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Chapter 22 NonlinearPartialDifferentialEquations Theultimatetopictobetouchedoninthisbookisthevastand activefieldofnonlinear partial differential equations. Leaving aside quantum mech anics, which remains to date an inherently linear theory, most real-world physical system s, including gas dynamics, fluid mechanics, elasticity, relativity, ecology, neurology, t hermodynamics, and many more, are modeled by nonlinear partial differential equations. Attempts to survey, in such a small space, even a tiny fraction of such an all-encompassing rang e of phenomena, methods, results, and mathematical developments, are doomed to fail ure. So we will be content to introduce a handful of prototypical, seminal examples that arise in the study of nonlinear waves and that serve to highlight some of the most significant physical and mathematical phenomena not encountered in simpler linear systems. We wil l only have space to look at simpleone-dimensionalmodels; thefarmorecomplicatedno nlinearsystemsthatgovernour three-dimensional dynamical universe quickly leadone to t hecutting edge of contemporary research. Historically, comparatively little was known about the ext raordinary range of behav- ior exhibited by the solutions to nonlinear partial differen tial equations. Many of the most fundamental phenomena that now drive modern-day resea rch, including solitons, chaos, stability, blow-up and singularity formation, asym ptotic properties, etc., remained undetected or at best dimly perceived in the pre-computer er a. The last sixty years has witnessed a remarkable blossoming in our understanding, du e in large part to the insight offered by the availability of high performance computers co upled with great advances in theunderstanding anddevelopment ofsuitablenumericalap proximationschemes. Newan- alytical methods, new mathematical theories, coupled with new computational algorithms have precipitated this revolution in our understanding and study of nonlinear systems, an activity that continues to grow in intensity and breadth. Ea ch leap in computing power coupled with theoretical advances has led to yet deeper unde rstanding of nonlinear phe- nomena, while simultaneously demonstrating how far we have yet to go. To make sense of this bewildering variety of methods, equations, and resu lts, it is essential build upon a firm foundation on, first of all, linear systems theory, and s econdly, nonlinear algebraic equations and nonlinear ordinary differential equations. Our presentation is arranged according to the order of the un derlying differential equation. First order nonlinear partial differential equat ions model nonlinear waves and arise in gas dynamics, water waves, elastodynamics, chemic al reactions, transport of pol- lutants, flood waves in rivers, chromatography, traffic flow, a nd a wide range of biological and ecological systems. One of the most important nonlinear phenomena, with no linear counterpart, is the break down of solutions in finite time, re sulting in the formation of 12/11/12 1169 c/circleco√yrt2012 Peter J. Olver discontinuous shock waves. A striking example is the supers onic boom produced by an airplane that breaks the sound barrier. As in the linear wave equation, the signals propa- gate along the characteristics, but in the nonlinear case th e characteristics can cross each other, precipitating the onset of a shock. The characteriza tion of the shock dynamics re- quires additional physical information, in the form of a con servation law, that supplements the original partial differential equation. Parabolic second order partial differential equations gove rn nonlinear diffusion pro- cesses, including thermodynamics, chemical reactions, di spersion of pollutants, and popu- lation dynamics. The simplest and most well understood is Bu rgers’ equation, which can, surprisingly, be linearized by transforming it to the heat e quation. This accident provides an essential glimpse into the world of nonlinear diffusion pr ocesses. In the limit, as the diffusion or viscosity tends to zero, the solutions to Burger s’ equation tend to the shock