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Introduction to Partial Differential Equations

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Introduction to Partial Differential Equations Introduction to Partial Differential Equations 1 HeatConductioninaSolidBody 1.1 Heat Transfer in One Spatial Direction Consider a homogeneous rod: length of rod is L lateral surface of rod is insulated rod has constant thermal conductivity k (k>0) rod has constant specificheatc (c>0) rod has constant density ρ (mass/volume) rod has congruent cross sections perpendicular to its lateral axis, the x-axis; each cross section has area A H1. Heat flows only parallel to the lateral axis of the rod. Consequently, the temperature must be the same at each point in the cross section at x. Let: u = u (x, t) be the temperature at cross section x at time t. (0 x L, t 0) ≤ ≤ ≥ q = q (x, t) be the heat flux (the time rate of heat transfer per unit area) across the cross section at x at time t. (Heat flux is a vector quantity whose direction is given by the sign of q (x, t) . Heat flows from left to right when q (x, t) is positive and flows from right to left when q (x, t) is negative.) e = e (x, t) be the specific internal energy (heat energy per unit mass) at cross section x at time t. f = f (x, t) , assumed known, be the time rate per unit volume at which heat is generated by processes within the rod. (When f>0 there is a heat source. When f<0 there is a heat sink.) 1 H2. Thermal Energy is Conserved: Thetimerateofchangeofthermal energy in any region V in a heat conducting solid is equal to the net rate at which heat flows into V across its boundary plus the net rate at which heat is produced by sources and/or sinks in V. Apply H2 to the region V of the rod with a x b to obtain: ≤ ≤ b b etρA dx = ( qx + f) Adx (1) − Za Za for all a<bin [0,L] and for all t 0. ≥ Lemma 1 If r (x) and s (x) are continuous functionsonanintervalI and if b b r (x) dx = s (x) dx for all a<bin I, Za Za then r (x)=s (x) for all x in I. The Lemma and (1) imply that etρ = qx + f (2) − Now, we need to relate the three unknowns e, q, and u through properties characteristic of heat flow. H3. (Fourier’s Law of Heat Conduction) Heat flows from regions of higher temperature to regions of lower temperature and the rate of flow is greater the more rapidly temperature changes with respect to distance. Therefore we assume that the rate of heat flow is proportional to the temperature gradient: q (x, t)= kux with k>0. − The thermal conductivity k is defined by this relation. Use of Fourier’s law in (2) give etρ = kuxx + f (3) H4. For many materials and over wide temperature ranges the specificinternal energy is a linear function of the temperature: e = cu + d (c>0,dconstants) The specificheatc is defined by this relation. The units of specificheat are [energy]/[mass][deg]. 2 From H4 et = cut. So (3) can be expressed as ρcut = kuxx + f, the inhomogeneous heat equation (in one spatial dimension). Let k 1 α = (thermal diffusivity) and F = f. scρ cρ Then the inhomogeneous heat equation can be expressed as 2 ut = α uxx + F. (IHE) If there are no sources/sinks present, then F =0and 2 ut = α uxx (HE) which is usually called the heat equation (AKA the diffusionequationinone spatial dimension). 1.2 Heat Transfer in Two and Three Spatial Directions Consider a homogeneous solid heat-conducting body: (Mathematically such a body is modeled as a bounded region D in xyz-space.) solid has constant thermal conductivity k (k>0) solid has constant specificheatc (c>0) solid has constant density ρ (mass/volume) In this context, u = u (x, y, z, t) , q = q (x, y, z, t) ,e= e (x, y, z, t) ,f = f (x, y, z, t) . As before assume: H2. Thermal energy is conserved H3. Fourier’s Law holds: q = k u − ∇ H4. Specific internal energy is a linear function of the temperature 3 The heat flux vector q is defined by the following property: Let S be any reasonable surface in the heat conducting medium and let n by a continuous unit normal vector field on S. (In mathematical lingo, the surfaced is orientable.) Then the heat per second flowing across S towardthesideofS specified by n is q dS = q n dS. · · ZZS ZZS A straightforward adjustment of the reasoning we used in two spatial dimension leads to