MAT-INF4300 - Mandatory Assignment
Sindre Froyn
Exercise 1
(a)
We are working with the boundary conditions
∆u = 0 on U
∂u
= g in ∂U.
∂ν
(*)
We assume that there is a function u ∈ C 2 (U) ∩ C(U ) that satisfies the
boundary conditions in (*) and show that
Z
g(x)dS(x) = 0,
∂U
proving it is a necessary condition.
Since we know by (*) that g = ∂u
on the boundary, we can exchange
∂ν
them in the integral.
Z
Z
∂u
(x)dS(x).
(1)
g(x)dS(x) =
∂U
∂U ∂ν
Now we can apply Green’s first identity and use that u ∈ C 2 (U) which means
the Laplacian exists:
Z
Z
∂u
(x)dS(x) =
∆u(x)dx.
(2)
∂U ∂ν
U
We integrate over the open set U and by (*) we know that the Laplacian ∆u
is 0 in U, and integrating the zero function is zero. Hence,
Z
∆u(x)dx = 0.
(3)
U
Combining (1),(2) and (3), we have shown that
Z
g(x)dS(x) = 0
∂U
which proves that it is a necessary condition if (*) is true.
1
(b)
We are going to show the equivalence between the solution u of (*) where we
have u ∈ C 2 (U) ∩ C(U ) and u minimizing the energy given by
Z
Z
1
2
I[w] =
|Dw| dx −
wgdS(x).
2 U
∂U
We begin by defining the set of admissible functions, i.e the set of all functions
w that satisfy the boundary condition.
A = w ∈ C 2 (U) w = g on ∂U .
⇒) We assume u solves (*). Using this assumption and choosing an
arbitrary, admissible w ∈ A, we get,
0=
Z
∆u(u − w)dx =
U
Z X
n
uxi xi (u − w)dx =
U i=1
n Z
X
i=1
uxixi (u − w)dx
U
For each integral i = 1, . . . , n we apply the integration by parts formula.
Z
Z
Z
uxixi (u − w)dx = − uxi (u − w)xi dx +
uxi (u − w)ν i dS
U
U
∂U
We sum these from i = 1 to n.
Z
Z X
n Z
n
X
uxi xi (u − w)dx = −
uxi (u − w)xi dx +
i=1
U
n
X
uxi (u − w)ν i dS
∂U i=1
U i=1
In the first integral on the right side, we see we have the dot product of
the gradients Du and D(u − w), and in the second integral
we have the
1
n
dot product between Du and the vector ν = ν , . . . , ν , and by definition,
Du · ν = ∂u
which by the assumption (*) is equal to g on the boundary ∂U.
∂ν
Exchanging this, we get:
Z
Z
= − Du · D(u − w)dx +
g(u − w)dS
U
=−
Z
∂U
|Du|2 − DuDwdx +
U
Z
gu − gwdS
∂U
So, we have established that
Z
Z
Z
2
0 = − |Du| dx +
DuDwdx +
U
U
∂U
2
gudS −
Z
∂U
gwdS
Z
2
|Du| dx −
U
Z
gudS =
∂U
Z
DuDwdx −
U
Z
(4)
gwdS
∂U
Applying absolute values, the Cauchy-Schwarz inequality and the Cauchy
inequality yields,
1
1
Du · Dw ≤ |DuDw| ≤ |Du||Dw| ≤ |Du|2 + |Dw|2,
2
2
so we can write
Z
Z
DuDwdx −
U
gwdS ≤
∂U
Z
U
1
|Du|2dx +
2
Z
U
1
|Dw|2dx −
2
Z
gwdS
∂U
and combining this with (4) gives:
Z
Z
Z
Z
Z
1
1
2
2
2
|Du| dx −
gudS ≤
|Du| dx +
|Dw| dx −
gwdS
U
∂U
U 2
U 2
∂U
Z
Z
Z
Z
1
1
2
2
|Du| dx −
gudS ≤
|Dw| dx −
gwdS.
∂U
U 2
∂U
U 2
This is true for u and since w was arbitrary, it is also true for all w ∈ A. We
have verified (*) ⇒ I[u] = minw∈A I[w].
