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Power and Justice

by: Boris Kuhn

Power and Justice POLI 13

Boris Kuhn

GPA 3.81


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This 32 page Class Notes was uploaded by Boris Kuhn on Thursday October 22, 2015. The Class Notes belongs to POLI 13 at University of California - San Diego taught by Staff in Fall. Since its upload, it has received 11 views. For similar materials see /class/226801/poli-13-university-of-california-san-diego in Political Science at University of California - San Diego.

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Date Created: 10/22/15
PDE LECTURE NOTES MATH 237AB BRUCE K DRIVERl ABSTRACT These are lecture notes from Math 237A7B See CdriverdatBruce CLASSFIL257AF94coursetex for notes on contraction semi7groups Need to add examples of using the Hille Yoshida theorem in PDE See See C driverdat Bruce DATA MATHFILE qftenotescoeareatex for co area material and applications to Sobolev inequalities CONTENTS 1 Some Examples 11 Some More Geometric Examples 2 First Order QuasieLinear Scalar PDE 21 Linear Evolution Equations 2 2 General Linear First Order PDE 23 QuasieLinear Equations 24 Distribution Solutions for Conservation Laws 25 Exercises 3 Fully nonlinear rst order PDE 31 An Introduction to Hamilton Jacobi Equations 4 Cauchy 7 Kovaleski Theorem 5 Test Functions and Partitions of Unity 5 1 Convolution 6 Cutoff functions and Partitions of Unity 7 Surfaces and Surface Integrals 8 Laplace s and Poisson s Equation 9 Laplacian in polar coordinates 91 Laplace s Equation Poisson Equation 10 Estimates on Harmonic functions 11 Green s Functions 12 A Little Distribution Theory 121 Max Principle a la Hamilton 13 Wave Equation 131 Corresponding rst order ODE 132 Spherical Means 14 Old Section Stuff Date October 23 2002 Fileszpdeiex 1 Department of Mathematics 0112 University of California San Diego La Jolla CA 9209370112 118 BRUCE K DRIVER39 141 Section 2 118 142 Section 3 120 15 Solutions to Exercises 121 151 Section 2 Solutions 121 PDE LECTURE NOTES MATH 237A7B 1 1 SOME EXAMPLES Example 11 Traffic Equation Consider cars travelling on a straight road ie R and let ut ac denote the density of cars on the road at time t and space at and vt at be the velocity of the cars at t Then for J 11 C R NJt futxdac is the number of cars in the set J at time t We must have b 12t acdac NJt ut avta 7 ut bvt b a 7 ib w mm mm a 8x 7 7 I Since this holds for all intervals 1 b we must have 390 d7 8W t 1 u 1 3577836 u 3512 35 To make life more interesting we may imagine that vt ac 7Fut acut 1 in which case we get an equation of the form 8 8 Eu EGW uf where Gu uf 7ut acFut ac ut A simple model might be that there is a constant maximum speed Um and maXi7 mum density um and the traffic interpolates linearly between 0 when u um to Um when u 0 ie v vml 7 in which case we get 8 8 Eu 711mg u1 7 Example 12 Burger s Equation Suppose we have a stream of particles travelling on R each of which has its own constant velocity and let ut ac denote the velocity of the particle at at at time i Let act denote the trajectory of the particle which is at 0 at time to We have C ut Differentiating this equation in t at t to implies 0 ut v 107 itto 1on 0 1007 90074107 300 which leads to Burger s equation 0 u u up Example 13 Minimal surface Equation Review Dominated convergence the7 orem and differentiation under the integral sign Let D C R2 be a bounded region with reasonable boundary uo 8D 7 R be a given function We wish to nd the function u D 7gt R such that u uo on 8D and the graph of u I u has least area Recall that the area of I u is given by An Amman D V1 Vu Zdac 2 BRUCE K DRIVERl Assuming u is a minimizer let 1 E 01D such that v 0 on 9D then 0 0Ausv dishD V1VU3 l2d d zng 1Vusv2dx 2 VuVU dx D 11Vu2 1 V Vu vdx D 1Vu2 from which it follows that V 4V1 0 ql W2 Example 14 Heat or Diffusion Equation Suppose that Q C R is a region of space lled with a material p is the density of the material at 96 E Q and C96 is the heat capacity Let uxt denote the temperature at time t 6 000 at the spatial point 96 E 9 Now suppose that B C R is a little volume in R 9B is the boundary of B and E B t is the heat energy contained in the volume B at time t FIGURE 1 A test volume B in 9 Then so on one hand 11 E305 B pxcxut xdx while on the other hand 12 E300 BltGltxgtvultaxgt7nltxgtgtdaltxx where G is a n Xnipositive de nite matrix representing the conduction properties of the material is the outward pointing normal to B at 96 6 9B and d0 denotes surface measure on 9B We are using lt to denote the standard dot product on w PDE