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# PARTIAL DIFF EQNS AMATH 353

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This 17 page Class Notes was uploaded by Mossie Pfannerstill V on Wednesday September 9, 2015. The Class Notes belongs to AMATH 353 at University of Washington taught by Eleftherios Kirkinis in Fall. Since its upload, it has received 14 views. For similar materials see /class/192273/amath-353-university-of-washington in Applied Mathematics at University of Washington.

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Date Created: 09/09/15

Amath 353 Partial Differential Equations Highlights Part ll Eleftherios Kirkinis December 7 2007 Abstract Review of solution techniques developed and the corresponding applied mathematical con structionsi 1 General solution of the wave equation in 1D Consider the wave equation in a in nite domain utt czum7 m E foo7 00 1 We can t solve it like this We introduce characteristic coordinates Emict nmct 2 Substituting into 1 for u 145 7 we obtain 1 67 039 3 Integrating twice7 uFEGn7 4 where the functions F7 G are arbitrary7 or uFm7ctGmct 5 which is the general solution to the wave equation We also used the sloppy notation of denoting a function of mt and E 7 with the same symbol u 2 D Alembert s solution for the wave equation Now assume we consider the eq 1 with initial data wemmeemm WWFJW7MMM o How can we use the general solution derived above to incorporate the initial data We just substi tute in 5 and we derive the two relations C95 g zwiggem rr 7 Fm g ai4mK a Substituting into 5 we obtain the D Alembert s solution to the wave equation 6 West fzctl i H yew lt9 1 iix7 t 5 2C 701 after some rearrangement 21 The WE in the presence of boundaries semiin nite domain The above general solution and D Alembert s solution can be used to nd the solution to a semi in nite domain for ex consider the case an 621i m E 07oo um0 utm0 9a 10 along with the boundary condition ii07 t O The above GS is ok if z gt ct but not ok in the region z lt ct because the argument of F becomes negative Thus we need to nd an expression that will replace F in the z lt ct region The answer comes from using the boundary condition ii07 t F7ct Cct O7 76172 11 Thus the GS for z lt ct can be written as iix7 t 7G7zict Gmct 12 and we can also write an expression that incorporates the initial conditions Understand these type of problems before you come to the exam 22 The WE in the presence of boundaries nite domain If the string is nite7 its vibrations are described by z e M um 0 M w 0 gltzgt lt13 with boundary conditions n07 t nL7 t 0 14 other types of BC also exist The solution is obtained with separation of variables M8 umt An sinwnt Bn coswnt sin ha 15 H n where 4 chm kn One then incorporates the initial conditions to nd An7 B What is the geometrical interpretation of the mode nnm7 t sin Aunt sin ham 16 Fixing x7 then an behaves as On sin aunt With time the string oscillates with frequency 4 radians per second and the amplitude of oscillation is xed at xed At other z the string oscillates with other amplitudes but the same frequency This is the frequency of the musical tone Rewriting the mode in terms of knziwnt and k zwnt leads to the usual pro le that propagates here the pro le is sinusoidal Understand the geometrical interpretation before you come to the exam 3 Elliptic PDEs Standard examples are the Laplace Poisson and Helmholtz equations Vzu 0 Vzu mi Vzu kzu 0 17 We brie y talked about the rst one as it represents the steady state temperature distribution in a given domain Normally as we did in the homework you solve these problems with separation of variables 4 First Order PDEs The method of characteristics 41 Linear lst order These are equations of the form aztut bz tum dxtu u uzt 18 and the coefficients a b d are known functions of their arguments Considering the rate of change of uzt t as measured by a moving observer z zt 1 8w 8w dm 109505 75 i E 3 19 then if the observer moves with