EXERCISES PDE 31.10.12-02.11.12

1. Exercise

Let U ∈ RN be a bounded open set. We say that v ∈ C2(U) is subharmonic iff −∆v ≤ 0 in U .

(a) Prove that subharmonic functions enjoy the following form of the mean-value property: forevery x ∈ U , for every r > 0 such that B(x, r) ⊂ U

v(x) ≤ B(x,r)

v(y) dy

(b) Use the previous property to prove the maximum principle for subharmonic functions:

maxU

v = max∂U

v .

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 withthe 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) :=

∂B(x,ρ)

v(ξ) dS(ξ) =

∂B(0,1)

v(x+ ρζ) dS(ζ)

for ρ ∈ (0, r). Observe that

limρ→0+

φ(ρ) = v(x) . (1.1)

Now using the divergence theorem one has

ψ′(ρ) =

∂B(0,1)

∇v(x+ ρζ) · ζ dS(ζ) =

∂B(x,ρ)

∇v(ξ) · ξ − xρ

dS(ξ) =

=

∂B(x,ρ)

∂v

∂ν(ξ) dS(ξ) =

r

N

B(x,ρ)

∆v(y) dy ≥ 0

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 B(x,r)

v(y) dy =1

αNrN

ˆB(x,r)

v(y) dy =1

αNrN

ˆ r

0

(ˆ∂B(x,ρ)

v(ξ) dS(ξ))dρ =

=1

rN

ˆ r

0

NρN−1ψ(ρ) dρ ≥ v(x)

rN

ˆ r

0

NρN−1 dρ = v(x) ,

as required.

(b) Assume that v takes its maximum at an interior point x0 . But for every r > 0 such thatB(x0, r) ⊂ U one has, by part (a)

B(x0,r)

v(y)− v(x0) dy ≥ 0 ;

1

2 EX1

since the integrand must be nonpositive this implies v(y) = v(x0) for every y ∈ B(x0, r). It followsthat v is constant in the connected component of U containing x0 (strong maximum principle) whichimplies the weaker conclusion. Alternative proof will appear here next week.

(c) By the usual rules of differentiation, one has

∆[φ(u)] = φ′(u)∆u+ φ′′(u)|∇u|2 .Since ∆u = 0 and φ′′ ≥ 0 by convexity, one gets ∆[φ(u)] ≥ 0 as required.

(d) For every i = 1, . . . , N the derivatives ∂u∂xi

are still harmonic. By the previous step with

φ(t) = t2 , the functions(∂u∂xi

)2are subharmonic. Then

|∇u|2 =

N∑i=1

(∂u∂xi

)2must be subharmonic as well.

2. Exercise

Let U ∈ RN be a bounded open set. Let v ∈ C2(U) satisfy−∆v = f in U

v = g on ∂U .

Prove that there exist a constant CU only depending on U such that

maxU|v| ≤ CU (max

∂U|g|+ max

U|f |) .

Solution: Let L := maxU |f | , and define v(x) := v(x) + L2N |x|

2 . One has

−∆v = f − L = f −maxU|f | ≤ 0

and therefore by the maximum principle for subharmonic functions and trivial estimates

maxU

v ≤ maxU

v = max∂U

v ≤ max∂U

g + L maxy∈∂U

1

2N|y|2 .

Therefore, setting MU := maxy∈∂U1

2N |y|2 and recalling the expression of L , we get

maxU

v ≤ max∂U

g +MU maxU|f | ≤ max

∂U|g|+MU max

U|f | (2.1)

where the last estimate is obvious. The same argument with −v in place of v gives

maxU

(−v) ≤ max∂U| − g|+MU max

U| − f | = max

∂U|g|+MU max

U|f | . (2.2)

Combining (2.1) and (2.2) gives

maxU|v| ≤ max

∂U|g|+MU max

U|f |

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

Exercise 1. Let U ⊂ RN be a bounded open set. Consider a sequence un of harmonic functionssuch 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 continuousfunctions; moreover, x ∈ U and r > 0 such that B(x, r) ⊂ U one has

u(x) = limn→+∞

un(x) = limn→+∞

∂B(x,r)

un(ξ) dS(ξ) =

∂B(x,r)

u(ξ) dS(ξ) ,

that is, it holds the mean-value property. This one implies first that u ∈ C∞(U) (remember theargument with the mollifiers!), and eventually that u is harmonic, since such is a C∞ function withthe mean value property.

Exercise 2. Fix R > 0 and consider a nonnegative function u ≥ 0 which is harmonic in the ballB(0, R). Prove the following Harnack’s inequality: for every x such that |x| < R one has

RN−2 R− |x|(R+ |x|)N−1

u(0) ≤ u(x) ≤ RN−2 R+ |x|(R− |x|)N−1

u(0) . (1.1)

Deduce that

supB(0,R2 )

u ≤ 3N infB(0,R2 )

u .

Solution: Denote by αN the volume of the unit sphere. By Poisson’s formula, for every x such that|x| < R one has

u(x) =

ˆ∂B(0,R)

KR(x, ξ)u(ξ) dS(ξ)

with KR the Poisson’s kernel

KR(x, ξ) :=R2 − |x|2

NαNR |ξ − x|N.

Since |ξ| = R > |x| by means of elementary triangle inequalities we get

R− |x|NαNR (R+ |x|)N−1

≤ KR(x, ξ) ≤ R+ |x|NαNR (R− |x|)N−1

and since u is nonnegative, substituting into the Poisson’s formula we arrive at

R− |x|NαNR (R+ |x|)N−1

ˆ∂B(0,R)

u(ξ) dS(ξ) ≤ u(x) ≤ R+ |x|NαNR (R− |x|)N−1

ˆ∂B(0,R)

u(ξ) dS(ξ) .

Since by the mean-value formulaˆ∂B(0,R)

u(ξ) dS(ξ) = NαNRN−1 u(0) ,

we eventually get (1.1).Now, for |x| = R

2 , Harnack’s inequality reads, after simple computations,

2N−2

3N−1u(0) ≤ u(x) ≤ 3 · 2N−2u(0) ,

1

2 EXERCISES PDE

which gives

max∂B(0,R2 )

u ≤ 3 · 2N−2u(0) and min∂B(0,R2 )

u ≥ 2N−2

3N−1u(0) .

Combining the two we arrive at

max∂B(0,R2 )

u ≤ 3N min∂B(0,R2 )

u

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(nπL y), with n ∈ N ;

(b) f is a generic C1 function satisfying f(0) = f(L) = 0 and thus can be developed in a Fourierseries

f(y) =

+∞∑n=1

an sin(nπ

Ly)

with+∞∑n=1

|an| < +∞ . (1.2)

Hint: first, try to see which functions of the form u(x, y) = v(x)w(y) satisfy the Laplace equationwith 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)

and

v(L) = 0 . (1.4)

Moreover, the Laplace equation givesv′′(x)

v(x)= −w

′′(y)

w(y)

for every x and y ; but this is possible only if both sides are equal to a constant, and thus there mustexist λ ∈ R such that

v′′(x)− λv(x) = w′′(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 aboundary-value problem of the form

w′′(y) + λw(y) = 0

w(0) = w(L) = 0 .(1.6)

Without entering the details of the related (beautiful) theory, we only observe that by solving explicitlyand imposing the boundary conditions, one sees that the problem (1.6) has solution only for a discreteset of values of λ (no wonder: this is not a Cauchy problem)! Precisely, it must be

λ = λn :=(nπL

)2

, n ∈ N (1.7)

EXERCISES PDE 3

and for given n the solutions of (1.6) are all multiples of the functions

wn(y) := sin(nπLy), n ∈ N .

For fixed n , now, and thus for fixed λn , by (1.5) and (1.4) we can recover the functions vn(x) assolutions of the problem

v′′n(y)− λnvn(y) = 0

vn(L) = 0 .

The solutions of this problem are all multiples of the functions

vn(x) := sinh(nπL

(x− 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

un(x, y) = µ sinh(nπL

(x− L))

sin(nπLy), µ ∈ R, n ∈ N . (1.8)

Now for f(y) = sin(nπL y) we immediately get that the solution can be recovered from (1.8) byimposing the boundary condition (namely, recover µ by imposing equality for x = 0) and we get thesolutions

un(x, y) = − 1

sinh(nπ)sinh

(nπL

(x− L))

sin(nπLy).

In the general case, we first proceed formally. Observe that finite linear combinations of the ungiven by (1.8) are still harmonic and 0 on the three sides of the square; but since f is given by aninfinite sum, we look for a candidate solution in the form of an infinite linear combination

u(x, y) =

+∞∑n=1

bn sinh(nπL

(x− L))

sin(nπLy).

By imposing equality for x = 0 with the Fourier development of f we get that our candidate solutionis

u(x, y) =

+∞∑n=1

− ansinh(nπ)

sinh(nπL

(x− L))

sin(nπLy). (1.9)

But since the solution is given by an infinite series, we are not done. We must first show that the seriesconverges, to have a well defined function! And actually, we want to show that it converges uniformlyon QL : thus, the limit will be a harmonic function by Exercise 1 (by construction the partial sumsare 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

0 ≤ −sinh

(nπL (x− L)

)sinh(nπ)

≤ 1

and thus, for every (x, y) ∈ QL∣∣∣− ansinh(nπ)

sinh(nπL

(x− L))

sin(nπLy)∣∣∣ ≤ |an| ;

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 harmonicfunctions, 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 positiveharmonic function. Thus, by Harnack’s inequality there exists a constant C(N, r,R) such that

supx∈B(0,r)

|uk(x)− uh(x)| = supx∈B(0,r)

(uk(x)− uh(x))

≤ C(N, r,R) infx∈B(0,r)

(uk(x)− uh(x)) ≤ C(N, r,R)(uk(0)− uh(0)) .

Since the last term of this chain of inequalities is by the hypothesis a Cauchy sequence, we get theconclusion.

Exercise 2. Let u : RN → R be a harmonic function. Prove that:

(a)

ˆRnu2(x) dx < +∞⇒ u ≡ 0;

(b)

ˆRn|∇u|2(x) dx < +∞⇒ u ≡ const.

Solution: For part (a), observe that u2 is subharmonic (Ex.1 (c), 31.10.12). Then by the mean-valueproperty one has, for every x ∈ Rn and every R > 0

u2(x) ≤ 1

αN RN

ˆB(x,R)

u2(x) dx ≤ 1

αN RN

ˆRnu2(x) dx =

C

αN RN.

