Then the variational problem J (v):= 1
2a(v, v)− (f, v)0,− (g, v)0, −→ min !
has exactly one solution u∈ H1(). The solution of the variational problem lies in C2()∩C1( ¯)if and only if there exists a classical solution of the boundary-value problem
Lu= f in ,
i,k
νiaik∂ku= g on , (3.4)
in which case the two solutions are identical. Here ν := ν(x) is the outward-pointing normal defined almost everywhere on .
Proof. Since a is an H1-elliptic bilinear form, the existence of a unique mini-mum u ∈ H1() follows from the Lax–Milgram Theorem. In particular, u is characterized by
a(u, v)= (f, v)0,+ (g, v)0, for all v ∈ H1(). (3.5) Now suppose (3.5) is satisfied for u ∈ C2()∩ C1( ¯). For v ∈ H01(), γ v= 0, and we deduce from (3.5) that
a(u, v) = (f, v)0 for all v∈ H01().
By (2.21), u is also a solution of the Dirichlet problem, where we define the boundary condition using u. Thus, in the interior we have
Lu= f in . (3.6)
For v ∈ H1(), Green’s formula (2.9) yields
v∂i(aik∂ku) dx = −
∂ivaik∂ku dx+
vaik∂ku νids.
46 II. Conforming Finite Elements
Now it follows from (3.5) and (3.6) that the second integral on the right-hand side of (3.7) vanishes. Suppose the function v0 := νiaik∂ku− g does not vanish.
Then
v20ds >0. Since C1( ¯)is dense in C0( ¯), there exists v ∈ C1( ¯)with
v0 · v ds > 0. This is a contradiction, and the boundary condition must be satisfied.
On the other hand, from (3.7) we can immediately see that every classical solution of (3.4) satisfies (3.5).
Neumann Boundary Conditions For the Helmholtz equation
−u + a0(x) u = f in , the natural boundary condition is
∂u
∂ν := ν · ∇u = g on ∂.
We call it the Neumann boundary condition. Here ∂u/∂ν is the normal derivative, i.e., the direction perpendicular to the tangent plane (if the boundary is smooth). [In the general case, the boundary condition in (3.4) also involves the normal direction if we define orthogonality w.r.t. the metric induced by the quadratic form with the matrix aik = aik(x).]
Clearly, the Poisson equation with Neumann boundary conditions,
−u = f in ,
∂u
∂ν = g on ∂, (3.8)
only determines a function up to an additive constant. This suggests that in formu-lating the weak version of this problem we should restrict ourselves to the subspace V := {v ∈ H1();
v dx = 0}. The bilinear form a(u, v) =
∇u · ∇v dx is not H1-elliptic, but in view of the variant (1.11) of Friedrichs’ inequality, it is V-elliptic.
We claim that the data of the boundary-value problem (3.8) must satisfy a certain compatibility condition. Indeed, with w := ∇u, equation (3.8) becomes
− div w = f in , νw= g on .
§3. The Neumann Boundary-Value Problem. A Trace Theorem 47
By the Gauss Integral Theorem,
div w dx=
∂wνds, and thus
f dx+
g ds = 0. (3.9)
This condition is not only necessary, but also sufficient. By the Lax–Milgram Theorem, we get u∈ V with
a(u, v)= (f, v)0,+ (g, v)0, (3.10) for all v ∈ V . Because of the compatibility condition, (3.10) also holds for v = const, and thus for all v ∈ H1(). As in Theorem 3.2, we now deduce that every classical solution of the variational problem satisfies the equation (3.8).
Another method to deal with the pure Neumann problem (3.8) will be dis-cussed in Problem III.4.21.
Mixed Boundary Conditions
In physical problems, we often encounter Neumann or natural boundary condi-tions whenever the flow over the boundary is prescribed. Sometimes a Neumann condition is prescribed on only part of the boundary.
3.3 Example. Suppose we want to determine the stationary temperature distribu-tion in an isotropic body ⊂ R3. On the part of the boundary where the body is mechanically clamped, the temperature is prescribed. We denote this part of the boundary by D. On the rest of the boundary N = \ D, we assume that the heat flux is so small that it can be considered to be 0. If there are no heat sources in , then we have to solve the elliptic boundary-value problem
u= 0 in , u= g on D,
∂u
∂ν = 0 on N.
This problem leads in a natural way to a Hilbert space which lies between H1()and H01(). Consider functions of the form
u∈ C∞()∩ H1(), u vanishes in a neighborhood of D.
Then the closure of this set w.r.t. the H1-norm leads to the desired space. This is a subspace of H1(), and by Remark 1.6, under very general hypotheses | · |1 is a norm which is equivalent to · 1.
48 II. Conforming Finite Elements
Proof of the Trace Theorem
We now present the proof of the trace theorem. For the sake of clarity, we restrict ourselves to domains inR2. The generalization to domains inRnis straightforward, and can be left to the reader as an exercise.
Suppose the boundary is piecewise smooth. In addition, suppose a cone condi-tion is satisfied at the (finitely many) points where the boundary is not smooth. Then we can divide the boundary into finitely many boundary pieces 1, 2, . . . , m so that for every piece i, after a rotation of the coordinate system, we have
1. For some function φ = φi ∈ C1[y1, y2],
i = {(x, y) ∈ R2; x = φ(y), y1≤ y ≤ y2}.