LECTURE NOTES MATH 237ArB 3 In order to see that we have the sign correct in 12 suppose that at 6 8B and Vuac gt 0 then the temperature for points near as outside of B are hotter than those points near as inside of B and hence contribute to a increase in the heat energy inside of B If we get the wrong sign then the resulting equation will have the property that heat ows from cold to hot Comparing Eqs 11 to 12 after an application of the divergence theorem shows that 13 paccx12txdac v GVut ac dag B B Since this holds for all volumes B C 9 we conclude that the temperature functions should satisfy the following partial differential equation 14 17000001207 90 V 39 GHVW 39 or equivalently that 1 900000 Setting gijac and 27W Z3GijPC3i 15 12t ac V G1Vut the above equation may be written as 16 12t ac Lut ac where 82 8 17 was Zg rmfx Zzwxwm The operator L is a prototypical example of a second order elliptic differential operator Example 15 Laplace and Poisson Equations Laplaces Equation is of the form Lu 0 and solutions may represent the steady state temperature distribution for the heat equation Equations like u 7 appear in electrostatics for example where u is the electric potential and p is the charge distribution Example 16 Shrodinger Equation and Quantum Mechanics 3 A 251W x EMWE VWWUJ Wlth ll 3970ll2 1 lnterpretation l t 352 dt the probability of nding the particle in A at time t A Notice similarities to the heat equation Example 17 Wave Equation Suppose that we have a stretched string supported at at 0 and ac L and y 0 Suppose that the string only undergoes vertical motion pretty bad assumption Let utac and Ttx denote the height and tension of the string at t at p0ac denote the density in equilibrium and To be the equilibrium string tension Let J 1 at Am C 0 L then 4 BRUCE K DRIVERl Alt x xAx FIGURE 2 A piece of displace string MJt z Juttxp0xdx is the momentum of the piece of string above J Notice that p0 xdx is the weight of the string above Newtonls equations state dM t d ut tt7p0d Force on String J Since the string is to only undergo vertical motion we require T057 96 Ax cosozmAm T057 96 cosozm 0 for all A96 and therefore that T05 96 cosozm To7 ie To cosozm T057 96 The vertical tension component is given by sinozmAm sinozm sinozmAm cosozm To un t7 x Ax un t T057 96 Ax sinozmAm T057 96 sinozm To Finally there may be a component due to gravity and air resistance7 say gravity 2 p0 xdx and resistance 2 l gcutt7 xdx J J So Newtonls equations become LITPALE uttt xp0xdx To um 157 x Ax um If7 LITPALE LITPALE p0xdx kxuttxdx and differentiating this in A96 gives W tt7 P0 Us At 96 P0676 kutt7 96 or equivalently that M96 P095 1 18 ut ttx mum 05796 1 utt7 PDE LECTURE NOTES MATH 237A7B 5 Example 18 Maxwell Equations in Free Space 8E V B a X 8B 7V E a X V E7VB Noticethat 82E 8B WVgtlt57VXVXEAE7VVEAE 81B and similarly at AB so that all the components of the electromagnetic elds satisfy the wave equation Example 19 Navier 7 Stokes Here utac denotes the velocity of a uid ad t at pt at is the pressure The Navier 7 Stokes equations state Bu 19 E Buu VAu 7 Vp f With u0 ac u0ac 110 V u 0 incompressibility Where f are the components of a given external force and uo is a given divergence free vector eld V is the viscosity constant he Euler equations are found by taking V 0 Equation 19 is Newton s law of motion again F ma See httpWWWclaymathorg for more information on this Million Dollar problem 11 Some More Geometric Examples Example 110 Einstein Equations Einstein s equations from general relativity are 1 Rlcg 7 5959 T Where T is the stress energy tensor Example 111 Yamabe Problem Does there exists a metric 91 u4n 2go in the conformal class of go so that 91 has constant scalar curvature This is equivalent to solving 77Agou Sgou kua Where y 42 a 3 k is a constant and 590 is the scalar curvature of go Example 112 Ricci Flow Hamilton introduced the Ricci 7 ow 39 E Rlcg as another method to create good metrics on manifolds This is a possible solution to the 3 dimensional Poincare conjecture again go to the Clay math web site for this problem 6 BRUCE K DRIVER1 2 FIRST ORDER QUASI LINEAR SCALAR PDE 21 Linear Evolution Equations Consider the rst order partial differential equation 21 8ut ac Zaix8iut at with u0 ac 21 where at E R and 35 are smooth functions on R Let 141 1111 anx and for u E 01R C let N d Auac 101400 tZac Vuac 141 ZZdiac8iuac ie is the rst order differential operator 21 aix8i With this notation we may write Eq 21 as 22 a Au with u0 f The following lemma contains the key observation needed to solve Eq 22 Lemma 21 Let A and 14 be as above and f E 01R R