wave solutions to the limiting first order dispersionless eq uation, and thus provides an alternate mechanism for unraveling shock dynamics. Third order partial differential equations arise in the stud y of dispersive wave motion, including water waves, plasma waves, waves in elastic media , and elsewhere. We first treat the basic linear dispersive model, comparing and contrasti ng it with the hyperbolic mod- els we encountered earlier in this text. The distinction bet ween group and wave velocity — observed when, for instance, surface waves propagate over water — is developed. Fi- nally, we introduce the remarkable Korteweg–deVries equat ion, which serves as a model for waves in shallow water, waves in plasmas, and elsewhere. Des pite its intrinsic nonlinearity, it supports stable localized traveling wave solutions, kno wn assolitons, that, remarkably, maintain their shape even under collision. The Korteweg–de Vries equation is the proto- typical example of an integrable system, and this discovery in the mid 1960’s inaugurated intense and ongoing research in the remarkable physical mod els that exhibit integrability, a development that has had many ramifications in both pure and applied mathematics. 22.1. Nonlinear Waves and Shocks. Before attempting to tackle any nonlinear partial different ial equations, we must care- fully review the solution to the simplest linear first order p artial differential equation. Linear Transport and Characteristics Thetransport equation ut+cux= 0, (22.1) is so named because it models the transport of, say, a polluta nt in a uniform fluid flow. Let us begin by assuming that the wave speed cis constant. According to Proposition 14.8, every solution is constant along the characteristic lines o f slope dx dt=c, namely x−ct= constant . (22.2) As a consequence, the solutions are traveling waves of the form u(t,x) =p(x−ct), (22.3) 12/11/12 1170 c/circleco√yrt2012 Peter J. Olver Figure 22.1. Traveling Wave. wherep(ξ) isan arbitrary function of the characteristic variable ξ=x−ct. To a stationary observer, the solution (22.3) appears as a wave of unchangin g form moving at velocity c. Whenc >0, the wave translates to the right, as illustrated in Figure 22.1. When c <0, the wave moves to the left, while c= 0 corresponds to a permanent wave form that remains fixed at its original location. Slightly more complicated, but still linear, is the non-uni form transport equation ut+c(x)ux= 0, (22.4) where the wave velocity c(x) depends upon the spatial position. This equation models unidirectional waves propagating through a non-uniform, b ut static medium. Generalizing the construction (22.2), we define a characteristic curve to be a solution to the autonomous ordinary differential equation dx dt=c(x). (22.5) Thus, unlike the constant velocity version, the characteri stics are no longer necessarily straight lines. Nevertheless, the preceding observation r emains valid: Proposition 22.1. Solutions to the linear transport equation (22.4)are constant on its characteristic curves. Proof: Letx(t) be a characteristic curve, i.e., a solution to (22.5), para metrized by the time t. Leth(t) =u(t,x(t)) be the value of the solution at the point ( t,x(t)) on the given characteristic curve. Our goal is to prove that h(t) is a constant function of t, and, as usual, this is done by proving that its derivative is ident ically zero. To differentiate h(t), we invoke the chain rule: dh dt=d dtu(t,x(t)) =∂u ∂t(t,x(t))+dx dt∂u ∂x(t,x(t)) =∂u ∂t(t,x(t))+c(x(t))∂u ∂x(t,x(t)) = 0. We replaced dx/dtbyc(x) since we are assuming that x(t) is a characteristic curve, and hence satisfies (22.5). The final combination of derivatives is zero whenever usolves the transport equation (22.4). Q.E.D. Since the characteristic curve differential equation (22.5 ) is autonomous, it can be immediately solved: b(x)≡/integraldisplaydx c(x)=t+k, (22.6) wherekis the constant of integration. Thus, the characteristic cu rves are “parallel”, each being a translate of the graph of t=b(x) in the direction of the taxis. The characteristic 12/11/12 1171 c/circleco√yrt2012 Peter J. Olver tx (t,x) (0,y) Figure 22.2. Characteristic