16.920J/SMA 5212 Numerical Methods for PDEs 4. Substituting the ax's into u, +2au +=S+S2+S,+S+.yields 2△x(1“ △x2·l,+ higher- order terms In other words, l--+0(△x2)+ dx 0(△x2),p i.e. the above representation is accurate up to O(4r) Some useful points to note These 4 steps are the general procedure used to obtain the representation of the spatial operator up to the order of accuracy O(ArP) d2u For other spatial operators, say we simply replace/du in(1)with dx d the said spatial operator 3. For one-sided representations, one can choose nodal points u,k,k>0. This may be important especially for representations on a boundary. For example +a+am+a2+…=0(△) dt One possibility is dx 2Ax which is also second-order accurate (We can also use a similar procedure to construct the finite difference scheme of Hermitian type for a spatial operator. This is not covered here) 6
16.920J/SMA 5212 Numerical Methods for PDEs 6 4. Substituting the αk ’s into 1 1 2 3 4 1 .... k j k j k k u α u S S S S = + =− ′ + = + + + + yields ( ) 2 1 1 1 1 2 6 j j j j u u u x u x + − ′ ′′′ − − = ∆ ⋅ + ∆ higher-order terms In other words, ( ) 1 1 2 .... 2 j j j j du u u u O x dx x + − − ✁ ✂ ′ = = + ∆ + ✄ ☎ ∆ ✆ ✝ i.e. the above representation is accurate up to ( ) 2 O ∆x . Some useful points to note: 1. These 4 steps are the general procedure used to obtain the representation of the spatial operator up to the order of accuracy ( ) p O ∆x . 2. For other spatial operators, say 2 2 j d u dx ✞ ✟ ✠ ✡ ☛ ☞ , we simply replace j du dx ✌ ✍ ✎ ✏ ✑ ✒ in (1) with the said spatial operator. 3. For one-sided representations, one can choose nodal points uj+k , k ≥ 0 . This may be important especially for representations on a boundary. For example 0 1 1 2 2 .... ( ) p j j j j du u u u O x dx α α + α + ✓ ✔ + + + + = ∆ ✕ ✖ ✗ ✘ One possibility is ( ) 1 2 2 3 4 2 j j j j du u u u O x dx x − + + + ✙ ✚ + = ∆ ✛ ✜ ∆ ✢ ✣ which is also second-order accurate. (We can also use a similar procedure to construct the finite difference scheme of Hermitian type for a spatial operator. This is not covered here). ( ), 2 p O ∆x p =
16.920J/SMA 5212 Numerical Methods for PDEs STABILITY ANALYSIS Discretization We obtain at x dt Ar(uo-2u+"2) dax2(4-2n2+n) 动=a2(1-21+1 N-1 (a2-2ux=1+lx) dt Note that we need not evaluate u at x=x, and x=XN since u, and u, are given as boundary conditions Side 6 STABILITY ANALYSIS Matrix Formulation Assembling the system of equations, we obtain dt 0 dhu,△x2 dt 0 1-2 DUN Slide 7 7
16.920J/SMA 5212 Numerical Methods for PDEs 7 STABILITY ANALYSIS Discretization We obtain at 1 1 2 1 2 : ( 2 ) o du x u u u dt x υ = − + ∆ 2 2 2 1 2 3 : ( 2 ) du x u u u dt x υ = − + ∆ 2 1 1 : ( 2 ) j j j j j du x u u u dt x υ = − − + + ∆ 1 1 2 2 1 : ( 2 ) N N N N N du x u u u dt x − υ − = − − − + ∆ 0 0 Note that we need not evaluate at and since and are given as boundary conditions. N N u x x x x u u = = Slide 6 STABILITY ANALYSIS Matrix Formulation Assembling the system of equations, we obtain Slide 7 1 1 2 2 2 2 1 1 2 2 1 0 1 2 1 0 1 2 1 1 0 1 2 o j j N N N du u u dt x du u dt du x u dt u du u x dt υ υ υ − − ✁ ✁ − ✂ ✄ ✁ ✁ ✂ ✄ ∆ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ − ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ = + ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ∆ − ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✂ ✄ ✂ ✄ ✂ ✄ ✂ ✂ ✄ ✂ ✄ − ✂ ✄ ☎ ✆✝☎ ✆ ✂ ✂ ∆☎ ✆ ✂☎ ✆✄ ✄ ✄ ✄ 0 ✄ 0 A