then d N N 23 Efoem c Afoemac 14foem Proof By de nition and so by the chain rule d N 3f 0 eWx VHEMWD 39 A m AHEMWD which proves the rst Equality in Eq For the second we will need to use the following two facts 1 e SA em a eSZ and 2 e Aac is smooth in ac Assuming this we nd if oetAac f oetsAx f Gem oeszx 14 f Gem dt d5 0 d5 0 which is the second equality in Eq l Theorem 22 The function u 6 Cl DAR de ned by 24 uUNE f m solves Eq Moreover this is the unique function de ned on DA which solves Eq 23 Proof Suppose that u 6 Cl DAR solves Eq 22 then d N warm meme 7 Aultte ltx o and hence 1 07 MW 1407 90 Hf Let to x0 6 DA and apply the previous computations with ac e Axo to nd ut0ac fetAac0 This proves the uniqueness assertion The veri cation that u de ned in Eq 24 solves Eq 22 is simply the second equality in Eq I PDE LECTURE NOTES MATH 237A7B 7 Notation 23 Let et afac ut at where u solves Eq 22 ie awn fe x The differential operator 41 01R R 7gt CRn R is no longer bounded so it is not possible in general to conclude 00 t m 7 Mn 25 e fiznlA f n0 lndeed to make sense out of the right side of Eq 25 we must know f is in nitely differentiable and that the sum is convergent This is typically not the case because if f is only Cl However there is still some truth to Eq 25 For example if f E CWR R then by Taylor s theorem with remainder k in N emf 7 Z EAnf 0tk 710 by which 1 mean for any at E R k in N t k emfac 7 7gt 0 as t 7gt 0 710 Example 24 Suppose n 1 and Aac 1 85 then e Aac at t and saw m 2 It is interesting to notice that ta 0 1 lt z 7 n 6 f0 E if 9 710 is simply the Taylor series expansion of fact centered at ac This series converges to the correct answer ie fact i f is real analytic For more details see the Cauchy 7 Kovaleski Theorem in Section 4 Example 25 Suppose n 1 and 1411 2 x2875 then e Aac 7 17m on DA tac 1 7 tr gt 0 and hence et afac 131 utx on DA where 26 ut 121 It may or may not be possible to extend this solution ut ac to a 01 solution on all R2 For example if limi goo does not exist then limmi ut as does not exist for any at gt 0 and so u can not be the restriction of Cl 7 function on R2 On the other hand if there are constants Ci and M gt 0 such that 0 for at gt M and c for at lt 7M then we may extend u to all R2 by de ning c if acgt0andtgt1ac 171 T if xlt0andtlt 1ac It is interesting to notice that act 1t solves 7352t 7Aact so any solution u E 01R2R to Eq 26 satis es ut 1t 0 ie u must be constant on the curves at 1t for t gt 0 and ac 1t for t lt 0 See Example 213 below for a more detailed study of Eq 26 Example 26 Suppose n 2 8 BRUCE K DRIVERl 1 lfAacy7yacie A 3 2 01 3 then em 1223 etAfacy faccost 7 ysint7ycost xsint at 1 0 at 2 lfAacy aty1e A y 0 1 y then t m at 7 e 0 at ellHo My emf yy 9663219 Theorem 27 Given A E CWRKR and h E 010R gtlt R R 1 Duhamel s Principle The unique solution u E 01DAR to 27 u Au h with u0 f and hence and hence is given by t ut7 emf eO TMhH7 d739 0 or more explicitly 28 utac fetAac 0t hTet TAacdT 2 The unique solution u E 01DAR to 29 u Au hu with u0 f is given by 210 um ef We Wife m which we abbreviate as 211 e lt Mgtfltx ef WM Wifewxn Proof We Will verify the uniqueness assertions7 leaving the routine check the Eqs 28 and 29 solve the desired PDE s to the reader Assuming u solves Eq 277 we nd d d a 52 x Enigma ut 7 Au lawn mama and therefore t ue as mam use hermenm 0 and so replacing an by e Aac in this equation implies utac fetAac 0 h7et TAacdT PDE LECTURE NOTES MATH 237ArB 9 Similarly ifu solves Eq 297 we nd with 2t E MML ute mx that 2m ultte ltx u 7 Au 076 ht7 e macut e mx ht7 e mxzt Solving this equation for 2t then implies utyeitAx Z efot hTe TAxdTZ0 efot hTe TAacdex Replacing at by em in this equation implies um 35 6f hTet TA1deetAxI I Remark 28 It is interesting to observe the key point to getting the simple expres sion in Eq 211 is the fact that veg f9 0 e f o e g o e emf e Ag That is to say Ed is an algebra homomorphism on functions This property does not happen for any other type of differential operator lndeed7 if L is some operator on functions such that e Lfg e Lf e Lg7 then differentiating at t 0 implies Mfg LfgfL97 ie L satis es the product rule One learns in differential geometry that this property implies L must be a vector eld Let us now use this result to nd the solution to the wave equation 212 um um with u07 f and u07 9 To this end7 let us notice the un um may be written as 8 7 8f 8 875 u 0 and therefore noting that at 375 Hay lt0 900 WW we have 3n 3x W x 6 8 9 f 90 9 f 1 i The solution to this equation is then a consequence of Duhamel7 s Principle which gives ut x equot5 fx 604 9 f as