Authors: Joe Benson, Denis Bashkirov, Minsu Kim, Helen Li, Alex Csar

Evans PDE Solutions, Chapter 2

Joe: 1, 2,11; Denis: 4, 6, 14, 18; Minsu: 2,3, 15; Helen: 5,8,13,17. Alex:10, 16

Problem 1. Write down an explicit formula for a function u solving the initial-value problem{ut + b · Du + cu = 0 on Rn × (0,∞)

u = g on Rn × {t = 0}

Here c ∈ R and b ∈ Rn are constants.

Sol: Fix x and t, and consider z(s) := u(x + bs, t + s)Then

z(s) = b · Du + ut

= −cu(x + bs, t + s)= −cz(s)

Therefore, z(s) = De−cs, for some constant D. We can solve for D by letting s = −t. Then,

z(−t) = u(x − bt, 0)= g(x − bt)

= Dect

i.e. D = g(x − bt)e−ct

Thus, u(x + bs, t + s) = g(x − bt)e−c(t+s)

and so when s = 0, we get u(x, t) = g(x − bt)e−ct . �

Problem 2. Prove that Laplace’s equation ∆u = 0 is rotation invariant; that is, if O is an orthogonaln × n matrix and we define

v(x) := u(Ox) (x ∈ R)then ∆v = 0.

Solution:Let y := Ox, and write O = (ai j). Thus,

v(x) = u(Ox)= u(y)

where y j =∑n

i=1 a jixi. This then gives that

∂v∂xi

=

n∑j=1

∂u∂y j

∂y j

∂xi

=

n∑j=1

∂u∂y j

a ji

1

2

Thus, ∂v∂x1...∂v∂xn

=

a11 . . . an1...

...a1n . . . ann

∂u∂y1...∂u∂yn

= OT

∂u∂y1...∂u∂yn

Dx · v = OT Dy · u

Now,

∆v = Dxv · Dxv

= (OT Dyu) · (OT Dyu)

= (OT Dyu)T OT Dyu

= (Dyu)T (OT )T OT Dyu

= (Dyu)T OOT Dyu

= (Dyu)T Dyu because O is orthogonal= (Dyu) · (Dyu)= ∆u(y)= 0

Problem 3. Modify the proof of the mean value formulas to show for n ≥ 3 that

u(0) =1

nα(n)rn−1

∫∂B(0,r)

gdS +1

n(n − 2)α(n)

∫B(0,r)

( 1|x|n−2 −

1rn−2

)f dx,

provided −∆u = f in B0(0, r)u = g on ∂B(0, r).

Solution: Set

φ(t) =1

nα(n)tn−1

∫∂B(0,t)

u(y)dS (y), 0 ≤ t < r,

and

φ(r) =1

nα(n)rn−1

∫∂B(0,r)

u(y)dS (y) =1

nα(n)rn−1

∫∂B(0,r)

gdS .

Then,

φ′(t) =tn

( 1α(n)tn

∫B(0,t)

∆u(y)dy)

=tn

( 1α(n)tn

∫B(0,t)− f dy

)=

−1α(n)tn−1

∫B(0,t)

f dy.

(See the proof of Thm2)

3

Let ε > 0 be given.

(1) φ(ε) = φ(r) −∫ r

ε

φ′(t)dt =1

nα(n)rn−1

∫∂B(0,r)

gdS −∫ r

ε

φ′(t)dt.

Using integration by parts, we compute

−

∫ r

ε

φ′(t)dt =

∫ r

ε

1nα(n)tn−1

∫B(0,t)

f dydt

=1

nα(n)

∫ r

ε

1tn−1

∫B(0,t)

f dydt

=1

nα(n)

([ 12 − n

1tn−2

∫B(0,t)

f dy]r

ε−

∫ r

ε

12 − n

1tn−2

∫∂B(0,t)

f dS dt)

=1

n(n − 2)α(n)

( ∫ r

ε

1tn−2

∫∂B(0,t)

f dS dt −1

rn−2

∫B(0,r)

f dy +1εn−2

∫B(0,ε)

f dy)

=:1

n(n − 2)α(n)

(I −

1rn−2

∫B(0,r)

f dy + J).

Observe thatJ :

1εn−2

∫B(0,ε)

f dy ≤ C · ε2, for some constant C > 0

and ∫B(0,ε)

1|x|n−2 f (x)dx =

∫ r

0dt

∫∂B(0,t)

1tn−2 f dS .

As ε → 0, I + J →∫

B(0,ε)1|x|n−2 f (x)dx. Thus,

limε→0−

∫ r

ε

φ′(t)dt =1

n(n − 2)α(n)

( ∫B(0,r)

1|x|n−2 f (x)dx −

1rn−2

∫B(0,r)

f dy)

=1

n(n − 2)α(n)

∫B(0,r)

( 1|x|n−2 −

1rn−2

)f dx.

Therefore, letting ε → 0, we have from (1)

u(0) = φ(0) =1

nα(n)rn−1

∫∂B(0,r)

gdS +1

n(n − 2)α(n)

∫B(0,r)

( 1|x|n−2 −

1rn−2

)f dx.

