When Point Boundary Conditions Are Meaningful and When They Are Not, - PowerPoint PPT Presentation

When Point Boundary Conditions Are Meaningful and When They Are Not, or, Why We Need Functional Analysis Jan Mandel University of Colorado Faculty Research Colloquium Denver, February 25, 2004

Contents 1. Energy minimization view of the Laplace equation and its discretization 2. Numerical examples 3. Analysis: Existence and uniqueness of solution, need for bounded trace operators 4. Explanation: The trace theorem and the Sobolev embedding theorem

Laplace equation in 2D: Formulation as minimization of energy � � E ( u ) = 1 �∇ u � 2 dxdy − fu dxdy → min 2 Ω Ω subject to u = 0 on a subset S of the boundary ∂ Ω equivalent to − ∆ u = f in Ω + constraints on u + natural boundary conditions ∂u ∂n = 0 on ∂ Ω where u itself not constrained already � 2 + � 2 , ∆ = ∂ 2 � ∂u � ∂u ∂x 2 + ∂ 2 here �∇ u � 2 = ∂y 2 is the Laplace operator ∂x ∂y discretization: u approximated by piecewise (multi)linear interpolation � � from values u ij on mesh points x ij , y ij , x ij and y ij multiples of step size h (divide each square of size h into two triangles, interpolate linearly on the triangles) the minimum is taken over a finite dimensional space V h

Laplace equation on unit square, zero constraint on one side − ∆ u = 1, u = 0 at the left edge, otherwise natural boundary conditions 0.5 value at the h square center 0.4 1 / 5 0.4 0.3 1 / 10 0.45 0.2 1 / 20 0.475 0.1 1 / 40 0.4875 0 20 15 20 1 / 80 0.49375 15 10 10 1 / 160 0.496875 5 5 0 0 The solution values at the right edge appear to converge to 0.5 when h ← h/ 2, the difference between two successive values decreases by a factor of 2. All is fine.

Laplace equation on unit square, zero constraints at corners − ∆ u = 1, u = 0 at the corners, otherwise natural boundary conditions 0.5 value at the h right edge 0.4 1 / 5 0.165 0.3 1 / 10 0.267045 0.2 1 / 20 0.376459 0.1 1 / 40 0.486533 0 40 30 40 1 / 80 0.596789 30 20 20 1 / 160 0.707091 10 10 0 0 The solution values at the center of the square appear to grow by about the same amount 0.11 when h ← h/ 2: the discrete solutions grow as log h and do not converge as h → 0.

Laplace equation on unit cube, zero constraint on one side − ∆ u = 1, u = 0 on the bottom, otherwise natural b.c. max value on h the top side 1 / 5 0.4 1 / 10 0.45 1 / 20 0.475 1 / 40 0.4875 The solution values at the top side appear to converge to 0.5 when h ← h/ 2, the difference between two successive values decreases by a factor of 2. All is fine.

Laplace equation on unit cube, zero constraint on one bottom edge − ∆ u = 1, u = 0 on one bottom edge, otherwise natural b.c. max value on h the top side 1 / 5 1.06818 1 / 10 1.50583 1 / 20 1.94613 1 / 40 2.38716 The solution values at the top side appear to grow by constant value about 0.53 when h ← h/ 2, the discrete solutions grow as log h and do not converge as h → 0.

Laplace equation on unit cube, zero constraint at bottom corners − ∆ u = 1, u = 0 at bottom corners, otherwise natural b.c. max value on h the top side 1 / 5 0.791113 1 / 10 1.68409 1 / 20 3.4825 1 / 40 7.08243 The solution values at the top side appears to grow by about a factor of 2 when h ← h/ 2. The discrete solutions grow as 1 /h and do not converge .

Examples when point constraints are OK Laplace equation on an interval: − u ′′ = f in [0 , 1], u = 0 at endpoints � u ′ � 2 dxdy − energy functional: E ( u ) = 1 � � fu dxdy → min 2 Ω Ω Biharmonic equation on an interval (beam bending): − u (4) = f in [0 , 1], u = 0 at endpoints + natural b.c. � u ′′ � 2 dx − energy functional: E ( u ) = 1 � � fudx → min 2 Ω Ω Biharmonic equation on a square (plate bending): � � ∂ 4 u ∂x 2 ∂y 2 + ∂ 4 u ∂ 4 u − ∂x 4 + 2 = f , u = 0 at corners + natural b.c. ∂y 4 � 2 � ∂ 2 u energy form: E ( u ) = 1 � � � dxdy − fu dxdy → min ∂x i ∂y j 2 Ω i + j =2 Ω

Common to all these problems: the energy minimization formulation E ( u ) = 1 2 a ( u, u ) − � f, u � → min subject to u ∈ V h , u | S = 0 where • a ( u, v ) is a positive semidefinite bilinear form, such as � a ( u, v ) = ∇ u · ∇ v dxdy , Ω • � f, u � is the L 2 inner product, � f, u � = � fu dxdy , Ω • v | S are the values of u restricted to a subset S of the boundary • V h is the space of piecewise linear functions on the mesh with step h