Curve. curves are therefore defined by the formula x=g(t+k), where g=b−1is the inverse function. (See Section 20.1 for full details.) Observe that the characteristic curves are the level sets of thecharacteristic variable ξ=b(x)−t. As a consequence, any function which is constant along the c haracteristic curves depends only on the value of the characteristic varia ble at each point, and hence takes the form u(t,x) =p(b(x)−t) (22 .7) for some function p(ξ). In other words, the characteristic curves are the common l evel curves of allsolutions to the transport equation. It is easy to check dire ctly that, provided b(x) is defined by (22.6), u(t,x) solves the partial differential equation (22.4) for anychoice of function p(ξ). To find the solution that satisfies the prescribed initial con ditions u(0,x) =f(x) (22 .8) we merely substitute the general solution formula (22.7). T his leads to the equation p(b(x)) =f(x),and, therefore, p(ξ) =f◦b−1(ξ) =f/parenleftbig g(ξ)/parenrightbig . Theresultingsolutionformulahasasimplegraphicalinter pretation: tofinditsvalue u(t,x) at a given point, we look at the characteristic curve passing through ( t,x). If this curve intersects the xaxis at the point (0 ,y), thenu(t,x) =u(0,y) =f(y). The construction is illustrated in Figure 22.2. Incidentally, if the charact eristic curve through ( t,x) doesn’t intersect the xaxis, the solution value u(t,x) is not prescribed by the initial data. Example 22.2. Let us solve the particular transport equation ∂u ∂t+1 x2+1∂u ∂x= 0 (22 .9) 12/11/12 1172 c/circleco√yrt2012 Peter J. Olver tx Figure 22.3. Characteristic Curves for ut+1 x2+1ux= 0. by the method of characteristics. According to (22.5), the c haracteristic curves satisfy the first order ordinary differential equation dx dt=1 x2+1. Separating variables and integrating, we find /integraldisplay (x2+1)dx=1 3x3+x=t+k, wherekis the integration constant. Some of the resulting characte ristic curves are plotted in Figure 22.3. The characteristic variable is ξ=1 3x3+x−t, and hence the general solution to the equation takes the form u=p/parenleftbig1 3x3+x−t/parenrightbig , wherep(ξ) is an arbitrary function. A typical solution, correspondi ng to initial data u(t,0) =1 1+(x+2.75)2, is plotted at times t= 0,2,5,10,25,50 in Figure 22.4. The fact that the characteristic curves are not straight means that, although the solution re mains constant along each individual curve, a stationary observer will witness a dyna mically changing profile as the wavemoves along throughthe non-uniform medium. The wavesp eeds up asit aapproaches the origin, and then slows back down once it passes and moves o ff to the right. As a result, we observe the wave spreading out as it approaches the origin , and then contracting as it moves off to the right. 12/11/12 1173 c/circleco√yrt2012 Peter J. Olver Figure 22.4. Solution to ut+1 x2+1ux= 0. tx Figure 22.5. Characteristic Curves for ut−xux= 0. Example 22.3. Consider the equation ut−xux= 0. (22.10) In this case, the characteristic curves are the solutions to dx dt=−x, and so xet=k, (22.11) wherekis the constant of integration; see Figure 22.5. It is easier to adopt ξ=xetas the characteristic variable here, noting that its level set s are the characteristic curves. The 12/11/12 1174 c/circleco√yrt2012 Peter J. Olver Figure 22.6. Solution to ut−xux= 0. solution therefore takes the form u=p(xet), (22.12) wherep(ξ) is an arbitrary function of ξ=xet. Given the initial data u(0,x) =f(x),the resulting solution is u=f(xet). For example, the solution u(t,x) =1 (xet)2+1=e−2t x2+e−2t corresponding to initial data u(t,0) =f(x) = (x2+1)−1is plotted in Figure 22.6 at times t= 0,1,2,3. Note that since the characteristic curves all converge on thetaxis, the solution becomes more and more concentrated at the origin. I n the limit, it converges to thefunction thatiszero everywhere except for thevalue u(t,0)≡1 attheorigin. Warning : The limit is nota delta function, since its value at x= 0 remains bounded. A Nonlinear Transport Equation Perhaps the simplest possible nonlinear partial differenti al equations is the nonlinear transport equation ut+uux= 0. (22.13) first systematically studied by Poisson and Riemann in the ea rly nineteenth century. Since it appears in so many applications, this equation appears in the literature under a variety of names, including the Riemann equation, the inviscid Burg ers’ equation, and the dis- persionless Korteweg–deVries equation. It and its multi-d imensional and multi-component generalizations play a crucial role in the modeling of gas dy namics, traffic flow, flood waves in rivers, chromatography, chemical reactions, and other a reas; see [ 188]. 