�

Problem 4. We say v ∈ C2(U) is subharmonic if

−∆v ≤ 0 in U.

(a) Prove for subharmonic v that

v(x) ≤?

B(x,r)v dy for all B(x, r) ⊂ U.

(b) Prove that therefore maxU v = max∂U v.(c) Let φ : R → R be smooth and convex. Assume u is harmonic and v := φ(u). Prove v is

subharmonic.

4

(d) Prove v := |Du|2 is subharmonic, whenever u is harmonic.

Solution.

(a) As in the proof of Theorem 2, set φ(r) :=>∂B(x,r)

v dS (y) and obtain

φ′(r) =rn

?B(x,r)

∆v(y)dy ≥ 0.

For 0 < ε < r, ∫ r

ε

φ′(s)ds = φ(r) − φ(ε) ≥ 0.

Hence, φ(r) ≥ limε→0

φ(ε) = v(x). Therefore,?B(x,r)

v dy =1

α(n)rn

∫B(x,r)

v dy =1

α(n)rn

∫ r

0

(∫∂B(x,s)

v(z) dS (z))

ds

=1

α(n)rn

∫ r

0nα(n)sn−1φ(s) ds ≥

1rn

∫ r

0nsn−1v(x) ds = v(x)

(b) We assume that U ⊂ Rn is open and bounded. For a moment, we assume also that U isconnected. Suppose that x0 ∈ U is such a point that v(x0) = M := maxU v. Then for0 < r < dist(x0, ∂U),

M = v(x0) ≤?

B(x0,r)v dy ≤ M.

Due to continuity of v, an equality holds only if v ≡ M within B(x0, r). Therefore, the setu−1({M}) ∩ U = {x ∈ U |u(x) = M} is both open and relatively closed in U. By the connect-edness of U, v is constant within the set U. Hence, it is constant within U and we concludethat maxU v = max∂U v.

Now let {Ui|i ∈ I} be the connected components of U. Pick any x ∈ U and find j ∈ Isuch that x ∈ U j. We obtain

v(x) ≤ maxU j

v = max∂U j

v ≤ max∂U

v

and conclude that maxU v = max∂U v.(c) For x = (x1, ..., xn) ∈ U and 1 ≤ i, j ≤ n,

∂2v∂xi∂x j

(x) =∂2

∂xi∂x jφ(u(x)) = φ′′(u(x)) ·

∂u∂xi

(x) ·∂u∂x j

(x) + φ′(u(x)) ·∂2u∂xi∂x j

(x).

Since φ is convex, then φ′′(x) ≥ 0 for any x ∈ R. Recall that u is harmonic and obtain

∆v = φ′′(u) ·n∑

i=1

(∂u∂xi

)2

+ ∆u = φ′′(u) ·n∑

i=1

(∂u∂xi

)2

≥ 0.

(d) We set v := |Du|2 =n∑

k=1

(∂u∂xk

)2. For x = (x1, ..., xn) ∈ U and 1 ≤ i, j ≤ n,

∂2v∂xi∂x j

(x) = 2n∑

k=1

[∂2u∂xi∂xk

(x) ·∂2u∂xi∂x j

(x) +∂u∂xk

(x) ·∂3u

∂xi∂x j∂xk(x)

].

5

Therefore,

∂2v∂xi

2 = 2n∑

k=1

( ∂2u∂xi∂xk

)2

+∂u∂xk·∂

∂xk

(∂2u∂xi

2

) ,∆v = 2

∑1≤i,k≤n

(∂2u∂xi∂xk

)2

+

n∑k=1

∂u∂xk·∂

∂xk

(∆u

)= 2

∑1≤i,k≤n

(∂2u∂xi∂xk

)2

≥ 0.

�

Problem 5: Prove that there exists a constant C, depending only on n, such that

maxB(0,1)|u| ≤ C

(max∂B(0,1)

|g| + maxB(0,1)| f |

)whenever u is a smooth solution of− 4 u = f in B0(0, 1)

u = g on ∂B(0, 1).

Proof: Let M := maxB(0,1) | f |, then we define v(x) = u(x) + M2n |x|

2 and w(x) = −u(x) + M2n |x|

2. Wefirst consider v(x) . Note that

− 4 v = − 4 u − M = f − M ≤ 0.

So, v(x) is a subharmonic funcion.From Problem 4 (b), we have

maxB(0,1)

v(x) = max∂B(0,1)

v(x) ≤ max∂B(0,1)

|g| +M2n.

That is

maxB(0,1)

u(x) ≤ maxB(0,1)

v(x) ≤ max∂B(0,1)

|g| +1

2nmaxB(0,1)| f |.

Then, for w(x), we have

− 4 w = 4u − M = − f − M ≤ 0.

Again, we can get

maxB(0,1)

w(x) = max∂B(0,1)

w(x) ≤ max∂B(0,1)

|g| +M2n.

i.e.

maxB(0,1)−u(x) ≤ max

B(0,1)w(x) ≤ max

∂B(0,1)|g| +

12n

maxB(0,1)| f |.

