Week_05 - ACM 100c - Methods of Applied Mathematics Week 5...

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Unformatted text preview: ACM 100c - Methods of Applied Mathematics Week 5 April 26, 2009 1 The heat equation in polar coordinates We next look at the heat equation in polar coordinates. We assume we have a circular plate given by < r < r < < 2 . At time t = 0 we assume there is an initial heat distribution given by u ( r,,t = 0) = f ( r, ) , and we assume that the boundary of the circular plate is kept at zero temperature for all time: u ( r ,,t ) = 0 . We would like to solve the heat equation u t = k 2 u in this domain with these initial and boundary conditions. It makes most sense to adopt polar coordinates since the boundary is easily expressed in these coordinates. The heat equation is u t = k 1 r r r u r n + 1 r 2 2 u 2 < r < r , < < 2 . We again try to use separation of variables to solve this problem. We set u ( r,,t ) = R ( r )( ) T ( t ) and try to separate the variables. We get T t T = k 1 rR r r R r + 1 r 2 . 1 We see that we can certainly separate the time variable and set T t T =- k 2 . This then leaves us with the equation- 2 = 1 rR r r R r + 1 r 2 . If we now add the first term on the left hand side and multiply both sides by r 2 we get- 2 r 2- r R r r R r = . We can see that we can now separate the r and variable by setting =- 2 . The choice of the sign of the separation constant is of course at our disposal but the choice made here makes particular sense as will be seen. The ODE for now becomes + 2 = 0 < < 2 . This is a Sturm-Liouville equation, but here the boundary conditions arise from the fact that we want the solution to be physically sensible. If we start at some point ( r, ) inside the plate and let +2 we expect that we get the same temperature since we have just gone around a circle to the same point. But in order for this to happen we must have that the solutions to + 2 = 0 < < 2 are periodic with period 2 . We therefore have a Sturm-Liouville equation but with peri- odic conditions. The solutions are = sin( m ) and = cos( m ) , m = 0 , 1 , 2 , , and the separation constant is the eigenvalue: 2 = m 2 . We next turn to the radial equation for R ( r ) and substitute what we have so far:- 2 r 2- r R r r R r =- m 2 , or, as an ODE, 1 r d dr r dR dr- m 2 r 2 R + 2 R = 0 . 2 We can also see that this can be written in the standard Sturm-Liouville form: d dr r dR dr- m 2 r R + 2 rR = 0 < r < r . Note, however, that this is a singular Sturm-Liouville problem because the leading term function (usually labeled p ( r ) ) in the Sturm-Liouville ODE vanishes at r = 0 . We also need to think about appropriate boundary conditions for R ( r ) . For r = r , we are on the boundary of the circle, so the temperature must vanish....
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This note was uploaded on 12/12/2009 for the course ACM 95c taught by Professor Nilesa.pierce during the Spring '09 term at Caltech.

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Week_05 - ACM 100c - Methods of Applied Mathematics Week 5...

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