Partial Differential Equations, 2nd Edition, L.C.Evans

Chapter 6 Second-Order Elliptic Equations

Shih-Hsin Chen∗, Yung-Hsiang Huang†

2017/03/27

1. (a) Direct computation. (b) c = ∆√a.

2. Proof. Define

B[u, v] =

∫U

n∑i,j=1

aijuxivxj + cuv dx, u, v ∈ H10 (U).

Clearly,

|B[u, v]| ≤ ‖aij‖∞‖Du‖L2(U)‖Dv‖L2(U) + ‖c‖∞‖u‖L2(U)‖v‖L2(U) ≤ α‖u‖H10 (U)‖v‖H1

0 (U),

for some positive constant α. For c(x) ≥ −µ, where µ is a fixed constant to be assigned later.

Using the uniform ellipticity, we have

θ

∫U

|Du|2 dx ≤∫U

n∑i,j=1

aijuxiuxj

= B[u, u]−∫U

cu2 dx

≤ B[u, u] + µ

∫U

u2 dx

≤ B[u, u] + Cµ

∫U

|Du|2 dx,

where the Poincare inequality is applied in the last inequality. Now we set µ =θ

2Cand see

that if c(x) ≥ θ

2C, then

θ

2

∫U

|Du|2 dx ≤ B[u, u].

∗Department of Math., National Taiwan University. Email: d03221002@ntu.edu.tw†Department of Math., National Taiwan University. Email: d04221001@ntu.edu.tw

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Therefore,

‖u‖2H1

0 (U) = ‖u‖2L2(U) + ‖Du‖2

L2(U)

≤ (1 + C)‖Du‖2L2(U)

≤ 2

θ(1 + C)B[u, u].

3. Proof. Define

B[u, v] =

∫U

n∑i,j=1

uxixjvxixj dx, u, v ∈ H20 (U).

Use the integration by part (divergence theorem),∫U

|D2u|2 dx =

∫U

n∑i,j=1

uxixjuxixj dx =

∫U

∑i,j=1

uxixiuxjxj dx =

∫U

|∆u|2 dx

for u, v ∈ H20 (U). (Fix i, consider F = uxiDuxi and then use

∫U

divF dx =∫∂U

F · ν dS.)

‖Du‖2L2(U) = −

∫U

u∆u dx

≤ ε‖u‖2L2(U) +

1

4ε‖∆u‖2

L2(U)

≤ εC‖Du‖2L2(U) +

1

4ε‖∆u‖2

L2(U)

the Poincar’e inequality is used in the last inequality. We choose ε =1

2Cso that

‖Du‖2L2(U) ≤ C‖∆u‖2

L2(U).

Therefore,

‖u‖2H2

0 (U) = ‖u‖2L2(U) + ‖Du‖2

L2(U) + ‖D2u‖2L2(U)

≤ (1 + C)‖Du‖2L2(U) + ‖∆u‖2

L2(U)

≤ (1 + C + C2)‖∆u‖2L2(U)

= (1 + C + C2)B[u, u].

Thus,1

1 + C + C2‖u‖2

H20 (U) ≤ B[u, u], u ∈ H2

0 (U).

Remark 1. The validity of the definition of weak solution to Biharmonic equation with zero

Dirichlet and zero Neumann boundary is based on the zero trace theorem for W 2,p0 function,

which is not proved in Evans’ book.

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4. Proof. ”⇒ ” take v = 1.

”⇐ ” Define

B[u, v] =

∫U

Du ·Dv dx, for u, v ∈ H1(U).

Consider a subspace of H1(U)

A = {u ∈ H1(U)∣∣(u)U = 0}

Let l(u) =∫Uu dx, then l is a continuous linear functional on H1(U) (the boundness of domain

U is used here.) Then A = l−1({0}) is closed in H1(U), so A is a Hilbert space with induced

inner product. Note that

|B[u, v]| ≤ ‖u‖A‖v‖A.

and by Poincare inequality,

‖u‖2L2(U) =

∫U

|u− (u)U |2 dx ≤ C

∫U

|Du|2 dx.