Combining these two together, we finally proved the problem. �

Problem 6. Use Poisson’s formula for the ball to prove

rn−2 r − |x|(r + |x|)n−1 u(0) ≤ u(x) ≤ rn−2 r + |x|

(r − |x|)n−1 u(0)

whenever u is positive and harmonic in B0(0, r). This is an explicit form of Harnack’s inequality.

6

Solution.Since y ∈ ∂B(0, r), then |x − y| ≤ |x| + r. Therefore,

u(x) =r2 − |x|2

nα(n)r

∫∂B(0,r)

g(y)|x − y|n

dS (y)

≥r2 − |x|2

nα(n)r

∫∂B(0,r)

g(y)(r + |x|)n dS (y) = rn−2 r − |x|

(r + |x|)n−1 ·1

nα(n)rn−1

∫∂B(0,r)

g(y)dS (y)

= rn−2 r − |x|(r + |x|)n−1

?∂B(0,r)

g(y)dS (y) = rn−2 r − |x|(r + |x|)n−1 u(0)

The inequality u(x) ≤ rn−2 r+|x|(r−|x|)n−1 u(0) can be proven in a similar way. �

Problem 7. Prove Poisson’s formula for a ball: Assume g ∈ C(∂B(0, r)) and let

u(x) =r2 − x2

nα(n)r

∫∂B(0,r)

g(y)|x − y|n

dS (y) for x ∈ B0(0, r).

Show that

Proof.

Problem 8.

Let u be the solution of 4u = 0 in Rn+

u = g on ∂Rn+

given by Poisson’s formula for the half-space. Assume g is bounded and g(x) = |x| for x ∈ ∂Rn+,

|x| le1. Show Du is not bounded near x = 0. (Hint: Estimate u(λen)−u(0)λ

.)

Proof: From formula (33) on page 37, we have

u(x) =2xn

nα(n)

∫∂Rn

+

g(y)|x − y|n

dy,

and u(0) = g(0) = 0. Thus, using hint, we get

u(λen) − u(0)λ

=2

nα(n)

∫∂Rn

+

g(y)|λen − y|n

dy

=2

nα(n)

∫|y|≤1

⋂∂Rn

+

g(y)|λen − y|n

dy +2

nα(n)

∫|y|>1

⋂∂Rn

+

g(y)|λen − y|n

dy

Taking absolute value on both sides, we have∣∣∣u(λen) − u(0)λ

∣∣∣ ≥ ∣∣∣ 2nα(n)

∫|y|≤1

⋂∂Rn

+

g(y)|λen − y|n

dy∣∣∣ − 2

nα(n)

∫|y|>1

⋂∂Rn

+

|g(y)||λen − y|n

dy

=I1 − I2.

7

Since g is bounded, so it is obvious that I2 is bounded and independent of λ. For I1, in this case,g(y) = |y|, so

I1 =2

nα(n)

∫|y|≤1

⋂∂Rn

+

|y||λen − y|n

dy

≥2

nα(n)

∫|y|≤1

⋂∂Rn

+

|y|(λ + |y|)n dy

Note that for fixed y, |y|(λ+|y|)n is increasing when λ is decreasing to 0, so by Monotone Convergence

theorem, we have

limλ→0

2nα(n)

∫|y|≤1

⋂∂Rn

+

|y|(λ + |y|)n dy

=

∫|y|≤1

⋂∂Rn

+

|y||y|n

dy

=

∫Bn−1(0,1)

|y||y|n

dy

=

∫ 1

0dr

∫∂Bn−1(0,r)

1|y|n−1 dS (y) = C

∫ 1

0

1rn−1 rn−2dr = ∞.

So, Du is unbounded near x = 0. �

Problem 10.

Suppose u is smootha nd solves ut − ∆u = 0 in Rn × (0,∞).(i) Show uλ(x, t) := u(λx, λ2t) also solves the heat equation for each λ ∈ R.

(ii) Use (i) to show v(x, t) := x · Du(x, t) + 2tut(x, t) solves the heat equation as well.

(i) uλt(x, t) = λ2ut(λx, λ2t) and uλxi(x, t) = λu(λx, λ2t) for each i. Then uλxi xi(x, t) = λ2uxi(λx, λ2t).Consequently, ∆uλ = λ2∆u and uλt − ∆uλ = λ2(ut − ∆u), so uλ solves the heat equation forall λ ∈ R.

(ii) We differentiate u(λx, λ2t) = u(λx1, . . . , λxn, λ2t) with respect to λ we get∑

k

xkuxk(λx1, . . . , λxk, λ2t) + 2λtut(λx1, . . . , λxn, λ

2t) = x · D(λx, λ2t) + 2tut(λx, λ2t).

