Numericl Methods for Prtil Differentil Equtions Eric de Sturler Deprtment of Computer Science University of Illinois t Urn-Chmpign 11/19/003 1 00 Eric de Sturler
Why More Generl Spces We now provide more forml frmewor for the pproximtion of solutions of PDEs nd the solution to certin minimiztion prolems. First we need to extend the spce of functions over which we wor (sy C [, ] for one-dimensionl second order equtions). 1. Mny useful prolems do not hve C solution. Approximtion over suspce of sis functions (sy, piecewise polynomils) my not esily give pproximtion tht is C. 3. Useful to define est pproximte solution using orthogonl projection, which in turn requires the spce to e complete. First we need to introduce numer of concepts. 11/19/003 00 Eric de Sturler
The Spce L { } Define the spce L [, ] = f :[, ] R f dx < with inner product uv, Here = uvdxnd norm ( ) 1 u u dx =. dx denotes Leesgue integrl, which is more generl thn the Riemnn integrl, ut yields the sme vlue when the Riemnn integrl is defined. A consequence of the ove choice for norm nd spce is tht ( u v) dx = 0 u = v even when u( x) = v( x) is not true for ll x [, ]. Such u nd v re memers of equivlence clss. 11/19/003 3 00 Eric de Sturler
The Spce Consider ( = 0, = 1), u : x x 0 if x 1 Then, u v = 1 1 if x =., nd v : x x if x 0 if x = 1. 1 1 1 1 However, ( u v ) dx = 0dx + 0dx = 0. 0 0 So, s elements of L we hve u We sy u = v lmost everywhere (.e.) 1 = v even if u( 1 ) ( 1 ) v. 11/19/003 4 00 Eric de Sturler
The Spce Functions re equl if they differ (only) on set of mesure zero. Any denumerle (countle) set hs mesure zero. Let u( x) = 1 x [, ] nd v( x) Then ( u v) dx = 0 x, x Q = 1, x [, ] \ Q. Agin, s elements of L we hve u = v even though u v for every x Q. The functions u nd v re lso sid to e memers of the sme equivlence clss. In generl, we represent n equivlence clss of functions y its smoothest memer. 11/19/003 5 00 Eric de Sturler
The Spce Verify tht uv, nd u for ll uv, L [, ] stisfy the properties for (rel) inner product nd norm, nd L [ ] is vector spce., For ny H C : H( x) > 0 we cn define the H-inner product uv, H = Huvdxnd H-norm ( ) 1 u = Hu dx. uv, = vu, u 0 αuv, = α uv, u = 0 u = 0 uu, 0 αu = α u uu, = 0 if nd only if u= 0 u + v u + v H 11/19/003 6 00 Eric de Sturler
Completeness Sequence ui L [, ], i = 1,,, converges in the men (L - norm) to u L [ ] if lim u u = 0., n This is not the sme s pointwise convergence: lim u ( ) ( ) n x = u x for ll x [, ]. n Sequence is Cuchy sequence if lim u u 0, or,, N() : m, n N um un n mn ε ε > < ε. m n If every Cuchy sequence converges to limit (tht is in the spce) we sy the spce is complete. 11/19/003 7 00 Eric de Sturler
Hilert Spces If vector spce with n inner product nd norm is complete with respect to this norm we cll such spce Hilert spce. L [, ] with inner product, Hilert spce. uvdx, nd norm ( ) 1 udx is { } As we will show lter V = w C [, ] w( ) = w( ) = 0 is not Hilert spce with respect to this norm or the energy norm. 11/19/003 8 00 Eric de Sturler
