SOLUTION OF Partial Differential Equations (PDEs)

1 1 SOLUTION OFPartial Differential Equations (PDEs) Mathematics is the Language of SciencePDEs are the expression of processes that occur across time & space: (x,t), (x,y), (x,y,z), or (x,y,z,t)2 Partial Differential Equations (PDE's)A PDEis an equation which includes derivatives of an unknown function with respect to 2 or moreindependentvariables3 Partial Differential Equations (PDE's)PDE's describe the behavior of many engineering phenomena: Wave propagation Fluid flow (air or liquid)Air around wings, helicopter blade, atmosphereWater in pipes or porous mediaMaterial transport and diffusion in air or waterWeather: large system of coupled PDE's for momentum, pressure, moisture, heat.

2 Vibration Mechanics of solids: stress-strain in material , machine part, structure Heat flow and distribution Electric fields and potentials Diffusion of chemicals in air or water Electromagnetism and quantum mechanics4 Partial Differential Equations (PDE's)Weather Prediction heat transport & cooling advection & dispersion of moisture radiation & solar heating evaporation air (movement, friction, momentum, coriolis forces) heat transfer at the surfaceTo predict weather one need "only" solve a very large systems ofcoupled PDE Equations for momentum, pressure, moisture, heat, n de energ a, masa, momento, vapor de agua,ecuaci n de estado de gases.

3 V= (u, v, w), T, p, = 1/ y q360x180x32x nvarresoluci n 1/ tDuplicar la resoluci n espacial supone incrementar el tiempo de c mputo en un factor 16 Modelizaci n Num rica del TiempoD+0D+1D+2 Condici n inicial H+0 H+24D+3D+4D+5s bado 15/12/2007 006 Partial Differential Equations (PDE's)Learning Objectives1) Be able to distinguish between the 3 classes of 2nd order, linear PDE's. Know the physical problems each class represents and the physical/mathematical characteristics of ) Be able to describe the differences between finite-difference and finite-element methods for solving ) Be able to solve Elliptical (Laplace/Poisson) PDEs using finite ) Be able to solve Parabolic (Heat/Diffusion) PDEs using finite Differential Equations (PDE's)Engrd 241 Focus:Linear 2nd-Order PDE's of the general formu(x,y), A(x,y), B(x,y), C(x,y), and D(x,y,u,,)The PDE is nonlinear if A, B or C include u, u/ x or u/ y, or if D is nonlinear in u and/or its first 4AC < 0 > Elliptic( Laplace Eq.

4 B2 4AC = 0 > Parabolic( Heat Eq.)B2 4AC > 0 > Hyperbolic( Wave Eq.) Each category describes different phenomena. Mathematical properties correspond to those CD0xyxy +++= 8 Partial Differential Equations (PDE's)Typical examples includeuuuu(x,y),(in terms ofand)xy Elliptic Equations (B2 4AC < 0) [steady-state in time] typically characterize steady-state systems (no time derivative) temperature torsion pressure membrane displacement electrical potential closed domain with boundary conditions expressed in terms ofA = 1, B = 0, C = 1 ==> B2 4AC = 4 < 0222220uuuuuD(x, y,u,,)xyxy + = Laplace Eq. Poisson PDEs Boundary Conditions for Elliptic PDE's:Dirichlet: u provided along all of edgeNeumann:provided along all of the edge (derivative in normal direction)Mixed:u provided for some of the edge and for the remainder of the edge Elliptic PDE's are analogous to Boundary Value ODE'su u 10 Elliptic PDEsEstimateu(x)fromLaplaceEquationsu(x= 1,y)or u/ xgivenonboundaryu(x,y =0) or u/ y given on boundaryyxu(x,y =1) or u/ y given on boundaryu(x=0,y)or u/ xgivenonboundary11 Parabolic PDEs Parabolic Equations (B2 4AC = 0) [first derivative in time ] variation in both space (x,y) and time, t typically provided are: initial values.

5 U(x,y,t = 0) boundary conditions: u(x = xo,y = yo, t) for all tu(x = xf,y = yf, t) for all t all changes are propagated forward in time, , nothing goes backward in time; changes are propagated across space at decreasing PDEs Parabolic Equations (B2 4AC = 0) [first derivative in time ] Typical example: Heat Conduction or Diffusion(the Advection-Diffusion Equation)22uxuu1D :kD(x, u,)tx =+ 2222uuxyuuu2D :kD(x, y, u,,)txy =++ A = k, B = 0, C = 0 > B2 4AC = 02kuD= +13 Parabolic PDEsEstimateu(x,t)fromthe HeatEquationu(x=1,t)givenonboundaryfor all tu(x,t =0) given on boundaryas initial conditions for u/ ttxu(x=0,t)givenonboundaryfor all t14 Parabolic PDEsx=L An elongated reactor with a single entry and exit point and a uniform cross-section of area A.