Taking λ = 1, we then have that v(x, t) = x · Du(x, t) + 2tut(x, t). u is smooth, so the secondderivatives of u(λx, λ2t) are continuous, meaning the mixed partials are equal. Therefore,vt −∆v = ∂

∂t∂λu(λx, λ2t)−∆ ∂∂λ

u(λx, λ2t) = ∂∂λ∂t u(λx, λ2t)− ∂

∂λ∆u(λx, λ2t) = ∂

∂λ(uλt −∆uλ) = 0,

since uλ satisfies the heat equation for all λ. Thus v does as well.

Problem 11: Assume n = 1 and u(x, t) = v( x2

t ).a) Show

ut = uxx

if and only if

(2) 4zv”(z) + (2 + z)v′(z) = 0 (z > 0)

8

b) Show that the general solution of (1) is

v(z) = c∫ z

0e−s/4s−1/2ds + d

c) Differentiate v( x2

t ) with respect to x and select the constant c properly, so as to obtain the funda-mental solution Φ for n = 1.

Solution:a) Assume that ut = uxx. Then

ut = −x2

t2 v′(

x2

t

)and

uxx = 2v′(

x2

t

)+ 4x2v′′

(x2

t

)So ut = uxx implies that

−x2

t2 v′(

x2

t

)= 2v′

(x2

t

)+ 4x2v′′

(x2

t

)or

4x2

t2 v′′(

x2

t

)+

(2t

+x2

t2

)v′

(x2

t

)= 0

If we let z = x2

t , we get4zt

v′′(z) +

(2t

+zt

)v′(z) = 0

Multiplying this equation by t gives the desired equality.For the other direction, reverse the steps, and hence our proof is done.

b)4zv′′ + (2 + z)v′ = 0

=⇒v′′

v′= −

12

1z−

14

=⇒

(by integrating) log(v′) = − log√

z −z4

+ c

=⇒

v′ = Cz−1/2e−z/4

=⇒

v = C∫ z

0e−s/4s−1/2ds + d

9

as is desired.

c)

v(z) = c∫ z

0e−s/4s−1/2ds + d

=⇒

v(

x2

t

)= c

∫ x2t

0e−s/4s−1/2ds + d

=⇒

v′(

x2

t

)= c

2xt

e−x24t

(x2

t

)−1/2

or

v′(

x2

t

)=

2c√

te−

x24t

Now we want to integrate over R and set the integral equal to 1. Thus we get

1 =2c√

t

∫ ∞

∞

e−x24t dx

Letting y = x√

4t, we get dy = (4t)−1/2dx and substituting, we get

1 =2c√

t

∫ ∞

∞

√4te−y2

dy

or

1 = 4c∫ ∞

∞

e−y2dy

Employing the identity∫ ∞∞

e−y2dy =

√π and solving for c, we get

c =1

4√π

Thus,

Φ(x, t) : = v′(

x2

t

)=

2c√

te−

x24t

=1

2√πt

e−x24t

is easily shown to solve the equationΦt = Φxx

�

Problem 12. Write down an explicit formula for a solution ofut − ∆u + cu = f in Rnx(0,∞)u = g on Rnx{t = 0},

where c∈ R.

10

Solution: Set v(x, t) = u(x, t)eCt. Then, vt = uteCt + CeCtu and vxi xi = uxi xieCt.

⇒

vt − ∆v = uteCt + CeCtu − eCt∆u= eCt(ut − ∆u + Cu)= eCt f .

So, v is a solution of vt − ∆v = eCt f in Rnx(0,∞)v = g on Rnx{t = 0},

By (17) (p.51),

v(x, t) =

∫Rn

Φ(x − y, t)g(y)dy +

∫ t

0

∫Rn

Φ(x − y, t − s)eCs f (y, s)dyds

where Φ is the fundamental solution of the hear equation. Since v(x, t) = u(x, t)eCt, we have

u(x, t) = eCt( ∫Rn

Φ(x − y, t)g(y)dy +

∫ t

0

∫Rn

Φ(x − y, t − s)eCs f (y, s)dyds).

�

Problem 13: Given g : [0,∞]→ R, with g(0) = 0, derive the formula

u(x, t) =x√

4π

∫ t

0

1(t − s)3/2 e

−x24(t−s) g(s)ds, x > 0

for a solution of the initial/boundary-value problemut − uxx = 0 inR+ × (0,∞)

u = 0 onR+ × {t = 0},u = g on{x = 0} × [0,∞).

Proof. We define

v(x, t) =

u(x, t) − g(t) x > 0,−u(−x, t) + g(t) x ≤ 0.

So, we have

vt(x, t) =

ut(x, t) − g′(t) x > 0,−ut(−x, t) + g′(t) x ≤ 0,

and

vxx(x, t) =

uxx(x, t) x > 0,−uxx(−x, t) x ≤ 0.

11

Hence, vt(x, t) − vxx(x, t) =

−g′(t) x > 0,g′(t) x ≤ 0.

v(x, 0) = 0,v(0, t) = 0.

