(x) dx =
f (x)v(x) dx
Ω
(11.15)
Ω
for every v ∈ H . Indeed, any smooth solution of (11.14) satisfies (11.15). Conversely, if u ∈ H
satisfies (11.15) and additionally u ∈ H 2 (Ω) , testing (11.15) on Cc∞ functions gives −∆u = f in
L2 (Ω) . Using this additional information, integrating by parts and using again (11.15) we get for
every v ∈ H
ˆ
f (x)v(x) dx =
Ω
ˆ
ˆ
ˆ
∂u
(ξ)v(ξ) dHn−1 (ξ) =
−
∆u(x)v(x) dx =
∇u(x) · ∇v(x) dx −
Ω
Ω
∂Ω ∂ν
ˆ
ˆ
∂u
(ξ)v(ξ) dHn−1 (ξ) ,
f (x)v(x) dx −
Ω
Γ2 ∂ν
2
since v = 0 on Γ1 . This gives ∂u
∂ν = 0 in L (Γ2 ) . Since u ∈ H we also have u = 0 a.e. on Γ1 , so
(11.15) is the weak formulation of (11.14).
Existence of a unique solution to (11.15) in H follows again by the Poincaré inequality and the
Lax-Milgram Theorem.
12. Exercises PDE 14.02.13
Exercise 1. Differentiation of convolutions revisited. Recall preliminarly the following form of the
Jensen inequality: let F ∈ L∞ (Rn ), F ≥ 0 with
ˆ
F (y) dy = 1 .
Rn
Then for every convex function ϕ : R → R and every g ∈ L1 (Rn ) it holds
ˆ
ˆ
ϕ
g(y)F (y) dy ≤
ϕ(g(y))F (y) dy .
Rn
Rn
∞
In the following %ε is a positive symmetric C mollifier with
ˆ
%ε (y) dy = 1
(12.1)
Rn
and supp%ε ⊂ B(0, ε) .
(a) Let 1 ≤ p < ∞ and f ∈ W 1,p (Rn ) . Prove that D(%ε ? f ) = %ε ? Df in Lp (Rn ) .
(b) Let 1 ≤ p < ∞ and f ∈ W 1,p (Rn ) . Prove that k%ε ? f kW 1,p (Rn ) ≤ kf kW 1,p (Rn ) .
(c) Let Ω be a bounded open subset of Rn and f ∈ W01,1 (Ω) . Prove that D(%ε ? f ) = %ε ? Df in
L1 (Ω) . Hint: extend f by setting it equal to 0 outside Ω . Is this extension in W 1,1 (Rn ) ?
(d) Let Ω be a bounded open subset of Rn and f ∈ W 1,1 (Ω) . Define
Ωε := {x ∈ Ω :
dist (x, ∂Ω) ≥ ε} .
Let Ω0 ⊂⊂ Ωε . Prove that D(%ε ? f ) = %ε ? Df in L1 (Ω0 ) . Hint: try first with C ∞ functions, then
approximate.
32
EXERCISES PDE
Solution: (a) Let ϕ ∈ Cc∞ (Rn ; Rn ) . Using the properties of the convolution with the mollifiers and
the definition of weak gradient we have
ˆ
ˆ
%ε ? f (x) div ϕ(x) dx =
f (x) %ε ? div ϕ(x) dx =
ˆ
Rn
ˆ
Rn
ˆ
f (x) div (%ε ? ϕ)(x) dx = −
Df (x) · %ε ? ϕ(x) dx = −
%ε ? Df (x) · ϕ(x) dx
Rn
Rn
Rn
as required.
(b) Applying Jensen’s inequality with F (y) = %ε (x − y) for fixed x and using Fubini’s Theorem
and (12.1), we get
ˆ
ˆ ˆ
p
p
%ε (x − y)f (y) dy dx ≤
|%ε ? f (x)| dx =
Rn
Rn ˆ
ˆ
Rn
ˆ
%ε (x − y)|f (y)|p dy dx =
|f (y)|p dy
Rn
Rn
Rn
which proves k%ε ? f kLp (Rn ) ≤ kf kLp (Rn ) . The same argument with Df in place of f gives the claim.
