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## PARTIAL DIFF EQNS

by: Mossie Pfannerstill V

25

0

6

# PARTIAL DIFF EQNS AMATH 353

Mossie Pfannerstill V
UW
GPA 3.59

Staff

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COURSE
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Staff
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6
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KARMA
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## Popular in Applied Mathematics

This 6 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 Staff in Fall. Since its upload, it has received 25 views. For similar materials see /class/192281/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 Eleftherios means free Kirkinis February 157 2008 Abstract Review of solution techniques developed and the corresponding applied mathemat ical constructions Not all subjects to be examined are included here 1 Midterm Review The Midterm will be comprehensive Everything we covered up to and including homework 6 No characteristics no applications just methodology to solve problems In particular give special attention to the following homework problems HW2 Problems 1273475 HW3 All problems HW4 Problems 24the Parseval Part7 6778 HW5 all are important but the essential ones are 476 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 DirichletBCssettemp NeumannBCsinsulated Periodic X00XL0 X 00X L0 X7LXLX 7LX L evaluesAn n 12 012 0123 efnan sin cos sin and cos 2 The solution is given in general by uzt Z AnTntXnx 3 nDL7r1 and for the above BCs Tnt e quot gt2t 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 2X0 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 In order to satisfy the le at t 0 we evaluate the solution 393 at t 0 and equate with f since the initial conditions given are ut 0 f f is a given known function to obtain 0 f Z AanWL 5 nDL7r1 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 A fOL Xmm 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 W 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 wmy7 2 through 61z62ycazw7y727 substituting into W 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 042 7 u 18 reduces to wt 04210 c with u7 t 3quotW M20 21UQif7 the equation ut 042 7 yum 19 reduces to wt 04210 5 Nonhomogeneous PDE s with source functions or external forcing Lesson 9 The general problem is Given a heat equation with a source function 9amp7 t g is known ut 042 9amp7 t 20 with BCs 04111407 75 511107 t 0 21 oagugAL7 t 8211L7 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 W 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

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