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Weak Form of Boundary Value Problems

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Weak Form of Boundary Value Problems Weak form of Boundary Value Problems Simulation Methods in Acoustics Note on finite dimensional description of functions I Approximation: N X f (x) ≈ f^(x) = qj φj (x) j=1 I Residual function: r(x) = f (x) − f^(x) I Collocation and Galerkin with test functions i (x) h i ; ri = 0 for all i = 1; 2 ::: N I Collocation: i (x) = δ(xi ) I Galerkin: i (x) = φi (x) I Other { φ choices are possible, but not used often in practice The boundary value problem I Given a BVP (PDE + BCs) Afu(x)g = f (x) x 2 Ω (1) [ Bi fu(x)g =g ¯i (x) x 2 Γi Γi = Γ (2) i I A and Bi are differential operators I Example (Poisson equation) @2u(x) − = 1 x 2 Ω = [0; 1] (3) @x2 u(x) = 0 x = 0 (Dirichlet) (4) @u(x) = −1 x = 1 (Neumann) (5) @x I In our example: 2 2 I A = −@ =@x , f (x) ≡ 1 I B1 = 1,g ¯1(x) ≡ 0 I B2 = @=@x,g ¯2 ≡ −1 Weak form construction I Construction steps: 1. Multiply PDE by a “sufficiently well behaved"1 function (x) 2. Integrate over the whole domain Ω 3. Integration by parts as much as possible (shift derivatives) 4. Apply BC 1. Multiply the PDE by the test function (x) (x)Afu(x)g = (x)f (x) 8x 2 Ω 2. If the equality holds, the integrals must be also equal Z Z (x)Afu(x)g dx = (x)f (x) dx 8 (x) Ω Ω I Statement: The solution of the weak form must also be the solution of the strong (original) form of the PDE I Proof: The weak form must hold for all (x). We can choose test functions that are nonzero only in Bδ(x0) for any x0 2 Ω. Thus, the original equation must hold for all x0 2 Ω. I Then, why it is a weak form? 1The definition well behaved depends on the type of PDE and BC Weak form example I. I Look at the Poisson equation example: @2u(x) − = f (x) x 2 Ω = [0; 1] ; f (x) ≡ 1 @x2 1. Multiply by the test function (x) @2u(x) − (x) = (x)f (x) @x2 2. The weak form is written as Z @2u(x) Z − (x) 2 dx = (x)f (x) dx Ω @x Ω 3. Apply integration by parts on the l.h.s. R x2 0 x2 R x2 0 I Recall that f g dx = [fg] − fg dx x1 x1 x1 I With f = @u=@x and g = (x) @u(x)1 Z 1 @ (x) @u(x) Z 1 − (x) + dx = (x)f (x) dx @x 0 0 @x @x 0 Weak form example II. 4. Apply boundary conditions I u(0) = 0 @u(x) I @x = −1 x=1 Z 1 Z 1 @u @ @u jx=1 + + dx = (x)1 dx @x x=0 0 @x @x 0 I Let's compare the strong and weak forms I Strong: 2nd derivative of u, Weak: 1st I Strong: f continuous on Ω, Weak: integrable (f integrable) I The weak form imposes less strict criteria on u and f ! But it requires some conditions for (x). I We can express these criteria using function space definitions: 2 0 I Strong: u 2 C (Ω), f 2 C (Ω) 1 −1 2 I Weak: u 2 C (Ω), f 2 C (Ω), u;@u=@x 2 L (Ω) Sobolev space definition I Square-integrable functions (in Lebesgue sense) Z 2 2 2 L (Ω) = f :Ω ! R kf k = f dx < 1 Ω p I Sobolev space ∼ the function and its derivatives are in L @f H1(Ω) = f 2 L2(Ω); 2 L2(Ω) @x I Note: derivatives are meant in a weak sense I g(x) is the weak derivative of u, i.e. g = @u=@x if and only if Z Z @ (x) 1 (x)g(x) dx = − u(x) dx 8 (x) 2 C0 (Ω) Ω Ω @x I The weak derivative is unique 1 I Note: C0 (Ω) means infinite times differentiable functions having a compact support on Ω Example: solve the weak form I Let's search the solution in the form: 2 I u(x) = Ax + Bx 2 I (x) = Ax^ + Bx^ I Find A, B such that the weak form is satisfied for all A^, B^ I Substitute: ((( 2 2 (( Ax^ + Bx^ + Ax^ + (Bx^ ((A(+ 2Bx + x=1 ((( x=0 Z 1 Z 1 A^ + 2Bx^ A + 2Bx dx = Ax^ + Bx^ 2 dx 0 0 I Rearrange (A and B on l.h.s.) Z 1 Z 1 1 A x 1 A^ B^ 1 2x dx = A^ B^ 2 dx − 0 2x B 0 x 1 I Evaluating the integrals 1 1 A −1=2 A^B^ = A^B^ 8 A^ B^ 1 4=3 B −2=3 I We obtain the solution: A = 0; B = −1=2, thus u(x) = −x2=2 How did we solve? 