Td7 fxi0tgf TFWdT fxiigf2fiidf faciigx2739itdffx2fitlg factfacit1tgxsds 10 BRUCE K DRIVERl The following theorem summarizes what we have proved Theorem 29 If f E 02R7R and g E 01R7R then Eq has a unique solution given by 213 was g m H m 71 1th 5mg Proof We have already proved uniqueness above The proof that u de ned in Eq 213 solves the wave equation is a routine computation Perhaps the most instructive way to verify that u solves um um is to observe7 letting y at s that g s gt9ydy At9ydytgydy A t wdyiOw wdy From this observation it follows that ut x Fac t Gac 7 t where Fm g M 75mm and Gov 5 we 7 Bandy Now clearly F and G are 02 7 functions and 8 78Fxt0and 88Gacit0 so that a 7 83 W x a 7 8 8 875 Fac t Gx 7 2 0 I Now let us formally apply Exercise 216 to the wave equation um um in which case we should let A2 783 and hence A 4783 Evidently we should take 1 cos i 78 5 t fac 7 t and Sine 783 1 lt d 1 d ac as s s 783 g 2 fig 2 757to 11 Thus with these definitions7 we can try to solve the equation 214 um um h with u07 f and u07 g by a formal application of Exercise 215 According to Eq 272 we should have t 7 14 COSUAV My Mhh d7 7 A 0 A ie 215 1 1 t 1 t t7T utx fxtfacit gacsds dT dy M7331 2 2 7 2 0 a 4M PDE LECTURE NOTES MATH 237A7B 11 An alternative way to get to this equation is to rewrite Eq 214 in rst order in time form by introducing v u to ltzgt7ltzgtlt2gtm me Where 0 1 A a W 0 A restatement of Theorem 29 is simply etAfx um 1 frtfrifftgrsds 9 WOW 2 f i7f xi9i9i 39 According to Du hamel s principle the solution to Eq 216 is given by 5517 l J View mg l The rst component of the last term is given by 1 t t7T 1 t t7T 7 h739ac sds d7 7 h739ydy 1739 2 0 T7t 2 0 7tT Which reproduces Eq 215 To check Eq 2157 it suf ces to assume f g 0 so that 1 t t7T um dT dy hm 2 0 447 so that um 7 um theorem Theorem 210 ff 6 02R7R and g E 01lR7lR7 and h E CR27R such that has exists and has 6 ClR27lR7 then Eq has a unique solution ut7 at given by Eq 214 Proof The only thing left to prove is the uniqueness assertion For this suppose that v is another solution7 then u 7 v solves the wave equation 212 With f g and hence by the uniqueness assertion in Theorem 297 u 7 v E 12 BRUCE K DRIVERl 211 A lidimensional wave equation with noniconstant coe icients Theorem 211 Let C35 gt 0 be a smooth function and CMY cac8 and fg E 020R Then the unique solution to the wave equation 217 um 2u Gui 5 c ui with u07 f and u07 g is 1 40 t0 1 l so 21 w x 5 me am 1 mm 5 glte sows it de ned for Lac 6 DC D7C Proof Uniqueness If u is a 02 7 solution of Eq 2177 then We mam and N N at a w some gm was Therefore N at 0 um 2 e g f as g f 6 Which has solution given by Duharnel7 s Principle as t N utac mm WC 9 Cf WWW 0 t N erce g Cf ear wmwf 0 eece t 9 60 wound eece in ow g Wound fequotcx fe 0x 1tgesc1ds EXiStence Let y 5300 50 d CESCdS Cyds in the integral in Eq 2187 then t etc etc 0 Lgesconds A m 20 A god j Ewgyd y iEiw Zm 0 C11 0 C11 From this observation7 it follows follows that 071 FWCWD GWWWD Where 1135 fxgyctz and G00 an 0 gy Now Clearly F and G are 02 7 functions and at 7 5 new 0 and at 5 Gowns 0 PDE LECTURE NOTES MATH 237ArB 13 so that a 7 5 um at i 5 8n 5 mm awmm 0 By Du hamel s principle7 we can similarly solve 219 um 5 h with 140 0 and Mo 0 Corollary 212 The solution to Eq 219 is Solution to Eq 217 1 t utac at timetif 0 with f 0 and g h7397 1 t tiT 17 h739escacds 2 0 Proof This is simply a matter of computing a number of derivatives 1 t u dT hTet Tcx h7eT tcac 0 un htx 10th ohme i cm 7 ohTeltHCx 2 5 1 id Him 30 d 1 id Hdh 30 d uig T it 75 35572J T it ds7 e 1 t dT h7 et cx 7 h7eT tcac and 2 o N 1 t N N CZu dT Ch7et Tcac 7 Ch7 eT tcac 0 Subtracting the second and last equation then shows um fpu h and it is clear that u07 0 and u07 0 l 22 General Linear First Order PDE In this section we consider the following PDE7 n 220 Zaix8iux Cmm 21 where 221 and C35 are given functions As above Eq 220 may be written simply as 221 The key observation to solving Eq 221 is that the chain rule implies d N 222 d ulte ltx mm s which we will write brie y as Eu 0 GSA flu 0 53A 15 Combining Eqs 221 and 222 implies d EMEMW CESAWDMESAW 14 BRUCE K DRIVERl which then gives 223 uesAx 2 eff CltE Altmgtgtd0ux Equation 222 shows that the values of u solving Eq 221 along any flow line of A7 are completely determined by the value of u at any point on this flow line Hence we can expect to construct solutions to Eq 221 by specifying u arbitrarily on any surface 2 which crosses the flow lines of A transversely7 see Figure 3 below FLOuO Lmes a A FIGURE 3 The flow lines of A through a noncharacteristic surface 2 Example 213 