Liner Functionls Liner functionl is mpping ϕ : L [, ] R such tht uv, L[, ] nd αβ, R : ϕ( αu+ βv) = αϕ( u) + βϕ( v). The functionl is ounded if α > 0: ϕ( u) α u, u L [, ]. Exmple: ϕ ( u) = u, u0 for ritrry (ut constnt) u0 L [, ]. Linerity follows from properties inner product nd oundedness from the Cuchy-Schwrz inequlity: ϕ ( u) = u, u u u. 0 0 Riesz Representtion Theorem: Let ϕ ( u) e ounded liner functionl over L [, ]. Then u0 L [, ] such tht ϕ ( u) = u, u0 for ll u L [, ]. (extends to ounded liner functionl on ny Hilert spce) 11/19/003 9 00 Eric de Sturler
The Liner Alger of PDEs f u = 1 1 p x ux + q x u g x udx with u V = w C [, ] w( ) = w( ) = 0. Note V = V. Consider ( ) ( ) ( ) ( ) { } Here p C 1 [, ], q, g C[, ], 0 < p0 p p1 nd 0 q q1 for x [, ]. Furthemore, p0, p1, q 1 re constnts. Let ( u, v) = puxvx + quvdx nd G( u) = gudx. ( u, v ) is symmetric iliner form, G( u ) is ounded liner functionl. 11/19/003 10 00 Eric de Sturler
The Liner Alger of PDEs Given ( u, v ) nd G( u ) we cn write f ( u) = 1 ( u, u) G( u). Let η = 1 nd τ R. Then 1 1 f ( u + τη) = ( u, u) + τ( u, η) + τ ( η, η) G( u) τg( η) ; 1 f ( u + τη) = f ( u) + τ ( ( u, η) G( η) ) + τ ( η, η). So, () 1 f ( u; η ) = ( u, η ) G ( η ), f u, η = η, η > 0 (verify), () ( ) ( ) ( 3 ) ( η) ( 4 ) ( η) f u, = f u, = 0 nd sme for higher derivtives. 11/19/003 11 00 Eric de Sturler
The Liner Alger of PDEs () 1 f ( u η) ( u η) G( η) η = ), implies ( u, η) G( η) The condition ( 1 ; =, = 0 for ll unit η V = for ll unit η V. This corresponds to pu η + qu η dx = g η dx x x for ll unit η V. (we form) Integrtion y prts gives ( pu x ) qu η dx g η dx + = x, pu + qu = g. which leds to the EL, ( ) x x 11/19/003 1 00 Eric de Sturler
The Liner Alger of PDEs Define the opertor L : V C[, ] Lu = pu + qu. : ( ) x x x x, Note tht (, ) = ( ) + = ( ) u η pu qu ηdx Lu ηdx nd tht L is symmetric nd liner (verify). Liewise G( η) ( ) ( ) = gηdx. u, η = G η Lu, η g, η = 0 Lu g, η = 0 Since Lu g is continuous we must hve Lu = g. (, ) ( ) x x for ll V u η = G η pu η + qudx = gηdx This is the we form of the PDE. η. 11/19/003 13 00 Eric de Sturler
The Liner Alger of PDEs Note tht ( u, η ) defines n inner product nd (hence) ( u, u ) 1 () defines norm. Therefore, f ( u; η) = ( η, η) > 0. From PDE theory we now tht the prolem Lu = g with u ( ) = u ( ) = 0 (nd under given conditions) hs unique solution û V. f ( uˆ+ η) = f ( uˆ ) + ( u, η) G( η) + 1 ( ηη, ) = f ( uˆ) + 1 ( ηη, ) > f ( uˆ) for ll η V, η 0. So, û is unique, strong, glol minimizer. 11/19/003 14 00 Eric de Sturler
The Liner Alger of PDEs { } The spce V = w C [, ] w( ) = w( ) = 0 is not complete with respect to energy- or L-norm; so, it s not Hilert spce. So, we cnnot formlly do in Prolems (formlly): C wht we hve done so fr. u g xx = g nd u ( ) = u ( ) = 0, (solution û ) g which is discontinuous: u u C [, ]. Best pproximtion in spce of sis functions with locl support is not well-defined. It requires orthogonl projection nd completeness: Hilert spce. 11/19/003 15 00 Eric de Sturler
Completion Process of completion, expnd the spce of dmissle functions V to lrger set V, such tht 1. We cn pproximte every element of V ritrrily closely y n element of V. The solution û of Lu = g with u ( ) = u ( ) = 0 is unique solution to vritionl prolem u V, s.t. ( u, η) = G( η) for ll η V 3. min f ( u) = min f ( u) = f ( uˆ ) u V u V We will see tht V contins much rougher functions thn V. 11/19/003 16 00 Eric de Sturler