Letting R to +∞ , we get u2(x) ≤ 0 for every x , which implies the conclusion.For part (b), apply part (a) to the harmonic functions ∂

∂xiu with i = 1, . . . , N to obtain ∇u ≡ 0

and thus the conclusion.

Exercise 3.

• Consider a holomorphic function f : C→ C and its real and imaginary part u : R2 → R andv : 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 holomorphicfunction. Is v unique?

Solution: We need to recall the following preliminaries about holomorphic functions. By the Cauchy-Riemann theorem, f is holomorphic in region of the complex plane D if and only if the followingconditions are satisfied

∂

∂xu =

∂

∂yv

∂

∂yu = − ∂

∂xv

(2.2)

and if this happens one has

f ′(z0) =∂

∂xu(x0, y0) + i

∂

∂xv(x0, y0) (2.3)

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 formulasˆΓ

f(z) dz = 0 (2.4)

EXERCISES PDE 5

Indeed, by (2.1), dz = dx+ i dy we getˆΓ

f(z) dz =

ˆΓ

P (x, y) · ds+ i

ˆΓ

Q(x, y) · ds ,

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

Γ

P (x, y) · ds =

ˆD

curlP (x, y) dxdy = 0

ˆΓ

Q(x, y) · ds =

ˆD

curlQ(x, y) dxdy = 0 ,

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

∂x2u+

∂2

∂y2u =

∂2

∂x∂yv − ∂2

∂y∂xv = 0

since we can exchange the order or derivations by smoothness of u . A similar proof shows that v isharmonic, 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 ′(z) =∂

∂xu(x, y)− i ∂

∂yu . (2.5)

Then, defining g(z) as the right-hand side, which depends only on u , it only suffices to show thatthere exists a holomorphic primitive f of g satisfying f(0) = u(0). If it exists, its imaginary part vwill 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

f(z) = u(0) +

ˆ[0,z]

g(w) dw

where [0, z] is the line segment connecting 0 and z , oriented from 0 to z . We observe that g satisfiesthe Cauchy-Riemann conditions; indeed, since u is smooth and harmonic

∂

∂xReg =

∂2

∂x2u = − ∂2

∂y2u =

∂

∂yImg

∂

∂xImg = − ∂2

∂x∂yu = − ∂2

∂y∂xu = − ∂

∂xReg .

From this, by (2.4) with Γ the union of the three segments [0, z0] , [z0, z] and [z, 0] all oriented fromthe first to the second endpoint (observe that this curve is the boundary of a domain interely containedin the ball, thus g is holomorphic inside!) we easily getˆ

[0,z]

g(w) dw −ˆ

[0,z0]

g(w) dw =

ˆ[z0,z]

g(w) dw ,

so that

f(z)− f(z0)

z − z0=

´[z0,z]

g(w) dw

z − z0

for every z and z0 in B(0, R); and now, only continuity of g suffices to say that the righ-hand sidehas a limit when z goes to z0 , which is equal to g(z0), as we wanted.

Remark 2.1. In Exercise 3 part (a) it was used the fact that when f is holomorphic, then u andv are C∞ . Actually, this is a consequence of the fact that the derivative of a holomorphic functionis still holomorphic, usually recovered as a consequence of Cauchy’s formula for the derivatives. It isinteresting to show that this fact can be also derived only using the Cauchy-Riemann conditions andthe 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 areholomorphic and ρn ? u and ρn ? v harmonic. As ρn ? u and ρn ? v converge locally uniformly to uand 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 ′ is holomorphic, as required.

3. Exercises PDE 21.11.12-23.11.12

Exercise 1. Assume N = 1 and u(x, t) = v( x√t).

(a) Show that ut = uxx if and only if v solves the differential equation

v”(z) +z

2v′(z) = 0

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 theinitial condition for u?)

Solution: (a) It holds, denoting with primes the derivatives of v with respect to its argument z :

∂

∂tu(x, t) =

−x2t√tv′(z) = −zv

′(z)

2t

∂2

∂x2u(x, t) =

1

2tv′′(z)

so that ut = uxx if and only if v solves the differential equation

v”(z) +z

2v′(z) = 0 .

(b) Setting w = v′ one has that it must be

w′(z) = ce−z24

for some arbitrary constant c ∈ R , whence

v(z) = c

ˆ z

0

e−s24 ds+ d

with d ∈ R again arbitrary. It is easy to check that

U(x, t) :=∂

∂xu(x, t) =

c√te−x24t

solves again the heat equation. Moreover we can impose to this function the “initial condition” forthe fundamental solution, that is it must approximate the Dirac delta when t tends to 0. This resultsin the coupling of the two conditions

limt→0

U(x, t) = 0 if |x| 6= 0

and ˆRU(x, t) dx = 1 (3.1)

for t in a neighborhood of 0. The first is automatically satisfied, while the second gives the requiredconstant c = 1

2√π

.

Also, observe that (3.1) asserts conservation of the total heat, a property that follows from theequation for a solution decaying at infinity with its derivative. It is actually, for t 6= 0

− d

dt

ˆRU(x, t) dx =

ˆRUxx(x, t) dx = Ux(+∞, t)− Ux(−∞, t) = 0

which implies (3.1) possibly with a constant different from 1, provided that the total heat keepsbounded 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 ofut −∆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 solvesvt −∆v = ectf(x, t) in Rn × (0,+∞)

v = g in Rn × t = 0 .

A solution of the previous problem is given via Duhamel’s principle by

v(x, t) =1

(4πt)N2

ˆRne−|x−y|2

4t g(y) dy +

ˆ t

0

ˆRn

ecs

(4π(t− s))N2e−|x−y|24(t−s) f(y, s) dsdy

Now, setting u(x, t) := e−ctv(t, x), with v as above, gives a solution to (3.2).

Exercise 3. Tychonov’s counterexample: Consider the holomorphic function g(z) = e−1z2 for z ∈

C\0 and denoting by g(k) its k -th derivative, define the function

u(x, t) =

+∞∑k=0

g(k)(t)

(2k)!x2k if t > 0 , x ∈ R

0 if t = 0 , x ∈ RRigorously justify that this is a solution of the heat equation with 0 Cauchy datum by showinguniform convergence of the series (and of the series of its time and space derivatives involved in theequation) 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

g(k)(t) =k!

2π i

ˆ∂B(t, t2 )

g(z)

(z − t)k+1dz (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

z = t(1 +1

2eiω)

whence1

z=

1

t

1

1 + 12eiω

=1

t

2(2 + e−iω)

5 + 4 cosω.

Therefore1

z2=

4

t2(2 + e−iω)2

(5 + 4 cosω)2.

so that

Re1

z2=

4

t24 + cos(2ω)− 4 cos(ω)

(5 + 4 cosω)2≥ 4

81t2,

where the last estimate is simply obtained by observing that 0 is a minimizer of the function at thenumerator, plus the trivial fact that (5 + 4 cosω)2 ≤ 81.

With this, we get

|e−1z2 | = e−Re

(1z2

)≤ e−

481t2

8 EXERCISES PDE

for every z ∈ Ct , therefore (3.3) gives

|g(k)(t)| ≤ k!

2π

ˆ∂B(t, t2 )

|g(z)||z − t|k+1

|dz| ≤ k!

2π

ˆ 2π

0

2k+1 e−Re

(1z2

)tk+1

t

2dθ ≤ k!

(2

t

)ke−

481t2 .

By recalling the easy estimate2kk!

(2k)!≤ 1

k!

for every k ∈ N , we conclude that ∣∣∣g(k)(t)

(2k)!x2k∣∣∣ ≤ e− 4

81t2

(x2

t

)k 1

k!

for every k ∈ N and every (x, t). Therefore the series defining u(x, t) is totally and thus uniformlyconvergent in [−a, a]× [δ,+∞); moreover, we have the estimate

|u(x, t)| ≤ e−4

81t2+ a2

t

for every x ∈ [−a, a] and t > 0. This implies

limt→0+

|u(x, t)| = 0

uniformly in x ∈ [−a, a] , which means that u ∈ C(R× [0,+∞)).A similar proof shows that the series

+∞∑k=0

g(k+1)(t)

(2k)!x2k , (3.4)

formally giving ut(x, t), as well as+∞∑k=1

g(k)(t)

(2k − 1)!x2k−1

and+∞∑k=1

g(k)(t)

(2(k − 1))!x2(k−1) (3.5)

formally giving ux(x, t), and uxx(x, t), respectively, uniformly converge in [−a, a]×[δ,+∞). Thereforeu(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 left-

moving wave?

Solution: (a) Clearly any function of the form (4.1) solves the equation, and it must be only shownthe converse. Take a solution u of the PDE uxy = 0 and, for fixed x , set fx(y) := ux(x, y). One hasthat

d

dyfx(y) = 0

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

dxF (x) .

We now set, for fixed y , gy(x) := u(x, y). We have

d

dxgy(x) = ux(x, y) =

d

dxF (x)

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 = 1

2 (ξ − η) and x = 12 (ξ + η). Consequently, let

v(ξ, η) := u( 12 (ξ − η), 1

2 (ξ + η)). denoting with ut differentiation with respect to the first argumentand with ux differentiation with respect to the second argument, we have

∂

∂ηv(ξ, η) = −1

2ut

(1

2(ξ − η),

1

2(ξ + η)

)+

1

2ux

(1

2(ξ − η),

1

2(ξ + η)

)and therefore

∂2

∂ξ∂ηv(ξ, η) = −1

4utt

(1

2(ξ − η),

1

2(ξ + η)

)− 1

4uxt

(1

2(ξ − η),

1

2(ξ + η)

)+

+1

4utx

(1

2(ξ − η),

1

2(ξ + η)

)+

1

4uxx

(1

2(ξ − η),

1

2(ξ + η)

)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 :

u(1

2(ξ − η),

1

2(ξ + η)

)= F (ξ) +G(η)

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 ′(x)−G′(x) = h(x)

for every x , that is F (x) +G(x) = g(x)

F (x)−G(x) =

ˆ x

0

h(s) ds+ const. .

We then have G(x) ≡ const. if and only if g(x) −´ x

0h(s) ds ≡ const. , which is to say that we have

a left-traveling wave if and only if g′ = h . Similarly we have a right-traveling wave if and only ifg′ = −h .

10 EXERCISES PDE

Exercise 2.

(a) Let E := (E1, E2, E3) and B := (B1, B2, B3) be solutions of the Maxwell equationsEt = curlB, Bt = −curlE

divE = divB = 0 .

Show

Ett −∆E = 0, Btt −∆B = 0 .