By formula (13) on page 49, we get

v(x, t) =

∫ t

0

1√

4π(t − s)

{∫ 0

−∞

e−(y−x)24(t−s) g′(s)dyds −

∫ ∞

0e−(y−x)24(t−s) g′(s)dyds

}Note that(page 46 Lemma) ∫ ∞

−∞

1√

4π(t − s)e−(y−x)24(t−s) dy = 1,

so when x > 0, we let y − x = −z and obtain

u(x, t) = v(x, t) + g(t)

= v(x, t) +

∫ t

0g′(s)ds

∫ ∞

−∞

1√

4π(t − s)e−(y−x)24(t−s) dy

= 2∫ t

0

1√

4π(t − s)−

12

∫ 0

−∞

e−(y−x)24(t−s) dy g′(s)ds

=

∫ t

0

1√π

(t − s)−12

∫ ∞

xe−z2

4(t−s) dz dg(s)

Integrating by parts, we get

u(x, t) =1√π

(t − s)−1/2∫ ∞

xe−z2

4(t−s) dz g(s)|s=ts=0

−

∫ t

0g(s)

1√π

12

(t − s)−3/2ds∫ ∞

xe−z2

4(t−s) dz

−

∫ t

0g(s)

1√π

(t − s)−1/2ds∫ ∞

xe−z2

4(t−s)−z2

4(t − s)2 dz

= I1 −

∫ t

0g(s)

1√π

12

(t − s)−3/2ds∫ ∞

xe−z2

4(t−s) dz

+

∫ t

0g(s)

1√π

(t − s)−1/2ds∫ ∞

x

−z2(t − s)

de−z2

4(t−s)

= I1 −

∫ t

0g(s)

1√π

12

(t − s)−3/2ds∫ ∞

xe−z2

4(t−s) dz

+

∫ t

0g(s)

1√

4π(t − s)−3/2ds (−z) e

−z24(t−s) |z=∞z=x

+

∫ t

0g(s)

1√π

12

(t − s)−3/2ds∫ ∞

xe−z2

4(t−s) dz

= I1 +x√

4π

∫ t

0

1(t − s)3/2 e

−x24(t−s) g(s)ds.

12

Now, we focus on I1 and define w2 to be z2

4ε ,

I1 = limε→0+

1√πε−1/2

∫ ∞

xe−z24ε dz g(t − ε)

= g(t) limε→0+

1√π

∫ ∞

x2/4ε2e−w2

dw = 0.

Thus, we proved

u(x, t) =x√

4π

∫ t

0

1(t − s)3/2 e

−x24(t−s) g(s)ds, x > 0.

Next, we need to show thatlimx→0+

u(x, t) = g(t).

Note that for any fixed δ > 0.

limx→0+

u(x, t) = limx→0+

x√

4π

∫ t

t−δ

1(t − s)3/2 e

−x24(t−s) g(s)ds

+ limx→0+

x√

4π

∫ t−δ

0

1(t − s)3/2 e

−x24(t−s) g(s)ds

= g(t) limx→0+

x√

4π

∫ t

t−δ

1(t − s)3/2 e

−x24(t−s) ds

= g(t) limx→0+

x√

4π

∫ δ

0

1s3/2 e

−x24s ds

For fixed x, we let s = x2/w2 and get

limx→0+

u(x, t) = g(t) limx→0+

x2√π

∫ x2/δ

∞

w3

x3 e−w2

4−2x2

w3 dw

= g(t) limx→0+

1√π

∫ ∞

x2/δ

e−w2

4 dw

= g(t)1√π

∫ ∞

0e−w2

4 dw = g(t).

Hence, we are done. �

Problem 14. We say v ∈ C21(UT ) is a subsolution of the heat equation if

vt − ∆v ≤ 0 in UT .

(a) Prove for a subsolution v that

v(x, t) ≤1

4rn

∫ ∫E(x,t;r)

v(y, s)|x − y|2

(t − s)2 dyds

for all E(x, t; r) ⊂ UT .(b) Prove that therefore maxUT v = maxΓT v

13

Solution.

(a) We may well assume upon translating the space and time coordinates that x = 0 and t = 0.As in the proof of Theorem 3, set

φ(r) :=1rn

∫ ∫E(r)

v(y, s)|y|2

s2 dyds,

ψ(y, s) := −n2

log(−4πs) +|y|2

4s+ n log r

and derive

φ′(r) ≥1

rn+1

∫ ∫E(r)−4n∆vψ −

2ns

n∑i=1

vyiyi dyds

=

n∑i=1

1rn+1

∫ ∫E(r)

4nvyiψyi −2ns

vyiyi dyds = 0.

For 0 < ε < r,r∫

ε

φ′(z)dz = φ(r) − φ(ε) ≥ 0.

Hence, φ(r) ≥ limε→0

φ(ε) = v(0, 0) · limε→0

1εn

∫ ∫E(ε)

|y|2

s2 dyds = 4v(0, 0), and the statement follows.(b) Suppose there exists a point (x0, t0) ∈ UT with u(x0, t0) = M := maxUT u. Then for all

sufficiently small r > 0, E(x0, t0; r) ⊂ UT . Using the result proved above, we deduce

M = v(x0, t0) ≤1

4rn

∫ ∫E(x0,t0;r)

v(y, s)|x − y|2

(t − s)2 dyds ≤ M,

since

1 =1

4rn

∫ ∫E(x0,t0;r)

|x0 − y|2

(t0 − s)2 dyds.