Preliminries L [, ] is complete (with given norm): lim u u = 0 u L : lim u u 0 nm, m n n n The right hnd side mens u n u (in terms of equivlence clss) Theorem: For ll u C 1 [, ], v( ) = 0: u', u ' uu, ( ). Define uv, = ( uv, ) nd u (, ) 1 L L = u u These re the energy inner product nd norm. 11/19/003 17 00 Eric de Sturler
Let { un } n 1 Completion V hve Cuchy property: lim u 0 = m un. mn, L u L = pu', u' + qu, u p0 u', u' = p0 ( u', u' + u', u' ) = p0 p0 u' + u ( ) ( u' u ) ρ + for ll u V. + for { 1 ρ = p ( ) } 0 min, nd So, ( u ' ) L ρ u u u V 11/19/003 18 00 Eric de Sturler
Completion Let { un } n 1 V hve Cuchy property: lim u u 0. =, mn Now, ( u ) m un ρ u m u n um un + nd L m n L ( ) lim u u 0 lim ρ u u + u u = 0 m n m n m n mn, L mn, Hence, lim u u = 0 nd mn, m n lim u u = 0 (Cuchy!). mn, m n such tht lim u u = 0 nd lim u u = 0 L complete implies there exist uu, L n n n We cll ũ the generlized derivtive of u. V is spce of ll functions u tht cn e otined in this wy. n 11/19/003 19 00 Eric de Sturler
Completion Since the energy inner product nd norm re defined over L, the existence of uu, L implies tht L is lso complete in this norm. Under ssumptions given, we now Lu = g with u ( ) = u ( ) = 0 hs solution u V V. So, there is no prolem stting u V s.t. u ( ) = u ( ) = 0 nd Lu g, η = 0 η V (prolem posed in V ). Using uniqueness of qudrtic prolem it is esy to see tht the solution in V is only solution. (ut now the underlying mthemtics is properly defined) In ddition we cn solve more generl prolems in we form. 11/19/003 0 00 Eric de Sturler
Let uu, L with ũ the generlized derivtive of u. In ddition, let u u n 0 nd u u n 0 ( L norms). Then u u n 0. L u = p ( u ') + qu dx p ( u ') + q u dx L 1 1 µ mx p, q u' dx + µ u dx = µ u' + µ u ( ) ( ) 1 1. So, n µ L n µ n u u u u + u u, nd since L is complete, we hve u u 0 nd n u u n 0, nd hence u u n 0. L 11/19/003 1 00 Eric de Sturler
We cn sy (intuitively) tht we creted the complete spce V from the incomplete spce V y dding the limit points of Cuchy sequences in V to V. Furthermore, we extend the relevnt inner product nd norm to the new elements (these limit points) to which they originlly did not pply, s those functions re not in V. 11/19/003 00 Eric de Sturler
We cn sy (intuitively) tht we creted the complete spce V from the incomplete spce V y dding the limit points of Cuchy sequences in V to V. Furthermore, we extend the relevnt inner product nd norm to the new elements (these limit points) to which they originlly did not pply, s those functions re not in V. The exmple in the oo, pp. 110 ff., shows tht there re V functions tht re not even once differentile. Wht ind of functions live in this spce? (V ) 11/19/003 3 00 Eric de Sturler
