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Unformatted text preview: 642:527 FALL 2009 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 by Greenberg, with the final result appearing in equation (18), which to avoid confusion with equations in these notes we will write as (G.16.7.18). 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 (G.16.7.18) 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 2009 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...
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This note was uploaded on 12/15/2009 for the course MATH 527 taught by Professor Staff during the Fall '08 term at Rutgers.
- Fall '08