⇐) We now assume u is the minimizer of the energy functionals I[w]
and show that this u solves (*). We fix a test function v ∈ Cc∞ (U) and let
ε ∈ R be an arbitrary, small number, then u + εv ∈ A, the admissible set of
functions, and we define
i(ε) = I[u + εv].
By assumption u is the minimizer of the functional I[·], so i(ε) is at its
minimum when ε = 0, and this means ∂i(0)
= i′ (0) = 0. This partial derivative
∂ε
can be quite easily calculated.
Z
Z
1
2
|D(u + εv)| dx −
(u + εv)gdS
i(ε) = I[u + εv] =
2 U
∂U
Z
Z
ε2
1
2
2
|Du| + εDuDv + |Dv| dx −
ug + εvgdS =⇒
=
2
∂U
U 2
Z
Z
′
2
i (ε) =
DuDv + ε|Dv| dx −
vgdS
U
∂U
and setting ε = 0 we get
′
0 = i (0) =
Z
DuDvdx −
U
Z
∂U
3
vgdS.
(5)
We can apply Green’s second identity on the integral over U:
Z
Z
Z
∂u
DuDvdx = − v∆udx +
vdS.
U
U
∂U ∂ν
Setting this into (5) yields,
Z
Z
0=
v∆udx −
Z
∂u
vdS +
vgdS
U
∂U ∂ν
∂U
Z
Z ∂u
0=
v∆udx +
− g vdS
∂ν
U
∂U
Since v is compactly contained in U it is 0 on the boundary and the boundary
integral vanishes. This leaves us with
Z
0=
v∆udx
U
which holds for all v ∈ Cc∞ (U), so ∆u = 0 in U, and the boundary condition
is satisfied since u ∈ A. We have shown ⇐ and completed the proof of the
equivalence.
Exercise 2
(a)
We have,
u(x) =
Z
∞
f (y)Φ(x − y)dy.
−∞
We differentiate with respect to x under the integral sign.
Z ∞
′
u (x) =
f (y)Φ′(x − y)dy
−∞
u′′ (x) =
′′
Z
∞
f (y)Φ′′ (x − y)dy
−∞
−u (x) = −
Z
∞
f (y)Φ′′(x − y)dy
−∞
Now we use the given condition that −Φ′′ = δ, the Dirac delta function, so
we have
Z ∞
′′
−u (x) =
δx f (y)dy = f (x),
−∞
where we used that f is continuous and compact, and the integral of the
Dirac delta function over R is 1.
4
(b)
We are going to determine the value C such that Φ(x) = Φ(−x). First, for
som x > 0 and using Φ(x) = −x+ − Cx − D, we get:
Φ(x) = −x − Cx − D,
Φ(−x) = Cx − D.
Setting these equal each other,
−x − Cx − D = Cx − D
x(−1 − C) = Cx
−1 − C = C
1
C=− .
2
This means we have the form
1
Φ(x) = x − x+ − D.
2
For some a ∈ R we have
Φ(a) =
a
a
−a−D = − −D
2
2
a
a
Φ(−a) = − − 0 − D = − − D,
2
2
so we can write
|x|
− D.
2
A function is homogenous if f (ax) = af (x), so obviously Φ(x) is only homogeneous for D = 0.
Φ(x) = −
(c)
We are going to find Green’s function for the region U = (0, l) for some l ∈ R.
Green’s function is defined as,
G(x, y) := Φ(y − x) − φx (y),
(x 6= y),
(6)
where Φ is the fundamental solution we found in b), and φx (y) is the corrector
function, which is the solution to the boundary-value problem
x
∆φ = 0
in (0, l)
x
φ = Φ(y − x) on ∂(0, l) = [0] ∪ [l].
5
We know that the Laplacian ∆φx (y) =
∂φx (y)
= A,
∂y
∂ 2 φx
(y)
∂y 2
= 0 in (0, l), so
φx (y) = Ay + B
, the boundary condition
for some constants A, B ∈ R. Since Φ(x) = − |x|
2
|y−x|
x
becomes φ (y) = Φ(y − x) = − 2 . We fix some x ∈ (0, l) and calculate the
boundary conditions.