Wht ind of functions live in this spce? (V ) Consider (1D) Soolev spces. Define inner product nd norm,, 1 () i () i =, =, + ', ' = 1 i= 0 + ' ' 1 1 = u, u, 1, uv u v uv u v uv u v dx u for ll uv, V= w C [, ] : w( ) = w ( ) = 0. { } Soolev norm. 1 nd energy norm. L re equivlent. Two norms. α nd. β re equivlent if constnts 0 < γ1 < γ exist, such tht γ u u γ u u V, 1 α β α where γ1, γ my depend on norms ut not on u. 11/19/003 4 00 Eric de Sturler
We ll show ρ u u σ u u V for some ρσ>, 0. 1 L 1 u = p ( u ') + qu dx p ( u ') dx = p u ', u ' = L 0 0 p0 p0 p0 p0 u', u' ', ' ', ', + u u u u u u + ( ) ρ ( ' ) ρ 1 u + u = u, where ρ p0 p 0 min,. ( ) 11/19/003 5 00 Eric de Sturler
We ll show ρ u u σ u u V for some ρσ>, 0. 1 L 1 u = p ( u ') + qu dx p ( u ') dx + q u dx L 1 1 ( ' ) 1 ' 1 σ σ 1 p u + q u u + u = u, where mx { p, q } σ. 1 1 Hence, we hve ρ u 1 u L σ u 1 from which we cn immeditely derive equivlence. 11/19/003 6 00 Eric de Sturler
Using this property of equivlence we cn show tht completing V with respect to the Soolev (1) norm rther thn then energy norm leds to the sme (complete) spce V. Becuse of this, V is lso clled Soolev spce. The Soolev inner product nd norm re defined on ll elements of V s long s we interpret derivtives re generlized derivtives. (in generl inner products nd norms re extended to the completion of spce with the proper interprettion) V is Hilert spce with respect to the Soolev inner product nd norm. 11/19/003 7 00 Eric de Sturler
We consider severl spces resulting from completion. Let C (, ) e the spce of times continuously differentile functions. () Let (, ) (, ) ( i C = w C w ) dx <, i = 1,,, { } () i () i () 1 () 1 ( ) ( ) = = + + + i= 0 1, nd uv, u, v uv u v u v dx u = u, u. The ltter re the Soolev inner product nd norm of order. Completion of the spce with respect to u proceeds nlogously to completion w.r.t. the energy norm. The resulting spce is referred to s H (, ) ( H for Hilert). 11/19/003 8 00 Eric de Sturler
We sy function u( x) defined on (, ) hs compct support if αβ, : < α< β< nd u( x ) = 0 if x (, α) ( β, ). The completion of C0 (, ), defined s ( ) { C ( ) ( )} 0, = w C, w hs compct support on,, w.r.t.. is denoted s H ( ). It is lso Soolev spce. 0, { } () () H ( ) ( ) ( ) 0, = w H, w i = w i () = 0, i = 0,1,, 1 Although V C 1 ( ) their completions w.r.t.. 1 re the sme: 1 V = H ( ) 0, 0, 11/19/003 9 00 Eric de Sturler
So, the prtil differentil eqution Lu = g pu + qu = g ( ) with oundry conditions u ( ) = u ( ) = 0, hs the vritionl formultion Find u H 1 0 (, ) such tht ( u, η) = G( η) η H 1 (, ) x x 0 11/19/003 30 00 Eric de Sturler
We proceed the sme wy in severl spce dimensions. Define L ( ) { } Ω = w : w dω< L. Ω ( Ω ) is Hilert spce w.r.t. the inner product nd norm uv, u = uvdω, Ω 1 = u, u. We denote prtil derivtives s follows. n = = 1 n i= 1 α α ( α) α =, = α1 αn x1 xn Define (, α ) α α, α,, α α, nd D u D u. 11/19/003 31 00 Eric de Sturler i
{ } ( ) Let C ( Ω ) = w C ( Ω) ( w α ) dω<, 0 α Ω The Soolev inner product nd norm on C ( Ω) re given y uv, u ( α), v ( α) = 0 α ( ) 1 u = u, u = u α 0 α ( ) So, in two dimensions we hve u = u + ux + uy + uxx + uxy + uyy dω. Ω The completion of C ( Ω) w.r.t.. is gin referred to s H ( Ω ). Agin H ( Ω ) is Hilert spce w.r.t. to the Soolev inner product nd norm. 11/19/003 3 00 Eric de Sturler.