Therefore

‖u‖2A = ‖u‖2

L2(U) + ‖Du‖2L2(U) ≤ (1 + C)‖Du‖2

L2(U) = (1 + C)B[u, u].

By Lax-Milgram, for all f ∈ L2(U) (with∫Uf dx = 0), there exists a unique uf ∈ A such that

B[uf , v] = (f, v)L2(U), ∀v ∈ A.

Let v ∈ H1(U), then v − (v)U ∈ A. Thus, we have

(f, v)L2(U) = (f, (v)U)L2(U) + (f, v − (v)U)L2(U)

= 0 +B[uf , v − (v)U ], (since

∫U

f dx = 0)

= B[uf , v] =

∫U

Duf ·Dv dx, ∀v ∈ H1(U).

Therefore, uf ∈ A ⊂ H1(U) is a weak solution. In general, uf + C is always a weak solution

for any constant C.

5. Motivation. If u ∈ C∞(U) solves the given boundary value problem, then for each v ∈ C∞(U),∫U

fv =

∫U

(−∆u)v =

∫∂U

−v ∂u∂n

+

∫U

∇u∇v =

∫∂U

vu+

∫U

∇u∇v.

By density theorem and trace theorem, we define B on (H1(U))2 by

B[u, v] =

∫∂U

(Tv)(Tu) +

∫U

∇u∇v,

where T is the trace operator with finite operator norm ‖T‖. We call u ∈ H1(U) is a weak

solution of the given Robin boundary value problem if B[u, v] =∫Ufv for all v ∈ H1(U).

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Proof. We prove uniqueness and existence by Lax-Milgram. Given u, v ∈ H1(U), it’s easy to see

|B[u, v]| ≤ max{1, ‖T‖2}‖u‖H1‖v‖H1 . Next, we prove B is strictly coercive by a contradiction

argument similar to the one for Poincare inequality.

Assume for each k ∈ N, there exists uk ∈ H1 such that ‖uk‖2H1 > kB[uk, uk].

Let vk = uk/‖uk‖H1 , then 1 = ‖vk‖2H1 > kB[vk, vk], so 1

k> B[vk, vk] ≥ ‖Dvk‖2

L2 ,

1k> ‖Tvk‖2

L2(∂U) and hence ‖vk‖2H1 < 1 + 1

k≤ 2. Rellich’s compactness theorem asserts there

exists a subsequence, still denoted by {vk}, converging to a function v in L2 sense. Note

‖Dvk‖L2 → 0, so for every φ ∈ C∞c (U),∫vDφ = lim

k→∞

∫vkDφ = lim

k→∞

∫−(Dvk)φ = 0

Hence v ∈ H1, Dv ≡ 0. Now we know vk → v in H1 sense, so ‖v‖H1 = 1. Moreover, v is

constant in each component of U . But

‖Tv‖L2(∂U) ≤ ‖T (v − vk)‖L2(∂U) + ‖Tvk‖L2(∂U) ≤ ‖T‖‖v − vk‖H1 +

√1

k→ 0, as k →∞,

which means v ≡ 0 on ∂U and then v ≡ 0 on U . This contradicts to ‖v‖H1 = 1. Therefore B

is strictly coercive.

6. (This boundary value problem is called Zaremba problem sometimes)

Definition. Similar to Exercise 5, we define B[u, v] =∫∇u∇v on H := H1 ∩ {u : Tu �Γ1= 0}.

The linearity and continuity of T ensures that H is a closed subspace of H1, so H is a Hilbert

space with induced inner product. u ∈ H is a weak solution of the given Robin boundary

value problem if B[u, v] =∫Ufv for all v ∈ H. (The zero Newmann boundary condition is

involved.)