(c) Let f˜ be the extension of f simply obtained by f˜ = 0 out of Ω . Trivially f˜ ∈ W 1,1 (Rn ) .
Indeed, since f ∈ W01,1 (Ω) there is a sequence fn ∈ Cc∞ (Ω) converging to f in W 1,1 . The extensions
f˜n simply obtained by f˜n = 0 out of Ω form now easily a Cauchy sequence in W 1,1 (Rn ) with limit
f˜. Step (a) gives now D(%ε ? f˜) = %ε ? Df˜. Since Df˜ = Df in Ω and Df˜ = 0 outside, we conclude.
(d) Let us only see the proof when f ∈ C ∞ (Ω) leaving to the reader the easy approximation
argument. Since supp%ε ⊂ B(0, ε) and by symmetry of the convolution
ˆ
ˆ
ˆ
%ε ? f (x) =
%ε (x − y)f (y) dy =
%ε (y)f (x − y) dy =
%ε (y)f (x − y) dy .
Rn
Rn
B(0,ε)
0
Now, when x ∈ Ω , the gradient Df (x − y) (in the variable x ) exists for every y ∈ B(0, ε) , since
in this case x − y ∈ Ω where f is differentiable. We can therefore derive under the sign of integral,
getting the required equality.
Exercise 2. Let N = 2 and B(0, 1) ⊂ R2 be the unit ball in the plane.
(a) For 0 < r < 1 let ur be the unique solution of the problem


∆ur = 0 in B(0, 1) \ B(0, r)
ur (x) = 1 if |x| = 1


ur (x) = 0 if |x| = r .
Compute ur .
(b) Define
(
vr (x) :=
ur (x) in B(0, 1) \ B(0, r)
0 if |x| ≤ r .
Prove that vr ∈ H 1 (B(0, 1)) and that
kvr − 1kH 1 → 0
when r → 0 .
(c) Let f ∈ C 1 (B(0, 1)) . Construct a sequence fr with suppfr ⊂ B(0, 1)\B(0, r) and kfr −f kH 1 →
0.
(d) Deduce that H01 (B(0, 1)\{0}) = H01 (B(0, 1)) .
(e) Let N = 1 . Is that true that H01 ((−1, 1)\{0}) = H01 ((−1, 1)) ? Hint: Sobolev convergence is
particularly strong in dimension 1 .
Solution: (a) The radial symmetry of the domain and of the data suggests a radial solution. Arguing
in the same way used to construct the fundamental solution and imposing the boundary conditions
one gets
log(|x|)
ur (x) =
+ 1.
| log r|
EXERCISES PDE
33
(b) vr are indeed Lipschitz continuous since obtained by gluing together two C 1 functions with
the same trace on the interface (see Exercise 2, 9-11.01.13). Furthermore, using polar coordinates
ˆ
ˆ 1
2π
2Cπ
|vr (x) − 1|2 dx = |B(0, r)| +
% log2 % d% ≤ |B(0, r)| +
.
2
log r r
log2 r
B(0,1)
Indeed, the function % → % log2 % is uniformly bounded in (0, 1] since it has a limit (namely 0 ) when
% → 0 . Now, the right-hand side vanishes when r → 0 . About the first derivatives, again by a direct
computation and using polar coordinates
ˆ 1
ˆ
ˆ
1
2π
2π
d% =
|D(vr (x) − 1)|2 dx =
|Dur (x)|2 dx =
2
| log r|
log r r %
B(0,1)
B(0,1)\B(0,r)
which again vanishes when r → 0 . Thus, the claim is proved.
(c) Set fr := vr f . Clearly suppfr ⊂ B(0, 1) \ B(0, r) . Furthermore, since f is C 1 up to the
boundary, we easily have
kfr − f kH 1 ≤ kf kC 1 kvr − 1kH 1
which gives the claim.
(d) We need to use a double approximation argument. First, of all, observe that clearly
Cc∞ (B(0, 1\{0}) ⊂ H01 (B(0, 1)) ;
taking the closure at the left-hand side, the inclusion H01 (B(0, 1)\{0}) ⊂ H01 (B(0, 1)) is trivial. Now,
let f ∈ H01 (B(0, 1)) and let fn ∈ Cc∞ (B(0, 1)) a sequence H 1 -converging to f . By step (c) applied to
each fn , for every n we can construct gn ∈ H01 (B(0, 1)\{0}) such that kgn − fn kH 1 ≤ n1 (construct a
sequence approximating fn and then choose a sufficiently close element of the sequence). Now, easily
gn converges to f in H 1 ! Since H01 (B(0, 1)\{0}) is a Banach space, f ∈ H01 (B(0, 1)\{0}) , and this
proves the reverse inclusion.
