The wave equation February 23, 2009ME 501B – Engineering Analysis 1Laplace Equation Conclusion and Laplace Equation Conclusion and The Wave EquationThe Wave EquationLarry CarettoMechanical Engineering 501BSeminar in Engineering Seminar in Engineering AnalysisAnalysisFebruary 23, 20092Overview• Review material to date– General approach for solving PDEs• Other ideas about Laplace’s Equation• Derivation and physical meaning of wave equation• Solution of the wave equation by separation of variables• Introduction to the D’Alambert solution of the wave equation• Midterm Exam, Wednesday, March 43Solving PDEs• Perform necessary operations if boundaries not homogenous– Diffusion: define u(x,t) = v(x,t) + w(x)• v satisfies diffusion equation with zero boundary conditions; w satisfies boundary– Laplace equation use superposition• Solution is sum of two or more solutions each of which has only one nonzero boundary• Get equation in appropriate coordinate system (rectangular or cylindrical)4Solving PDEs II• Get appropriate separation of variables solution– Want eigenfunctions to express initial or boundary condition• Use homogenous boundary conditions to determine constants in solutions and eigenvalues• General, sum of all eigenfunctions, used to fit initial or boundary condition5Midterm Exam• Wednesday, March 4• Covers material on diffusion and Laplace equations• Includes material up to and including tonight’s lecture and homework for Monday, March 2• Open book and notes, including homework solutions6Review Spherical Laplace• Laplace’s equation in a sphere has solutions in Legendre polynomials, Pn(x)0sinsin12=⎟⎟⎠⎞⎜⎜⎝⎛∂∂∂∂+⎟⎠⎞⎜⎝⎛∂∂∂∂φφφφururr()∑∞=−−+φ=ϕ01)(cos),(nnnnnnrBrAPru∑=−−−−−=)2/int(02)!2()!(!2)!22()1()(nmmnnnnxmnmnmmnxPThe wave equation February 23, 2009ME 501B – Engineering Analysis 27Review Hollow Cylinder• Consider various boundary conditions • Nonzero conditions on upper or lower surface only gives Bessel eignefunctions• Nonzero conditions on inner or outer surface gives sine or cosine eigenfunctions8Review Hollow Cylinder II• Laplace’s Equation in two-dimensional cylindrical region 0 ≤ z ≤ L and Ri≤ r ≤ Ro– u(r,0) = u(r,L) = u(Ri,z) = 0)(),(0122zuzRuzururrrRo==∂∂+∂∂∂∂LmrKRKRIrIzCzrummmimimmmmπλλλλλλ=⎥⎦⎤⎢⎣⎡−=∑∞=10000)()()()()sin(),([]∫λλλ−λλλ=LRmmimimmimmdzzuzRKRIRKRIRKC00000000)()sin()()()()()(29Review Hollow Cylinder III• Laplace’s Equation in two-dimensional cylindrical region 0 ≤ z ≤ L and Ri≤ r ≤ Ro– u(r,0) = u(Ri,z) = u(Ro,z) = 0)(),(0122ruLruzururrrN==∂∂+∂∂∂∂• Eigenvalues, λm= αm/R0and eigenfunction0)()(000000=⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛RRYJYRRJimmmimαααα[]∑∞=λλ−λλλ=1000000)()()()()sinh(),(mmmmmmmrYRJrJRYzCzruEigenfunction, P0(λmr)10Review Eigenvalues = f(Ri/R0)-0.35-0.25-0.15-0.050.050.150.250 20406080100radius ratio = 0.1radius ratio = 0.5radius ratio = 0.90)()(000000=⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛RRYJYRRJimmmimαααα11Review Hollow Cylinder IV• Eigenfunction expansion in P0(λmr) = Y0(λmR0)J0(λmr) – J0(λmR0) · Y0(λmr)[])]()([2)sinh()()()(020200200RJRJLdrrPrruRJCmimmRRmNimmmiλ−λλλλπλ=∫∑∞=+=100000)()()()()sinh()sinh(),(mmimmimmmRJRJrPRJLzUzruλλλλλλπ• Solution for uN(r) = U, a constant12Review Conclusions• Approach to solving Laplace equation is similar to that of diffusion equation– Main difference is that second dimension (y or r) in Laplace equation gives closed boundary instead of open boundary in time– Use separation of variables– Have eigenfunction solution (sine/cosine, Bessel or other) in one dimension– Use eigenfunction expansion to fit condition at one boundaryThe wave equation February 23, 2009ME 501B – Engineering Analysis 313Review Conclusions II• Use superposition to solve Laplace equation with more than one nonzero boundary• Additional cylindrical geometry considerations– Complex Bessel functions when radial boundary is not eigenfunction solution– Must include both Y0and J0when radial coordinate does not start at zero (must have zero boundary at inner radius)14Vector Calculus• Important results for Laplace equation• See notes on vector calculus or chapters nine and ten in Kreyszig for background details not given here• Results are independent of coordinate system, but Cartesian used for examples• Introduce gradient and divergence which are vector/scalar functions15Gradients• Gradient is a vector in written here in Cartesian space where we have f(x,y,z)zfyfxfffgrad∂∂+∂∂+∂∂=∇= kjizyx ∂∂+∂∂+∂∂=∇ kji• Definition of gradient• Del operator• grad f is magnitude and direction of maximum gradient, df/ds• Grad f is perpendicular to line of constant f16Physical Gradients• Gradients of Laplace equation solutions often proportional to flux terms– Heat flux and temperature gradient– Diffusion flux and mass fraction gradient– Velocity and velocity potential in ideal flow– Current and electrostatic potential• If we have a plot of constant potential the lines perpendicular to the potential are flux lines17 18Divergence• Divergence converts vector, v = vxi + vyj+ vzk, into a scalar written as div vzvyvxvdivzyx∂∂+∂∂+∂∂=⋅∇= vvzyx ∂∂+∂∂+∂∂=∇ kji• Definition of divergence• Del operator• Gauss divergence theorem (n is vector normal to surface, pointing outward)dAdVdivSurfaceVolumeEnclosednvv ⋅=∫∫∫∫∫The wave equation February 23, 2009ME 501B – Engineering Analysis 419• Example of heat flux vector, q, (W/m2)• q·n is component of q, normal to surface, dA, flowing outward• Integrand in surface integral, q·ndA is heat flow (watts) flowing out through infinitesimal area, dA• Surface integral gives total heat flow through surface in outward directiondAdVdivSurfaceVolumeEnclosednqq ⋅=∫∫∫∫∫20Relation to Laplace Equation• The vector, v, may be gradient of a scalar, representing a flux: v = -k grad u()dAdVugradkdivdVdivSurfaceVolumeEnclosedVolumeEnclosednvv ⋅=−=∫∫∫∫∫∫∫∫• For constant k ()dAkudVdVugraddivSurfaceVolumeEnclosedVolumeEnclosednv⋅−=∇=∫∫∫∫∫∫∫∫1221Interpretation of 02=∇ u• When v = -k grad u is a flux that is the gradient of a scalar, Laplace’s equation for u says that