6 A mass balance is developed for a finite segment xalong the tank's longitudinal axis in order to derive a Differential equation for concentration (V = A x). xx=0c(x,t) = concentrationat time, t, and distance, PDEsx=L xx=0cc(x)c(x)VQc(x)Qc(x)xDAtxx = + Flow inFlow outDispersion inc(x)c(x)DAxkVc(x)xxx ++ Dispersion outDecayreaction22cc cQcDkcttAxx ==> = As t and x ==> 016 Hyperbolic PDEs Hyperbolic Equations (B2 4AC > 0) [2nd derivative in time ] variation in both space (x, y) and time, t requires: initial values: u(x,y,t=0), u/ t (x,y,t = 0) "initial velocity" boundary conditions: u(x = xo,y = yo, t) for all tu(x = xf,y = yf, t) for all t all changes are propagated forward in time, , nothing goes backward in PDEsHyperbolic Equations (B2 4AC > 0) [2nd derivative in time] Typical example: Wave Equation A = 1, B = 0, C = -1/c2 ==> B2 4AC = 4/c2> 0 Models vibrating string water waves voltage change in a wire22222uuxtu1 u1D.

7 D ( x , y, u ,,)0xct += 18 Hyperbolic PDEsEstimateu(x,t)fromthe WaveEquationu(x=1,t)givenonboundaryfor all tu(x,t =0) and u(x,t=0)/ t given on boundaryas initial conditions for 2u/ t2 Now PDE is second order in timetxu(x=0,t)givenonboundaryfor all t19 Numerical Methods for Solving PDEsNumerical methods for solving different types of PDE's reflect the different character of the problems. Laplace- solve all at once for steady state conditions Parabolic(heat) and Hyperbolic(wave) Equations . Integrate initial conditions forward through Finite Difference (FD) Approaches (C&C Chs. 29 & 30)Based on approximating SOLUTION at a finite # of points, usuallyarranged in a regular grid.

8 Finite Element (FE) Method (C&C Ch. 31)Based on approximating SOLUTION on an assemblage of simply shaped (triangular, quadrilateral) finite pieces or "elements" which together make up (perhaps complexly shaped) this course, we concentrate on FD applied to elliptic and parabolic Difference for Solving Elliptic PDE's Solving Elliptic PDE's: Solve all at once Liebmann Method: Based on Boundary Conditions (BCs) and finite difference approximation to formulate system of Equations Use Gauss-Seidel to solve the system 2222y0uuuuD(x, y, u,,)xxy += Laplace Difference Methods for Solving Elliptic PDE's1. Discretize domain into grid of evenly spaced points2.

9 For nodes where u is unknown: w/ x = y = h, substitute into main equation3. Using Boundary Conditions, write, n*m Equations for u(xi=1:m,yj=1:n) or n*m Solve this banded system with an efficient scheme. Using Gauss-Seidel iteratively yields the Liebmann ,ji,ji1,j2222u2uuxO( x )ux() + + =+222222i 1,ji 1,ji,j 1i,j 1i,juuuu4uuuO(h )xyh + ++++ +=+i, j 1i,ji, j 12222u2uuyO( y )uy() + + =+22 Elliptical PDEs The Laplace Equation2222yuu0x += The Laplace moleculeii+1i-1j-1j+1ji 1, ji 1, ji, j 1i, j 1i, jTTTT4T0+ + +++ =If x= ythen23 Elliptical PDEs The Laplace molecule:T = 0o CT = 100o CT = 100o CT = 0o C111213212223313233 TTTT T T TT T111213212223313233-100T-TTTT=TTTT-41010 00001-4101000001-4001000100-4101000101-4 101000101-4001000100-4100000101-41000001 01-4 100-20000-10000-100 1,11,21,32,12,22,33,13,23,3 The temperature distribution can be estimated by discretizingthe Laplace equation at 9 points and solving the system of linear 1, ji 1, ji, j 1i, j 1i, jTTTT4T0+ + +++ =Excel24 SOLUTION of Elliptic PDE's: Additional Factors Primary (solve for first): u(x,y) = T(x,y) = temperature distribution Secondary (solve for second):heat flux.

10 Obtain by employing:xyTTqk andqkxy = = i1,ji1,jTTTx2x+ i, j 1i, j 1 TTTy2y + then obtain resultant flux and direction:22nxyqqq=+y1xxqtanq0q => y1xxqtanq0q =+ < (with in radians)25 Elliptical PDEs: additional factorsi1,ji1,jxTTTqkkx2x+ = i, j 1i, j 1yTTTqkky2y+ = k' = cal/s cm CAt point 2,1 (middle left):qx~ (50-0)/(2 10cm) = cal/(cm2 s)qy~ ( )/(2 10cm) = (cm2 s) T = 0o CT = 100o CT = 100o CT = 0o cal /(cm s)=+= + = ===+ = ooo26 SOLUTION of Elliptic PDE's: Additional FactorsNeumann Boundary Conditions(derivatives at edges) employ phantom points outside of domain use FD to obtain information at phantom point,T1,j+ T-1,j+ T0,j+1+ T0,j-1 4T0,j= 0 [*]If given then use to obtainSubstituting [*]:Irregular boundaries use unevenly spaced molecules close to edge use finer meshTx 1, ji 1, jTTTx2x = 1, j1, jTTT2xx = 1,j0,j 10,j 10,jT2T2 xTT4 T0x+ ++ = 27 Elliptical PDEs: Derivative Boundary Conditions The Laplace molecule:Insulated ==> T/ y = 0T = 50o CT = 100o CT = 75o C1,11,21,32,12,22,33,13,23,3i 1, ji 1, ji, j 1i, j 1i, jTTTT4T0+ + +++ =5,13,1 TTT2yy = 0 Derivative (Neumann) BC at (4,1).