(b) Assume that u := (u1, u2.u3) solves the evolution equation of linear elasticity

utt − µ∆u− (λ+ µ)∇(divu) = 0 in R3 × (0,+∞) . (4.3)

Show that w := divu and w := curlu each solve wave equation, but with different speed ofpropagation.

Solution: (a) We use the following vector calculus identity:

curl(curlF) = ∇(divF)−∆F (4.4)

for every C2 vector field F . Taking into account that divE = 0, (4.4) yields

curl(curlE) = −∆E .

Observing that by Maxwell’s equations Ett = curlBt = −curl(curlE) we get that Ett−∆E = 0. Theproof of the other case is similar.

(b) Using (4.3), one has

wtt = curlutt = µ curl∆u + (λ+ µ) curl(∇(divu)) .

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 propagationµ .

Analogously

wtt = div (utt) = µdiv (∆u) + (λ+ µ) div (∇(divu)) = µ∆w + (λ+ µ) div (∇w)

again exchanging the order of differentiation and recalling that w = divu . Since div (∇w) = ∆w weget

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 thereexists 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

dtw = Aw, w(0) = g (4.5)

and show that

w(t) :=

n∑i=1

〈g, vi〉eλitvi

is its only solution.(b) Give an analogous statement for this other initial value problem

d2

dt2w = Aw, w(0) = g,

d

dtw(0) = h . (4.6)

EXERCISES PDE 11

We now want to generalise this method to an infinite-dimensional situation where we look to thesolution of the wave equation with periodic boundary conditions, that is

utt − uxx = 0 (x, t) ∈ (0, 2π)× (0,+∞)

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 .

(4.7)

To this end, let

V := v ∈ C2([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 withthe usual L2 scalar product

〈u, v〉 :=1

2π

ˆ 2π

0

u(x)v(x) dx ,

where the operator A is given by

A :=d2

dx2: V ⊂ L2((0, 2π))→ L2((0, 2π)) . (4.9)

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 begeneralised to an infinite-dimensional situation, and using the basis (eikx)k∈Z . Then showthat you actually found a solution, in this way!

Solution: (a) Uniqueness is a well-known fact. By a direct calculation, since λivi = Avi we get

d

dtw(t) :=

n∑i=1

〈g, vi〉eλitλivi =

n∑i=1

〈g, vi〉eλitAvi = A( n∑i=1

〈g, vi〉eλitvi),

so that ddtw = Aw as we wanted. On the other hand w(0) =

∑ni=1〈g, vi〉vi = 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 , thesolution to (4.5) is given by

w(t) :=

n∑i=1

gieλitvi ; (4.10)

only, gi 6= 〈g, vi〉 , in general!(b) Setting v(t) := d

dtw and W := (v, w) ∈ V × V the problem is equivalent to the following one

d

dtW = BW, W (0) = (g, h)

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 diagonalisablein 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 bythe set of vectors

(W1)+, (W1)−, . . . , (Wn)+, (Wn)−

12 EXERCISES PDE

where for every i = 1, . . . , n we have set

(Wi)+ =1√

1 + |λi|(vi,√λivi)

and

(Wi)− =1√

1 + |λi|(vi,−

√λivi) .

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∑i=1

(wi)+e√λit(Wi)+ +

n∑i=1

(wi)−e−√λit(Wi)− .

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

w(t) =

n∑i=1

aie√λitvi +

n∑i=1

bie−√λitvi (4.11)

with ai = (wi)+√1+|λi|

and bi = (wi)+√1+|λi|

. But the key point is that now we can easily determine

ai and bi by imposing the initial conditions, since vi is an orthonormal basis! Precisely, itmust be

ai + bi = 〈g, vi〉√λiai −

√λibi = 〈h, vi〉

for every i . Solving the system gives that the coefficients ai and bi in (4.11) are given by

ai =1

2

(〈g, vi〉+

〈h, vi〉√λi

), bi =

1

2

(〈g, vi〉 −

〈h, vi〉√λi

). (4.12)

(b2) dim ker A = m 6= 0. Denoting with vin−mi=1 the eigenvectors of A relative to the nonzeroeigenvalues λin−mi=1 we see that the analogous of (4.11), that is the function

w1(t) =

n−m∑i=1

aie√λitvi +

n−m∑i=1

bie−√λitvi , (4.13)

wtih ai and bi given by (4.12) still solves the equation in (4.6), but with a different initialdatum: precisely w1(0) = g − Pker A g and d

dtw1(0) = h− Pker A h where Pker A denotes the

orthogonal projection onto ker A . But now, given an orthonormal basis vker1 , . . . , vkerm ofker 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:

z(t) :=

m∑j=1

[〈g, vkerj 〉+ 〈h, vkerj 〉t]vkerj . (4.14)

Indeed z(t) ∈ ker A for every t , d2

dt2 z(t) ≡ 0 = Az(t), and z(0) = Pker A g as well asddtz(0) = Pker A h so that we conclude by linearity.

(c) Simply interpret (4.7) as a generalised Cauchy problem

d2

dt2w = Aw, w(0) = g,

d

dtw(0) = h .

with w : R→ V , where this one is given by (4.8), and A is given by (4.9). Setting u(t, ·) = w(t) theboundary 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 Fourierseries; since (eikx)k∈Z ⊂ V for, it constitutes an orthonormal basis of V , too. A direct calculationshows that for every k ∈ Z

A(eikx) = −k2eikx

so that they are eigenvectors of A with eigenvalues −k2 .(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 := 〈g, eikx〉we get that the coefficients ak and bk in (4.12) satisfy ak = bk = 1

2αk . Therefore the function w1(t)defined in (4.13) reduces in our case to

w1(t) =∑

k∈Z, k 6=0

1

2αk[ei|k|t + e−i|k|t]eikx =

∑k∈Z, k 6=0

αk cos(|k|t)eikx .

By a direct computation, αk = − 24k4 for every k 6= 0, so that

w1(t) = −∑

k∈Z, k 6=0

24

k4cos(|k|t)eikx = −48

∞∑k=1

1

k4cos(kt) cos(kx) .

The eigenvector corresponding to the eigenvalue 0 is the constant 1, so that in our case

z(t) = 〈g, 1〉 =8π4

15

therefore we get to our candidate solution

w(t) =8π4

15− 48

∞∑k=1

1

k4cos(kt) cos(kx) . (4.15)

The verification that (4.15) is actually the solution of (4.7) is an easy exercise in differentiation ofseries 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

v(t, x) :=

ˆ +∞

−∞

e−s2

4t

√tu(s, x) ds

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 variableF and G . With the change of variable z = x+ s one has

ˆ +∞

−∞

e−s2

4t

√tF (x+ s) ds =

ˆ +∞

−∞

e−(x−z)2

4t

√t

F (z) dz

and similarly, with z = x− s we get

ˆ +∞

−∞

e−s2

4t

√tG(x− s) ds =

ˆ +∞

−∞

e−(x−z)2

4t

√t

G(z) dz

so that

v(t, x) =

ˆ +∞

−∞

e−(x−z)2

4t

√t

(F (z) +G(z)) dz

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 thereexists δ > 0 such that if 0 ≤ t0 < s0 < t1 < s1 < . . . tn < sn ≤ 1 are points in [0, 1] satisfyng

n∑i=1

(si − ti) ≤ δ

thenn∑i=1

|u(ti)− u(si)| ≤ ε

An absolutely continuous function u is continuous and u(t) (the pointwise derivative) exists for almostevery t ∈ [0, 1].

Show that if u ∈ W 1,p(0, 1) for 1 ≤ p then there exists v absolutely continuous such that u = valmost everywhere; moreover the weak derivative of u is given exactly by the pointwise derivative vand v ∈ Lp(0, 1). Moreover, if p > 1:

|u(x)− u(y)| ≤ |x− y|1−1p (

ˆ 1

0

|u′(t)|p dt)1p (5.1)

for every x and y in [0, 1].

Solution: The key of the exercise, important in itself, is that u satisfies the fundamental Theoremof calculus, that is

u(x) = u(0) +

ˆ x

0

u′(t) dt (5.2)

for every x ∈ [0, 1], with u′ 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 absolutelycontinuous function is needed, but only Fubini’s theorem and exercise 4 below. Let us see how. Givena function v ∈ L1([0, 1]), define V (x) :=

´ x0v(y) dy and we claim that

V ′ = v , (5.3)

that is, the weak derivative of V is v . To this aim, take ϕ ∈ C∞c ([0, 1]) and observe that by Fubini’sTheorem

−ˆ 1

0

V (x)ϕ′(x) dx = −ˆ 1

0

( ˆ x

0

v(y) dy)ϕ′(x) dx = −

ˆ 1

0

( ˆ 1

y

ϕ′(x) dx)v(y) dy =

ˆ 1

0

v(y)ϕ(y) dy .

This proves (5.3). Now, if you define U(x) := u(0) +´ x

0u′(y) dy , (5.3) gives that (U − u)′ = 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’sinequality), when p > 1; it is indeed for every 0 ≤ x < y ≤ 1

|u(y)− u(x)| =∣∣∣ˆ y

x

u′(t) dt∣∣∣ = |y − x|

(∣∣∣ 1

|y − x|

ˆ y

x

u′(t) dt∣∣∣p) 1

p ≤ |y − x|1−1p (

ˆ 1

0

|u′(t)|p dt)1p .

Absolute continuity of u follows from (5.2) with the only observation that for every v ∈ L1([0, 1]),V (x) :=

´ x0v(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) :=minv(x), N , one has ˆ 1

0

|v(x)− vN (x)| dx ≤ ε

2. (5.4)

This is because the sequence vk(x) := minv(x), k is L1 convergent to v . Now, fix δ = ε2N ; for

every finite sequence of points 0 ≤ t0 < s0 < t1 < s1 < . . . tn < sn ≤ 1 satisfyngn∑i=1

(si − ti) ≤ δ

EXERCISES PDE 15

one hasn∑i=1

|V (ti)− V (si)| =ˆBn

v(y) dy

with Bn := ∪ni=0[ti, si] . Since now, using (5.4) and our choice of δˆBn

v(y) dy ≤ˆBn

|v(y)− vN (y)| dy +

ˆBn

vN (y) dy ≤ ε

2+N |Bn| ≤

ε

2+Nδ = ε ,

our claim is proved.Finally, define

Uh(t) :=1

h(u(t+ h)− u(t)) .