12/11/12 1175 c/circleco√yrt2012 Peter J. Olver The first order partial differential equation (22.13) has the form of a transport equa- tion, whose wavevelocity c=udepends, not onthe position x, but rather onthe sizeof the disturbance. Larger waves move faster, and overtake smalle r waves. Waves of elevation, whereu >0, move to the right, while waves of depression, where u <0, move to the left. Fortunately,themethodofcharacteristicsthatwasdevelo pedforlinearwaveequations also works in the present nonlinear situation and leads to a c omplete solution. Mimicking our previous construction, (22.5), let us define a characteristic curve of the nonlinear wave equation (22.13) to be a solution to the ordinary differentia l equation dx dt=u(t,x). (22.14) As such, the characteristics depend upon the solution u, which, in turn, is based on the characteristic variable. So we appear to be trapped in a circ ular argument. The resolution of the apparent conundrum is to observe that, as in the linear case, the solution u(t,x) remains constant along its characteristics, and this fact w ill allow us to simultaneously specify both. To prove this claim, suppose that x=x(t) parametrizes a characteristic curve. We need to show that the function h(t) =u(t,x(t)), which is obtained by evaluating the solution along the curve, is constant. As usual, constancy i s proved by checking that its derivative is identically zero. Invoking the chain rule, an d then (22.14), we deduce that dh dt=d dtu(t,x(t)) =∂u ∂t(t,x(t))+dx dt∂u ∂x(t,x(t)) =∂u ∂t(t,x(t))+u(t,x(t))∂u ∂x(t,x(t)) = 0. The final expression vanishes because uis assumed to solve the wave equation (22.13) at all values of ( t,x), including those on the curve ( t,x(t)). This verifies our claim that h(t) is constant, and so the solution uis constant on the characteristic curve. This has the implication that the right hand side of equation (22.14) is a constant whenever x=x(t) defines a characteristic curve, and so the derivative dx/dtis a constant — namely the value ofuon the curve. In this manner, we arrive at the key deduction th at the characteristic curve must be a straight line x=ut+k, (22.15) whosecharacteristic slope uequals the value assumed by the solution uon it. The larger uis, the steeper the characteristic line, and the faster that part of the wave travels. The corresponding characteristic variable ξ=x−tudepends upon the solution, which can now be written in implicit form u=f(x−tu), (22.16) wheref(ξ) is an arbitrary function of the characteristic variable. T he solution u(t,x) can be found by solving the algebraic equation (22.16). For exam ple, if f(ξ) =αξ+β is an affine function, with α,βconstant, then u=α(x−tu)+β,and hence u(t,x) =αx+β 1+αt(22.17) 12/11/12 1176 c/circleco√yrt2012 Peter J. Olver Figure 22.7. Two Solutions to ut+uux= 0. is the corresponding solution to the nonlinear transport eq uation. At each fixed t, the graph of the solution is a straight line. If α >0, the solution flattens out as t→ ∞. On the other hand, if α <0, the straight line rapidly steepens to vertical as tapproaches the critical time t⋆=−1/α, at which point the solution ceases to exist — it is said to “bl ow up”. In Figure 22.7, we graph the solution with α= 1, β=.5, whent= 0,1,5,20 on the top row, and α=−.2, β=.1, at times t= 0,3,4,4.9 on the bottom row. In the second case, the solution becomes vertical as t→5 and then ceases to exist. In general, to construct the solution u(t,x) to the initial value problem u(0,x) =f(x), (22.18) we note that, at t= 0, the implicit solution formula (22.16) reduces to u(0,x) =f(x). Thus, thefunction fcoincideswiththeinitialdata. However, because (22.16)i sanimplicit equation for u(t,x), it is not immediately evident (a) whether it can be solved to give a well-defined value for u(t,x), and, (b) even granted this, how to describe the solution’s qualitat ive features and dynamical behavior. A more instructive and revealing strategy is based on the fol lowing geometrical con- struction, inspired by the linear version appearing in Figu re 22.2. Through each point (0,y) on the xaxis, draw the characteristic line x=tf(y)+y (22.19) whose slope, namely f(y) =u(0,y), equals the value of the initial data at that point. According to the preceding argument, the solution will have the same value on the entire characteristic line (22.19), and so u(t,tf(y)+y) =f(y). (22.20) For example, if f(y) =y, thenu(t,x) =ywhenever x=ty+y; eliminating y, we recover u(t,x) =x/(t+1), which agrees with one of our straight line solutions (22 .17). 12/11/12 1177 c/circleco√yrt2012 Peter J. Olver tx Figure 22.8. Characteristic Lines for f(x) =1 4sin(1.8x−.8). tx Figure 22.9. Characteristic Lines for a Rarefaction Wave. Now, the trouble with our construction is immediately appar ent from the illustrative Figure 22.8. Any two characteristic lines that are not paral lel must cross each other somewhere. The value of the solution must equal to the slope o f the characteristic line, and so, at the crossing point, the solution is required to ass ume two different values, one corresponding to each line. Something is clearly amiss, and we need to study the resulting solutions in more depth. It turns out that there are three basic scenarios. The first, t rivial case is when all the 12/11/12 1178 c/circleco√yrt2012 Peter J. Olver Figure 22.10. Rarefaction Wave. characteristic lines are parallel and so the difficulty does n ot arise. In this case, they all have the same slope, say c, which means that the solution has the same value on each one. Therefore, u(t,x)≡cis a trivial constant solution. The next simplest case occurs when the initial data f(x) is everywhere increasing, so f(x)≤f(y) whenever x≤y, which is assured if its derivative is never negative: f′(x)≥0. In this case, as in sketched in Figure 22.9, the characterist ic lines emanating from the x axis fan out into the right half plane, and so never cross each other when t≥0. Each point (t,x) fort≥0 lies on a unique characteristic line, and the value of the so lution at ( t,x) is equal to the slope of the line. Consequently, the solution is well-defined at all future times. Physically, such solutions represent rarefaction waves , which gradually spread out as time progresses. A typical example, corresponding to initial da ta u(0,x) = tan−13x+π 2, is plotted in Figure 22.10 at successive times t= 0,1,2,3. Note how the slope of the solution gradually diminishes as the rarefaction wave spre ads out. The more interesting case is when f′(x)<0. Now some of the characteristic lines starting at t= 0 will cross at some point in the future. If ( t,x) lies on two or more distinct characteristic lines, the value of the solution u(t,x), which should equal the characteristic slope, is no longer uniquely determined. Although one might be tempted to deal with such multiply-valued solutions in a purely mathematical framew ork, from a physical standpoint this is unacceptable. The solution u(t,x) is supposed to represent a physical quantity, e.g., density, velocity, pressure, etc., and must therefore assu me a unique value at each point. The mathematical model has broken down, and fails to agree wi th the physical reality. Before confronting this difficulty, let us first, from a theore tical standpoint, try to understand what happens if we were to continue the solution a sa multiply-valuedfunction. To be specific, consider the initial data u(0,x) =π 6−1 3tan−1x, (22.21) appearing in the first plot in Figure 22.12. The correspondin g characteristic lines are sketched in Figure 22.11. Initially, they do not cross, and t he solution remains a well- defined, single-valued function. However, eventually one r eaches a critical time, t=t⋆>0, when the first two characteristic lines cross each other. Sub sequently, a wedge-shaped region appears