Uniqueness and Existence. They are established by Lax-Milgram again, the coercivity of B is

established by almost the same argument as Exercise 5.

7. Proof. By definition, we know for each v ∈ H1(Rn),∫∇u∇v + c(u)v =

∫fv.

Take v = −D−hi Dhi u, 0 < |h| � 1. Then v ∈ H1 and has compact support. Hence∫

|DhiDu|2 +Dh

i c(u)Dhi u = −

∫f(D−hi Dh

i u).

By mean value theorem,

Dhi c(u)(x)Dh

i u(x) =c(u(x+ hei))− c(u(x))

hDhi u(x) = c′(· · · )|Dh

i u(x)|2 ≥ 0.

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Hence ∫|Dh

iDu|2 ≤ |∫f(D−hi Dh

i u)| ≤ 1

2

∫|f |2 +

1

2

∫|D−hi Dh

i u|2

≤ 1

2

∫|f |2 +

1

2

∫|DDh

i u|2

The last inequality follows easily from mean value theorem. Hence∫|Dh

iDu|2 ≤∫|f |2, for all

small enough h. This implies∫|D2u|2 ≤

∫|f |2. (Evans, Page 292)

8. Proof. Differentiate the equation Lu = 0, we get (Daij)uxixj + aij(Du)xixj = 0. Multiply this

equation by 2Du, we have 2Du · (Daij)uxixj = −2aijDu · (Du)xixj . Adding −2aij(Du)xi(Du)xj

to both sides, we find

2Du · (Daij)uxixj − 2aij(Du)xi(Du)xj = −aij(|Du|2)xixj (1)

On the other hand, multiply the origin equation by 2u and add 2aijuxiuxj to both sides, we

have

2aijuxiuxj = 2aijuxiuxj + 2aijuuxixj = aij(u2)xixj (2)

(1)− λ(2) implies

−2λaijuxiuxj + 2Du · (Daij)uxixj − 2aij(Du)xi(Du)xj = −aij(λu2 + |Du|2)xixj

By uniform ellipticity, boundedness of Daij and Young’s inequality, the left hand side is non-

positive for large λ. The second assertion is a consequence of weak maximum principle to

L(|Du|2 + λu2) ≤ 0

9. Proof. By weak maximum principle, minUw = min

∂Uw = w(x0). Let v1 = u + ‖f‖∞w and

v2 = u− ‖f‖∞w, then Lv1 ≥ 0, Lv2 ≤ 0.

By weak maximum principle, minUv1 = min

∂Uv1 = ‖f‖∞min

∂Uw = ‖f‖∞w(x0) = v1(x0) and

maxU

v2 = max∂U

v2 = −‖f‖∞min∂U

w = −‖f‖∞w(x0) = v2(x0). Therefore,

0 ≥ ∂v1

∂ν(x0) =

∂u

∂ν(x0) + ‖f‖∞

∂w

∂ν(x0),

0 ≤ ∂v2

∂ν(x0) =

∂u

∂ν(x0)− ‖f‖∞

∂w

∂ν(x0).

Since u = 0 on ∂U , ∇u//ν. So

|∇u(x0)| = |∂u∂ν

(x0)| ≤ ‖f‖∞|∂w

∂ν(x0)|.

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10. Proof. We omit (a) since is standard. For (b), if u attains an interior maximum, then the

conclusion follows from strong maximum principle.

If not, then for some x0 ∈ ∂U, u(x0) > u(x) ∀x ∈ U . Then Hopf’s lemma implies ∂u∂ν

(x0) > 0,

which is a contradiction.

Remark 2. A generalization of this problem to mixed boundary conditions is recorded in

Gilbarg-Trudinger, Elliptic PDEs of second order, Problem 3.1.

11. Proof. Define

B[u, v] =

∫U

∑i,j

aijuxivxj dx for u ∈ H1(U), v ∈ H10 (U).

