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## PART DIFF EQUATIONS

by: Estevan Champlin Sr.

74

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3

# PART DIFF EQUATIONS MATH 124A

Estevan Champlin Sr.
UCSB
GPA 3.66

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
3
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 3 page Class Notes was uploaded by Estevan Champlin Sr. on Thursday October 22, 2015. The Class Notes belongs to MATH 124A at University of California Santa Barbara taught by Staff in Fall. Since its upload, it has received 74 views. For similar materials see /class/226881/math-124a-university-of-california-santa-barbara in Mathematics (M) at University of California Santa Barbara.

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Date Created: 10/22/15
Math 124A PDEs Solving the diffusion heat equation Fix a positive number k and consider the diffusion equation on the Whole line DE utkum on0lttltoo7 ooltoltoo Properties of the diffusion equation o If 3133715 and 32337 t are both solutions of DE so 3 kSlm and Sgt 1632m7 then the function uo7 t 1131337 t 1232337 t Where a1 and a2 are any two numbers is a solution of You know this because the di usion equation is linear You can also check it by taking partial derivatives Find au 3 it E 131 232 Wu 32 um w 1131 1232 Check that ut kum 0 o If 333715 is a solution of DE7 then the function uo7t 333 y7 t is a solution of DE Where y is any xed number Check 455715 393 y7tlt 5W1 yy 15 Then ut17 t kum27 t 513 y7 t kSmo y7 t 0 since S solues DE o If 333715 is a solution of DE7 then uo7t ff 333 y7tgydy is a solution of DE for any function g of one variable This is also easy to check simply by di erentiating u If you can pull deriva tives under the integral sign Intuitiuely we can think of this integral as an in nite sumi In the integral above we are summing solutions of DE so by linearity u should solve More rigorously if we partition the line into intervals of size Ay for each integer k is just some number and by the previous property gAy 333 g t solues By linearity 00 k k 3 t S t A 507 2009 6 n7 y is a solution of Of course as n gets large Sn2t gt uat This implies that u also solues DE to see this take limits of both sides of the equation 3 0 Finally ifuo t is a solution of DE the new function uo t uao at is also a solution of DE for any positive number a Check by di erentiating use the chain rule 3 do 3 uto t ampuax at x atampat Withax at umot a aumao at g 63 Then ut hum O at every point with t gt O We will start by solving DE with special choice for the initial condition The problem will be 5116 Olttltoo ooltoltoo Sa0 6a 00 lt a lt 00 The solution 333 t that we nd is alternately called the source function for the diffusion equation Green s function for the diffusion equation or the fun damental solution of the diffusion equation This solution gives us all others If we want to solve the problem utkum Olttltoo ooltoltoo 00 lt x lt 007 the solution is uaarfSa y wami Why Well by our properties of the diffusion equation this solves DE ie ut Also we can check whether the initial condition holds 0 um ww Wi wwwww 0 OO Since ua7 t solves DE and u27 O qb27 the function u solves the problem a Intuitiuely this is true by linearity of the di usion equation We can write as the integral of the functions qby6a y and since the equation is linear the solution u should be the integral of the solutions qbySa y7 t each of which has initial condition qby6a y All we have to do is gure out what the function So7t is lnstead7 it will be easier to solve for Q where S g and Q solves the problem Qtka0lttltoo7 ooltxltoo mammal 23 S will solve DE with 33370 Hx 6a Again7 from our properties of the diffusion equation we know that Qa27 at solves But this function has initial condition Qa27 O Check that Because of this special property of H7 we expect that 0071 6205957 at since we expect that two solutions of DE that start off with same initial condition are the same in other words7 we expect uniqueness Then the value of the function Q depends only on the ratio Q27 t will be constant on the curves 1 mt This means that oat 2 g for some function g of one variable later7 we ll call the variable We know Q solves DE7 and we will use that to gure out what ODE g solves Start differentiating Q remember to use the chain rule For the derivative with respect to t7 hold a constant7 and notice 2t27 so t W g t amt g and omega g We know Qt ka Plugging in our cornputations above7 this means that eae Also7

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