2 I By choosing the trial functions (u(x) = Ax + Bx ) the trial function space was discretized (x and x2 are its two elements). Thus, we arrived at a finite dimensional problem (number of dimensions is d = 2 in this example) 2 I By letting (x) = Ax^ + Bx^ we used the Galerkin method. I Note that we cannot apply the collocation method 2 ( i = δ(xi )) as @ =@x must be in L I By means of the discretization the BVP was reduced to an algebraic set of linear equations! Thus, in the end a system of linear equations was solved. I In this case, we \luckily" got the analytical solution, however, in general we only get an approximate solution Another solution I Let's solve with a different right hand side: @2u(x) 1 − = − u(0) = 0; u0(1) = 0 @x2 1 + x2 I All the same until we get Z 1 " −x # 1 1 A 1+x2 A^ B^ = A^ B^ 2 dx 8 A^ B^ 1 4=3 B −x 0 1+x2 I By \dropping" A^ B^ 1 1 A − log(2)=2 A −0:7425 = ! = 1 4=3 B π=4 − 1 B 0:3959 I The discretized weak form gives an approximate solution Visualization of the example I The approximation gets better if we increase the dimensionality of our function space, e.g. by adding higher order polynomials Order = 2 Order = 5 0 0 Weak solution u(x) -d2u(x)/dx2 -0.2 -0.2 f = -1/(x2 + 1) -0.4 -0.4 -0.6 u(x) u(x) -0.6 -0.8 Weak solution u(x) -0.8 -1 -d2u(x)/dx2 f = -1/(x2 + 1) -1 -1.2 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 x x Beam example I. I Solve the following BVP (static deflection of a clamped{free Euler{Bernoulli beam) EIu0000(x) = f (x) u(0) = 0 u0(0) = 0 (Dirichlet) u00(L) = 0 u000(L) = 0 (Neumann) I Construct the weak form Z L Z L EI (x)u0000(x) dx = (x)f (x) dx 0 0 I Integration by parts once . Z L ! Z L 000 L 0 000 EI [ u ]0 − u dx = f dx 0 0 I . and twice Z L ! Z L 000 L 0 00 L 00 00 EI [ u ]0 − [ u ]0 + u dx = f dx 0 0 Beam example II. I Choose the test and trial function spaces as iπ φ = = sin i = 1 ::: n i i 2L (i − n)π φ = = cos i = n + 1 ::: 2n = N i i 2L I Note: by letting i = φi we use the Galerkin method I The approximations using the above spaces reads as N X u(x) ≈ u^(x) = qj φj (x) = φ(x)q j=1 N X T T (x) ≈ ^(x) = wj φj (x) = w φ (x) j=1 I Note: φ is a row vector, q and w are column vectors Beam example III. I Notice that the expressions Z L 000L 0 00L 00 00 u 0 u 0 u dx 0 I Can all be written in a similar form 2 (α) (β) (α) (β)3 8 9 φ φ ··· φ φ q1 1 1 1 N <> => T 6 . .. 7 . w 4 . 5 . (α) (β) (α) (β) > > :qN ; φN φ1 ··· φN φN R L (α) (β) I Integration 0 dx or evaluation of φi φj (product of shape functions and their derivatives) at a fixed point is needed. Each term gives an N × N matrix in the square brackets. I On the r.h.s. we have a column vector in the brackets 2φ (x)f (x)3 Z L 1 T . T w 6 . 7 dx = w b 0 4 5 φN (x)f (x) Beam example IV. I Write the system in matrix notation: T T EI w (A1 + A2 + A3) q = w b Note that this is a scalar equation T I This must hold for all vectors w , which is only possible if EI Aq = b I Thus, we arrive at a system of linear algebraic equations Beam example V. I Satisfying the BCs I Contrary to the previous example, our trial (shape) functions 0 do not satisfy the Dirichlet BCs. (φi (0) 6= 0, φi (0) 6= 0) I Constraints must be imposed: 8 9 q1 > > φ (0) φ (0) ··· φ (0) <>q2 => 0 1 2 N = 0 0 0 . φ1(0) φ2(0) ··· φN (0) > . > 0 > > :qN ; I The same in matrix form: Acq = 0 I We already know some methods for constraints .
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