Let us again consider the PDE in Eq 26 above but now with initial data being given on the line 96 t ie ut 962 with u7 A for some f E C1 R R The characteristic equations are given by 224 tS l and x s x2s and the flow lines of this equations must live on the solution curves to x2 ie on curves of the form t for C E R and x 07 see Figure 213 PDE LECTURE NOTES MATH 237A7B 15 Any solution to u 121 must be constant on these characteristic curves Notice the line at t crosses each characteristic curve exactly once while the line i 0 crosses some but not all of the characteristic curves Solving Eqs 224 with t0 A 10 gives A 225 ts S A and 355 H S and hence A us A7H SA fA for all A and s gt 71A for a plot of some of the integral curves of Eq 224 Let A 226 t A lt 1 5 WW and solve for A A A27 271 A7 0 ac 1t7orac ac at which gives act 7 l i act 7 l2 4x2 2x I Now to determine the sign7 notice that when 5 0 in Eq 226 we have t A ac So taking t at in the right side of Eq 227 implies 227 A 3271 t x271242 7 271 ix21 7 x with 230 7 2x 7 721 with 7 39 Therefore7 we must take the plus sing in Eq 227 to nd A act 7 1 act 7 12 4x2 7 2x and hence act 7 l act 7 l2 4352 228 utx f 2f 16 BRUCE K DRIVERl When at is srnall7 A xt7117xt41f11 liacti11 x 2x 2x 1 7 act so that ut7 at g f when at is small Thus we see that ut0 f0 and ut7 at is 01 if f is Cl Equation 228 sets up a one to one correspondence between solution u to u 12mg and f E 010R R Example 214 To solve 229 gouge guy Aacyu with u f on 517 let Aacy 17y 875 gay The equations for acsys eSAac7 y are Ms as and Ms as from which we learn 53AM 11 5390721 Then by Eq 2237 Mm a y elt9 1gt ultxy Letting 3531 A 640531 in this equation gives um mm and then choosing s so that 1 6406711 2 6723062 12 ie so that s 1nac2 212 We then nd 7 A 1 96711 1490721 7 exp 5 1 i 901 NW Notice that this solution always has a singularity at 731 00 unless f is constant Characteristic curves for Eq 229 along with the plot of 51 PDE LECTURE NOTES MATH 237A7B Example 215 The PDE7 230 efuf uy u With uac0 gac7 has characteristic curves determined by ac 5 and y 1 and along these curves solutions u to Eq 230 satisfy 1 231 EMMA Way Solving these characteristic equations gives 5 s 232 76453 671 af xd5 ld5 5 0 0 so that 233 355 7 lne 7 5 and 315 yo 5 From Eqs 232 and one shows 315 yo 671 7 67 so the characteristic curves are contained in the graphs of the functions y C 7 67 for some constant C Some characteristic curves for Eq 230 Notice that the line y 0 intersects some but not all of the characteristic curves Therefore Eq 230 does not uniquely determine a function u de ned on all of R2 On the otherhand if the intial condition were u0y gy the method would produce an everywhere de ned solution Since the initial condition is at y 07 set yo 0 in Eq and notice from Eq231 that 234 u7 lne 7 57 5 uac5y5 53141070 639350 Setting Ly 7 lne 0 7 57 5 and solving for are7 5 implies 5 y and 0 7lne 31 and using this in Eq 234 then implies uacy eyg 7 lny 5775 as This solution is only de ned for y gt 75 18 BRUCE K DRIVER39 Example 216 In this example we will use the method of characteristics to solve the following non7linear equation7 235 121 312 uy u2 with u 1 on y 235 As usual let 17y solve the characteristic equations7 ac 2 and y 12 so that 0 210 as s s Hm linenisyo Now let 0310 A72A be a point on line y 2x and supposing u solves Eq 235 Then 275 uacsys solves d z 121 312 uy u2acy 22 with 270 u2 1 and hence 236 u 7 menSn 7 as 7 17 Let 237 W v and solve the resulting equations A71 7 s x71 and Ail2 7 s 3171 for 5 gives 5 x 1 7 231 1 and hence 238 17s12y 17x71x 1y 1xy2x7y Combining Eqs 236 7 238 gives lt my u as 7 y 35y 2x 7 y Notice that the characteristic curves here lie on the trajectories determined by GE 7 39 i 7 i C 39 1 t1 1 7 y 1e y 7 as or equ1va en y as y 1 Car Some characteristic curves PDE LECTURE NOTES MATH 237A7B 19 23 QuasiLinear Equations In this section we consider the following PDE 239 Axz Wag25x Zaixux82 ux cxux 21 where aixz and cxz are given functions on Lg G R X R and Axz a1x z a1 95 Assume u is a solution to Eq 239 and suppose 955 solves x s AxsuxsThen from Eq 239 we nd diiw s Zaimsuxs32uxs Cxs7uxsa see Figure 4 below We have proved the following Lemma t 23 70 mm L 7 9 20 Crosswe foun FIGURE 4 Determining the values of u by solving ODE7s Notice that potential problem though where the projection of character istics cross in 95 7 space Lemma 217 Letw Lg 7T1w x 7T2w z andYw Axzcx Ifu is a solution to Eq 239 then u7T1 o eSY x0ux0 7T2 o esYx0ux0 Let E be a surface in R 957 space ie E U Co R 1 a R such that 20 x0 and DE is injective for all y E U Now suppose