(e) No. If the equality holds, any f ∈ H01 ((−1, 1)) could be approximated by a sequence fn ∈
∞
Cc ((−1, 1) \ {0}) in the H 1 topology, and in particular, by Morrey’s imbedding in dimension 1 ,
uniformly. This would imply f (0) = 0 which has no general reason to be.
Exercise 3. Let Ω ⊂ Rn be a smooth bounded connected open set, and u ∈ C ∞ (Ω) a solution of the
PN
2
elliptic equation Lu = − i,j=1 aij (x) ∂x∂i ∂xj u = 0 in Ω . Assume that the coefficients aij ∈ C 1 (Ω) .
Show that for λ sufficiently large not depending on u the function v := |∇u|2 + λu2 satisfies Lv ≤ 0
in Ω . Deduce that
(12.2)
k∇ukL∞ (Ω) ≤ C k∇ukL∞ (∂Ω) + kukL∞ (∂Ω) .
Solution: We called A(x) the matrix whose coefficients are aij (x) and we denote with h·, ·i the
scalar product between vectors in Rn . The norm we consider on matrices is the Frobenius norm.
We start by observing that by a direct computation, for any w ∈ C ∞ (Ω) one has
2
Lw (x) = −
For w =
√
N
X
i,j=1
aij (x)
∂2
w2 (x) = −hA(x)∇w(x), ∇w(x)i + 2w(x)Lw(x) .
∂xi ∂xj
λu, since u is a solution and by ellipticity, we get
L(λu2 (x)) ≤ −λα|∇u(x)|2
(12.3)
∂u
∂xk
with α the coercivity constant of A. For k = 1, . . . , N and w =
we have
∂u 2 2
∂u
∂u
∂u
L
(x) ≤ −α∇(
)(x) + 2
(x)L
(x) .
∂xk
∂xk
∂xk
∂xk
By a direct computation and since Lu = 0 we have
N
N
∂u
X
X
∂
∂
∂2
∂
∂2
L
(x) =
Lu(x) +
aij (x)
u(x) =
aij (x)
u(x) .
∂xk
∂xk
∂xk
∂xi ∂xj
∂xk
∂xi ∂xj
i,j=1
i,j=1
34
EXERCISES PDE
Since aij ∈ C 1 (Ω) we get easily
∂u
∂u
(x)L
(x) ≤ CN |∇u(x)||D2 u(x)|
∂xk
∂xk
2
with D u the Hessian matrix and CN a constant depending on the C 1 norm of the aij ’s and on the
dimension N . Summing over k from 1 to N we have
2
L(|∇u|2 (x)) ≤ −α|D2 u(x)|2 + CN |∇u(x)||D2 u(x)| .
(12.4)
We now recall Young’s inequality in the following form: for every a, b ∈ R and ε > 0 one has
ab ≤
1
ε2 2
a + 2 b2 .
2
2ε
(12.5)
Using now (12.3), (12.4) and (12.5) with a = |D2 u(x)|, b = |∇u(x)| and ε =
L(|∇u|2 (x) + λu2 (x)) ≤ −λα|∇u(x)|2 +
q
α
2CN
we obtain
2
4CN
|∇u(x)|2 .
α
2
the right-hand side is nonpositive so that setting v := |∇u|2 + λu2 one has Lv ≤ 0
For λ ≥ 2CαN
in Ω . By the weak maximum principle
kvkL∞ (Ω) ≤ kvkL∞ (∂Ω) .
k∇uk2L∞ (Ω)
≤ kvkL∞ (Ω) . It is also easy to se that there is a constant C depending on
2
N and λ, thus on N , CN and α such that kvkL∞ (∂Ω) ≤ C k∇ukL∞ (∂Ω) + kukL∞ (∂Ω) . From this,
Now, trivially
(12.2) follows.