By (5.2) and the Lebesgue differentiation Theorem we have that

Uh(t) =1

h

ˆ t+h

t

u′(y) dy → u′(t)

in L1([0, 1]) when h → 0, and therefore almost everywhere. Being Uh(t) exactly the differencequotients of u at t , we have shown that u is pointwise almost everywhere differentiable, and that itspointwise derivative coincides a.e. with u′ .

Exercise 3. Let U := (−1, 1)× (−1, 1). Define u as follows

u(x1, x2) :=

1− x1 in T1 := x1 > 0, |x2| < x11 + x1 in T2 := x1 < 0, |x2| < −x11− x2 in T3 := x2 > 0, |x1| < x21 + x2 in T4 := x2 < 0, |x1| < −x2

For which 1 ≤ p ≤ ∞ does u belong to W 1,p(U)?

Solution: Let us take ϕ := (ϕ1, ϕ2) ∈ C∞c (U ;R2) and define the vectors ν+ := (− 1√2, 1√

2) (unit

normal to the line x1 = x2) and ν− := ( 1√2, 1√

2) (unit normal to the line x1 = −x2). First, we

compute ˆT1

x1 divϕ(x1, x2) dx1dx2 .

Since div (x1ϕ)− (1, 0) ·ϕ = x1divϕ , by the divergence theorem (taking into account the orientationof the exterior normal!) we getˆ

T1

x1 divϕ(x1, x2) dx1dx2 = −ˆT1

(1, 0) · ϕ(x1, x2) dx1dx2 +ˆx1=x2; x1>0

ξ1 ϕ(ξ) · ν+ ds(ξ)−ˆx1=−x2; x1>0

ξ1 ϕ(ξ) · ν− ds(ξ) .(5.5)

Here we took into account that ϕ(1, x2) ≡ (0, 0). In a similar way we getˆT4

x2 divϕ(x1, x2) dx1dx2 = −ˆT4

(0, 1) · ϕ(x1, x2) dx1dx2 +ˆx1=−x2; x2<0

ξ2 ϕ(ξ) · ν− ds(ξ)−ˆx1=x2; x2<0

ξ2 ϕ(ξ) · ν+ ds(ξ) .

But this is clearly equivalent toˆT4

x2 divϕ(x1, x2) dx1dx2 = −ˆT4

(0, 1) · ϕ(x1, x2) dx1dx2 +

−ˆx1=−x2; x1>0

ξ1 ϕ(ξ) · ν− ds(ξ) +

ˆx1=x2; x1<0

ξ1 ϕ(ξ) · ν+ ds(ξ) .(5.6)

16 EXERCISES PDE

The same reasonings lead toˆT2

x1 divϕ(x1, x2) dx1dx2 = −ˆT2

(1, 0) · ϕ(x1, x2) dx1dx2 +ˆx1=−x2; x1<0

ξ1 ϕ(ξ) · ν− ds(ξ)−ˆx1=x2; x1<0

ξ1 ϕ(ξ) · ν+ ds(ξ)(5.7)

and ˆT3

x2 divϕ(x1, x2) dx1dx2 = −ˆT3

(0, 1) · ϕ(x1, x2) dx1dx2 +ˆx1=−x2; x1<0

ξ1 ϕ(ξ) · ν− ds(ξ)−ˆx1=x2; x1>0

ξ1 ϕ(ξ) · ν+ ds(ξ)(5.8)

Since ˆU

divϕ(x1, x2) dx1dx2 = 0

again by the divergence Theorem, we have that

−ˆU

udivϕ(x1, x2) dx1dx2 =

ˆT1

x1 divϕ(x1, x2) dx1dx2 −ˆT2

x1 divϕ(x1, x2) dx1dx2+ˆT3

x2 divϕ(x1, x2) dx1dx2 −ˆT4

x2 divϕ(x1, x2) dx1dx2 .

Using (5.5), (5.7), (5.8), and (5.6) we conclude that

−ˆU

udivϕ(x1, x2) dx1dx2 =

ˆU

v · ϕ(x1, x2) dx1dx2

where v ∈ L∞(U ;R2) is defined as

v(x1, x2) :=

(−1, 0) in T1

(1, 0) in T2

(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 ′ of U ; it suffices to show u constant in U ′ . For εsufficiently small, now, the convolutions with the simmetric mollifiers ρε are well defined; furthermore,for every ϕ ∈ C∞(U ′;Rn), one has, by simmetry of mollifiers and since ρε ? ϕ ∈ C∞(U ;Rn) for εsmall enough, thatˆ

U ′divϕ(x)(ρε ? u)(x) dx =

ˆU

u(x)(ρε ? divϕ)(x) dx =

ˆU

u(x)div (ρε ? ϕ)(x) dx .

By the hypothesis, the right-hand side is 0, so that D(ρε ? u) = 0 on U ′ . By smoothness of ρε ? uthis implies that ρε ? u is constant in U ′ ; by Lp convergence, when ε is going to 0 we get that u isalmost 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 ∈ C∞c (Ω), prove that thefollowing interpolation inequality holds:(ˆ

Ω

|Du(x)|2 dx)2

≤ C(ˆ

Ω

u(x)2 dx)( ˆ

Ω

|D2u(x)|2 dx), (6.1)

with D2u the Hessian matrix.(b) Assume ∂Ω ∈ C1 , let u ∈W 2,2(Ω) ∩W 1,2

0 (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 getˆΩ

|Du(x)|2 dx = −ˆ

Ω

u(x)∆u(x) dx

since u = 0 on ∂Ω. By the Cauchy-Schwarz inequality we then have(ˆΩ

|Du(x)|2 dx)2

≤(ˆ

Ω

u(x)2 dx)( ˆ

Ω

|∆u(x)|2 dx)≤ C

(ˆΩ

u(x)2 dx)( ˆ

Ω

|D2u(x)|2 dx),

where the last estimate is obvious. This proves (6.1).(b) By (6.2) and the divergence Theorem, we getˆ

Ω

|Dvn(x)|2 dx = −ˆ

Ω

vn(x)∆vn(x) dx+

ˆ∂Ω

vn(ξ)∂

∂νvn(ξ) dS(ξ) . (6.3)

Now, considering wn as in the hint, we obviously haveˆ∂Ω

vn(ξ)∂

∂νvn(ξ) dS(ξ) =

ˆ∂Ω

(vn − wn)(ξ)∂

∂νvn(ξ) dS(ξ)

since wn(ξ) ≡ 0. By the divergence Theorem, we then getˆ∂Ω

vn(ξ)∂

∂νvn(ξ) dS(ξ) =

ˆΩ

div [(vn(x)− wn(x))Dvn(x)] dx =ˆΩ

(vn(x)− wn(x))∆vn(x) dx+

ˆΩ

D(vn(x)− wn(x)) ·Dvn(x) dx .

Now, by W 2,2 convergence of vn to u , ∆vn and Dvn are bounded in L2 , while ‖vn−wn‖W 1,2(Ω) →‖u− u‖W 1,2(Ω) = 0 as n goes to +∞ , therefore

limn→+∞

ˆ∂Ω

vn(ξ)∂

∂νvn(ξ) dS(ξ) = 0 . (6.4)

By (6.4) and the W 2,2 convergence of vn to u , taking the limit in (6.3) we getˆΩ

|Du(x)|2 dx = −ˆ

Ω

u(x)∆u(x) dx

and we conclude as in part (a).

Exercise 2. Poincare’s inequality:

• Let u ∈W 1,20 ((0, d)). Prove that

ˆ d

0

u(x)2 dx ≤ d2

ˆ d

0

u′(x)2 dx . (6.5)

• Let u ∈W 1,20 ((0, d)× RN−1). Prove thatˆ

[0,d]×RN−1

u(x)2 dx ≤ d2

ˆ[0,d]×RN−1

|Du(x)|2 dx . (6.6)

Hint: Use density of C∞c 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

|u(x)− u(y)| ≤ |x− y| 12 (

ˆ d

0

|u′(t)|p dt) 12

18 EXERCISES PDE

and that u is absolutely continuous (in particular, continuous). Therefore u ∈ W 1,20 ([0, d]) implies

u(0) = 0 and then for y = 0 we have

|u(x)| ≤ |x| 12 (

ˆ d

0

|u′(t)|p dt) 12 ≤ d 1

2 (

ˆ d

0

|u′(t)|p dt) 12 ,

where the last estimate is trivial. Taking squares and integrating in [0, d] with respect to x , we get(6.5).

Now, let u ∈ C∞c ((0, d)×RN−1). Writing x = (x1, x) for every x ∈ [0, d]×RN−1 , with x1 ∈ [0, d]and x ∈ RN−1 , we have that for fixed x the function u(·, x) belongs to C∞c ((0, d)). We thereforehave, by (6.5), thatˆ d

0

u(x1, x)2 dx1 ≤ d2

ˆ d

0

( d

dx1u(x1, x)

)2

dx1 = d2

ˆ d

0

( ∂

∂x1u(x)

)2

dx1

for every x ∈ RN−1 . Now, by Fubini’s Theoremˆ[0,d]×RN−1

u(x)2 dx =

ˆRN−1

(ˆ d

0

u(x1, x)2 dx1

)dx ≤

d2

ˆRN−1

( ˆ d

0

( ∂

∂x1u(x)

)2

dx1

)dx ≤ d2

ˆ[0,d]×RN−1

|Du(x)|2 dx .

This proves (6.6) when u ∈ C∞c ((0, d) × RN−1). In the general case, there exists un ∈ C∞c (Ω) suchthat un → u in W 1,2(Ω). In particular

limn→∞

ˆ[0,d]×RN−1

un(x)2 dx =

ˆ[0,d]×RN−1

u(x)2 dx

and

limn→∞

ˆ[0,d]×RN−1

|Dun(x)|2 dx =

ˆ[0,d]×RN−1

|Du(x)|2 dx .

Since ˆ[0,d]×RN−1

un(x)2 dx ≤ d2

ˆ[0,d]×RN−1

|Dun(x)|2 dx ,

(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.

Exercise 3. Let Ω be an open subset of Rn . We denote by C∞(Ω)W 1,∞(Ω)

the set of functions vin 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)W 1,∞(Ω)

6= W 1,∞(Ω), find a function v ∈ W 1,∞(Ω) that does not belong

to C∞(Ω)W 1,∞(Ω)

.

Solution: W 1,∞ convergence corresponds to uniform convergence of the sequence of functions andof the sequence of gradients. Therefore, since the limit of continuous functions must be continuous,one easily has that

C∞(Ω)W 1,∞(Ω)

⊆ C1(Ω) .