Conclude that u|E(x0,t0;r) = M. The argument used in the proof of Theorem 4 will finish theproof.

�

Problem 15.(a) Show the general solution of the PDE uxy = 0 is

u(x, y) = F(x) + G(y)

for arbitrary functions F,G.(b) Using the change of variables ξ = x + t, η = x − t, show utt − uxx = 0 if and only if uξη = 0.(c) Use (a),(b) to rederive d’Alembert’s formula.

Solution:(a)uxy = 0⇒ ux = f (x)⇒ u(x, y) =

∫f (x)dx + G(y)

uyx = 0⇒ uy = g(y)⇒ u(x, y) =∫

g(y)dy + F(x)

14

This implies u(x, y) = F(x) + G(y).

(b)x =

ξ+η

2 , y =ξ−η

2Define u := u

( ξ+η2 ,

ξ−η

2

)uξ =

12

ux +12

ut and uξη =14

uxx −14

uxt +14

utx −14

utt =14

(uxx − utt)

Hence, uξη = 0⇔ utt − uxx = 0.

(c)By (b), utt − uxx = 0⇒ uξη = 0, and u(ξ, η) = F(ξ) + G(η) by (a),i.e, u(x, y) = F(x + t) + G(x − t).Since u(x, 0) = g, ut(x, 0) = h,

(3) u(x, 0) = F(x) + G(x) = g(x),

ut(x, 0) = F′(x) −G′(x) = h(x)Integration⇒

(4) F(x) −G(x) =

∫ x

0h(y)dy + C, C:constant.

(2) + (3); F(x) =12(g(x) +

∫ x

0h(y)dy + C

)(2) − (3); G(x) =

12(g(x) −

∫ x

0h(y)dy −C

)Thus,

u(x, y) = F(x + t) + G(x − t) =12(g(x + t) +

∫ x+t

0h(y)dy + C

)+

12(g(x − t) −

∫ x−t

0h(y)dy −C

)=

12(g(x + t) +

∫ x+t

0h(y)dy + C + g(x − t) +

∫ 0

x−th(y)dy −C

)=

12[g(x + t) + g(x − t)

]+

12

∫ x+t

x−th(y)dy (x ∈ R, t ≥ 0).

�

Problem 16.

Assume E = (E1, E2, E3) and B = (B1, B2, B3) solve Maxwell’s equations:

Et = curl BBt = − curl E

div B = div E = 0

Show that utt − ∆u = 0 where u = Bi or Ei for i = 1, 2, 3.

Solution.

15

curl(curl E) = curl(−Bt)

=

(−∂2B3

∂y∂t+∂2B2

∂z∂t,−∂2B3

∂x∂t+∂2B1

∂z∂t,−∂2B2

∂x∂t+∂B1

∂y∂t

)= −

∂

∂tcurl B

= −∂

∂tEt

= −∂2E∂t2

However, we also know that curl(curl E) = ∇(div E) − ∇2E = −∇2E. Then Ei satisfies utt − ∆u = 0for i = 1, 2, 3.

Similarly, curl(curl B) = curl Et = −∂2B∂t2 , and curl(curl B) = ∇(div B) − ∇2B = −∇2B, so Bi satisfies

utt − ∆u = 0 for i = 1, 2, 3.

Problem 17.(Equipartition of energy) Let u ∈ C2 (R × [0,∞)) solve the initial value problem forthe wave equation in one dimension:utt − uxx = 0 in R × (0,∞)

u = g; ut = h on R × {t = 0}.

Suppose g, h have compact support. The kinetic energy is k(t) := 12

∫ ∞−∞

u2t (x, t)dx and the potential

energy is p(t) := 12

∫ ∞−∞

u2x(x, t)dx. Prove

(i) k(t) + p(t) is constant in t.(ii) k(t) = p(t) for all large enough times t.

Proof. (i.) We define e(t) = k(t) + p(t) = 12

∫ ∞−∞

(u2

t + u2x

)dx. Since g, h have compact support, so

we haved e(t)

dt=

12

∫ ∞

−∞

2ututt + 2uxuxt dx∫ ∞

−∞

ututt dx −∫ ∞

−∞

uxxut dx

=

∫ ∞

−∞

ut (utt − uxx) dx = 0.

Hence, e(t) ≡ e(0).(ii.)By d’Alembert’s formula on page 68, we have

u(x, t) =12

[g(x + t) + g(x − t)

]+

12

∫ x+t

x−th(y)dy.

So,

ut =12

[g′(x + t) − g′(x − t)

]+

12

[h(x + t) + h(x − t)] ,

andux =

12

[g′(x + t) + g′(x − t)

]+

12

[h(x + t) − h(x − t)] .