the inhomogeneous heat equation 2 ut = α (uxx + uyy + uzz)+F (IHE) in three spacial dimensions, and, if there are no sources/sinks present to the heat equation 2 ut = α (uxx + uyy + uzz) (HE) in three spacial dimensions. Here α and F are defined as in the one spatial dimension case. The derivation referred to above is most easily carried out using the (Gauss) Divergence Theorem: If D is a region in space with bounding surface S oriented by its outer normal n, then under quite general assumptions (see a book on vector analysis or multivariable calculus or Kellog’s Potential Theory) div F dV = F dS · ZZZD ZZS for any smooth vector field F defined on D S. If the thermal situation is such that there is∪ not heat flow in say the z-direction, then the 3-D heat equations reduce to their 2-D versions in which all functions are independent of z : 2 ut = α (uxx + uyy)+F (IHE) in two spacial dimensions, and, if there are no sources/sinks present to the heat equation 2 ut = α (uxx + uyy) . (HE) 4 1.3 Initial and Boundary Conditions for Heat Conduction Problems and Separation of Variables The foregoing heat equations do not, by themselves, determine the temperature within the solid. In addition, we need to know the initial temperature in the solid and the manner in which heat moves across the boundary of the solid. For example, consider heat flow in the homogeneous rod with no sources or sinks, whose initial temperature profile is given by f (x) , and whose ends are kept at temperature zero. Intuition suggests that the temperature u = u (x, t) in the rod is determined by the initial boundary value problem: 2 ut α uxx =0 for 0 <x<L,t>0, (HE) u (−x, 0) = f (x) for 0 x L, (IC) ⎧ u (0,t)=0,u(L, t)=0 for t ≤ 0. ≤ (BC) ⎨ ≥ ⎩ Let’s try to solve this problem. Grand strategy: (HE) and (BC) are linear and homogeneous in u. By the su- perposition principle, (finite) linear combinations of solutions to (HE) and (BC) are again solutions to (HE) and (BC). Maybe we can adjust the coefficients in the linear combination so that the sum also satisfies (IC). Key Step 1: Guess some solutions to the (HE). What is a reasonable educated guess? Key Step 2: Refine the guess in Step 1 so that the solutions in Step 1 also satisfy (BC). Key Step 3: Superpose the solutions in Step 2 so that the resulting linear combination will satisfy (IC). If the grand strategy works we will have found a solution of the given initial boundary value problem for the rod. Steps 1 and 2: There are no sources or sinks in our problem and the ends of the rod are kept at temperature zero. It is reasonable to expect that the temper- ature u (x, t) 0 as t . Since solutions to DEs often involve exponential → →∞ 5 functions, we look for solutions to the heat equation that decay exponentially in time — say λt u (x, t)=e− X (x) λ>0. For convenience replace λ by α2λ to get α2λt u (x, t)=e− X (x) λ>0. α2λt Our hope is to find X (x) so that u (x, t)=e− X (x) satisfies (HE) and (BC). Let’s see if there is such an X... Summary: We found an infinite number of functions 2 2 2 2 nπx u (x, t)=e α n π t/L sin n =1, 2, 3, ... (ss) n − L that satisfy (are solutions of) (HE) and (BC). Along the way we reviewed or learned the following: z2 zn ez =1+z + + + + for all complex z, 2! ··· n! ··· z3 z5 z7 sin z = z + + for all complex z, − 3! 5! − 7! − ··· z2 z4 z6 cos z =1 + + for all complex z. − 2! 4! − 6! − ··· Thus, setting z = iθ with θ real in the first series and doing some algebra leads to the basic Euler identity eiθ =cosθ + i sin θ. Each of the following four equations, also called Euler identities, implies the other three: iθ iθ e =cosθ + i sin θ, e− =cosθ i sin θ, iθ iθ iθ −iθ e + e− e e− cos θ = , sin θ = − . 2 2i Note that eiθ =1for any real θ. Theta is used here because¯ it often¯ is a polar angle. The Euler identities are also true with θ is complex in¯ which¯ case it is usually replace by z. For the record and later use: If z = x + iy, with x and y real, then ez = exeiy = ex (cos y + i sin y)=ex cos y + iex sin y, 6 which expresses exp (z) in terms of its real and imaginary parts. Easy conse- quences are: ez =0for any real or complex number z. 6 ez =1 z =2πin for n an integer. ⇐⇒ cos z =0 z = nπ/2 for n an odd integer.
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