velocity ba then comparing the above equations dm 1 7 7 20 dt a du d 7 7 21 dt a u This system is called the characteristic system and the curves z mt base characteristic curves Consider the simple PDE ut Citm 0 IO um0 22 Labeling a position on the x axis at t0 by E the characteristic system gives zct57 23 where F is an arbitrary function Thus the general solution to 22 is ux t Fx 7 ct 24 Incorporating the initial data we obtain ux t 7 ct 25 ie a translation of the initial pro le to the right at speed 0 Notes The initial data propagate along characteristics For the above problem it is constant as we walk along one characteristic but it might change if we jump onto another characteristic 42 Quasilinear lst order PDEs These are equations of the form ax tuut bmt Mum dx t u u ux t 26 and the coefficients a bd are known functions of their arguments The steps to derive the char acteristic system are analogous But now the phase speed ba depends on how high the pro le is Characteristics are not parallel anymore and they might intersect If they do the solution at the point of intersection becomes multi valued There are two important cases of waves that may occur as such expansion or rarefaction waves fans and shock or compression waves 421 Rarefact ion waves Consider for example the equation 3 xlt0 pt2ppz0 px04 300 27 This implies that waves on the negative z axis propagate with velocity 2 3 thus slower than the waves that emanate from the positive z axis For each p value between 3 and 4 we thus must introduce a fan that takes care of the expansion region This is given by m2pt 3ltplt4 28 or z 7 6t 8t 29 0 2t lt lt Thus the solution is 3 m lt 675 px t 4 m gt St 30 625 lt z lt 82 422 Shock waves Consider for example the equation 4 xlt0 pt2ppz0 px03 300 31 Now waves emanating from the negative x axis immediately overtake those from the positive x axis since the former travel at speed 2 4 and the latter slower at speed 2 3 The solution becomes multi valued and is no longer valid Instead we expect that there will be a shock wave 505 joining the characteristics from the different regions To nd the position of the shock wave at any time t we consider the RankineHugoniot conditions for the shock velocity 1 qright of ms 7 qleft of ms dt pright of 5 7 pleft of 5 32 where qp is the ux of the conservation law pt qz 0 33 For this particular problem q p2 and the RankineHugoniot conditions become dx d 7 x9t 7t 34 because the shock starts at X0 and t0 Thus the solution is 4 x lt 7t 10x7t 3 zgt775 35 As this is one of the most important topics you will ever learn in PDEs try to understand these type of problems before you come to the exam 5 Solution of PDES in curvilinear coordinates The equations remain the same but because the boundaries are given in curved domains circles ellipses spheres cylinders we need to rewrite the differential operators and notable V2 in these coordinates 51 Laplacels equation in planepolar coordinates 1 1 urr l Ur l E1499 With separation of variables the solution is 00 u A0 Bo lnr Emma Bnr cos n0 Cmn Dnr sin nd 37 n1 We distinguish various cases i The steady state temperature distribution inside a disk 0 lt r lt 1 with the temperature speci ed on the boundary ur 1 0 90 9 given function In this case we discard the ln and the r terms in 37 since they become unbounded at the origin The left over should be used to determine the constants An 0 through Fourier series In the exam you will get one of these problems where gr 1 0 cos30 for example Solve this problem before the exam date ii Annulus a lt r lt I Here nothing becomes unbounded and we have to retain all terms and thus determine through FS all constants To do so we are equipped with the temperature distribution at r a ur a 0 910 and the temperature distribution at r b ur b 0 920 9192 known functions iii The steady state temperature distribution outside a disk r gt 1 with the temperature speci ed on the boundary ur 10 90 9 given function In this case we discard the ln and