But also the converse inclusion holds! To prove this, it simply suffices to follow exactly the steps ofthe Meyers-Serrin Theorem (Theorem 2, Section 5.3.2 in the book of Evans); observe actually that instep 2, since now the functions ζiu are C1

c (U), by mollification one can find ui ∈ C∞c (Ω) such thatsuppui ⊂Wi and ‖ui − ζiu‖W 1,∞(Ω) ≤ δ

2i (why?). Therefore

C∞(Ω)W 1,∞(Ω)

= C1(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 C1(B(0, 1)). Consider for instance v(x) := |x| . This is clearly no C1

function in B(0, 1), since it is not differntiable at 0. On the other hand, for every ε > 0 one has

v ∈ C1(B(0, 1) \B(0, ε)), so that, given ϕ ∈ C∞c (B(0, 1);Rn) one hasˆB(0,1)\B(0,ε)

v(x) divϕ(x) dx = −ˆB(0,1)\B(0,ε)

x

|x|· ϕ(x) dx−

ˆ∂B(0,ε)

|ξ|ϕ(ξ) · ν(ξ) dS(ξ) .

Since ∣∣∣ ˆ∂B(0,ε)

|ξ|ϕ(ξ) · ν(ξ) dS(ξ)∣∣∣ ≤ Nα(N)εN‖ϕ‖L∞(B(0,1);Rn)

and x|x| ∈ L

∞(B(0, 1);Rn), by taking the limit as ε→ 0 we get by dominated convergenceˆB(0,1)

v(x) divϕ(x) dx = −ˆB(0,1)

x

|x|· ϕ(x) dx ,

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 thefunction

u(x1, x2, . . . , xN ) :=

1 if xN > 0

0 if xN < 0

belongs to W 1,p(Ω) for every 1 ≤ p ≤ ∞ , but there exists no sequence vn of C∞(Ω) functions suchthat 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 actuallybelongs to L∞(B(0, 1)), but does not belong to W 1,1(B(0, 1))! Indeed, suppose by contradiction thatW 1,1(B(0, 1)). Then, it exists g ∈ L1(B(0, 1);Rn) such thatˆ

B(0,1)

u(x)divϕ(x) dx = −ˆB(0,1)

g(x) · ϕ(x) dx

for every ϕ ∈ C∞c (B(0, 1);Rn). Now, let us define B(0, 1)+ := B(0, 1) ∩ xN > 0 and similarlyB(0, 1)− := B(0, 1)∩xN < 0 . Since Du = 0 in B(0, 1)+ , and obviously every ϕ ∈ C∞c (B(0, 1)+;Rn)also belongs to C∞c (B(0, 1);Rn), it must beˆ

B(0,1)+g(x) · ϕ(x) dx = 0

for every ϕ ∈ C∞c (B(0, 1)+;Rn). Thus, g = 0 a.e. in B(0, 1)+ . Arguing similarly in B(0, 1)− , weget g = 0 a.e. in B(0, 1). We should therefore haveˆ

B(0,1)

u(x)divϕ(x) dx = 0 (7.1)

for every ϕ ∈ C∞c (B(0, 1);Rn). But on the other hand, by definition of u and the divergence theoremwe have ˆ

B(0,1)

u(x)divϕ(x) dx =

ˆB(0,1)+

divϕ(x) dx = −ˆB(0,1)∩xN=0

ϕ(ξ) · eN (ξ) dS(ξ)

for every ϕ ∈ C∞c (B(0, 1);Rn), and this has clearly no reason of being 0! So, u /∈W 1,1(B(0, 1)).Now, if a sequence vn of C∞(Ω) functions such that vn → u in W 1,p(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 bya set of Lebesgue measure 0, we also have that vn and Dvn are Cauchy sequences in Lp(B(0, 1)) andLp(B(0, 1);Rn), respectively. Thus, vn has a limit u in W 1,p(B(0, 1)), and it must be u = u almost

20 EXERCISES PDE

everywhere in B(0, 1). But since the definition of weak derivative does not depend on the Lebesguerepresentative, 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 fg ∈W 1,1(Ω) and compute the weak gradient D(fg).(b) Show that when N = 1 it suffices to take p = q = 1.(c) Kettenregel: assume that F : R → R ∈ C1 ∩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 1p + 1

q = 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

‖fngn − fmgm‖1 = ‖fn(gn − gm) + (fn − fm)gm‖1 ≤ ‖fn‖p‖gn − gm‖q + ‖gm‖q‖fn − fm‖pfor every n and m in N . Thus, fngn is a Cauchy sequence in L1(Ω), and similarly we can provethat D(fngn) = fnDgn + gnDfn is a Cauchy sequence in L1(Ω;Rn). So, fngn is Cauchy in W 1,1(Ω)and is converging to fg in L1 . It follows that fg is in W 1,1(Ω) and that fngn is converging to fgin W 1,1(Ω). In particular,

D(fg) = fDg + 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 ∈ W 1,10 ((0, 1)). Take fn and gn sequences of C∞c functions

converging to f , and g , respectively, in W 1,1((0, 1)). Since for every x ∈ [0, 1], one has

|fn(x)− f(x)| ≤ˆ 1

0

|f ′n(y)− f ′(y)| dy

we get that actually fn converges to f uniformly in [0, 1], and clearly, also gn converges to g uniformlyin [0, 1]! Thus, fngn converges to fg uniformly in [0, 1]. On the other hand

(fngn)′ = fng′n + gnf

′n ;

by the uniform convergences of fn and gn , and the L1 convergences of f ′n and g′n , we easily get that

(fngn)′ → fg′ + gf ′

in L1((0, 1)). This proves that fg belongs to W 1,10 ((0, 1)).

In the general case, we only can take fn and gn sequences of C∞([0, 1]) functions converging tof , and g , respectively, in W 1,1((0, 1)). By the fundamental Theorem of calculus, we easily get, forevery m and n ∈ N

|fn(0)− fm(0)| ≤ |fn(x)− fm(x)|+ˆ 1

0

|f ′n(y)− f ′m(y)| dy

so that integrating from 0 to 1, it is

|fn(0)− fm(0)| ≤ˆ 1

0

|fn(x)− fm(x)| dx+

ˆ 1

0

|f ′n(y)− f ′m(y)| dy

which yields that fn(0) is a Cauchy sequence. Now, again by the fundamental theorem of calculus,we have

|fn(x)− fm(x)| ≤ |fn(0)− fm(0)|+ˆ 1

0

|f ′n(y)− f ′m(y)| dy

for every x ∈ (0, 1), so that fn is uniromly Cauchy. It follows that fn converges to f uniformlyin [0, 1], and similarly gn converges to g uniformly in [0, 1]. Arguing as in the previous step, theconclusion follows.

EXERCISES PDE 21

(c) Since F : R → R ∈ C1 ∩W 1,∞(R), for every u ∈ W 1,p(Ω) the functions F (u), F ′(u), andF ′(u)Du are a.e. well defined and belong to Lp(Ω), Lp(Ω), and Lp(Ω;Rn), respectively. Now, takinga sequence un of C∞(Ω) functions such that un → u in W 1,p(Ω), we first have

‖F (un)− F (u)‖Lp(Ω) ≤ ‖F ′‖L∞(R)‖un − u‖Lp(Ω)

which gives F (un)→ F (u) in Lp(Ω). By boundedness of F ′ , the convergence

F ′(un)Du→ F ′(u)Du in Lp(Ω;Rn)

can be proved by dominated convergence. Finally, one has

limn‖F ′(un)Dun − F ′(u)Du‖Lp(Ω;Rn) = lim

n‖F ′(un)Dun − F ′(un)Du‖Lp(Ω;Rn)

≤ ‖F ′‖L∞(R) limn‖Dun −Du‖Lp(Ω;Rn) = 0 ,

whence it follows that F (u) ∈W 1,p(Ω), and that the weak gradient is given by F ′(u)Du .

Exercise 3. Let Ω be a bounded open subset of Rn , and u ∈ W 1,p(Ω). Define as usual u+(x) :=maxu(x), 0 , and u−(x) := max−u(x), 0 . Show that u+ , u− , and |u| ∈ W 1,p(Ω), too, andcompute the corresponding weak gradients.

Hint: u+ = limε→0 Fε(u), with

Fε(z) :=

(z2 + ε2)

12 − ε if z ≥ 0

0 if z < 0 .

Solution: Take Fε as in the hint, and observe that Fε ∈ C1 ∩W 1,∞(R). Therefore, the previousexercise gives that Fε(u) ∈W 1,p(Ω) and that

D(Fε(u)) = F ′ε(u)Du = u+(ε2 + u2)−12Du ,

where the last equality follows by a direct computation. Since u+(ε2 + u2)−12 → χu>0 a.e. in Ω,

by dominated convergence we get that

D(Fε(u))→ χu>0Du

in Lp(Ω;Rn). It follows that u+ belongs to W 1,p(Ω), and that

D(u+) = χu>0Du .

Since u− = u+ − u and |u| = u+ − u− we get that u− , and |u| ∈ W 1,p(Ω), too, and the weakgradients are given by

D(u−) = −χu≤0Du and D(|u|) = χu>0Du− χu≤0Du .

8. Exercises PDE 9-11.01.13

Exercise 1. Prove the following multiplicative form of the trace inequality: if u ∈W 1,2(Rn+), denot-

ing with Tu the trace of u on the hyperplane RN−1 one has

‖Tu‖2L2(RN−1) ≤ C‖u‖L2(RN+ )‖Du‖L2(RN+ ) .

Solution: Being Rn+ an extension domain, for every u ∈ W 1,2(Rn+) there exists a sequence un of

functions in C∞c (Rn) converging to u in W 1,2(Rn+). Now, for every x ∈ RN−1 by the fundamentalTheorem of calculus one has

u2n(x, 0) = −

ˆ +∞

0

∂

∂xNu2n(x, xN ) dxN = −2

ˆ +∞

0

un(x, xN )∂

∂xNun(x, xN ) dxN

for every n ∈ N . This implies by Fubini’s TheoremˆRN−1

u2n(x, 0) dx = −2

ˆRn+un(x)

∂

∂xNun(x) dx .

22 EXERCISES PDE

Since for every n and every x the trace Tun(x) of un on the hyperplane RN−1 is simply equal toun(x, 0), the previous equality and the Cauchy-Schwarz inequality yield

‖Tun‖2L2(RN−1) ≤ 2 ‖un‖L2(Rn+) ‖Dun‖L2(Rn+) .

