527heatcyl

527heatcyl - 642:527 FALL 2007 THE HEAT EQUATION IN A DISK...

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Unformatted text preview: 642:527 FALL 2007 THE HEAT EQUATION IN A DISK Periodic and singular Sturm-Liouville problems In these notes we study the two-dimensional heat equation in a disk of radius a : α 2 ∇ 2 u ( x, y, t ) = ∂ ∂t u ( x, y, t ) , x 2 + y 2 ≤ a 2 . Here ∇ 2 , also written as △ , is the two-dimensional Laplacian, that is, ∇ 2 u = u xx + u yy . We will impose a homogeneous Dirichlet boundary condition at the boundary of the disk, i.e., require that u ( x, y, t ) = 0 if x 2 + y 2 = a 2 ; this could easily be replaced by a Neumann or Robin boundary condition. We will also impose an initial condition specifying the temperature distribution at time t = 0. Since the domain and the PDE have rotational symmetry it is natural to work in polar coordinates ( r, θ ), where x = r cos θ and y = r sin θ . We must then rewrite the x and y derivatives contained in ∇ 2 as derivatives with respect to r and θ . The result is derived in Section 16.7 of our text, with the final result appearing in equation (18); that equation refers to cylindrical coordinates, but to pass to polar coordinates one simply neglects all derivatives with respect to z . The r derivatives appearing in (18) can conveniently be rewritten as ∂ 2 u ∂r 2 + 1 r ∂u ∂r ≡ 1 r ∂ ∂r parenleftbigg r ∂u ∂r parenrightbigg . Thus our problem becomes 1 r ∂ ∂r parenleftbigg r ∂u ∂r parenrightbigg + 1 r 2 ∂ 2 u ∂θ 2 = 1 α 2 ∂u ∂t , ≤ r < a, ≤ θ < 2 π, t > 0; (1) u ( a, θ, t ) = 0 , ≤ θ < 2 π, t > 0; (2) u ( r, θ, 0) = f ( r, θ ) , ≤ r < a, ≤ θ < 2 π. (3) To attack this problem we will first find, by separation of variables, solutions of the PDE (1) that also satisfy the boundary conditions (2); then we will form a solution that also satisfies the initial condition (3) as a superposition of these product solutions. To apply the separation of variables method we look for solutions of (1) in the form u ( r, θ, t ) = R ( r )Θ( θ ) T ( t ) . Inserting this proposed solution into (1) leads to Θ T r d dr parenleftbigg r dR dr parenrightbigg + RT r 2 d 2 Θ dθ 2 = R Θ α 2 dT dt . We divide this equation by R Θ T to obtain 1 r ( rR ′ ) ′ R + 1 r 2 Θ ′′ Θ = 1 α 2 T ′ T . (4) 1 640:527 FALL 2007 THE HEAT EQUATION IN A DISK 2 The left hand side of (4) depends only on r and θ , while the right hand side depends only on t ; this is possible only if both sides are constant. Let us call this constant − λ ; then we have 1 r ( rR ′ ) ′ R + 1 r 2 Θ ′′ Θ = − λ and 1 α 2 T ′ T = − λ. (5) The second equation in (5) is easily solved for T ( t ): T ( t ) = e − λα 2 t . (6) Remark 1: Let us discuss more carefully why (5) follows from (4). Equation (4) has the form G ( r, θ ) = H ( t ), and this equality is supposed to hold for all r , θ , and t . Now choose any fixed values r 1 and θ 1 of the variables r and θ , and define λ by G ( r 1 , θ 1 ) = − λ . But now for any t , H ( t ) = G ( r 1 , θ 1 ) = − λ , which is just the second equation in (5). The first, which is just the second equation in (5)....
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This note was uploaded on 09/29/2009 for the course 642 527 taught by Professor Speer during the Fall '07 term at Rutgers.

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527heatcyl - 642:527 FALL 2007 THE HEAT EQUATION IN A DISK...

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