the r terms in 37 since they become unbounded at in nity The left over should be used to determine the constants B Dn through Fourier series In the exam you will get one of these problems where gr 1 0 sin3 9 for example Solve this problem before the exam date Amath 353 Partial Differential Equations Highlights Part I Eleftherios Kirkinis February 37 2009 Abstract Review of solution techniques developed and the corresponding applied mathemat ical constructions Not all subjects to be examined are included here 1 Exam I Review Exam I will be comprehensive Everything we covered from hw2 up to and including home work 6 No characteristics no applications just methodology to solve problems In particular give special attention to the following homework problems HW2 Problems 1727374 HW3 1273677 HW4 Problems 12735the Parseval Part7 78 HW5 All but 6 HW6 All Full solution to some of the problems are in the three Fourier series hand outs The hand out on average values and Parseval should also be reviewed Review the accompanying document with possible exam questions Last year exam is similar to your midterm If you solve and understand the problems of 2007 midterm you most likely be in the 70 range This document is not very well proof read so there might be some typos below 2 Solution methods for homogeneous no source heat equation in a nite domain We are solving ut 042 m E 07 L 1 B03 X0 0XL 0 evaluesAn n 1 2 efnan sin L The solution is given in general by ut Z AnTntXnv nDL7r1 and for the above BCs Tnt e quot gt2t DirichletBCs settemp NeumannBCs insulated Periodic X 0 0X L 0 7r 2 mr 2 T T 012 0123 WWI WWI WWI cos T sin Tami cos T 2 3 Method Separation of variables ut XTt Determine XT by substituting into Separating variables in an partial di erential equation gives rise to two applied mathe matical problems 0 As described in the table above the function X that describes the spatial variation of the temperature will satisfy a Sturrn Liouville eigenvalue problem Lesson 7 of the form X AZX 0 X00 XL0 4 these are Dirichlet BC7s depending on the boundary conditions employed Neverthe less a S L problem is characterized by the following properties i There is an in nity of discrete eigenvalues An they are quantized7 and an in nity of corresponding eigenfunctions ii Eigenfunctions corresponding to di erent eigenvalues are orthogonal iii They are complete 0 In order to satisfy the le at t 0 we evaluate the solution 3 at t 0 and equate with f since the initial conditions given are ut 0 known function to obtain f Z ATLXWltgt7 f f is a given 5 nDL7r1 XL XL7X L which is a Fourier series representation see hand outs of the given function Thus to nd the solution we need to determine the 7integration constants7 An and these are determined by the usual integrals constructed by calculating average values at both sides L f0 Xnfd An L L X72d 6 plot of fx gtlt its periodic extension and 5 term Fourier Series iii Mii Figure 1 Approximation of f m by its Fourier Series Example 1 in the matlab code fourierserieslm we show the graph of f m on 71717 its periodic extension and the Fourier series 2 1 1 fFS 7sin7m7isin27mgsin37mnj 7 7r In gure 1 we see that increasing the number of terms in the FS leads to a better approxi mation of Example2 Consider the solution of the heat equation with Neumann insulated boundary condition ut 04211ng x e 0712 uz0t 11412775 0 8 and initial condition uif0f27 6 0712 9 We know that the soln to the PDE is an in nite sum of cosines This implies that the function to be expanded is f 2 on 7127127 ie expand as an even function This leads to 1 1 1 1 f N FS E 7 cos 27m 7 cos lwm c0s67m 10 from which we can read o the solution of the PDE I Figure 2 we show how this solution decays to the value 112 all over the interval 0712 Why does it not decay to zero and what is the physical meaning of this behavior plot of the SOlUUOH aiiime 0 005 0 5 0 0 5 plot of the SOlUUOH at time 0 02 1 0 5 0 0 5 1 1 5 plot of the SOlUUOH aiiime 1 715 04 02k 1 715 1 05 