Existence of ordinry derivtives of u H depends on nd sptil dimension (nd smoothness of oundry). 1 H Ω C Ω 1d: ( ) ( ) n Ω R nd oundry sufficiently smooth: s H ( Ω) C ( Ω ) where s < n nd ( α) C( Ω, ) : mx u ( x) C u α : α s. x Ω d/3d: H ( Ω) C ( Ω ) 0 Anlogous to the hierrchy ( ) 1( ) C C C ( ) Ω Ω Ω We hve L ( Ω ) = H 0 ( Ω) H 1 ( Ω) H ( Ω) 11/19/003 33 00 Eric de Sturler
The finite element method typiclly pproximtes the solution using (continuous) piecewise polynomils. Let u 1 C ( Ω ) nd u C ( ) Then u H ( Ω ) nd u H + 1 ( Ω ). Ω nd u e piecewise polynomil. A function u defined on Ω hs compct support on Ω if Ω Ω, Ω is closed : u = 0 for x Ω \ Ω. 0 0 0 The completion of C 0 ( Ω ) is denoted y H 0 ( Ω ). If oundry is sufficiently smooth: i w H0 ( Ω ) = w H ( Ω ) = 0 on Γ, i = 0,1,, 1 i n 11/19/003 34 00 Eric de Sturler
Lx-Milgrm Lemm Let V e Hilert spce with IP.,. V nd norm 1. = uu,. V Furthermore, : V V R such tht 1. ( αu+ βvw, ) = α( uw, ) + β( vw, ), α, β R, uvw,, V,. ( w, αu + βv) = α( w, u) + β( w, v), α, β R, u, v, w V, 3. β : ( u, v) β u v, u, v V, 4. ρ > 0: ( u, u) ρ u, u V. V V V We see from (1)+() tht () is iliner form, (3) sys tht () is ounded, nd (4) sys tht () is coercive. Further, (, ) is symmetric if ( u, v) = ( v, u), u, v V. 11/19/003 35 00 Eric de Sturler
Furthermore, let G : V R such tht 5. G( αu + βv) = αg( u) + βg( v) α, β R, u, v V, δ : G u δ u, u V. 6. ( ) V These properties indicte tht G ( ) is ounded liner functionl. Lx-Milgrm lemm: If properties (1)-(6) hold, then there exists unique û ( uˆ, u) = G( u), u V. V s.t. Furthermore, if (, ) is symmetric, then û V is the unique 1 glol minimizer of f ( u) = ( u, u) G( u), u V. 11/19/003 36 00 Eric de Sturler
Proof (symmetric cse): (1)+(4)+symmetry: (, ) is inner product nd ( u, u ) 1 is norm. (3)+(4): u V nd ( u, u ) 1 re equivlent norms. V complete w.r.t.. V nd equivlence of the norms implies tht V is complete w.r.t... So, V is Hilert spce w.r.t. (, ) nd u. 1 ρ Further, G (.) ounded w.r.t.., since G( u) δ u V u. Hence, the Riesz representtion theorem sys: There exists unique uˆ V : ( uˆ, u) = G( u), u V. 1 ρ ( ) ( ) ( ) ( ) f uˆ+ u = f uˆ + uˆ, u G u + u, u f uˆ + u V > f uˆ [ ] ( ) ( ) ( ) 11/19/003 37 00 Eric de Sturler
Consider PDE Lu = g of order with oundry conditions. To wor with Hilert spce the oundry conditions must e nturl, giving H ( Ω ), or homogeneous essentil oundry conditions, giving H ( Ω ), or comintion of these. 0 In the cse of inhomogeneous oundry conditions we must dpt the prolem (s we sw efore) to go from woring with n ffine vector spce to vector spce. Since ( u, v ) is (Leesgue) integrl over Ω it must contin (generlized) derivtives of order t most. 11/19/003 38 00 Eric de Sturler
Consider PDE Lu = g of order with oundry conditions. To wor with Hilert spce we need either nturl, giving H ( Ω ), or homogeneous essentil oundry conditions, giving H 0 ( Ω ), or comintion of these. For inhomogeneous oundry conditions we must dpt the prolem to go from n ffine vector spce to vector spce. Since ( u, v ) is (Leesgue) integrl over Ω it must contin (generlized) derivtives of order t most, which is gurnteed y the spces H ( Ω ) or H ( Ω ). 0 11/19/003 39 00 Eric de Sturler