φx (0) = Φ(0 − x) =⇒ A(0) + B = −
| − x|
x
=⇒ B = −
2
2
|l − x| l>x (l − x)
== −
2
2
We know l > x since x ∈ (0, l). We know what B is, and replace it.
φx (l) = Φ(l − x) =⇒ A(l) + B = −
Al −
x
l
x
l
1 x
=− +
=⇒ Al = − + x =⇒ A = − +
2
2 2
2
2
l
We have found an expression for the corrector function.
1 x
x
y xy x
y−
= − +
−
φx (y) = − +
2
l
2
2
l
2
Using this and the known Φ(y − x) we have found Green’s function (6) for
the region U = (0, l).
G(x, y) = −
|y − x| y xy x
+ −
+
2
2
l
2
(d) By the Representation formula using Green’s function (Theorem 12 in
Evans), the solution u for the boundary value problem
−∆u = f
on U
u=g
in ∂U.
can be expressed with Green’s function.
Z
Z
∂G
(x, y)dS(y) +
f (y)G(x, y)dy
u(x) = −
g(y)
∂ν
U
∂U
(x ∈ U)
In our case, g = 0 so the boundary integral is 0, and we are left with:
Z
Z l
|y − x| y xy x u(x) =
f (y)G(x, y)dy =
f (y) −
dy
+ −
+
2
2
l
2
U
0
We have reduced the problem to an integral we can calculate once we have
an explicit expression for f .
6
Exercise 3
We are going to find Green’s function for the positive half ball B+ , as illustrated by the following image.
Setting U = B+ we are going to find Green’s function as specified in
equation (6) where the corrector function satisfies the conditions
x
∆φ = 0
in B+
φx = Φ(y − x) on ∂B+ = Γ1 ∪ Γ2 .
We apply the hint, using Green’s function for the unit ball:
G(x, y) = Φ(y − x) − Φ(|x|(y − x
e))
and reflection across xn = 0 which we denote by:
x′ = x1 , . . . , xn−1 , −xn .
For y ∈ Γ2 we can apply the same idea as we did for Green’s function for the
unit ball: we define φx (y) = Φ(y − x′ ), and φx (y) = Φ(y − x) for y ∈ Γ2 since
x = x′ .
From the expression for the normal unit ball, we already have the appropriate expression for y ∈ Γ1 :
φx (y) = Φ(|x|(y − x
e))
and to compensate for this when we have points on y ∈ Γ3 we introduce the
canceling term −Φ(|x|(y − x
e′ )). Collecting all these we get:
φx (y) = Φ(y − x′ ) + Φ(|x|(y − x
e)) − Φ(|x|(y − x
e′ ))
which leaves us with an expression, the reflection over xn = 0 of Green’s
function over the unit ball,
G(x, y) = Φ(y − x) − Φ(y − x′ ) − Φ(|x|(y − x
e)) + Φ(|x|(y − x
e′ )).
7
Exercise 4
In order for u ∈ W 1,p (B), B = B(0, 1), the underlying Lp norm must be
finite for the function and its weak derivatives.
First we write out the definition of the Sobolev norm.
X
kukW 1,p (B) =
kD α ukLp (B) = kukLp (B) + kDukLp (B) ≤ CkDukLp (B)
|α|≤1
where the last step follows by Poincare’s inequality.
From here I outline my ideas and an attempted solution. With the function u given explicitly as u(x) = ln |x| we can define the gradient Du(x) =
|x|−1 . Ignoring the constant C we arrived at, we are left with showing the
following integral is finite.
Z
Z
Z
Z
−p
p
−p
|x|−p dx
|x| dx +
kDukLp (B) =⇒
|Du| dx =
|x| dx =
B
B
Bε
B\Bε
where Bε = B(0, ε). The function |x|−p blows up at at the point |x| = 0,
so the second integral where we removed this is finite. With a test function
φ ∈ Cc∞ and after applying one of Green’s identities, we should be able to
show that the first integral, containing |x| = 0, is bounded for certain values.
Using this we should find that the function is contained in values of p with
some dependance on n, much like example 3, page 246 in Evans.
End
8