16.920J/SMA 5212 Numerical Methods for PDEs STABILITY ANALYSIS PDE to Coupled ODEs Or in compact form du lu+b where u=[a1u2…… △x2 We have reduced the I-D PDe to a set of Coupled OdEs! Slide 8 STABILITY ANALYSIS Eigenvalue and Eigenvector of Matrix A If A is a nonsingular matrix, as in this case, it is then possible to find a set of eigenvalues 元={4,2,…,…,不- A-11=0 For each eigenvalue 1. we can evaluate the eigenvector p consisting of a set of mesh point values v/,i.e y=Y vy STABILITY ANALYSIS Eigenvalue and Eigenvector of Matrix A The(N-1)x(N-1)matrix E formed by the(N-1)columns r diagonalizes the matrix A by E-AE=A
16.920J/SMA 5212 Numerical Methods for PDEs 8 STABILITY ANALYSIS PDE to Coupled ODEs Or in compact form We have reduced the 1-D PDE to a set of Coupled ODEs! Slide 8 STABILITY ANALYSIS Eigenvalue and Eigenvector of Matrix A If A is a nonsingular matrix, as in this case, it is then possible to find a set of eigenvalues λ = {λ1 ,λ2 ,....,λ j ,....,λN −1} from det( A− λI ) = 0. For each eigenvalue , we can evaluate the eigenvector consisting of a set of mesh point values , i.e. j j j i V v λ Slide 9 STABILITY ANALYSIS Eigenvalue and Eigenvector of Matrix A The ( 1) ( 1) matrix formed by the ( 1) columns diagonalizes the matrix by j N N E N V A − × − − 1 E AE − = Λ where [ 1 2 1 ] T N u u u u = − 2 2 0 0 0 T o N u u b x x υ υ ✁ ✂ = ✄ ☎ ∆ ∆ ✆ ✝ ✞ du Au b dt = + ✟ ✟✠✟ 1 2 1 Tj j j j V N v v v − ✡ ☛ = ☞ ✌
16.920J/SMA 5212 Numerical Methods for PDEs 0 0 Slide 10 STABILITY ANALYSIS Coupled oDEs to Uncoupled ODEs Starting fre Au+b Premultiplication by E yield e-ld e-l4u+E-b AEE)b+E-b E=(EAEE-b+E-b STABILITY ANALYSIS Coupled oDEs to Uncoupled ODEs Continuing from di AE-u+E-6 Letu=E-i and F=E-B,we have
16.920J/SMA 5212 Numerical Methods for PDEs 9 Slide 10 STABILITY ANALYSIS Coupled ODEs to Uncoupled ODEs Starting from du Au b dt = + ✟ ✟✠✟ 1 Premultiplication by E yields − 1 du 1 1 E E Au E b dt − − − = + Slide 11 STABILITY ANALYSIS Coupled ODEs to Uncoupled ODEs Continuing from 1 du 1 1 E E u E b dt − − − = Λ + 1 1 Let U E u and F E b, we have − − = = ✁ ✁ ✁ ✁ 1 2 1 where N λ λ λ − ✂ ✄ ☎ ✆ ☎ ✆ ☎ ✆ Λ = ☎ ✆ ☎ ✆ ☎ ✆ ✝ ✞ 0 0 ( ) 1 du 1 1 1 E E A EE u E b dt − − − − = + ✟ ✟ ✟ I Λ ( ) 1 du 1 1 1 E E AE E u E b dt − − − − = + ✠ ✠ ✠
16.920J/SMA 5212 Numerical Methods for PDEs di=au+F which is a set of Uncoupled ODEs STABILITY ANALYSIS Coupled oDEs to Uncoupled ODEs Expanding yields dU=U1+F due=au, +F2 d=4+F Since the equations are independent of one another, they be solved The idea then is to solve for u and determine i= eU STABILITY ANALYSIS Coupled oDEs to Uncoupled ODEs Considering the case of b independent of time, for the general j"equation is the solution forj=1, 2, . N-l
16.920J/SMA 5212 Numerical Methods for PDEs 10 d U U F dt = Λ + ✁ ✁ ✁ which is a set of Uncoupled ODEs! Slide 12 STABILITY ANALYSIS Coupled ODEs to Uncoupled ODEs Expanding yields 1 1 1 1 dU U F dt = λ + 2 2 2 2 dU U F dt = λ + j j j j dU U F dt = λ + 1 1 1 1 N N N N dU U F dt λ − = − − + − Since the equations are independent of one another, they can be solved separately. The idea then is to solve for U and determine u = EU ✂ ✂ ✂ Slide 13 STABILITY ANALYSIS Coupled ODEs to Uncoupled ODEs Considering the case of independent of time, for the general equation, th b j ✄ jt 1 j j j j U c e F λ λ = − is the solution for j = 1,2,….,N−1