Since un is converging to u in W 1,2(Rn+), by continuity of the trace operator Tun is converging

to Tu in L2(RN−1), so that we get

‖Tu‖2L2(RN−1) ≤ 2 ‖u‖L2(Rn+) ‖Du‖L2(Rn+) .

by letting n going to +∞ , as required.

Exercise 2. Let Ω be an open set of Rn , and Γ a C1 hypersurface such that Ω = Ω1 ∪Ω2 ∪Γ withΩ1 and Ω2 open connected disjoint.

(a) Let f1 ∈ C1(Ω1) and f2 ∈ C1(Ω2). Under which conditions the function f ∈ L∞(Ω) definedby 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 . Bythe Divergence Theorem we have, for every ϕ ∈ C∞c (Ω),ˆ

Ω

f(x)divϕ(x) dx =

ˆΩ1

f1(x)divϕ(x) dx+

ˆΩ2

f2(x)divϕ(x) dx =

−ˆ

Ω1

Df1(x) · ϕ(x) dx−ˆ

Ω2

Df2(x) · ϕ(x) dx+

ˆΓ

(f1(ξ)− f2(ξ))ϕ(ξ) · νΓ(ξ) dS(ξ)

therefore f ∈W 1,∞(Ω) if and only if f1 = f2 on Γ. In that case one also has

Df =

Df1 in Ω1

Df2 in Ω2 .

(b) It suffices to repeat the previous reasonings with suitable using of the notion of trace to deducethat the required condition is Tf1 = Tf2 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) suchthat 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| ≤ 1

2 , which is connected, and let for instance

f(x) :=

(|x|2 − 1

4 )2 if xN > 0, |x| ≤ 12

0 elsewhere .

Denoting with B(0, 1

2

)+the upper hemisphere B

(0, 1

2

)∩ xN > 0 , one has by the Divergence

theorem that for any ϕ ∈ C∞c (Ω;Rn)ˆΩ

f(x)divϕ(x) dx =

ˆB(

0,12

)+ f(x)divϕ(x) dx = −4

ˆB(

0,12

)+(|x|2 − 14 )x · ϕ(x) dx ;

indeed, no boundary term appears since either f or ϕ are 0 on ∂B(0, 1

2

)+. It follows that f ∈

W 1,∞(Ω). On the other hand, for every 0 < ε < 12 , one has

|f(0, . . . , 0, ε)− f(0, . . . , 0,−ε)||(0, . . . , 0, ε)− (0, . . . , 0,−ε)|

=(ε2 − 1

4 )2

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 thatDu ∈ 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 theproof 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 Ω′ ⊂⊂ Ω. For n large, depending onΩ′ , it holds Dun = %n ?Du in Ω′ (the formula Dun = D%n ?u is instead true on the whole Rn : whythis difference?). We therefore deduce by Jensen’s inequality (how?) that

‖Dun‖Lp(Ω′) ≤ ‖Du‖Lp(Ω) . (9.1)

By Poincare-Wirtinger and (9.1), if un,Ω′ denotes the integral mean of un on Ω′ we get that thereexists a constant CΩ′ depending on Ω′ such thatˆ

Ω′|un(x)− un,Ω′ |p dx ≤ CΩ′‖Du‖pLp(Ω) .

Since un is converging to u in L1(Ω′), by Fatou’s Lemma, denoting with uΩ′ the integral mean of uon Ω′ , we arrive at ˆ

Ω′|u(x)− uΩ′ |p dx ≤ CΩ′‖Du‖pLp(Ω)

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 suchthat the ball B(x + λ

neN ,1n ) ⊂ Ω for all x ∈ V . Therefore, if one defines un(x) = u(x + λ

neN ) one

has un ∈W 1,1(V ) and since we only did a translation of the argument

‖Dun‖Lp(V ) ≤ ‖Du‖Lp(Ω) . (9.2)

Now, defining vn = %n ? un ∈ C∞(Rn), by condition B(x + λneN ,

1n ) ⊂ 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 thefirst step plus (9.2) proves that u in Lp(V ).

Third step: Ω can be written as a finite union Ω = ∪mi=1Vi where V0 ⊂⊂ Ω and each Vi , i ≥ 1 issuch that the second step can be applied. Sinceˆ

Ω

|u(x)|p dx ≤m∑i=1

ˆVi

|u(x)|p dx

the finiteness of the right-hand side gives u ∈ Lp(Ω). Since Du ∈ Lp(Ω) by the hypothesis, theexercise 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 thatˆ

Rn|f(x)| dx = sup

ˆRnf(x) · ϕ(x) dx : ϕ ∈ C∞c (Rn;Rm), ‖ϕ‖∞ ≤ 1

. (9.3)

(b) Let C1,N the Sobolev constant in Rn , i.e. the least possible constant such that the Sobolevinequality

‖u‖ NN−1≤ C‖Du‖1

holds for every u ∈W 1,1(Rn). Prove the following isoperimetric inequality: for every smooth boundedopen set Ω ⊂ Rn one has

|Ω|N−1N ≤ C1,NHn−1(∂Ω) , (9.4)

where |Ω| is the Lebesgue measure of Ω and Hn−1(∂Ω) is the (Hausdorff, or surface) measure of theboundary. Hint: approximate the characteristic function χΩ by convolution and use (9.3) and thedivergence theorem to estimate the L1 norm of the gradients.

24 EXERCISES PDE

Solution: (a) Define

f1(x) :=

f(x)|f(x)| if f(x) 6= 0

0 otherwise

Let %n be the usual mollifiers, and, given a sequence ϕn of C∞c (Rn) functions with 0 ≤ ϕn ≤ 1 withϕn(x)→ 1 for every x , set gn := ϕn(%n ?f1). It is clear that gn ∈ C∞c (Rn;Rm) and that ‖gn‖∞ ≤ 1.Furthermore gn converges to f1 a.e. Observing that f(x) · f1(x) = |f(x)| for every x , by dominatedconvergence one hasˆ

Rn|f(x)| dx = lim

n

ˆRnf(x) · gn(x) dx ≤ sup

ˆRnf(x) · ϕ(x) dx : ϕ ∈ C∞c (Rn;Rm), ‖ϕ‖∞ ≤ 1

.

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

‖Du‖1 = sup ˆ

Rnudivϕ : ϕ ∈ C∞c (Rn;Rn), ‖ϕ‖∞ ≤ 1

. (9.5)

Now, if %n are the usual mollifiers, we have for every ϕ ∈ C∞c (Rn;Rn) with ‖ϕ‖∞ ≤ 1 that also‖%n ? ϕ‖∞ ≤ 1. Therefore, by symmetry of the mollifiers, standard properties of the convolution andalso using the divergence theorem, we haveˆ

Rn(%n ? χΩ)(x) divϕ(x) dx =

ˆRnχΩ(x) div (%n ? ϕ)(x) dx =

ˆΩ

div (%n ? ϕ)(x) dx =ˆ∂Ω

(%n ? ϕ)(ξ) · ν(ξ) dHn−1(ξ) ≤ Hn−1(∂Ω)

for every smooth bounded open set Ω ⊂ Rn . It follows from this and (9.5) that for every n ∈ N onehas

‖D(%n ? χΩ)‖1 ≤ Hn−1(∂Ω) (9.6)

for every smooth bounded open set Ω ⊂ Rn .Now, combining (9.6) with the Sobolev inequality gives

‖%n ? χΩ‖ NN−1≤ C1,NHn−1(∂Ω)

for every smooth bounded open set Ω ⊂ Rn and every n ∈ N . Letting n to +∞ we get

‖χΩ‖ NN−1≤ C1,NHn−1(∂Ω)

which is exactly (9.4).

Exercise 3. Let p > 2. Show with a counterexample that the Morrey’s Imbedding Theorem is notverified 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+1γ+1 > p , which is equivalent to

α− (γ + 1)p > −1 . (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 Holder functions. On theother hand by Fubini’s theorem one hasˆ

Dα

|uγ(x, y)|p dxdy =

ˆ 1

0

xα−γp dx < +∞

EXERCISES PDE 25

since by (9.7) one gets α− γp > −1. Similarly, being Du(x, y) = (−γx−(1+γ), 0) one getsˆDα

|Duγ(x, y)|p dxdy = γpˆ 1

0

xα−(γ+1)p dx < +∞

again using Fubini’s theorem and (9.7). Therefore uγ ∈ W 1,p(Dα) but it cannot satisfy Morrey’simbedding 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

fN (t) :=2

π

N∑n=1

an sin(nt)

where an =´ π

0f(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 theoremto the odd extension of f to [−π, π]).

(a) Assume that f ∈ H10 ((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 ∈ H1

0 ((0, π)) ∩H2((0, π)) prove that there exists a constant C such that

‖fN − f‖H1 ≤ C

N‖f‖H2 . (10.1)

Solution: (a) Since f(0) = f(π) = 0 integrating by parts we obtain

nan = −ˆ π

0

f ′(t) cos(nt) dt (10.2)

for every n ∈ N . Since √

2π cos(nt) : n ∈ N is an orthonormal system of L2(0, π), Bessel’s inequality

givesN∑n=1

n2a2n =

N∑n=1

( ˆ π

0

f ′(t) cos(nt) dt)2

≤ π

2‖f ′‖22

for every N ∈ N . Being a convergent series,∑Nn=1 n

2a2n is a Cauchy sequence.

On the other hand, since √

2π 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

‖f ′M − f ′N‖22 = ‖M∑n=N

nan cos(n·)‖22 =

M∑n=N

n2a2n‖ cos(n·)‖22 =

M∑n=N

n2a2n . (10.3)

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 H1

0 ((0, π)), and therefore uniformly by Morrey’simbedding Theorem.

(b) If f ∈ H2((0, π)) we can once more integrate by parts in (10.2), obtainingˆ π

0

f ′(t) cos(nt) dt = − 1

n

ˆ π

0

f ′′(t) sin(nt) dt

so that

n2an =

ˆ π

0

f ′′(t) sin(nt) dt

and, again using Bessel’s inequality,N∑n=1

n4a2n ≤

π

2‖f ′′‖22

26 EXERCISES PDE

for every N ∈ N . Combining this with (10.3) for every N ∈ N and every M ∈ N with M ≥ N , onehas

‖f ′M − f ′N‖22 =

M∑n=N

n2a2n ≤

1

N2

M∑n=N

n4a2n ≤

π

2N2‖f ′′‖22 .

When M goes to +∞ we therefore obtain

‖f ′ − f ′N‖2 ≤√π

2

1

N‖f ′′‖2

Since by Poincare’s inequality

‖f − fN‖2 ≤ π‖f ′ − f ′N‖2we get that

‖f − fN‖H10 ((0,π)) ≤

√π

2(1 + π)

1

N‖f ′′‖2

as required.

