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by: Estevan Champlin Sr.


Estevan Champlin Sr.
GPA 3.66


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This 4 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 85 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 The Wave Equation with a Source Consider the problem of the wave equation with a source utt cgumfxt on ooltxltoo 149570 x Int1570 1W The solution u is given by the formula 20 um lt ltxctgt ltx ctgtgti 11 ydy2 Af lt1 where the last term is the double integral of f over the domain of dependence for the point 0571 i f 2 t ijj ywwyl d5 Notice that we could check directly that this formula satis es utt cgu x If using the fundamental theorem of calculus carefullyl If u is a solu tion it is also easy to show that the formula must hold by calculating the integral A f Matt cm um tgt lt ltxctgt ltx ctgtgt jam dy This can be done directly by changing variables to the characteristic coor dinates E and 77 or by using Green s theorem See 34 for details For the wave equation with a source on the half line we will consider the case of a Dirichlet boundary condition at x 0 If we have the boundary condition u0 t 0 we can use the formula for the solution on the whole line taking the odd extensions of the initial conditions q and 11 as well as the odd extension in the x Variable of the source f This works since if all of these functions are odd every term in the formula 1 equals 0 when x 0 By fODD the odd extension of f in the x variable we mean f 35715 ifxgt0 fooDltx7t Given a point x t the domain of dependence is entirely in the rst quadrant if x ct gt 0 so in this region the boundary condition does not affect the solution But in the region 05 lt ct the boundary condition has an effect and the odd extensions of q 11 and f can be used in the formula 1 to determine the solution Now we wish to consider a general Dirichlet boundary condition In other words we want to solve the problem utt cgumfxt on0ltxltoo TIC157 0 05 M9670 W95 u0t ht We will prove that the formula for the solution is given by newWmcwww cwa 5f we nd mm 1435715 OD1D T Ct 0DD 35 Ct T 21 5 ODD y dy if 35 lt Ct 5 ffA form W 2 Notice that in the region 05 gt ct the formula is identical to the solution for the problem on the whole line The only effect of changing the boundary condition from 0 to Mt is to add a term that depends on h to the solution in the region 05 lt ct The function ht is constant along the characteristics with positive slope so this function is basically a wave propogating to the right To solve this problem on the half line with the general Dirichlet boundary condition u0 t Mt we will start by letting vxt ux t Mt For simplicity let s assume q 11 0 Then 1 solves the problem vii 02v fxt h t on0ltxltoo var 0 hltogt 114060 hltogt v0 t 0 and we know how to solve for 111 If x gt ct we have mm gem hltogtgt h m dy A f h h0 th0Ah liAf We can calculate the double integral of h by rst doing the integral with respect to y and then integrating by parts 1 t 1 zct t h 1d h d h t d 201 1120 l gt8 0 lt8gtlt 8 t t 1hsds hst 5 0 30 ht h0 thO Plugging this into the formula for v we have vxt ht iffA f Therefore for x gt ct 1 t 7 7 2C As expected for this region the boundary condition has no effect For the second region 05 lt ct we have that m t gem h0 310le h0ODDdy Af mm ict Notice that since the initial condition 1195 0 h0 is a constant we used the odd extension h0 at the left endpoint x ct which is less than 0 in this region Similarly for the extension of the initial velocity we use h0 to the left of 0 and h0 to the right Therefore vxt10 h0d 1IClh0d ih if 7 i20 11a y 20 0 y 20 A ODD 20 A ODD hO hO 1 1 7 t7 15 7 h 7 ODD 20mm 20mm macAr x 1 1 7h 0 7 h 7 ODD c 20M ODD 2c 3 Now we only need to carefully compute the double integral of thD on A One way to do this is to break up the region A into four parts see the picture below Region I is the only one where the odd extension is used on this region thDy 5 h 5 Everywhere else on A thDy 5 h s m x ct xlcl Since for each xed 5 the function h 5 is a constant by symmetry we have that 1UUthDysdgds Ih sdy U h s 0 Then on the square region 1 if 21 x 757 h 5d5 h 5dsdy h 5ds III 20 0 0 C 0 a a a h0 ht c lt gt c lt c On the top triangle 1 1 t zctis 7 h sd5 7 h 5dyds 20 IV 20 779 170t78 t Ahabads ti 1hsds t sWs SamC ht g ht ght g t Adding the integrals over each of these four regions we nd 21C hgm ht g ht ghO


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