in the ( t,x) plane, consisting of points which lie on the intersection o f three 12/11/12 1179 c/circleco√yrt2012 Peter J. Olver tx Figure 22.11. Characteristics for a Shock Wave. different characteristic lines with different slopes; at suc h points, the solution achieves three distinct values. Outside the wedge, the points only be long to a single characteristic line, and the solution remains single-valued. (The boundar y of the wedge consists of points where only two characteristic lines cross.) To fully appreciate what is going on, look now at the sequence of pictures of the multiply-valued solution at successive times in Figure 22. 12. Since the initial data is positive, f(x)>0, all the characteristic slopes are positive. As a conseque nce, all the points on the solution curve will move to the right, at a speed equal to their height. Since the initial data is a decreasing function, points lying to th e left will move faster than those on the right, and eventually overtake them. Thus, as time pas ses, the solution steepens. At the critical time t⋆when the first two characteristic lines cross, say at x⋆, the tangent to the solution curve has become vertical: ∂u ∂x(t,x⋆)−→ ∞ ast−→t⋆. Afterwards, thesolutiongraphnolongerrepresentsasingl e-valuedfunction; itsoverlapping lobes lie over points ( t,x) in the aforementioned wedge. The critical time t⋆can be determined from the implicit solution formula (22.16 ). Indeed, if we differentiate with respect to x, we find ∂u ∂x=∂ ∂xf(x−tu) =f′(ξ)/parenleftbigg 1−t∂u ∂x/parenrightbigg ,where ξ=x−tu is the characteristic variable, which is constant along the characteristic lines. Solving, ∂u ∂x=f′(ξ) 1+tf′(ξ). 12/11/12 1180 c/circleco√yrt2012 Peter J. Olver Figure 22.12. Multiply–Valued Solution. Therefore, the slope blows up, ∂u ∂x−→ ∞,as t−→ −1 f′(ξ). In other words, if the initial data has negative slope at posi tionx, sof′(x)<0, then the solution along the characteristic line emanating from the p oint (0,x) will break down at the time −1/f′(x). As a consequence, the earliest critical time is t⋆= min/braceleftbigg −1 f′(x)/vextendsingle/vextendsingle/vextendsingle/vextendsinglef′(x)<0/bracerightbigg . (22.22) For instance, for the particular initial configuration (22. 21) represented by the pictures, f′(x) =−1 3(1+x2),and so the critical time is t⋆= min(3(1+ x2)) = 3. Now, while mathematically plausible, such a multiply-valu ed solution is physically untenable. So what happens after the critical time t⋆? One needs to choose which of the possiblesolutionvaluesateachpoint( t,x)containedinthewedgeisphysicallyappropriate. Indeed, the mathematics by itself is incapable of specifyin g how to continue the solution past the critical time at which the characteristics begin to cross. We therefore must return to the underlying physics, and ask what sort of phenomenon ar e we trying to model. The most instructive is to view the differential equation as a sim ple model of compressible fluid flow in a single space variable, e.g., motion of gas in a long pi pe. If we push a piston down the end of a long pipe then the gas will move ahead of the pi ston and thereby be compressed. However, if the piston moves too rapidly, the ga s piles up on top of itself, and a shock wave forms and propagates down the pipe. Mathemat ically, the shock is represented by a discontinuity where the solution abruptly changes value. 12/11/12 1181 c/circleco√yrt2012 Peter J. Olver Conservation Laws and Shocks One way to resolve our mathematical dilemma relies on the fac t that the partial differential equation takes the form of a conservation law, i n accordance with the following definition†. Definition 22.4. Aconservation law is an equation of the form ∂T ∂t+∂X ∂x= 0. (22.23) The functions TandXare known, respectively, as the conserved density and associated flux. In the simplest situations, the conserved density T(t,x,u) and flux X(t,x,u) depend on the time t, the position x, and the solution u(t,x) to the physical system. (Higher order conservation laws, which also depend upon derivative s ofu, will appear in the final section.) We can clearly rewrite the nonlinear transport eq uation (22.13) in the following conservation law form:∂u ∂t+∂ ∂x/parenleftbig1 2u2/parenrightbig = 0, (22.24) where the conserved density and flux are, respectively, T=u, X =1 2u2. Thereason for calling(22.23)aconservationlaw comesfrom thefollowingobservation. Proposition 22.5. Given a conservation law (22.23), d dt/integraldisplayb aT dx=−X/vextendsingle/vextendsingle/vextendsingleb x=a. (22.25) The proof of (22.25)isimmediate— assuming sufficient smooth ness that allowsoneto bring the derivative inside the integral sign, and then invo king the Fundamental Theorem of Calculus: d dt/integraldisplayb aT dx=/integraldisplayb a∂T ∂tdx=−/integraldisplayb a∂X ∂xdx=−X/vextendsingle/vextendsingle/vextendsingleb x=a. Formula (22.25) says that the rate of change of the integrate d density over an interval depends only on the flux through its endpoints. In particular , if there is no net flux into or out of the interval, then the integrated density is conserved , meaning that it remains constant over time. All physical conservation laws — mass, m omentum, energy, and so on — for systems governed by partial differential equation s are of this form. (For ordinary differential equations, conservation laws coinci de with first integrals, as discussed in Section 20.3.) †Here we describe the one-dimensional situation. See Exercise for conservation laws for n-dimensional dynamics. 12/11/12 1182 c/circleco√yrt2012 Peter J. Olver t t +∆ta=s(t)b=s(t+∆t) u+ u−x Figure 22.13. Conservation of Mass Near a Shock. For the transport equation (22.24), the integrated conserv ation law (22.25) takes the specific form d dt/integraldisplayb au(t,x)dx=1 2/bracketleftbig u(t,a)2−u(t,b)2/bracketrightbig . (22.26) Viewing the equation as a model for, say, compressible fluid fl ow in a pipe, the integral on the left hand side represents the total mass of the fluid con tained in the interval [ a,b]. The right hand side represents the mass flux intothe interval through its two endpoints, and thus the conservation equation (22.26) is the mathemati cal formalization of basic mass conservation — mass is neither created nor destroyed, but ca n only enter a region as a flux through its boundary. In particular, if there is zero mass flu x, then we deduce conservation of the total mass. With this in hand, let us return to the physical context of the nonlinear transport equation. We will assume that mass conservation continues t o hold even within a shock, which, from a purely molecular standpoint, makes eminent ph ysical sense. By definition, ashockis a jump discontinuity in the solution u(t,x). Suppose that, at time t, a shock occurs at position x=s(t). We require†that both the left and right hand limits u−(t) =u(t,s(t)−) = lim x→s(t)−u(t,x), u+(t) =u(t,s(t)+) = lim x→s(t)+u(t,x), of the solution on either side of the shock discontinuity are well defined. Let us further assume that, in time, the shock x=s(t) follows a smooth — meaning C1— path. Now, referring to Figure 22.13, Consider a small time interval, f romttot+ ∆t. During this †With more analytical work, [ 188], the listed assumptions can all be rigorously justified. 12/11/12 1183 c/circleco√yrt2012 Peter J. Olver time, the shock moves from position a=s(t) to position b=s(t+ ∆t). The total mass contained in the interval [ a,b] at time t, before the shock has passed through, is m(t) =/integraldisplayb au(t,x)dx≈u+(t)(b−a) =u+(t)/bracketleftbig s(t+∆t)−s(t)/bracketrightbig , whereu+(t) is the average value of u(t,x) over the interval. After the shock has passed, the total mass has become m(t+∆t) =/integraldisplayb au(t+∆t,x)dx≈u−(t)(b−a) =u−(t)/bracketleftbig s(t+∆t)−s(t)/bracketrightbig , whereu−(t) refers to the average value of u(t+∆t,x) over the same interval. In the limit as ∆t→0, the point b=s(t+∆t)−→s(t) =a, and hence the averages lim ∆t→0u+(t) =u+(t), lim ∆t→0u−(t) =u−(t), tend to the limiting solution values on the right and left han d sides of the shock disconti- nuity. Thus, the limiting rate of change in mass across the sh ock at time tis dm dt= lim ∆t→0m(t+∆t)−m(t) ∆t = lim ∆t→0/bracketleftbig u−(t)−u+(t)/bracketrightbigs(t+∆t)−s(t) ∆t=/bracketleftbig u−(t)−u+(t)/bracketrightbigds