0 05 1 15 Figure 2 Evolution of the solution to eq 8 with initial pro le f 2 for the rod in the interval 0712 3 Nonhomogeneous Boundary conditions Lesson 6 In general the bc7s imposed on u7t are 04111407 75 511107 t 9175 211ng7 t 8211L7 t 9275 In the exam we will consider the simpler case u07 t k1 uL7 t k2 To eliminate the non homogeneity7 we introduce a new function U7 75 through uzt k1 gm 7 k1 Wm Then U satis es a heat equation with homogeneous BCs However the le change 4 Elimination of lower order derivatives in the PDE Lesson 8 Given a general PDE an buyy Cu duz euy fuz u 07 4 16 a7b707d767f given7 it is possible to eliminate the lower order derivatives by introducing a new function w y7 2 through 61z62ycazw7y727 substituting into 16 and determining the unknown 01702703 so that lower derivatives do not occur in the equation for 10 Special cases from Farlow o with u7 t e f wm f7 the equation ut 04211 7 u 18 reduces to wt 04210 c with u7 t equotg 2gt20 2wm7if7 the equation ut 04211 7 yum 19 reduces to wt 04210 5 Nonhomogeneous PDE s with source functions or external forcing Lesson 9Win 2009 nonexaminable The general problem is Given a heat equation with a source function g7 t g is known ut 04211 g7 t 20 with BCs 0411140775 811107 t 0 21 042L7 t 5211L t 0 22 and le u7t 0 f7 f known7 use an assumed form of the solution ut ZTntXnv 23 The Xn are known and are the eigenfunctions of the corresponding homogeneous no 9 problem with the same BCs Assuming an expansion for g in the form 9 5 Z gntXWt7 24 where the 9 are known7 we need to determine the Tnt7s We substitute the assumed solution into the non homogeneous pde 20 and this leads to an equation for Tnt in the form 0451305 97205 25 6 Other applied mathematical techniques Parseval relates the average of the square of a function over a period with the Fourier coef cients of the series that represents the function 1 L Llfrvl2drv 2 MP 26 where on are the Fourier coef cients of the complex Fourier series representing The bottom line is that the average total energy7 is proportional to the sum of the energies of each individual mode 7 Fourier Transforms highlights An important by product of the Fourier Transform technique is the calculation of integrals and the integral representation of certain functions For example in class we showed that the seemingly intractable integral fxw1kzekdx 27 can easily be calculated by resorting to the actual form of f that we used to derive it ie 6 7 gt0 m 64 x lt 0 lt28 Furthermore7 the Fourier lntegral Theorem guarantees that at a point of jump of r7 the integral will converge to the mid point of the jump7 and thus giving us zero in the above problem While in Fourier k space7 given F1k 172k where f 1F1k f1 and f 1F2k f2 known functions7 what is the inverse Fourier transform of the product F1k F2k ie what is f71F1k 39 F200 29 The answer is the convolution of f1 with f2 1 0 13 as f2c E w m e uf2udu lt30 Using this property of the FT7s we can now solve many PDEs in the in nite domain ie all the problems in homework 6 8 Important functions for solving PDEs These are the Dirac 6 function and the Gaussian or Normal distribution function Consider the latter with the following forms in con guration and Fourier space respectively 1 2 1 k2a2 M fwgmv FUC 7677 31 and it is given that the Gaussian function f satis es the condition X 1 i 00 We 2a2d 1 32 and thus F satis es a similar equation Below we graph f and F as we vary the parameter a This is called the width parameter and measures the distance from the center axis to the in ection point of f Also the Dirac delta function and solution of PDEs as these are Figure 3 Gaussian function f in 31 Notice that as a a 07 the graph becomes narrower and the amplitude increases described in the corresponding hw Figure 4 The FT of the Gaussian function f is again a Gaussian in 31 Notice that as 0 a 07 the graph becomes Wider and the amplitude remains constant The former happens because the Width parameter for F is 10 