5.4. THE HEAT EQUATION AND CONVECTION-DIFFUSIONc2006 Gilbert Strang5.4 The Heat Equation and Convection-DiffusionThe wave equation conserves energy. The heat equation ut= uxxdissipates energy.The starting conditions for the wave equation can be recovered by going backward intime. The starting conditions for the heat equation can never be recovered. Compareut= cuxwith ut= uxx, and look for pure exponential solutions u(x, t) = G(t) eikx:Wave equation: G0= ickG G(t) = eickthas |G| = 1 (conserving energy)Heat equation: G0= −k2G G(t) = e−k2thas G < 1 (dissipating energy)Discontinuities are immediately smoothed out by the heat equation, since G is ex-ponentially small when k is large. This section solves ut= uxxfirst analytically andthen by finite differences. The key to the analysis is the beautiful fundamentalsolution starting from a point source (delta function). We will show in equation (7)that this special solution is a bell-shaped curve:u(x, t) =1√4πte−x2/4tcomes from the initial conditionu(x, 0) = δ(x) . (1)Notice that ut= cux+ duxxhas convection and diffusion at the same time. Thewave is smoothed out as it travels. This is a much simplified linear model of thenonlinear Navier-Stokes equations for fluid flow. The relative strength of convectionby cuxand diffusion by duxxwill be given below by the Peclet number.The Black-Scholes equation for option pricing in mathematical finance also hasthis form. So do the key equations of environmental and chemical engineering.For difference equations, explicit methods have stability conditions like ∆t ≤12(∆x)2. This very short time step is more expensive than c∆t ≤ ∆x. Implicitmethods can avoid that stability condition by computing the space difference ∆2Uat the new time level n + 1. This requires solving a linear system at each time step.We can already see two major differences between the heat equation and the waveequation (and also one conservation law that applies to both):1. Infinite signal speed. The initial condition at a single point immediatelyaffects the solution at all points. The effect far away is not large, because of thevery small exponential e−x2/4tin the fundamental solution. But it is not zero.(A wave produces no effect at all until the signal arrives, with speed c.)2. Dissipation of energy. The energy12R(u(x, t))2dx is a decreasing functionof t. For proof, multiply the heat equation ut= uxxby u. Integrate uuxxbyparts with u(∞) = u(−∞) = 0 to produce the integral of −(ux)2:Energy decayddtZ∞−∞12u2dx =Z∞−∞uuxxdx = −Z∞−∞(ux)2dx ≤ 0 . (2)c2006 Gilbert Strang3. Conservation of heat (analogous to conservation of mass):Heat is conservedddtZ∞−∞u(x, t) dx =Z∞−∞uxxdx =hux(x, t)i∞x=−∞= 0 . (3)Analytic Solution of the Heat EquationStart with separation of variables to find solutions to the heat equation:Assume u(x, t) = G(t)E(x). Then ut= uxxgives G0E = GE00andG0G=E00E.(4)The ratio G0/G depends only on t. The ratio E00/E depends only on x. Sinceequation (4) says they are equal, they must be constant. This produces a usefulfamily of solutions to ut= uxx:E00E=G0Gis solved by E(x) = eikxand G(t) = e−k2t.Two x-derivatives produce the same −k2as one t-derivative. We are led to exponentialsolutions of eikxe−k2tand to their linear combinations (integrals over different k):General solution u(x, t) =12πZZZ∞−∞bu0(k)eikxe−k2tdx. (5)At t = 0, formula (5) recovers the initial condition u(x, 0) because it inverts theFourier transform bu0(Section 4.4.) So we have the analytical solution to the heatequation—not necessarily in an easily computable form ! This form usually requirestwo integrals, one to find the transform bu0(k) of u(x, 0), and the other to find theinverse transform of bu0(k)e−k2tin (5).Example 1 Suppose the initial function is a bell-shaped Gaussian u(x, 0) = e−x2/2σ.Then the solution remains a Gaussian. The number σ that measures the width of thebell increases to σ + 2t at time t, as heat spreads out. This is one of the few integralsinvolving e−x2that we can do exactly. Actually, we don’t have to do the integral.That function e−x2/2σis the impulse response (fundamental solution) at time t = 0to a delta function δ(x) that occurred earlier at t = −12σ. So the answer we want (attime t) is the result of starting from that δ(x) and going forward a total time12σ + t:Widening Gaussian u(x, t) =pπ(2σ)pπ(2σ + 4t)e−x2/(2σ + 4t). (6)This has the right start at t = 0 and it satisfies the heat equation.5.4. THE HEAT EQUATION AND CONVECTION-DIFFUSIONc2006 Gilbert StrangThe Fundamental SolutionFor a delta function u(x, 0) = δ(x) at t = 0, the Fourier transform is bu0(k) = 1. Thenthe inverse transform in (5) produces u(x, t) =12πReikxe−k2tdk One computation ofthis u uses a neat integration by parts for ∂u/∂x. It has three −1’s, from the integralof ke−k2tand the derivative of ieikxand integration by parts itself:∂u∂x=12πZ∞−∞(e−k2tk)(ieikx) dk = −14πtZ∞−∞(e−k2t)(xeikx) dk = −xu2t. (7)This linear equation ∂u/∂x = −xu/2t is solved by u = ce−x2/4t. The constantc = 1/√4πt is determined by the requirementRu(x, t) dx = 1. (This conservesthe heatRu(x, 0) dx =Rδ(x) dx = 1 that we started with. It is the area under abell-shaped curve.) The solution (1) for diffusion from a point source is confirmed:Fundamental solution fromu(x, 0) = δ(x)u(x, t) =1√4πte−x2/4t. (8)In two dimensions, we can separate x from y and solve ut= uxx+ uyy:Fundamental solution fromu(x, y, 0) = δ(x)δ(y)u(x, y, t) =1√4πt2e−x2/4te−y2/4t. (9)With patience you can verify that u(x, t) and u(x, y, t) do solve the 1D and 2D heatequations (Problem). The zero initial conditions away from the origin arecorrect as t → 0, because e−c/tgoes to zero much faster than 1/√t blows up. Andsince the total heat remains atRu dx = 1 orRRu dx dy = 1, we have a valid solution.If the source is at another point x = s, then the response just shifts by s. Theexponent becomes −(x−s)2/4t instead of −x2/4t. If the initial u(x, 0) is a combinationof delta functions, then by linearity the solution is the same combination of responses.But every u(x, 0) is an integralRδ(x−s) u(s, 0) ds of point sources ! So the solution tout= uxxis an integral of the responses to δ(x −s). Those responses are fundamentalsolutions starting from all points x = s:Solution from any u(x, 0) u(x, t) =1√4πtZ∞−∞u(s, 0) e−(x