uo E gt R is given we wish to solve for 2 such that 239 holds and u no on 2 Let 240 5731 1 7H O esy SQLUMEQW then 0 0 7T1 o Yx0u0x0 Ax0u0x0 and Dy 0 0 Dy20 20 BRUCE K DRIVER39 Assume Z is nonCharacteristic at are that is Aac0 u0aco Ran 30 where 30 R714 A R is de ned by d 2 0v 620 d l02sv for all 1 6 R714 8 Then g 8351 Wig are all linearly independent vectors at 0 0 E RX Rn l So R X R714 A R has an invertible differential at 00 and so the inverse function theorem gives the existence of open neighborhood 0 E W C U and 0 E J C R such that aleXw is a homeomorphism onto an open set V J X W C R see Figure 5 Because of Lemma 217 if we are going to have a C1 7 solution u to A W4Fwa Lgt I a IW KW a FIGURE 5 Constructing a neighborhood of the surface 2 near 10 where we can solve the quasielinear PDE Eq 239 with u uo on X it would have to satisfy 241 1490 7T2 0 6 2y7u02y with 821 1 W100 ie ac sy Proposition 218 The function u in Eq 241 solves Eq on V with u uo on 2 Proof By de nition of u in Eq 241 and in Eq 240 MM MY 0 e EQUJMQQIW A syy7u syy and 242 dis sv 772Y syy7u syy C Syy7u syy On the other hand by the chain rule 243 uwwy Vu 57u 4548711 Vu 57u 39 A Syy7u syy Comparing Eqs 242 and 243 implies Vu syy A syy7u syy C syy7u syy Since J X W V u solves Eq 239 on V Clearly uqgt0y u02y so u uo on E l PDE LECTURE NOTES MATH 237ArB 21 Example 219 Conservation Laws Let F R A R be a smooth function we wish to consider the PDE for u ut 1 244 0 u 8Fu u F uu with u0 1 The characteristic equations are given by d 245 t 5 1 15 F z5 and 0 5 The solution to Eqs 245 with t0 0 10 1 and hence 210 ui07 960 14071 WE are given by t5 5 275 91 and 15 1 5F 91 So we conclude that any solution to Eq 244 must satisfy 1457 x 5F 9 900 This implies letting 1931 1 5F 91 that um x 905102 ln order to nd 11571 we need to know 115 is invertible ie that 115 is monotonic in 1 This becomes the condition 0 lt 11400 1 iF 99 If this holds then we will have a solution Example 220 Conservation Laws in Higher Dimensions Let F R A R be a smooth function we wish to consider the PDE for u ut 1 246 0 u V u F u Vu with u0 1 The characteristic equations are given by 247 t 5 1 15 11125 and 1125 0 5 The solution to Eqs 247 with t0 0 10 1 and hence 20 ui07 960 140796 WE are given by t5 5 275 91 and 15 1 5F 91 So we conclude that any solution to Eq 246 must satisfy 24 ms x mm gm This implies letting 1931 1 5F 91 that 14MB 9015400 ln order to nd 11571 we need to know 11 is invertible Locally by the implicit function theorem it suf ces to know MW v tF gav3u9av I tF gavV9avl v is invertible Alternatively let y 1 5F 91 so 1 y 7 5F 91 in Eq 248 to learn using Eq 248 which asserts 91 u5 1 5F 91 u5y 145711 911 7 sF g 911 7 5174145711 This equation describes the solution u implicitly 22 BRUCE K DRIVER39 Example 221 Burger s Equation Recall Burger s equation is the PDE 249 u uui 0 with u0 35 g35 where g is a given function Also recall that if we view ut 35 as a time dependent vector eld on R and let 355 solve W W 10 then u 15535 u uiu 0 Therefore 35 has constant acceleration and 355 350 350t 350 g350t This equation contains the same information as the characteristic equations In deed the characteristic equations are 250 t 5 l 35 5 25 and 2 5 0 Taking initial condition 50 0 350 350 and 20 u0 350 g350 we nd 55 5 25 g350 and 355 350 5g350 According to Proposition 218 we must have 251 145ny 59000 1457 905 1407940 9 900 Letting 11350 350 tg350 the solution to t 35 5 350 59350 is given by 5 t and 350 Therefore we nd from Eq 251 that 252 14MB g M100 This gives the desired solution provided 11571 is well de ned Example 222 Burger s Equation Continued Continuing Example 221 Supe pose that g 2 0 is an increasing function ie the faster cars start to the right then 11 is strictly increasing and for any i 2 0 and therefore Eq 252 gives a solution for all t 2 0 For a speci c example take g35 max350 then 7 lt35 if 3520 w x H 130 1 7 1t 135 if 3520 x ifxgo and therefore 7 1 7 1t 135 if 3520 mwm so 7 0 if x 0 Notice that ut 35 A 0 as t A 00 since all the fast cars move off to the right leaving only slower and slower cars passing 35 E R Example 223 Now suppose g 2 0 and that g 350 lt 0 at some point 350 E R ie there are faster cars to the left of 350 then there are to the right of 350 see Figure 6 Without loss of generality we may assume that 350 0 The projection of a number of characteristics to the 535 plane for this velocity pro le are given in Figure 7 below Since any 02 7 solution to Eq249 must be constant on these lines with the value given by the slope it is impossible to get a 02 7 solution on all of R2 with this initial