16

We assume that there exists a positive constant M so that [−M,M] ⊇ supp(g′) and [−M,M] ⊇supp(h).Note that for a fixed t > M, −M ≤ x − t ≤ M ⇔ 0 < t − M ≤ x ≤ t + M and −M ≤ x + t ≤ M ⇔−t − M ≤ x ≤ −t + M < 0.Thus, when t > M :

(a) 0 < t − M ≤ x ≤ t + M.Then we have

h(x + t) = g(x + t) = 0.

So,

u2t =

14

g′(x − t)2 +14

h(x − t)2 −12

g′(x − t)h(x − t) = u2x.

(b) −t − M ≤ x ≤ −t + M < 0.Then,

u2t =

14

g′(x + t)2 +14

h(x + t)2 +12

g′(x + t)h(x + t) = u2x.

(c) Otherwise

g′(x + t) = g′(x − t) = h(x + t) = h(x − t) = 0.

So, combining all the cases, it is obvious that when t > M, k(t) = p(t). �

Problem 18. Let u solve {utt − ∆u = 0 in R3 × (0,∞)u = g, ut = h on R3 × {t = 0},

where g, h are smooth and have compact support. Show there exists a constant C such that

|u(x, t)| ≤ C/t (x ∈ R3, t > 0).

Solution.

From the conditions it follows that there exist R,M > 0 such that spt g, spt h ⊂ B(0,R) andg(y) ≤ M, |Dg(y)| ≤ M, h(y) ≤ M for any y ∈ R3. Kirchhoff’s formula gives the solution of theinitial-value problem:

u(x, t) =

?∂B(x,t)

th(y) + g(y) + Dg(y) · (y − x) dS (y).

Denote by Σ the intersection ∂B(x, t) ∩ B(0,R). Observe that the area of Σ is not greater than thearea of the sphere ∂B(0,R). Then, for t > 0, we obtain∣∣∣∣∣∣

?∂B(x,t)

th(y) + Dg(y) · (y − x) dS (y)

∣∣∣∣∣∣ =1

4πt2

∣∣∣∣∣∣∫∂B(x,t)∩B(0,R)

th(y) + Dg(y) · (y − x) dS (y)

∣∣∣∣∣∣≤

14πt2

∫∂B(x,t)∩B(0,R)

t · |h(y)| + |Dg(y)| · |y − x| dS (y)

≤1

4πt2 · 4πR2 · (tM + tM) =2R2M

t.

17

For t > 1, using the same argument, we get∣∣∣∣∣∣?∂B(x,t)

g(y) dS (y)

∣∣∣∣∣∣ =1

4πt2

∣∣∣∣∣∣∫∂B(x,t)∩B(0,R)

g(y) dS (y)

∣∣∣∣∣∣ ≤ 14πt2 · 4πR2 · M =

R2Mt2 ≤

R2Mt.

Notice now that the area Σ is not greater than the area of the sphere ∂B(x, t). Then for 0 < t ≤ 1,∣∣∣∣∣∣?∂B(x,t)

g(y) dS (y)

∣∣∣∣∣∣ =1

4πt2

∣∣∣∣∣∣∫∂B(x,t)∩B(0,R)

g(y) dS (y)

∣∣∣∣∣∣ ≤ 14πt2 · 4πt2 · M ≤

Mt.

Without loss of generality, we can take R > 1. Then, combining the estimates obtained above, weconclude |u(x, t)| ≤ 3R2 M

t . �

Evans PDE Solutions, Chapter 5

Alex: 4, Helen: 5, Rob H.: 1

Problem 1.

Suppose k ∈ {0, 1, . . .}, 0 < γ < 1. Prove Ck,γ(U) is a Banach space.

Solution:

1. First we show that || · ||Ck,γ(U) is a norm, where we recall that

||u||Ck,γ(U) =∑|α|≤k

||Dαu||C(U) +∑|α|=k

[Dαu]C0,γ(U) ,

and

[u]C0,γ(U) = supx,y∈U

{|u(x) − u(y)||x − y|γ

}.

For the sake of opaqueness we now omit subscripts on all norms unless it is unclear from context.

2. For any λ ∈ R we have first

[λu] = supx,y∈U

|λu(x) − λu(y)||x − y|γ

= |λ| supx,y∈U

|u(x) − u(y)||x − y|γ

= |λ| [u] ,

and certainly||Dα(λu)||C(U) = ||λDαu|| = |λ| · ||Dαu|| .

So

||λu|| =∑|α|≤k

||Dα(λu)|| +∑|α|=k

[Dα(λu)]

= |λ|∑|α|≤k

||Dαu|| + |λ|∑|α|=k

[Dαu]

= |λ| · ||u|| .

3. If u = 0 it is obvious that ||u|| = 0. On the other hand, ||u|| = 0 implies that

||Dαu||C(U) = 0

18

for every |α| ≤ k. In particular this is true for α = 0 so that the supremum of D0u = u on U is 0, i.e.u ≡ 0.

4. Finally we must prove the triangle inequality. We know the triangle inequality is true for the supnorm || · ||C(U). We can also see that for any α which makes sense

[Dα(u + v)] = [Dαu + Dαv] ≤ [Dαu] + [Dαv] .