By Exercise 5.17, φ(u) ∈ H1(U). Then, for all v ∈ C∞c (U), v ≥ 0,

B[φ(u), v] =

∫U

∑i,j

aij(φ(u))xivxj dx

=

∫U

∑i,j

aijφ′(u)uxivxj dx, (φ′(u) is bounded since u is bounded)

=

∫U

∑i,j

aijuxi(φ′(u)v)xj −

∑i,j

aijφ′′(u)uxiuxjv dx

≤ 0−∫U

φ′′(u)v|Du|2 dx ≤ 0, by convexity of φ.

(We don’t know whether the product of two H1 functions is weakly differentiable. This is why

we do not take v ∈ H10 .) Now we complete the proof with the standard density argument.

12. Proof. Given u ∈ C2(U) ∩ C(U) with Lu ≤ 0 in U and u ≤ 0 on ∂U . Since U is compact and

v ∈ C(U), v ≥ c > 0. So w := uv∈ C2(U) ∩ C(U). Brutal computation gives us

−aijwxixj =−aijuxixjv + aijvxixju

v2+aijvxiuxj − aijuxivxj

v2− aij 2

vvxj

vxiu− vuxiv2

=(Lu− biuxi − cu)v + (−Lv + bivxi + cv)u

v2+ 0 + aij

2

vvxjwxi , since aij = aji.

=Lu

v− uLv

v2− biwxi + aij

2

vvxjwxi

Therefore,

Mw := −aijwxixj + wxi[bi − aij 2

vvxj]

=Lu

v− uLv

v2≤ 0 on {x ∈ U : u > 0} ⊆ U

If {x ∈ U : u > 0} is not empty, Weak maximum principle to the operator M with bounded

coefficeints (since v ∈ C1(U)) will lead a contradiction that

0 < max{u>0}

w = max∂{u>0}

w =0

v= 0

Hence u ≤ 0 in U .

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13. (Courant-Fischer’s principle)

Proof. First we need to note the same proof of Theorem 2 on Page 356 leads the following

theorem:

Theorem 3. For the symmetric operator L = −(aijuxi)xj on a bounded smooth domain U where

(aij) is symmetric and satisfies the uniform ellipticity. Denote k-th zero Dirichlet eigenvalue

and eigenfunction by λk and ωk respectively, then

λk = min{B[u, u] | u ∈ H10 , ‖u‖L2 = 1, (u, ωi)L2 = 0, 1 ≤ i ≤ k − 1}

Through this theorem, we know the desired identity is true if we replace equality by ≤. Now

we prove the converse:

Given k − 1 dimensional subspace S. By spectral theory for compact self-adjoint operators,

there exists y = Σk1ciωi ∈ span{ω1 · · ·ωk} with ‖y‖L2 = 1, such that y ∈ S⊥, so

min{B[u, u] | u ∈ S⊥, ‖u‖L2 = 1} ≤ B[y, y] = Σk1|ci|2λi ≤ Σk

1|ci|2λk ≤ λk.

Since S is arbitrary chosen, we are done.

14. Proof. Let Y := {u ∈ C∞(U) | u > 0 in U, u = 0 on ∂U}. Let u1 > 0 be the zero Dirichlet

eigenfunction with respect to the eigenvalue λ1 which exists by Krein-Rutman theorem (Page

361, Theorme 3). By classical Schauder’s theory, we know u1 ∈ Y and therefore supu∈Y

infx∈U

Luu≥ λ1.

Given v ∈ Y , then w = v − u1 ∈ Y . Note that

Lv

v=L(u1 + w)

u1 + w= λ1 +

L(w)− λ1w

u1 + w

Consider the adjoint operator

L∗w = ∂ij(aijw)− ∂i(biw) + cw = aij∂ijw + [∂j(a

ij + aji)− ∂ibi]∂iw + (c+ ∂ijaij − ∂ibi)w,

the Krein-Rutman theorem tells us that, for m > supU |c+∂ijaij−∂ibi|, there exists λ∗1, such that

λ∗1 +m is the simple principle zero Dirichlet eigenvalue of L∗+m with positive eigenfunction u∗1.