( u, v) = G( v), v V corresponds to we form. The PDE is otined (formlly) y integrtion y prts ( u, v) = Lu g, v +nd. integrl G( v) g, v This requires sufficient derivtives of the solution to exist. The coefficient functions must lso collorte. This corresponds to existence of the clssicl solution. We usully don t cre. If clssicl solution exists, the we solution is the sme. If non clssicl solution exists, the we solution is ll we hve. If clssicl solution importnt, we my require, e.g., u H Ω C Ω (ut my not exist depending on eq.) ( ) ( ) 11/19/003 40 00 Eric de Sturler
If the iliner form (.,.) is symmetric it defines n inner product, the energy inner product (nd energy norm). The spce V (.,.,. V V ) we consider to solve PDEs is the Soolev spce of required order with pproprite constrints on oundry. Hence, we re concerned with estlishing the equivlence of the Soolev norm nd the energy norm. G() v = gvdω ounded follows from Cuchy-Schwrz Ω Boundedness of (.,.): usully esy to show Coercivity of (.,.): some equivlent of u', u' u, u ( ) 11/19/003 41 00 Eric de Sturler
Useful Inequlities Friedrichs first inequlity: 1 Let V = { u H ( Ω ) u = 0 on Γ 0 } nd. V =. 1 Ω u dω α u dω for some 0 Friedrichs second inequlity: u dω+ u dγ α u 0 Ω α > nd ll u V. 1, for some 0 Ω Γ Trce inequlity: 1 udγ α u 1, for some α > 0, ll u H ( Ω ). Γ 1 α >, ll u H ( Ω ). 11/19/003 4 00 Eric de Sturler
Exmples PDE: u = g on Ω u = 0 on Γ Proceed s efore: v udω= gvdω. Ω Intgrtion y prts: uxxdω= uvnd x 1 Γ+ uvd x x Ω 11/19/003 43 00 Eric de Sturler Ω Ω Γ Ω Sme for u yy nd some further mnipultion gives uv + uv dω+ vudγ = gvdω (we form) x x y y n Ω Γ Ω 1 Now we clerly wnt u V = H 0 ( Ω ) nd v V lso. Hence, the nd. integrl is zero. ( u, v) = uxvxdω (symmetric, iliner) G() v Ω = gvdω Ω
Exmples ( u, v) = u vdω (symmetric, iliner) Ω G() v = gvdω (liner functionl, ounded y CS for L IP) Ω Using F1 we get coercivity: 1 α ( u, u) = u dω= u dω+ u dω= ρ u 1 Ω Ω Ω To get oundedness: We hve ( u, u) = u dω u dω+ u dω= u 1 Ω Ω Ω 1 1 1 1. This gives ( u, v) ( u, u) ( v, v) u v. Lx-Milgrm stisfied. 11/19/003 44 00 Eric de Sturler
Exmples PDE: u = g on Ω u = 0 on Γ where u V = H 1 0 ( Ω ) Let ( u, v) = u vdω nd G() v Ω = gvdω. Ω Then the PDE corresponds to we form: 1 For ll v H 0 ( Ω ): ( u, v) = G() v This hs unique solution, which is lso the minimum of corresponding qudrtic functionl. 11/19/003 45 00 Eric de Sturler
Exmples PDE: u = g on Ω u = 0 on Γ 0 u n = 0 on Γ Γ 0 Integrtion y prts gives (formlly) uv + uv dω+ vudγ = gvdω x x y y n Ω Γ Ω 1 { } Here we wnt uv, V= w H ( Ω ) w= 0 on Γ We form ecomes uv + uv dω= gvdω x x y y Ω Ω The rest is the sme s the previous exmple. 0 11/19/003 46 00 Eric de Sturler
Exmples PDE: u = g on Ω u + σu = 0 on Γ, where 0 < s σ( x) σ n This gives ( ) ( σ ) 0 1 u vdω + v u + u = gvdω, n Ω Γ Ω nd integrtion y prts gives (formlly) u vdω+ σuv = gvdω Ω Γ Ω 1 Here we wnt uv, V= H ( Ω ) ( u, v) = u vdω+ σuv G() v (symmetric, iliner) Ω = gvdω (liner nd ounded) Ω Γ 11/19/003 47 00 Eric de Sturler
Exmples Coercive: ( ) { }(, = Ω+ σ Γ min 1, σ ) 0 Ω+ Γ u u u d u d u d u d Ω Γ Ω Γ σα ˆ 1 Γ So, (F): (, ) σˆ( ) u u u dω+ u dγ u Ω Bounded: ( u, u) u dω+ σ u dγ u + σ u dγ 1 1 1 Ω Γ Γ Now trce inequlity gives: ( u, u) u + σ u dγ u + σ α u ( 1+ σ α) u 1 1 1 1 1 1 1 Γ 1 1 σα 1 1 1 u v u u v v u v Now Lx-Milgrm cn e pplied. So, (, ) (, ) (, ) ( 1+ ) 11/19/003 48 00 Eric de Sturler