Therefore we can easily conclude

||u + v|| =∑|α|≤k

||Dα(u + v)|| +∑|α|=k

[Dα(u + v)]

≤∑|α|≤k

(||Dαu|| + ||Dαv||) +∑|α|=k

([Dαu] + [Dαv])

= ||u|| + ||v||.

5. We need only show that Ck,γ(U) is complete. So let {um} be a Cauchy sequence. Then {um(x){is a Cauchy sequence for every x, so define u to be the pointwise limit of the um. Now if V is anybounded subset of U, then V is compact, so that um ⇒ u uniformly on any V . Since the um areuniformly continuous on V by assumption, this implies that u is uniformly continuous on V as well(and so, a fortiori u ∈ C(U)). Therefore u ∈ C(U).

What we would really like would be to have u ∈ Ck(U). But similar arguments show that u hasderivatives Dαu for all |α| ≤ k on U by restricting first to bounded subsets of U to find the derivativesand then using uniform convergence on these subsets to show the derivatives must also be uniformlycontinuous on bounded subsets since the Dαum were.

This leaves us with only showing that the norm of u is finite, so that in fact u ∈ Ck,γ(U). But forevery n we have

||un − u|| =∑|α|≤k

supx∈U|Dαun(x) − Dαu(x)| +

∑|α|=k

supx,y∈U

|Dαun(x) − Dαun(y) − Dαu(x) + Dαu(y)||x − y|γ

= limm⇒∞

∑|α|≤k

supx∈U|Dαun(x) − Dαum(x)| +

∑|α|=k

supx,y∈U

|Dαun(x) − Dαun(y) − Dαum(x) + Dαum(y)||x − y|γ

= lim

m⇒∞||un − um||.

In particular, since {um} is Cauchy there is some N so that n,m ≥ N implies ||un − um|| ≤ 1. Lettingm approach∞, this implies that ||uN − u|| < 1. Now the triangle inequality applies to give

||u|| ≤ ||uN − u|| + ||uN || < 1 + ||uN || < ∞.

�

Problem 4.

Assume U is bounded and U ⊂⊂⋃N

i=1 Vi. Show there exist C∞ functions ζi (i = 1, . . . ,N) such that0 ≤ ζ1 ≤ 1, supp ζi ⊂ Vi i = 1, . . . ,N∑Ni=1 ζi = 1 on U.

The functions {ζi}N1 for a partition of unity.

19

Solution. Assume U is bounded and U ⊂⊂⋃N

i=1 Vi. Without loss of generality, we may assumethat the Vi are open, for if they are not, we can replace Vi by its interior. We note that, since U isbounded, U is compact. Each x ∈ U has a compact neighbourhood Nx contained in Vi for some i.Then {N◦x } is an open cover of U, which then has a finite subcover N◦x1

, . . . ,N◦xn. We now let Fi be

the union of the Nxk contained in Vi. Fi is the compact since it is the finite union of compact sets.The C∞ version of Urysohn’s Lemma (Folland, p.245) allows us to find smooth functions ξ1, . . . , ξN

such that ξi = 1 on Fi and supp(ξi) ⊂ Vi. Since the Fi cover U, U ⊂ {x :∑n

1 ξi(x) > 0} and we canuse Urysohn again to find ζ ∈ C∞ with ζ = 1 on U and supp(ζ) ⊂ {x :

∑n1 ξi(x) > 0}. Now, we let

ξN1 = 1 − ζ, so∑N+1

1 ξi > 0 everywhere. We then take

ζi =ξi∑N+1

1 ξ j

as our partition of unity.

Problem 5 (Helen) Prove that if n = 1 and u ∈ W1,p(0, 1) for some 1 ≤ p < ∞, then u is equal a.e.to an absolutely continuous function, and u′ which exists a.e. belongs to Lp(0, 1).Proof. Since u ∈ W1,p(0, 1), so by definition on page 242 and 244, we have some function v ∈Lp(0, 1) such that ∫

(0,1)u Dφdx = −

∫(0,1)

vφdx, ∀φ ∈ C∞c ((0, 1)) .

Note that v ∈ Lp(0, 1), so by Holder’s inequality, we have ‖v‖L1 ≤ ‖v‖Lp ‖1‖Lq < ∞, which meansv ∈ L1(0, 1). Thus, we can define function f (x) on (0, 1) by the following formula

f (x) = u(12

) +

∫ x

12

v(t)dt, ∀x ∈ (0, 1).

According to the Fundamental Theorem of Calcalus, f is absolutely continuous. Now we willprove u = f a.e.By the definition of f , we have f ′ = v a.e. So for any φ ∈ C∞c ((0, 1)) we get∫

(0,1)f Dφdx = −

∫(0,1)

f ′φdx = −

∫(0,1)

vφdx.

Therefore, ∫(0,1)

( f − u) Dφdx = 0 ∀φ ∈ C∞c ((0, 1)) ,

which means u = f + const. And note that u(12 ) = f ( 1

2 ), hence u = f a.e. So u′ exists a.e. andsatisfy u′ = v a.e., so u′ ∈ Lp(0, 1). �