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Week_05 - ACM 100c Methods of Applied Mathematics Week 5 1...

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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 < θ < 2 π. At time t = 0 we assume there is an initial heat distribution given by u ( r,θ,t = 0) = f ( r,θ ) , and we assume that the boundary of the circular plate is kept at zero temperature for all time: u ( r ,θ,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 ∂θ 2 < r < r , < θ < 2 π. We again try to use separation of variables to solve this problem. We set u ( r,θ,t ) = R ( r )Θ( θ ) T ( t ) and try to separate the variables. We get T t T = k 1 rR ∂ ∂r r ∂R ∂r + 1 r 2 Θ θθ Θ . 1 We see that we can certainly separate the time variable and set T t T =- kλ 2 . This then leaves us with the equation- λ 2 = 1 rR ∂ ∂r r ∂R ∂r + 1 r 2 Θ θθ Θ . If we now add the first term on the left hand side and multiply both sides by r 2 we get- λ 2 r 2- r R ∂ ∂r r ∂R ∂r = Θ θθ Θ . We can see that we can now separate the r and θ variable by setting Θ θθ Θ =- μ 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 Θ now becomes Θ θθ + μ 2 Θ = 0 < θ < 2 π. 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,θ ) inside the plate and let θ → θ +2 π 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 Θ θθ + μ 2 Θ = 0 < θ < 2 π are periodic with period 2 π . We therefore have a Sturm-Liouville equation but with peri- odic conditions. The solutions are Θ = sin( mθ ) and Θ = cos( mθ ) , m = 0 , 1 , 2 , ··· , and the separation constant is the eigenvalue: μ 2 = m 2 . We next turn to the radial equation for R ( r ) and substitute what we have so far:- λ 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 + λ 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 + λ 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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Week_05 - ACM 100c Methods of Applied Mathematics Week 5 1...

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