So what went wrong in some of the problems? E ( u ) = 1 2 a ( u, u ) − � f, u � → min subject to u ∈ V h , v | S = 0 In the bad cases there exist functions v such that v | S = 1 and a ( v, v ) and � v, v � are small for small h we had f = 1 = ⇒ ( f − v ) | S = 0 and a ( f, f ) = a ( f, v ) = 0 = ⇒ functions u = t ( f − v ) satisfy the boundary conditions u | S = 0, large negative for the optimal t � �� � t 2 E ( t ( f − v )) = 2 a ( v, v ) − t � f, f � + t � f, v � � �� � � �� � � �� � small small O (1) = ⇒ ∃ u h ∈ V h : E ( u h ) → −∞ as h → 0 = ⇒ The continuous problem E ( u ) → min in a space V ⊃ V h ∀ h has no solution

How do the bad functions look like? We have seen them before: for the Laplace equation in 2D, for example 1 0.5 0.8 0.4 0.6 0.3 0.4 0.2 0.2 0.1 0 0 80 40 60 80 30 40 60 30 40 20 40 20 20 10 20 10 0 0 0 0 E ( u h ) = 1 a ( v h , v h )+ � v h , v h � → 0 2 a ( u h , u h ) + � 1 , u h � → −∞ v h | S = 1

So how can we tell when things will go right? It is enough if there are no functions v such that v | S is large and a ( v, v ) and � v, v � are small. Use Applied Functional Analysis to construct • a complete space V with the inner product a ( u, v ) + � u, v � , such that V ⊃ V h , and also V ⊃ C ∞ (Ω). These are Sobolev spaces. • a bounded linear operator T : V → W (some other normed linear space), such that Tu = u | S when u ∈ V is also continous on Ω. These are trace operators . Then the above is made precise as � Tv � W ≤ const � u � V and everything falls into place nicely, except when a trace operator T does not exist for the Sobolev space V of functions on Ω and the subset S of the boundary of Ω for the given problem!

How exactly will everything fall into place nicely, or, existence and uniqueness of the solution ( ∗ ): E ( u ) = 1 2 a ( u, u ) − � f, u � → min subject to u ∈ V, Tu = 0 a ( u, v ) is inner product on V , V is complete T is bounded = ⇒ ker T is closed and so it is also complete the map u �→ � f, u � is bounded from V to R the derivative of E ( u ) in every direction v ∈ ker T should be zero = ⇒ ( ∗ ) ⇐ ⇒ u ∈ ker T : a ( u, v ) = � f, u � ∀ v ∈ ker T Riesz representation theorem: a ( · , · ) inner product on complete space ker T , map u �→ � f, u � bounded = ⇒ ∃ ! u What fails when a bounded trace operator does not exist? the very existence of the space ker T the problem is posed in or, if T exists but is not bounded then ker T is not complete such constraints are effectively invisible to the problem!

Sobolev spaces Elements of Sobolev spaces are generalized functions defined only “almost everywhere” – pointwise values do not exist , but they can be integrated, and their derivatives are again generalized functions. For the Laplace equation in 2D: � ∂u ∂x + ∂u ∂v ∂v V = H 1 (Ω) , � u, v � H 1 (Ω) = ∂y + uv dxdy ∂x ∂y Ω For the biharmonic equation in 2D: ∂ 2 u ∂ 2 v � � V = H 2 (Ω) , � u, v � H 2 (Ω) = ∂x i ∂y j + uv dxdy ∂x i ∂y j Ω i + j =2 In general, the inner product (and so the norm) in Sobolev space H m (Ω), Ω ⊂ R n , involve the partial derivatives of order m in a similar manner as above.

Trace and embedding theorems Theorem (Trace theorem) Let Ω ⊂ R n and S be a subset of the boundary of Ω of positive boundary measure (+technical assumptions) . Then the restriction operator u �→ u | S can be extended to a bounded linear operator from H m (Ω) to some other space if and only if m > 1 2 . Theorem (Sobolev embedding theorem) Let Ω ⊂ R n (+technical assumptions) and x ∈ Ω. Then the mapping u �→ u ( x ) can be extended to a bounded mapping from H m (Ω) to R if and only if m > n 2 . (Yes, there are Sobolev spaces of fractional order... a different talk.)

Existence of traces in H m (Ω), m = 1, on S ⊂ Ω (for the Laplace equation) dim Ω ⊂ R n dim S = 0 dim S = 1 dim S = 2 yes 1 , 2 n = 1 no 2 yes 1 n = 2 no 2 no 3 yes 1 n = 3 ⇒ m = 1 > 1 1 Trace theorem: dim S = n − 1 , trace exists ⇐ 2 ⇒ m = 1 > n 2 Embedding theorem: dim S = 0 , trace exists ⇐ 2 3 The case dim S = 1 (edge), n = 3 requires more work Note: point and edge loads are another talk.

Existence of traces in H m (Ω), m = 2, on S ⊂ Ω (for the biharmonic equation) dim Ω ⊂ R n dim S = 0 dim S = 1 dim S = 2 yes 1 , 2 n = 1 yes 2 yes 1 n = 2 yes 2 yes 3 yes 1 n = 3 ⇒ m = 2 > 1 1 Trace theorem: dim S = n − 1 , trace exists ⇐ 2 ⇒ m = 2 > n 2 Embedding theorem: dim S = 0 , trace exists ⇐ 2 3 The case dim S = 1 (edge), n = 3 requires more work Note: point and edge loads are another talk.