Amath 353 Partial Differential Equations Sample midterm questions Eleftherios Kirkinis October 267 2007 Abstract Questions 1 State the conditions of the Dirichletls theoremi 2 What is the value of the Fourier series representation of the function 071ltzlt0 fI1 0ltzlt3 1 at the point z 0 At the point z 7 9 Does the function 13 have a Fourier cosine series in 77rJr 9 State Parsevalls theoremi It is used to solve What kind of problems 5 Classify the following equations parabolic7 elliptic7 hyperboliciz7 y E 007 00 15mm um 710uyy z 7 3 2 um zyuyy 07 3 by rst constructing a suitable quadratic form and examining the sign of its determinant on i Farlow problem 74 For the Sturm Liouville eigenvalue problem y z AMI 07 yO 07 yL 07 4 find the 39 and col 1 quot 39 f 39 examine all possi ble values of LThen verify that i There is an infinite number of eigenvalues With a smallest but no largesti ii The nth eigenfunction has n 7 l zerosi iii The eigenfunctions are orthogonal 7 9 Heat equation with convection and lateral surface heat loss Reduce the parabolic pde um 41 7 2ut Su 0 5 to the onedimensional heat equation Solution 1 uzt 67121721114177 wt Eu 6 Solve the nonhomogeneous equation u0t lu7rt 07 uz0 0 7 u um 6721 sin 51 Solution To get rid of the nonhomogeneous bcs we introduce the new variable Uz7 t through 7r 7 Wm I Uzt 8 W The U satis es UL Um emsinsz U0t 0U7rt 0 Uz0 E 9 7r the 39 f 39 of the col 1 quot system are Xn sin nzi Thus with the usual ansatz for the solution of the nonhomogeneous equation7 we nd the solution W I linzt i722 iii 725 uzt7 7r Zgne s1nnz236 10 23e s1n51 5 10 Given lltzlt2 11 Sketch the even function fC of period 4 the odd function f5 of period 4 and the function fp of period 2 each of which equals on 072 EX pand each one in an appropriate Fourier series Expand fp in a sinecosine series and a complex exponential seriesi fx 0ltzlt1 Solution 2 l 3 ltsingisinwz sinmgt 12 l 2 l 3 l 5 fcz ltco 7 cosigcosmgt 13 l 2 l l 239 0 1 fpz E 7 g E gsinmrz E g E gem l4 Dddn 71700 10 What are the properties characterizing a S L eigenvalue problem H H E0 H 9 i There is an in nity of discrete eigenvalues An they are quantized7 and an in nity of corresponding eigenfunctions Xn ii Eigenfunctions corresponding to different eigenvalues are orthogonal iii They are complete 1 Consider the nonhomogeneous boundary conditions are imposed on the heat equation u a um7 u07 t uL7 t 161 762 15 16 How is the inhomogeneity eliminated If you introduce a new function U7 state what equation7 boundary and initial conditions it satis es To eliminate the nonhomogeneity7 we introduce a new function Uzt through I uxt k1 Z062 7 k1 Uz7 t 17 Then U satis es a heat equation with homogeneous BCsi However the le changer Given a general PDE Sum 411 Buzz 2 2uy 2uz 100u 07 18 eliminate the lower order derivatives Solution Introduce a new function wz7 y 2 through Wayyz ECIHMHWMI oz 19 substituting into 18 and determining the unknown Cl 52 Cg so that lower derivatives do not occur in the equation for wi Given the heat equation 20 ut 1211 7 u nd the solution up to arbitrary constantsi Solution First perform the change of vars uzt e wzt7 to reduce the pde to wt a wmi Then the BCs on w will be simple enough so that we can use the table that gives the L t and t Given the heat equation ut 04211 7 vugc 21 and some BCs nd the solution up to some arbitrary constantsi Solution 2 with uzt 6 47 th wzt7 the equation reduces to wt 1 Then apply the table to nd the solution 2 w i 14 Find the Fourier transform of the following function H 539 z 6 711 otherwise fltzgt 3 22gt Then nd an integral representation for the function What is the value of the integral at z il a Solve the diffusion equation With convection u kum cum7 z E 700007 uz0 23 Use the convolution and the shift theoremi b Consider the initial condition to be Sketch the corresponding uzt for various positive timesi Comment on the signi cance of the convection term cu

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