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WEBSITE : http://maths-minhthe.violet.vn NUMERICAL ANALYSIS PROBLEM DIFFERENTIAL EQUATION II 6200 1. Answer : From the figure above, we have : T T T T T T T min T min min T h T 1 1 c s (c s) cot cot ,c h , : the radius of the inscribed circle 2 2 2 2 Let the three int erior angles of T be a b c h 1 1 cot cot 2 2 2 2 We have : by a cons tant for all h h T then for som α β = + − = ρ + ρ = ρ θ = α ≤ β ≤ γ ⇒ ≤ ≤ α β ⇒ = + ρ θ ≤ θ ⇔ θ ≤ α ≤ β ≤ γ θ ∀ ∈ Τ ≤ σ ρ T T min T T min min e h 1 1 We have : cot cot 2 2 2 2 Because and cot is a decrea sing function so h 1 1 cot cot 2 2 2 2 cot cot where cot 2 2 2 σ α β = + ρ θ ≤ α ≤ β α α ≤ + ρ α θ θ ≤ ≤ σ = 2. Answer : a) Difference between regular and uniform grid T T T h T T T h Regular : , such that h h h,where h maxh ,T h Uniform grid : , h max h , K 0 K The regular grid is stronger than the uniform grid b)The mesh in the figure includes both uniform and regular grid ≤ σ ∃α α ≤ ≤ = ∈ Τ ρ ≤ σ = > 3. Answer : j j j 1 j 1 j 1 1 j z j 1 z z j h a)Show that tan x 2 h tan x 2 h h tan dx tan .x 2 2 We know that : 1 h h tan .x 1 x tan 2 2 h This is true when we have known that z tan , 0 2 − − − − − ∂ φ = θ ∂ ∂ φ = θ ∂ ⇒ φ = θ = θ φ = ⇔ θ = ⇒ = θ = θ ∫ ( ) 2 2 j I b)Consider the function u(x,y) y . Its linear int erpo tant vanishes at z and equals h / 4 at the two other vertices Show that h u u x 2tan = ∂ − = ∂ θ ( ) ( ) j j j j j I 2 2 z I z 2 z 2 z z j Answer : h u u x 2tan h With u(x,y) y (u) 4 h h u u dx .x 2tan 2tan We know that : h (u) 4 h h h .x x tan 2tan 4 2 h This is true when we have known that z tan , 0 2 From the parts above, we can see that Genera ∂ − = ∂ θ = ⇒ φ = ⇒ − = = θ θ φ = ⇔ = ⇒ = θ θ = θ ∫ lly, then the are of traingle T,which have creat ed by (0,h / 2);(0, h / 2) and (h / 2tan ,0) is− θ j j 2 z 2 0 z 1 h h 1 h h h (u) . . tan . . tan tan 2 2 2 2 2 2 4 h Hence, if (u) then tan 1 45 4 But, we have the shape regularity condition : φ = θ + θ = θ φ = θ = ⇒ θ = T min T min h cot cot 2 2 1 tan 2 θ θ = ≤ ρ ≤ θ min T T 0 min min tan tan 2 h 2 45 tan 1 Thus,the tan may be cause large error θ ρ θ ⇔ ≤ = θ ≤ θ = θ ≤ θ 4. " Consider the BVP problem : find a periodic funct ion u such that u (x) f(x) on ( 1,1 )− = Ω = − a) State the weak formulation a(u,v) (v) = Answer : First. We multiply the equation by test function v(x) [ ] " 1 1 1 " 1 1 ' " ' 1 1 " ' ' ' 1 1 1 ' ' ' ' 1 u v fv on ( 1,1 ), v H ( 1,1 ) We compute the weak form by int egrating over 1,1 u v dx fv dx Set p v dp v dx dq u dx q u 1 So : u v dx u .v u v dx 1 u (1).v(1) u ( 1).v( 1) u v dx Where we have used t − − − − − − = Ω = − ∈ − − − = = ⇒ = = ⇒ = − = − + = − = − + − − + ∫ ∫ ∫ ∫ ∫ ' ' 1 1 ' ' 1 1 1 2 1 hat u( 1) u(1) 0 u ( 1) u (1) 0 Hence, a(u,v) u v dx , (v) fv dx u( 1) u(1) 0 f L and f dx 0 so u c is a solution − − − − = = ⇒ − = = = = − = = ∈ = + ∫ ∫ ∫ b) Use the Lax-Milgram theorem to prove the existence and stability of the solution Recall : The Lax–Milgram theorem This is a formulation of the Lax–Milgram theorem which relies on properties of the symmetric part of the bilinear form. It is not the most general form. Let V be a Hilbert space and a bilinear form on V , which is 1. bounded: and 2. coercive: there is a unique solution to the equation a ( u , v ) = l ( v ) Answer : + We will prove the property 1 : bounded: We have ( ) 1 1 ' ' 1 1 1 a(u,v) u v dx , (v) fv dx ,H H − − = = = Ω ∫ ∫ 1 1 ' ' ' ' 1 1 1/2 1/2 1 1 2 2 ' ' 1 1 H H a(u,v) u v dx u v dx u dx v dx (applying to Cauchy Schazt inequality) C. u . v ,C 1 − − − − = ≤ ≤ − ≤ = ∫ ∫ ∫ ∫ 1/2 1/2 1 1 1 2 2 1 1 1 (u) fv dx f dx v dx − − − = ≤ ∫ ∫ ∫ 2 2 L . H L . H f v ,C f C. v ≤ = ≤ + We continue to prove the property 2 : coercive: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 1 L L 2 x x 2 2 ' 2 ' 1 1 2 2 2 2 2 2 ' L L We will prove the Poincare inequality : u C. u , u H We see that u (x) u( 1) u (t)dt 2 u ( 1) u (t) dt for 1 x 1 ( We have used Cauchy 's inequality : a b 2 a b ) We have : u 2. u ( 1) u For Ω Ω − − Ω Ω ≤ ∇ ∀ ∈ Ω = − + ≤ − + − < < + ≤ + ≤ − + ∫ ∫ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 2 2 2 2 L L L L L 2 L 1 1 L 2 2 ' L L 2 2 2 2 2 2 ' ' ' ' H 2 ' ' ' 2 H 1 1 u H with u( 1) u(1) 0 u 0 , is boundary u 2 u Poincare inequality means : u u u u 2 u 3 u 3a u,u Where a(u,u) uu dx u 1 Thus, a u,u u 3 ! u H such that a u,u (u), v H Ω Ω Ω Ω Ω Ω Γ Ω Ω Ω Ω Ω ∈ Ω − = = ⇒ = Γ ⇒ ≤ ≤ + ≤ + = = = = ≥ ⇒ ∃ ∈ Ω = ∀ ∈ ∫ ( ) Ω c) 0o 1 2 n n h n n n n n ij ji j i i j n n h i i i i n n i i i 1 i 1 i n n n 1 n i i i 1 i n h 1 n 2 Partition as 1 x x x . . . x 1 u H H a(u ,v ) (v ) , v H The stiffness matrix is symmetric if A A a( , ) a( , ) We have u u u u u u ( ) u So H span , , . = ≠ ≠ ≠ ≠ Ω − = < < < < = ∈ ⊂ = ∀ ∈ = ⇔ φ φ = φ φ = ϕ = ϕ + ϕ + ϕ = ϕ + ϕ + ϕ = ϕ + ϕ ϕ ∑ ∑ ∑ { } n 1 . . , − ϕ ij ij ij i 1,j ij i 1,j i,j 1 ij i,j 1 u u f u 4u u u 4u u 6 6 + + − + + = + + + + ⇔ + Thus, 2 ij 1 4 1 h A 4 16 4 36 1 4 1 = . This stiffness matrix is symmetric d) Set up a set equations based on finite differenc es 1 0 1 1 2 3 4 j j i i ij From the figure above ,we have the equations : (1 y)(1 x) ; x(1 y) ; xy ; (1 x)y Using the Intergrating to compute K h ( )dx dy ; [0,h] [0,h] x x y y ϕ = − − ϕ = − ϕ = ϕ = − ∂ϕ ∂ϕ ∂ϕ ∂ϕ = + × ∂ ∂ ∂ ∂ ∫ 1 0 1 1 1 1 1 11 3 3 1 1 2 2 0 0 K ( )dxdy ; [0,1] [0,1] x x y y 1 1 (1 y) (1 x) 2 h (1 y) (1 x) dxdy h h 3 0 3 0 3 ∂ϕ ∂ϕ ∂ϕ ∂ϕ = + = × ∂ ∂ ∂ ∂ − − = − + − = + = ∫ ∫ ∫ 1 0 1 2 2 2 2 22 3 3 1 1 2 2 0 0 K h ( )dxdy x x y y 1 1 (1 y) x 2 h (1 y) x dxdy h h 3 0 3 0 3 ∂ϕ ∂ϕ ∂ϕ ∂ϕ = + ∂ ∂ ∂ ∂ − = − + = + + = ∫ ∫ ∫ 1 0 1 3 3 3 3 33 3 3 1 1 2 2 0 0 K h ( )dxdy x x y y 1 1 (1 y) x 2 h (1 y) x dx dy h h 3 0 3 0 3 ∧ ∧ ∂ϕ ∂ϕ ∂ϕ ∂ϕ = + ∂ ∂ ∂ ∂ − = − + = + + = ∫ ∫ ∫ 1 0 1 1 2 1 2 12 21 3 2 3 1 1 2 0 0 K K ( )dxdy ; [0,1] [0,1] x x y y 1 1 (1 y) x x 1 h (1 y) (1 x)x dxdy h h 3 0 2 3 0 2 ∂ϕ ∂ϕ ∂ϕ ∂ϕ = = + = × ∂ ∂ ∂ ∂ − = − − + − = − + − = ∫ ∫ ∫ [ ] 1 0 1 3 3 1 1 13 31 2 3 2 3 1 1 0 0 K K h ( )dxdy x x y y 1 1 1 1 y y x x 1 h (1 y)y (1 x)x dx dy h h 2 0 3 0 2 0 3 0 3 ∂ϕ ∂ϕ ∂ϕ ∂ϕ = = + ∂ ∂ ∂ ∂ = − − − − = − + − + = − ∫ ∫ ∫ 1 0 1 3 3 2 2 23 32 2 3 3 1 1 2 0 0 K K h ( )dxdy x x y y 1 1 y y x 1 h (1 y)y x dxdy h h 2 3 0 3 0 2 ∂ϕ ∂ϕ ∂ϕ ∂ϕ = = + ∂ ∂ ∂ ∂ = − − = − + + = ∫ ∫ ∫ The element stiffness matrix 2 1 1 3 2 3 1 2 1 K h 2 3 2 1 1 2 3 2 3 − ⇒ = − We have the matrix four the aparts so : ( ) ( ) ( ) ( ) ( ) ( ) ij i j 1 1 1 2 1 1 2 3 1 3 3 1 A a , 2 8 a , 4 3 3 1 a , . d x d y 2 1 a , 2 1 a , a , 3 1 3 1 1 A 3 8 3 3 1 3 1 = ϕ ϕ ⇒ ϕ ϕ = × = ⇒ ϕ ϕ = ∇ ϕ ∇ ϕ = × = = ϕ ϕ ϕ ϕ = ϕ ϕ = − − − ⇒ = − − ∫ e) We have 1 ' ' 1 1 1 a(u,v) u (x)v (x)dx (v) f(x)v(x)dx a(.,.) is binear form − − = = ∫ ∫ 0 1 n 1 n 1 x x . . . x x 1 − − = < < < < < [...]... u" L2 ( −1,1) Thus,Ce'a ' s lemma can be applied along the same lines to devive error estimates for finite element and u sin g higher order polynomials of the subspace Vh Bonus Problem 2 : From the figure above , we have the equations : N1(x,y) = (1 − y)(1 − x) ; N2 (x, y) = x(1 − y) ; N3 (x,y) = xy ;N4 (x, y) = (1 − x)y The stiffness matrix form : ∂N ∂Nj ∂Ni ∂Nj + K ij = ∫ i dxdy i, j = 1 ,2... the local stencil 2 8 ×4 = 3 3 −1 −1 At B ,C,D,E : ×2 = 6 3 Thus, the local stencil for ∆ u u sin g the bilinears is : At A : −1 1 K = −1 3 −1 −1 8 −1 − 1 − 1 − 1 And computing as the problem 4c) we have the local stencil for the mass matrix is 1 4 1 h2 Thus, M = 36 4 16 4 1 4 1 . WEBSITE : http://maths-minhthe.violet.vn NUMERICAL ANALYSIS PROBLEM DIFFERENTIAL EQUATION II 6200 1. Answer : From the figure above, we have. matrix is symmetric d) Set up a set equations based on finite differenc es 1 0 1 1 2 3 4 j j i i ij From the figure above ,we have the equations : (1 y)(1 x) ; x(1 y) ; xy. polynomials of the subspace V + − ⇒ − ≤ Bonus Problem 2 : ( ) 1 2 3 4 1 j j i i ij 0 1 2 2 11 0 13 From the figure above ,we have the equations : N (x,y) (1 y)(1 x) ; N (x,y) x(1 y)