Since λ∗1(u∗1, u1)L2 = (L∗u∗1, u1)L2 = (u∗1, Lu1)L2 = λ1(u∗1, u1)L2 , the fact that u∗1, u1 > 0 implies

λ1 = λ∗1. Consider

(L(w)− λ1w, u∗1) = λ∗1(w, u∗1)− λ1(w, u∗1) = 0

Due to u∗1 > 0, there are only two possibilities:

(1) If L(w)− λ1w ≡ 0. Since λ1 is simple, w = u1. Then Lvv

= L(2u1)2u1

= λ1.

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(2) If L(w) − λ1w changes sign in U . Then there exist x ∈ U such that (L(w) − λ1w)(x) < 0

and hence L(v)(x)v(x)

= λ1 + (L(w)−λ1w)(x)(u1+w)(x)

< λ1. Since v is arbitrary, supu∈Y

infx∈U

Luu≤ λ1.

Hence supu∈Y

infx∈U

Luu

= λ1.

15. Proof. (We use Reynolds transport theorem without mentions. )

λ(τ) = λ(τ)

∫U(τ)

|ω(x, τ)|2 =

∫U(τ)

−(∆xω)(x, τ)ω(x, τ)

=

∫U(τ)

|∇xω(x, τ)|2.

Differentiate the original equation and ‖w‖L2(Uτ ) = 1 with respect to τ , we get

−∆xωτ = λωτ + λω

0 =

∫∂U(τ)

ω2v · ν +

∫U(τ)

(ω2)τ = 0 +

∫U(τ)

(ω2)τ ,

Hence,

λ(τ) =

∫∂U(τ)

|∇xω|2v · ν +

∫U(τ)

(|∇xω|2)τ

=

∫∂U(τ)

|∂ω∂ν|2v · ν +

∫U(τ)

2∇xω∇xωτ , Since u = 0 on ∂U.

=

∫∂U(τ)

|∂ω∂ν|2v · ν +

∫U(τ)

2ω(−∆xωτ )

=

∫∂U(τ)

|∂ω∂ν|2v · ν +

∫U(τ)

2ω(λωτ + λω)

=

∫∂U(τ)

|∂ω∂ν|2v · ν + λ · 0 + 2λ

16. Proof. Direct computations tell us e−iσtv := eiσ(ω·x−t) satisfies the wave equation utt−σ2∆u = 0

and r(∂v∂r− iσv) = iσ(x · ω − |x|)v which did not tend to 0 as |x| → 0. On the other hand,

r(∂Φ∂r−iσΦ) = −Φ which tends to 0 as |x| → ∞. The proof to show Φ is a fundamental solution

is almost the same as usual Laplacian case. Try to mimic the proof Evans gives on page 24.

17. I think we need to assume w = O(R−1). and the second expression Evans gives in this problem

is wrong, the integrand on the left-hand side should be |wr|2 + σ2|w|2 − 2σIm(wwr).

Proof. Given x0 ∈ R3, let BR denote the ball centered at x0 with radius R and Φx0(x) :=

Φ(x− x0), where Φ is defined in Exercise 16 (b). Hence

w(x0) =

∫BR

[(∆ + σ2)Φx0

]w −

[(∆ + σ2)w

]Φx0 =

∫∂BR

∂Φx0

∂rw − ∂w

∂rΦx0 (3)

=

∫∂BR

[∂Φx0

∂r− iσΦx0

]w −

[∂w∂r− iσw

]Φx0 (4)

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the magnitude of the integrand is o(R−1) ∗O(R−1), so its integral tends to 0 as R→∞.

Remark 4. Further results for Radiation condition on general dimension N ≥ 2 can be found

in G. Eskin, Lectures on Linear PDEs, Section 19. (Need some knowledges on stationary phase

formula.)

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