2. The domain i = {(x, y) ⊂ R2; φ(y) < x < φ(y) + r, y1 < y < y2} is contained in , where r > 0.
We now apply an argument used earlier for the Poincar´e–Friedrichs inequality. For v∈ C1( ¯)and (x, y)∈ ,
We take the square of this equation, and use Young’s inequality (a+b)2 ≤ 2a2+2b2. Applying the Cauchy–Schwarz inequality to the squares of the integrals gives
r2v2(φ (y), y)≤ 2
We now insert the values
1dt = r and
t2dt = r3/3. Dividing by r2 and integrating over y, we get
y2
The arc length differential on is given by ds =
1+ φ2dy. Thus, we have a dense set. Because of the completeness of L2(), it can be extended to all of H1()without enlarging the bound.
§3. The Neumann Boundary-Value Problem. A Trace Theorem 49
Note that the cone condition excludes cusps in the domain. The domain
:= {(x, y) ∈ R2; 0 < y < x5 <1} has a cusp at the origin, and H1()contains the function
u(x, y) = x−1, whose trace is not square-integrable over .
We would like to point out that Green’s formula (2.9) also holds for functions u, w ∈ H1(), provided that satisfies the hypotheses of the trace theorem.
The space H1() is isomorphic to a direct sum H1() ∼ H01()⊕ γ (H1()).
Specifically, every u∈ H1() can be decomposed as u= v + w,
according to the following rule. Let w be the solution of the variational problem
|w|21→ min ! More exactly, suppose
(∇w, ∇v)0, = 0 for all v∈ H01(), w− u ∈ H01().
Let v := u − w ∈ H01(). Here γ is an injective mapping on the set of functions w which appear in the decomposition.
We now consider the connection with continuous functions. As usual, the norm u∞ = u∞, is based on the essential supremum of |u| over . [It is not a Sobolev norm.]
3.4 Remarks. (1) Let ⊂ R2 be a convex polygonal domain, or a domain with Lipschitz continuous boundary. Then H2()is compactly imbedded in C( ¯), and
v∞ ≤ cv2 for all v ∈ H2(), (3.11) for some number c= c().
(2) For every open connected domain ⊂ R2, H2()is compactly imbedded in C().
The above results are not the sharpest possible in this framework. Because of their importance, and because they follow simply from the trace theorem, we now give their proofs.
50 II. Conforming Finite Elements
Choose an angle ϕ and a radius r with the following property: for every two points x, y ∈ ¯ with x − y < r, there exists a cone K with angle ϕ, diameter r, and tip at x such that y ∈ ∂K. We now rotate the coordinate system so that x and y differ in only the first coordinate. By the trace theorem, we deduce that
v0,∂K ≤ c(r, ϕ)v1,K.
Then for v ∈ H2(), ∂1v ∈ H1(), and thus ∂1v0,∂K ≤ c(r, ϕ)v2,. By Remark 1.8,|v(x) − v(y)| ≤ c(r, ϕ)√
x − y · v2,. Thus, v is H¨older con-tinuous with exponent 1/2. Using v0,K ≤ v2,, we get a bound for |v| in K, and (3.11) follows. The Arzel`a–Ascoli Theorem now establishes the compactness assertion.
For every m≥ 1, we can find a polygonal domain m which contains all of the points x ∈ whose distance from 0 is at most m, and whose distance from
∂is at least 1/m. Since =
m>0m, (2) follows from (1).
Practical Consequences of the Trace Theorem
For practical applications, it is tempting to believe that only classical solutions are of importance, and that weak solutions with singularities are nothing more than interesting mathematical objects. However, this is far from the truth, as the following example shows.
3.5 Example. Suppose we erect a tent over a disk with radius R such that its height at the center is 1. Find the shape of the tent which has the minimal surface area.
Suppose u(x) is the height of the tent at the point x. Then it is well known that the surface area is given by
BR
1+ (∇u)2dx.
Here BRis the disk with radius R and center at 0. Now
1+ (∇u)2≤ 1+12(∇u)2, and for small gradients the difference between the two sides of the inequality is small. Thus, we are led to the variational problem
1 2
BR
(∇v)2dx −→ min ! (3.12)
subject to the constraints
v(0)= 1,
v = 0 on ∂BR.
§3. The Neumann Boundary-Value Problem. A Trace Theorem 51
We now show that the constraint v(0) = 1 will be ignored by the solution of the variational problem. The singular function
w0(x)= log logeR
r with r = r(x) = x
has a finite Dirichlet integral (3.12). We smooth it to get
wε(x) =
w0(x) for r(x)≥ ε, log logeR
ε for 0≤ r(x) < ε.
Now|wε|1,BR ≤ |w0|1,BR, and wε(0) tends to∞ as ε → 0. Thus, uε = wε(x)
wε(0)
for ε= 1, 1/2, 1/3, . . . provides a minimizing sequence for J (v) which converges almost everywhere to the zero function. This means that the requirement u(0)= 1 was ignored.
The situation is different if we require that u = 1 on a curve segment. This would be the case if the tent were attached to a ring on the tent pole or if the tent is put over a rope so that it assumes the shape of a roof. While the evaluation of an H1 function at a point does not make any sense, its evaluation on a line in the L2 sense is possible. A condition on a function which is defined on a curve segment will be respected almost everywhere.
Fig. 8. Tent attached to a loop of a rope to prevent an extreme force concentration at the tip
For most larger tents, the boundary of the tent at the tip is a ring instead of a single point. Or (see Fig. 8) the tent may be attached to a loop of rope. This avoids very high forces, since the force applies to the ring or loop, rather than at a single point. The trace theorem explains why the point is to be replaced by a one-dimensional curve.
52 II. Conforming Finite Elements
Problems