Exercise 2. Cea’s Lemma: Consider a symmetric, bounded, coercive bilinear form B : H ×H → Ron a Hilbert space H , that is

B(u, v) = B(v, u) , B(u, v) ≤ C1‖u‖H‖v‖H , B(u, u) ≥ c0‖u‖2H .

(a) Fix a linear continuous functional F : H → R . Justify that there exists a unique u ∈ H suchthat

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 ) (10.4)

for every vN ∈ HN . Deduce Cea’s estimate:

‖u− uN‖H ≤√

C1

c0inf‖u− vN‖ : vN ∈ HN . (10.5)

(d) Consider now H = H10 ((0, π)) and HN := spant 7→ sin(nt) : n = 1, . . . , N . For a(x) ∈

C1[0, π] with 1 ≤ a(x) ≤ 2 for every x and f ∈ L2((0, π)) set

B(u, v) :=

ˆ π

0

a(t)u′(t) v′(t) dt (10.6)

and F (v) :=´ π

0f(t) v(t) dt . Check that u and uN as in (a) and (b), respectively, exist, and show

that uN → u in H10 ((0, π)) as N goes to +∞ . If additionally u ∈ H2((0, π)) show that there exists

a constant C such that

‖u− uN‖H1 ≤ C

N‖u‖H2 .

Solution: (a) follows from the Lax-Milgram theorem. The same holds for (b) since the restriction ofB 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‖u− uN‖2H ≤ B(u− uN , u− uN ) ≤ B(u− uN , u− uN ) +B(vN − uN , vN − uN ) =

B(u− vN , u− vN ) ≤ C1‖u− vN‖2Hwhence

‖u− uN‖H ≤√

C1

c0‖u− vN‖H

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 Lax-

Milgram theorem can be applied. The only relevant point to this end is to prove coercivity of B ,which follows from the Poincare inequality. Namely, one has

B(u, u) ≥ ‖u′‖22 ≥1

1 + π2‖u‖2H1 .

It follows now from the previous exercise that

inf‖u− vN‖H10 ((0,π)) : vN ∈ HN → 0

as N goes to +∞ . In particular, by (10.1), if u ∈ H2 one has for every N ∈ N

inf‖u− vN‖H10 ((0,π)) : vN ∈ HN ≤

C

N‖u‖H2 .

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

H1ΓD (Ω) := u ∈ H1(Ω), Tu = 0 on ΓD ,

with T the trace operator. For λ > 0 define

Bλ(u, v) =

ˆΩ

∇u(x) · ∇v(x) dx− 1

λ

ˆ∂Ω

Tu(ξ)Tv(ξ) dHn−1(ξ)

for u and v ∈ H1ΓD

(Ω). Prove that, when λ is sufficiently large, for every f ∈ L2(Ω) there exists a

unique uf ∈ H1ΓD

(Ω) such that

Bλ(uf , v) =

ˆΩ

f(x) v(x) dx

for every v ∈ H1ΓD

(Ω). Hint: does the Poincare inequality apply in H1ΓD

(Ω)?

Solution: Following the hint, let us first prove that there exists a constant C such that

‖u‖L2(Ω) ≤ C‖∇u‖L2(Ω) (10.7)

for every u ∈ H1ΓD

(Ω). To prove this, it suffices to argue as in the proof of the Poincare-Wirtingerinequality (Theorem 1, Section 5.8.1 in Evans’book): contradicting (10.7) would produce the existenceof a sequence un in H1

ΓD(Ω) with ‖un‖|L2(Ω) = 1 for every n which is converging to a constant c in

H1(Ω). By continuity of the trace operator, H1ΓD

(Ω) is a Hilbert subspace of H1(Ω), so it must be

c = 0, since this one is the only constant in H1ΓD

(Ω). On the other hand, by strong L2 convergence

we must have ‖c‖L2(Ω) = c|Ω| 12 = 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. Assumethat λ > CT (1 + C2). Then for every u ∈ H1

ΓD(Ω) one has, by the continuity of the trace and by

(10.7), that

Bλ(u, u) = ‖∇u‖2L2(Ω) −1

λ‖Tu‖2L2(∂Ω) ≥

1

1 + C2‖u‖2H1(Ω) −

CTλ‖u‖2H1(Ω) = c0‖u‖2H1(Ω)

with c0 := 11+C2 − CT

λ > 0 due to the assumption on λ . Therefore for λ > CT (1 + C2) the bilinear

form Bλ is coercive on H1ΓD

(Ω). For any λ > 0 one has also by the Cauchy-Schwarz inequality andthe continuity of the trace operator that

Bλ(u, v) ≤ (1 +CTλ

)‖u‖H1(Ω)‖v‖H1(Ω)

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, andcontinuous 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

1

2a(u, u)− L(u) = min

1

2a(v, v)− L(v) : v ∈ H

. (11.2)

Hint: if u is a minimizer, fix v ∈ H and consider the function F (t) := 12a(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 functionF (t) := 1

2a(u + tv, u + tv) − L(u + tv), t ∈ R . The function F attains then its minimum value fort = 0. Since by a direct computation

F (t) =1

2a(u, u)− L(u) + t(a(u, v)− L(v)) +

t2

2a(v, v) (11.3)

imposing F ′(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) := 12a(u+ tv, u+ tv)− L(u+ tv), t ∈ R . By (11.3) and (11.1) we obtain

F (t) =1

2a(u, u)− L(u) +

t2

2a(v, v) .

By coerciveness of a , a(v, v) ≥ 0 which implies that F attains its minimum at 0. In particularF (0) ≤ F (1), therefore

1

2a(u, u)− L(u) ≤ 1

2a(u+ v, u+ v)− L(u+ v) (11.4)

for every v ∈ H . Now, for an arbitrary w ∈ H , applying (11.4) with v = w − u one gets

1

2a(u, u)− L(u) ≤ 1

2a(w,w)− L(w)

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 saythat u ∈ H1(Ω) is a weak solution of the Neumann problem

−∆u = f in Ω∂u∂ν = 0 on ∂Ω

(11.5)

if ˆΩ

∇u(x) · ∇v(x) dx =

ˆΩ

f(x)v(x) dx (11.6)

for any v ∈ H1(Ω).(a) Justify that actually (11.6) is a weak formulation of the problem (11.5) by showing that if u

solves (11.6) and in addition u ∈ H2(Ω) then −∆u = f in L2(Ω) and ∂u∂ν = 0 in L2(∂Ω) in the sense

of traces.(b) Show that a necessary and sufficient condition for the existence of a solution of (11.6) isˆ

Ω

f(x) dx = 0 . (11.7)

Hint: use Lax-Milgram’s theorem in the Hilbert space

H := u ∈ H1(Ω) :

ˆΩ

u(x) dx = 0 . (11.8)

(c) Let g ∈ L2(∂Ω). Generalise part (a) and (b) to the nonhomogeneuous Neumann problem−∆u = f in Ω∂u∂ν = g on ∂Ω

(11.9)

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 ∈ H2(Ω), for every v ∈ C∞c (Ω) integrating byparts and using (11.6) we have

−ˆ

Ω

∆u(x)v(x) dx =

ˆΩ

∇u(x) · ∇v(x) dx =

ˆΩ

f(x)v(x) dx

which implies −∆u = f in L2(Ω). Using this additional information, integrating by parts and usingagain (11.6) we get for every v ∈ H1(Ω) ˆ

Ω

f(x)v(x) dx =

−ˆ

Ω

∆u(x)v(x) dx =

ˆΩ

∇u(x) · ∇v(x) dx−ˆ∂Ω

∂u

∂ν(ξ)v(ξ) dHn−1(ξ) =ˆ

Ω

f(x)v(x) dx−ˆ∂Ω

∂u

∂ν(ξ)v(ξ) dHn−1(ξ) ,

which gives ∂u∂ν = 0 in L2(∂Ω).

Observe in addition that via integration by parts also the converse holds: if u ∈ H2(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 thesubspace H as in (11.8), we first prove that, if u ∈ H satisfiesˆ

Ω

∇u(x) · ∇v(x) dx =

ˆΩ

f(x)v(x) dx (11.10)

for every v ∈ H , then (11.6) holds. Indeed, when v ∈ H1(Ω), then v = v − 1|Ω|

´Ωv(x) dx belongs to

H ; furthermore, ∇v = ∇v and, by (11.7),ˆΩ

f(x)v(x) dx =

ˆΩ

f(x)v(x) dx .

30 EXERCISES PDE

Therefore (11.10) implies (11.6). Now, existence of a unique solution to (11.10) in H follows by theLax-Milgram theorem once it is checked that the bilinear form B : H ×H → R defined by

B(u, v) :=

ˆΩ

∇u(x) · ∇v(x) dx

is continuous and coercive. Continuity is obvious, while coercivity follows by the Poincare-Wirtingerinequality

‖∇u‖2 ≥ CΩ‖u‖2for every u ∈ H (add the details!).

(c) The solution is very similar to the previous case, so I only sketch it. Integrating by parts, oneeasily gets that a smooth solution of (11.9) satisfiesˆ

Ω

∇u(x) · ∇v(x) dx =

ˆΩ

f(x)v(x) dx+

ˆ∂Ω

g(ξ)v(ξ) dHn−1(ξ) . (11.11)

This is therefore our candidate weak formulation. Testing (11.11) on C∞c functions we get that if usolves it and in addition u ∈ H2(Ω) then −∆u = f in L2(Ω). Using this additional information,integrating by parts and using again (11.11) we get for every v ∈ H1(Ω)ˆ

Ω

f(x)v(x) dx =

−ˆ

Ω

∆u(x)v(x) dx =

ˆΩ

∇u(x) · ∇v(x) dx−ˆ∂Ω

∂u

∂ν(ξ)v(ξ) dHn−1(ξ) =ˆ

Ω

f(x)v(x) dx−ˆ∂Ω

(∂u∂ν

(ξ)− g(ξ))v(ξ) dHn−1(ξ) ,

which gives ∂u∂ν = g in L2(∂Ω).