condition Physically there are collisions taking place which causes the formation of a shock wave in the system PDE LECTURE NOTES MATH 237ArB 23 FIGURE 6 An intial velocity pro le Where collisions are going to occur This is the graph 01391 1 l at DZ FIGURE 7 Crossing of projected characteristics for Burger s equation 24 Distribution Solutions for Conservation Laws Let us again consider the conservation law in Example 219 above We Will now restrict our attention to nonenegative times Suppose that u is a Cl 7 solution to 253 u 0 With u07 at 91 and ab 6 0307 00 X R Then by integration by parts7 0 dagmm Fltugtea 7R W E Rd1t20 1th Fuq5 990940xdacAdr zod u wwt wFut7x t7x 24 BRUCE K DRIVERT De nition 224 A bounded measurable function ut7 v is a distributional so lution to Eq 253 iff o 9rr0rdw dm Zodtowwm Fltulttwgtgt lttwgtgt for all test functions 15 E 03D where D 0 00 X R Proposition 225 Ifu is a distributional solution of Eq and u is Cl then u is a solution in the classical sense More generally ifu E 01R where R is an open region contained in D0 000 X R and 254 doc dtut mm m Fan some 90 0 R 720 for all 15 E 03R then at 0 on R Proof Undo the integration by parts argument to show Eq 254 implies u wounded 0 R for all 15 E 03R This then implies at 0 on R l Theorem 226 Rankine Hugoniot Condition Let R be a region in D0 and r ct for t 6 ab be a 01 curve in R as pictured below in Figure 8 rxchtl n AM 0 m V l l o g h gt g l i no 4 FIGURE 8 A curve of discontinuities of u Suppose u E 01R cab and u is bounded and has limiting values uJr and u on r ct when approaching from above and below respectively Then u is a distributional solution of at 0 in R if and only if 255 at 0 on R cab and 256 ct ntct 7 u tct Futct 7 Fu tct for all t 6 ab PDE LECTURE NOTES MATH 237ArB 25 Proof The fact that Equation 255 holds has already been proved in the pre vious proposition For 256 let 9 be a region as pictured in Figure 8 above and ab 6 039 Then 257 0 J5 Fuq5dt dac uq5 Fuq5dt dac uq5 Fuq5dt dac where at lt Ct 9itx 2 awe Now the outward normal to 9i along 0 is I t 71 W i lt4 7 1 c39t2 and the surface measure along 0 is given by da39t 1 c39t2dt Therefore nt dat 1W 71dt where the sign is chosen according to the sign in 9i Hence by the divergence theorem7 uq5 Fuq5dt dac q5q5dt dac nt da39t r21 91 6m 7 i 07 coma t7 Ct t 7 Fut7 Ctdt Putting these results into Eq 257 gives 0 7610 WW CW 7 u it C0 7 Fut7 C0 7 Fu it Ct t7 Ctdt for all qb which implies c39t utct 7 u t Ct Futct 7 Fu tct I Example 227 In this example we will nd an integral solution to Burger s Equae tion7 u uui 0 with initial condition 0 121 14035 171 03131 1 30 The characteristics are given from above by act 1 7 xot am ace 6 07 1 1t xotifaco 0 and 1t 10 if 10 2 1 BRUCE K DRIVERT Projected characteristics For the region bounded determined by t g x g 1 and t g 1 we have ut x is equal to the slope of the line through 75 x and 1 1 i e x i 1 t i 1 39 de ne in the region where characteristics ut x Notice that the solution is not well cross i e in the shock zone R2 2 tm 17521 m21andxgt see Figure 9 Let us now look for a distributional solution of the equation valid for FIGURE 9 The schock zone and the values of u away from the shock zone all mt by looking for a curve Ct in R2 such that above C75 u 0 while below C75 u 1 To this end we will employ the Rankine Hugoniot Condition of Theorem 2126 To do this observe that Burger s Equation may be written as at 0 where So the Jump condition is em i u Fu i Fm PDE LECTURE NOTES7 MATH 237A7B 27 0 1c39 0 22 1 22 Hence ct and therefore Ct t l for t Z 0 So we nd a distributional solution given by the values in shown in Figure 10 that is FIGURE 10 A distributional solution to Burgerls equation 25 Exercises Exercise 21 For A E IX7 let 258 6 z A Show directly that l 6 is convergent in LX When equipped With the operator norm 2 6 is differentiable in t and that e A6 Exercise 22 Suppose that A E LX and v E X is an eigenvector of A With eigenvalue A ie that A1 A1 Show emu emu Also show that X R and A is a diagonalizable n X n matrix With A SDS l with D diag1n then 6 Set3871 Where em diagetgt 1e quot Exercise 23 Suppose that AB 6 LX and A7 B E AB BA 2 0 Show that 6AB eAeB Exercise 24 Suppose A E ClR7 LX satis es AtAs 0 for all 813 6 R Show I ATd739 e is the unique solution to Atyt With y0 76 Exercise 25 Compute 6 When H315 28 BRUCE K DRIVERl and use the result to prove the formula coss t cos 5 cost 7 sin ssin t Hint Sum the series and use ewe etSA Exercise 26 Compute em when A DOC C309 1 c 0 with a7 17 c E R Use your result to compute e IA where A E R and I is the 3 X 3 identity matrix Hint