The 1-D Heat Equation 18.303 Linear Partial Differential Equations Matthew J. Hancock Fall 2006 1 The 1-D Heat Equation 1.1 Physical derivation Reference: Guenther & Lee §1.3-1.4, Myint-U & Debnath §2.1 and §2.5 [Sept. 8, 2006] In a metal rod with non-uniform temperature, heat (thermal energy) is transferred from regions of higher temperature to regions of lower temperature. Three physical principles are used here. 1. Heat (or thermal) energy of a body with uniform properties: Heat energy = cmu, where m is the body mass, u is the temperature, c is the specific heat, units [c] = L2T−2U−1 (basic units are M mass, L length, T time, U temperature). c is the energy required to raise a unit mass of the substance 1 unit in temperature. 2. Fourier’s law of heat transfer: rate of heat transfer proportional to negative temperature gradient, Rate of heat transfer ∂u = (1) −K0 area ∂x where K0 is the thermal condu ctivity, units [K0] = MLT−3U−1 . In other words, heat is transferred from areas of high temp to low temp. 3. Conservation of energy. Consider a uniform rod of length l with non-uniform temperature lying on the x-axis from x = 0 to x = l. By uniform rod, we mean the density ρ, specific heat c, thermal conductivity K0, cross-sectional area A are ALL constant. Assume the sides 1� � � � of the rod are in sulated and only the ends may be exposed. Also assume there is no heat source within the rod. Consider an arbitrary thin slice of the rod of width Δx between x and x + Δx. The slice is so thin that the temperature throughout the slice is u (x, t) . Thus, Heat energy of segment = c × ρAΔx × u = cρAΔxu (x, t) . By conservation of energy, change of heat in from heat out from heat energy of = left boundary − right boundary . segment in time Δt From Fourier’s Law (1), ∂u ∂u cρAΔxu (x, t + Δt) − cρAΔxu (x, t) = ΔtA −K0 − ΔtA −K0∂x ∂x x x+Δx Rearranging yields (recall ρ, c, A, K0 are constant), u (x, t + Δt) −Δt u (x, t) = K0 cρ ��∂u ∂x � x+Δx − Δx �∂u ∂x � x � Taking the limit Δt, Δx → 0 gives the Heat Equation, ∂u ∂2u ∂t = κ ∂x2 (2) where κ = K0 (3) cρ is called the thermal diffusivity, units [κ] = L2/T . Since the slice was chosen arbi-trarily, the Heat Equation (2) applies throughout the rod. 1.2 Initial condition and boundary conditions To make use of the Heat Equation, we need more information: 1. Initial Condition (IC): in this case, the initial temperature distribution in the rod u (x, 0). 2. Boundary Conditions (BC): in this case, the temperature of t he rod is affected by what happens at th e ends, x = 0, l. What happens to the temperature at the end of the rod must be specified. In reality, the BCs can be complicated. Here we consider three simple cases for the boundary at x = 0. 2(I) Temperature prescribed at a boundary. For t > 0, u (0, t) = u1 (t) . (II) Insulated boundary. The heat flow can be prescribed at the boundaries, ∂u (0, t) = φ1 (t)−K0∂x (III) Mixed condition: an equation involving u (0, t), ∂u/∂x (0, t), etc. Example 1. Consider a rod of length l with insulated sides is given an initial temperature distribution of f (x) degree C, for 0 < x < l. Find u (x, t) at subsequent times t > 0 if end of rod are kept at 0o C. The Heat Eqn and corresponding IC and BCs are thus PDE: ut = κuxx, 0 < x < l, (4) IC: u (x, 0) = f (x) , 0 < x < l, (5) BC: u (0, t) = u (L, t) = 0, t > 0. (6) Physical intuition: we expect u 0 as t → ∞.→1.3 Non-dimensionalization Dimensional ( or physical) terms in the PDE (2): k, l , x, t, u. Others could be introduced in IC and BCs. To make the solution more meaningful and simpler, we group as many physical constants together as possible. Let the characteristic length, time and temperature be L∗, T∗ and U∗, respectively, with dimensions [L∗] = L, [T∗] = T , [U∗] = U. Introduce dimensionless variables via xt � � u (x, t) f (x) xˆ = , tˆ= , ˆ ˆ t = , x) = .u x, ˆfˆ(ˆ (7) L∗ T∗ U∗ U∗ The variables ˆx, tˆ, ˆu are dimensionless (i.e. no units, [ˆx] = 1). The sensible choice for the characteristic length is L∗ = l, the length of the rod. While x is in the range 0 < x < l, ˆ < ˆx is in the range 0 x < 1. The choice of dimensionless variables is an ART. Sometimes the statement of the problem gives hints: e.g. the length l of the rod (1 is nicer to deal with than l, an unspecified quantity). Often you have to solve the problem first, look at the solution, and try to simplify the notation. 3� � � � From th e chain rule, ∂u ∂uˆ∂tˆU∗ ∂uˆ ut = = U∗ = ∂t ∂tˆ∂t T∗ ∂tˆ, ∂u ∂uˆ ∂xˆ U∗ ∂uˆux = = U∗ = ∂x ∂xˆ ∂x L∗ ∂xˆU∗ ∂2uˆuxx = L2 ∗ ∂xˆ2 Substituting these into the Heat Eqn (4) gives ∂uˆT∗κ∂2uˆ=ut = κuxx ⇒ ∂tˆL2 ∗ ∂xˆ2 To make the PDE simpler, we choose T∗ = L2/κ = l2/κ, so that ∗∂uˆ∂2uˆ= , 0 x < 1, ˆ< ˆ t > 0. ∂tˆ∂xˆ2The characteristic (diffusive) time scale in the problem is T∗ = l2/κ. For different substances, this gives time scale over which diffusion takes place in the problem. The IC (5) and BC (6) must also be non-dimen sionalized: IC: uˆ (ˆx, 0) = fˆ(ˆx) , 0 x < 1,< ˆBC: uˆ 0, tˆ= uˆ 1, tˆ= 0, t > ˆ0. 1.4 Dimensionless problem Dropping hats, we have the dimensionless problem PDE: ut = uxx, 0 < x < 1, (8) IC: u (x, 0) = f (x) , 0 < x < 1, (9) BC: u (0, t) = u (1, t) = 0, t > 0, (10) where x, t are dimensionless scalings of physical position and time. 2 Separation of variables Ref: Guenther & Lee, §4.2 and 5.1, Myint-U & Debnath §6.4 [Sept 12, 2006] We look for a solution t o the dimensionless Heat Equation (8) – (10) of the form u (x, t) = X (x) T (t) (11) 4Take the relevant partial derivatives: ′′ ′ uxx = X (x) T