We then show that (11.11) has a solution if and only ifˆΩ

f(x) dx+

ˆ∂Ω

g(ξ) dHn−1(ξ) = 0 . (11.12)

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 ∈ Hsatisfies ˆ

Ω

∇u(x) · ∇v(x) dx =

ˆΩ

f(x)v(x) dx+

ˆ∂Ω

g(ξ)v(ξ) dHn−1(ξ) . (11.13)

for every v ∈ H , then (11.11) holds. Existence of a solution to (11.13) in H follows again by theLax-Milgram Theorem; indeed, the bilinear form B : H ×H → R defined by

B(u, v) :=

ˆΩ

∇u(x) · ∇v(x) dx

is continuous and coercive, while the linear functional

L(v) :=

ˆΩ

f(x)v(x) dx+

ˆ∂Ω

g(ξ)v(ξ) dHn−1(ξ)

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 , andassume that Hn−1(Γ1) > 0. Let f ∈ L2(Ω). Find the weak formulation of the mixed boundary-valueproblem

−∆u = f in Ω

u = 0 on Γ1

∂u∂ν = 0 on Γ2

(11.14)

and prove existence of a weak solution by using Lax-Milgram’s theorem in a suitable Hilbert subspaceof H1(Ω).

EXERCISES PDE 31

Solution: The presence of the Dirichlet boundary condition suggests to operate in the Hilbert sub-space

H := u ∈ H1(Ω), u = 0 on Γ1 ,where we have already seen that the Poincare inequality

‖∇u‖2 ≥ CΩ‖u‖2holds (short reminder: the only constant function in H is 0 and ∂Ω is smooth).

The weak formulation is: finding u ∈ H such thatˆΩ

∇u(x) · ∇v(x) dx =

ˆΩ

f(x)v(x) dx (11.15)

for every v ∈ H . Indeed, any smooth solution of (11.14) satisfies (11.15). Conversely, if u ∈ Hsatisfies (11.15) and additionally u ∈ H2(Ω), testing (11.15) on C∞c functions gives −∆u = f inL2(Ω). Using this additional information, integrating by parts and using again (11.15) we get forevery v ∈ H ˆ

Ω

f(x)v(x) dx =

−ˆ

Ω

∆u(x)v(x) dx =

ˆΩ

∇u(x) · ∇v(x) dx−ˆ∂Ω

∂u

∂ν(ξ)v(ξ) dHn−1(ξ) =ˆ

Ω

f(x)v(x) dx−ˆ

Γ2

∂u

∂ν(ξ)v(ξ) dHn−1(ξ) ,

since v = 0 on Γ1 . This gives ∂u∂ν = 0 in L2(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 Poincare inequality and the

Lax-Milgram Theorem.

12. Exercises PDE 14.02.13

Exercise 1. Differentiation of convolutions revisited. Recall preliminarly the following form of theJensen inequality: let F ∈ L∞(Rn), F ≥ 0 withˆ

RnF (y) dy = 1 .

Then for every convex function ϕ : R→ R and every g ∈ L1(Rn) it holds

ϕ(ˆ

Rng(y)F (y) dy

)≤ˆRnϕ(g(y))F (y) dy .

In the following %ε is a positive symmetric C∞ mollifier withˆRn%ε(y) dy = 1 (12.1)

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 ‖%ε ? f‖W 1,p(Rn) ≤ ‖f‖W 1,p(Rn) .

(c) Let Ω be a bounded open subset of Rn and f ∈ W 1,10 (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 Ω′ ⊂⊂ Ωε . Prove that D(%ε ? f) = %ε ? Df in L1(Ω′). Hint: try first with C∞ functions, thenapproximate.

32 EXERCISES PDE

Solution: (a) Let ϕ ∈ C∞c (Rn;Rn). Using the properties of the convolution with the mollifiers andthe definition of weak gradient we haveˆ

Rn%ε ? f(x) divϕ(x) dx =

ˆRnf(x) %ε ? divϕ(x) dx =ˆ

Rnf(x) div (%ε ? ϕ)(x) dx = −

ˆRnDf(x) · %ε ? ϕ(x) dx = −

ˆRn%ε ? Df(x) · ϕ(x) dx

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 ˆRn|%ε ? f(x)|p dx =

ˆRn

∣∣∣ ˆRn%ε(x− y)f(y) dy

∣∣∣p dx ≤ˆRn

ˆRn%ε(x− y)|f(y)|p dy dx =

ˆRn|f(y)|p dy

which proves ‖%ε ? f‖Lp(Rn) ≤ ‖f‖Lp(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 ∈W 1,10 (Ω) there is a sequence fn ∈ C∞c (Ω) converging to f in W 1,1 . The extensions

fn simply obtained by fn = 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

%ε ? f(x) =

ˆRn%ε(x− y)f(y) dy =

ˆRn%ε(y)f(x− y) dy =

ˆB(0,ε)

%ε(y)f(x− y) dy .

Now, when x ∈ Ω′ , the gradient Df(x − y) (in the variable x) exists for every y ∈ B(0, ε), sincein 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 ∈ H1(B(0, 1)) and that

‖vr − 1‖H1 → 0

when r → 0.(c) Let f ∈ C1(B(0, 1)). Construct a sequence fr with suppfr ⊂ B(0, 1)\B(0, r) and ‖fr−f‖H1 →

0.(d) Deduce that H1

0 (B(0, 1)\0) = H10 (B(0, 1)).

(e) Let N = 1. Is that true that H10 ((−1, 1)\0) = H1

0 ((−1, 1))? Hint: Sobolev convergence isparticularly strong in dimension 1.

Solution: (a) The radial symmetry of the domain and of the data suggests a radial solution. Arguingin the same way used to construct the fundamental solution and imposing the boundary conditionsone gets

ur(x) =log(|x|)| log r|

+ 1 .

EXERCISES PDE 33

(b) vr are indeed Lipschitz continuous since obtained by gluing together two C1 functions withthe same trace on the interface (see Exercise 2, 9-11.01.13). Furthermore, using polar coordinatesˆ

B(0,1)

|vr(x)− 1|2 dx = |B(0, r)|+ 2π

log2 r

ˆ 1

r

% log2 % d% ≤ |B(0, r)|+ 2Cπ

log2 r.

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 directcomputation and using polar coordinatesˆ

B(0,1)

|D(vr(x)− 1)|2 dx =

ˆB(0,1)\B(0,r)

|Dur(x)|2 dx =2π

log2 r

ˆ 1

r

1

%d% =

2π

| log r|which again vanishes when r → 0. Thus, the claim is proved.

(c) Set fr := vrf . Clearly suppfr ⊂ B(0, 1) \ B(0, r). Furthermore, since f is C1 up to theboundary, we easily have

‖fr − f‖H1 ≤ ‖f‖C1‖vr − 1‖H1

which gives the claim.(d) We need to use a double approximation argument. First, of all, observe that clearly

C∞c (B(0, 1\0) ⊂ H10 (B(0, 1)) ;

taking the closure at the left-hand side, the inclusion H10 (B(0, 1)\0) ⊂ H1

0 (B(0, 1)) is trivial. Now,let f ∈ H1

0 (B(0, 1)) and let fn ∈ C∞c (B(0, 1)) a sequence H1 -converging to f . By step (c) applied toeach fn , for every n we can construct gn ∈ H1

0 (B(0, 1)\0) such that ‖gn−fn‖H1 ≤ 1n (construct a

sequence approximating fn and then choose a sufficiently close element of the sequence). Now, easilygn converges to f in H1 ! Since H1

0 (B(0, 1)\0) is a Banach space, f ∈ H10 (B(0, 1)\0), and this

proves the reverse inclusion.(e) No. If the equality holds, any f ∈ H1

0 ((−1, 1)) could be approximated by a sequence fn ∈C∞c ((−1, 1)\0) in the H1 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

elliptic equation Lu = −∑Ni,j=1 aij(x) ∂2

∂xi∂xju = 0 in Ω. Assume that the coefficients aij ∈ C1(Ω).

Show that for λ sufficiently large not depending on u the function v := |∇u|2 + λu2 satisfies Lv ≤ 0in Ω. Deduce that

‖∇u‖L∞(Ω) ≤ C(‖∇u‖L∞(∂Ω) + ‖u‖L∞(∂Ω)

). (12.2)

Solution: We called A(x) the matrix whose coefficients are aij(x) and we denote with 〈·, ·〉 thescalar 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

Lw2(x) = −N∑

i,j=1

aij(x)∂2

∂xi∂xjw2(x) = −〈A(x)∇w(x),∇w(x)〉+ 2w(x)Lw(x) .

For w =√λu , since u is a solution and by ellipticity, we get

L(λu2(x)) ≤ −λα|∇u(x)|2 (12.3)

with α the coercivity constant of A . For k = 1, . . . , N and w = ∂u∂xk

we have

L(( ∂u

∂xk

)2

(x))≤ −α

∣∣∣∇(∂u

∂xk)(x)

∣∣∣2 + 2∂u

∂xk(x)L

( ∂u∂xk

(x)).

By a direct computation and since Lu = 0 we have

L( ∂u∂xk

(x))

=∂

∂xkLu(x) +

N∑i,j=1

∂

∂xkaij(x)

∂2

∂xi∂xju(x) =

N∑i,j=1

∂

∂xkaij(x)

∂2

∂xi∂xju(x) .

34 EXERCISES PDE

Since aij ∈ C1(Ω) we get easily

2∂u

∂xk(x)L

( ∂u∂xk

(x))≤ CN |∇u(x)||D2u(x)|

with D2u the Hessian matrix and CN a constant depending on the C1 norm of the aij ’s and on thedimension N . Summing over k from 1 to N we have

L(|∇u|2(x)) ≤ −α|D2u(x)|2 + CN |∇u(x)||D2u(x)| . (12.4)

We now recall Young’s inequality in the following form: for every a, b ∈ R and ε > 0 one has

ab ≤ ε2

2a2 +

1

2ε2b2 . (12.5)

Using now (12.3), (12.4) and (12.5) with a = |D2u(x)| , b = |∇u(x)| and ε =√

α2CN

we obtain

L(|∇u|2(x) + λu2(x)) ≤ −λα|∇u(x)|2 +4C2

N

α|∇u(x)|2 .

For λ ≥(

2CNα

)2the right-hand side is nonpositive so that setting v := |∇u|2 + λu2 one has Lv ≤ 0

in Ω. By the weak maximum principle

‖v‖L∞(Ω) ≤ ‖v‖L∞(∂Ω) .

Now, trivially ‖∇u‖2L∞(Ω) ≤ ‖v‖L∞(Ω) . It is also easy to se that there is a constant C depending on

N and λ , thus on N , CN and α such that ‖v‖L∞(∂Ω) ≤ C(‖∇u‖L∞(∂Ω) + ‖u‖L∞(∂Ω)

)2

. From this,

(12.2) follows.