Sum the series Theorem 228 Suppose that T E LX fort 2 0 satis es 1 Semiigroup property T0 Idx and TTS TS for all st 2 0 2 Norm Continuity t A T is continuous at 07 ie HT ilHLw gt 0 as t l 0 Then there exists A E LX such that T e where e is de ned in Eq 258 Exercise 27 Prove Theorem 228 using the following outline 1 First show t E 000 A T E LX is continuous 2 For 6 gt 07 let S6 foe TTdT E LX Show S6 A I as e l 0 and conclude from this that 56 is invertible when 6 gt 0 is sufficiently small For the remainder of the proof x such a small 6 gt 0 3 Show 1 te TSe TTdT 5 t and conclude from this that 1 1 7 7 ltlfgt T 56 6Te Idx Using the fact that 56 is invertible7 conclude A limio t 1 T 7 1 exists in LX and that E AlTeilS1 6 5 Now show using the semigroup property and step 4 that T AT for all t gt 0 6 Using step 57 show dilte mT 0 for all t gt 0 and therefore e mT e OATO 1 Exercise 28 Higher Order ODE Let X be a Banach space7 Z1 Co X and f E C J gtlt LLX be a Locally Lipschitz function in x 3517 Show the nth ordinary differential equation7 259 110 ftytytylt 1t with 21000 a for k 012 n 7 1 where 318 711371 is given in Z1 has a unique solution for small t E J Hint let yt ytgt y 1t and rewrite Eq 259 as a rst order ODE of the Ztyt with yO 31831371 PDE LECTURE NOTES MATH QSYAVB 29 Exercise 29 Use the results of Exercises 26 and 28 to solve 7 2gt yt 0 with 310 a and b Hint The 2 X 2 matrix associated to this system7 A7 has only one eigenvalue 1 and may be written as A I B where 32 0 Exercise 210 Suppose that A R A LX is a continuous function and UV R A LX are the unique solution to the linear differential equations Vt Atvt with V0 I and 260 Ut 7UtAt with U0 I Prove that Vt is invertible and that V 1t Ut Hint 1 show dig UtVt 0 which is suf cient if dimX lt 00 and 2 show compute yt VtUt solves a linear differential ordinary differential equation that has y E 0 as an obvious solution Then use the uniqueness of solutions to ODEs The fact that Ut must be de ned as in Eq 260 is the content of Exercise 7 below Exercise 211 Duhamel7 s Principle 1 Suppose that A R A LX is a contine uous function and V R A LX is the unique solution to the linear differential equation in Eq 7 Let at E X and h E CRX be given Show that the unique solution to the differential equation 261 gt Atyt ht with y0 at is given by 262 yt Vtac Vt Ot Vt1ht dT Hint compute V 1tyt when y solves Eq 261 Exercise 212 Duhamel7 s Principle 11 Suppose that A R A LX is a con tinuous function and V R A LX is the unique solution to the linear differential equation in Eq 7 Let W0 6 LX and H E CR7 LX be given Show that the unique solution to the differential equation 263 Wt AtWt Ht with W0 W0 is given by 264 Wt VtW0 Vt1VT 1HTdm Exercise 213 NoneHomogeneous ODE Suppose that U Co X is open and Z R X U gt X is a continuous function Let J 21 be an interval and to E J Suppose that y E 01JU is a solution to the nonehomogeneous differential equation 265 Ztyt with yto at E U De ne Y E 01J 7 t0R gtlt U by YO E t t0yt t0 Show that Y solves the homogeneous differential equation 266 Yt AYt with Y0 to yo SO BRUCE K DRIVERl where Atx E 1Zac Conversely suppose that Y E 01J 7 t0lR gtlt U is a solution to Eq 266 Show that Yt t t0yt to for some y E 01J U satisfying Eq 265 In this way the theory of nonehomogeneous ode s may be reduced to the theory of homogeneous ode s Exercise 214 Differential Equations with Parameters Let W be another Bae nach space U gtlt V Co X X W and Z 6 CU gtlt V X be a locally Lipschitz function on U gtlt V For each 3510 6 U gtlt V let i 6 J75 A q5t 110 denote the maximal solution to the ODE 267 Zytw with 310 ac Prove 268 D txw ERX Ugtlt Vt Jf w is open in R X U gtlt V and ab and are continuous functions on D Hint lf yt solves the differential equation in 267 then vt E ytw solves the differential equation 269 w 2mm with 120 110 where Aacw E Zacw0 E X X W and let t vt Now apply the Theorem 7 to the differential equation 269 Exercise 215 Abstract Wave Equation For A E LX and t E R let 7 00 42 2n 2n costA 7 Z t A and n0 2n 11014 i0 12n1t2n1A2n A 7 2n 1 7170 Show that the unique solution y E 02 RX to 270 gi A2110 0 with 110 yo and 120 220 e X is given by sintA ya 0050140210 A Remark 229 Exercise 215 can be done by direct veri cation Alternatively and more instructively rewrite Eq 270 as a rst order ODE using Exercise 28 In doing so you will be lead to compute etB where B E LX X X is given by 0 I B o v where we are writing elements of X X X as column vectors You should 961 902 e costA w 7A sintA costA then show where 1 2n1 n n As1ntA 2 g 1A2 1 110


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