<p> Advective - Diffusive Equation (Steady - State)</p><p>SS : not accumulating or generating any more source effects</p><p>Conservation + Constitutive = PDE  2   q  0 q  D u  V u D  u  V u  0 (D, V constant) (V, D > 0) 2 d u d u 1  D : D 2  V  0 d x d x </p><p>Dimensionless Form : x / L</p><p> 1        1     x L x  x  L </p><p> d 2 u d u  P e  0 V L  d  P e  "Peclet #" d  D ratio of advection to diffusion effects</p><p>FD form :</p><p>2 x U i 2  P e  ?   0 h    x / L h </p><p>ME 515 - Numerical Solution of PDEs - Sullivan, Lec # 9 1 U  U  i) Centered : P i 1 i 1 ; multiplying the FD form e 2 h by h 2 :</p><p>1 + P e h - 2 1 - P e h = 0 2 2 P h Note: Diag. Dominance when e  1 i.e. P h  2 2 e (Weak Form) A 2nd order linear constant coeff. soln => exp( )</p><p>Exact Solution of Difference Equations :</p><p> P e h   P e h  U i 1  1    2 U i  U i 1  1    2   2 </p><p>U i  U i 1 Try i ; solve for  U i   2 U i 1   U i 1</p><p> P e h   P e h   1    2    1   2  0  2   2  1 2   P e h   P e h  2   4  4  1    1   2  2        P e h  2  1    2 </p><p>ME 515 - Numerical Solution of PDEs - Sullivan, Lec # 9 2  2     P e h   1   1   1        2     P h 1  e 2</p><p>P h 1  e 2   P h 1  e 2 1  1 C o n s t a n t </p><p> P e h   1    2  2   P e h   1    2 </p><p> i U i  A  B 2 Exact Soln to PDE : 2  e x p P e h  U  C  D e x p P e </p><p>------</p><p>The following soln satisfies the governing equation: V L  V x  e x p    e x p    D   D  U  V L e x p    1  D </p><p>ME 515 - Numerical Solution of PDEs - Sullivan, Lec # 9 3 V L e x p    D   1 C  D  V L  V L  let e x p    1 e x p    1  D   D  V x L U  C  D e x p    C  D e x p P   D L  e</p><p> i U i  A  B 2  C  D e x p P e  or 2  e x p P e h  i 2  e x p P e i h   e x p P e </p><p>------</p><p>P h 1  e 2 P e h   l e t   2 P h 2 1  e 2 1 2  1  1  </p><p>Binomial Expansion</p><p>Centered Formualtion gives: 2 3  2 3  2  1  1          1  2   2   2   </p><p>2 3 2  2  4 e x p 2   1  2       1  2   2 2  3   2 ! 3 ! 3</p><p>ME 515 - Numerical Solution of PDEs - Sullivan, Lec # 9 4 Ex. BC's U = 1  = 1 , U = 0</p><p>C o n s t .</p><p>U e x p ( P e  P e</p><p>0 </p><p>Numerical : Accuracy depends on agreement between and exp (Peh)</p><p>As h  0 : convergence with 0( ? ) (student try it)</p><p>As h  large : becomes negative! " S p u r i o u s U i O s c i l l a t i o n " w h e n F D P e h > 1 A n a l y t i c 2 </p><p> i . e . P e h > 2</p><p>0 1 </p><p>ME 515 - Numerical Solution of PDEs - Sullivan, Lec # 9 5 V x P h  "Cell Peclet #" e D</p><p> d u ii) Alternative : "Upstream Weighting" of d  i.e. backward F.D. </p><p>2 x U i  U i  U i 1   P e    0 h 2  h </p><p>1 + P e h - 2 - P e h 1 = 0</p><p>Note: Always - (Weak) Diagonal Dominance</p><p>Difference Equations :</p><p>2 1  P e h   2  P e h      0</p><p>2 2  P e h  2 P e h   4 1  P e h    2</p><p>2  P h  P h   e e 2</p><p>ME 515 - Numerical Solution of PDEs - Sullivan, Lec # 9 6 1  1 (Constant) and 2  1  P e h (Never Negative)</p><p>Or in terms of : upstream 2  1  2 </p><p> Upstream Weighting  no spurious oscillation Accuracy : 2 v e r s u s e x p P e h </p><p> d u iii) Alternative : "Downstream Weighting" of d  i.e forward Finite difference  x U i  U i 1  U i   P e    0 h 2  h </p><p>1 - 2 + P e h 1 - P e h = 0</p><p>2 1  2  P e h    1  P e h    0</p><p>1 1  1 and 2  ---> 1  P e h</p><p>Same order approx. to e x p P e h  as case ii (upstream weighting) again in terms of : 2 3  downstream 2  1  2   4   8   </p><p>ME 515 - Numerical Solution of PDEs - Sullivan, Lec # 9 7 But : P e h  1  Oscillations</p><p>Summary:</p><p>Upstream P e h   ; 0 h  2 Centered P e h  2 ; 0 h  Downstream P e h  1 ; 0 h </p><p>ME 515 - Numerical Solution of PDEs - Sullivan, Lec # 9 8 Wave Equation 2 U 2 U   0 t 2  E l l i p t i c i n S p a c e   H y p e r b o l i c O v e r a l l</p><p>U x , t   V x   e j t Periodic Solution:  A m p l i t u d e</p><p>Helmholtz Equation : will always work for a linear hyperbolic eqn.</p><p>2 2 2 2 2 2  V   V  0  x V i , j  y V i , j   h V i , j  0 E l l i p t i c</p><p>2 - D Molecule :</p><p>1</p><p>- 4 1 2 h 2 1</p><p>1 = 0</p><p>ME 515 - Numerical Solution of PDEs - Sullivan, Lec # 9 9</p><p>Diagonal Dominance:</p><p>- lost as h  0 Cannot converge to PDE w/ D.D. - available only when </p><p>2 h 2  4  4 2 h 2  8</p><p> h  Poor accuracy ; for ~ 1% , need 1 0 2  L    s o t h i s i s h    L 2 0 </p><p>Recall a U x x  b U x y  c U y y  d U x  e U y  f U  g a U  c U  f U  g x x y y and assumed f  0 & for f < 0 speed convergence Telegraph Equation</p><p>2 U U 2 U    c 2  0 t 2 t x 2</p><p>is dampening or frictions factor which is necessary It allows initial conditions to decay</p><p>ME 515 - Numerical Solution of PDEs - Sullivan, Lec # 9 10</p>