dt, which is the product of the shock speed times minus the jump ma gnitude at the shock discontinuity. On the other hand, at any t < τ < t +∆t, the mass flux into the interval [a,b] is, according to the right hand side of (22.26), 1 2/bracketleftbig u(τ,a)2−u(τ,b)2/bracketrightbig −→1 2/bracketleftbig u−(t)2−u+(t)2/bracketrightbig as ∆t−→0. For conservation of mass to hold across the shock, the limiti ng value of the rate of change in mass must equal the limiting mass flux, /bracketleftbig u−(t)−u+(t)/bracketrightbigds dt=1 2/bracketleftbig u−(t)2−u+(t)2/bracketrightbig , from which we discover the Rankine–Hugoniot condition ds dt=1 2u−(t)2−u+(t)2 u−(t)−u+(t)=u−(t)+u+(t) 2. (22.27) So, to maintain conservation of mass, the speed of the shock m ust equal the average of the solution values on either side. A shock appears when one or more characteristic lines cross. For this to occur, charac- teristics to the left of the shock must have larger slope (or s peed), while those to the right must have smaller slope. Since the shock speed is the average of the two characteristic slopes, this means u−(t)>ds dt=u−(t)+u+(t) 2> u+(t). (22.28) 12/11/12 1184 c/circleco√yrt2012 Peter J. Olver xu Figure 22.14. Equal Area Rule. While it is theoretically possible to construct a shock solu tion to (22.13)that maintains the Rankine–Hugoniot constraint (22.27) but violates (22.28) , such solutions are excluded on physical grounds, in that they violate causality, [ 105], which requires that characteristics are only allowed to enter shocks, not leave, and, furthermor e, are not stable under small perturbations, [ 188]. The dynamics of shock wave solutions is then prescribed by the Rankine–Hugoniot and causality conditions (22.27,28). Howdoesonedeterminethemotionoftheshock inpractice? Th eanswer isbeautifully simple. Since the total mass, which at time tis the area under the curve u(t,x), must be conserved, one merely draws the vertical shock line where th e areas of the two lobes in the multiply-valued solution are equal, as in Figure 22.14. Thi sEqual Area Rule ensures that the total mass of the shock solution matches that of the origi nal (why?), as required by the physical conservation law. Example 22.6. An illuminating special case is when the initial data has the form of a step function with a single jump discontinuity at the ori gin: u(0,x) =f(x) =a+bσ(x) =/braceleftbigga, x < 0, b, x > 0.(22.29) If†a > b >0, then the initial data is already in the form of a shock wave. Fort >0, the mathematical solution constructed by continuing along the characteristic lines is multiply- valued in the region bt < x < at , where it assumes both values aandb; see Figure 22.15 . The Equal Area Rule tells us to draw the shock line halfway al ong, atx=1 2(a+b)t, in order that the two triangles have the same area. Therefore, t he shock moves with speed c=1 2(a+b) equal to the average of the two speeds at the jump, and so this particular shock wave solution is u(t,x) =a+bσ(x−ct) =/braceleftbigga, x < ct, b, x > ct,where a > c=a+b 2> b.(22.30) †Cases where aorbare negative are left to the exercises. 12/11/12 1185 c/circleco√yrt2012 Peter J. Olver xu a b Figure 22.15. Multiply–Valued Step Wave. tx Figure 22.16. Characteristic Lines for the Step Wave Shock. A graph of the characteristic lines appears in Figure 22.16. By way of contrast, suppose 0 < a < b, so the initial data has a jump upwards. In this case, the characteristic lines diverge from the initial dis continuity, and the mathematical solution is not specified at all in the wedge-shaped region at < x < bt . Now our task is to decide how to connect the two regions where the solution is well-defined. The simplest connection is an affine function, i.e., a straight line. Indee d, a simple modification of the rational solution (22.17) produces the function u(t,x) =x t, whichnotonlysolvesthedifferentialequation, but alsohas therequiredvalues u(t,at) =a, 12/11/12 1186 c/circleco√yrt2012 Peter J. Olver Figure 22.17. Piecewise Affine Rarefaction Wave. andu(t,bt) =bat the two edges of the wedge. The resulting solution is the pi ecewise affine rarefaction wave u(t,x) =  a, x ≤at, x/t, at ≤x≤bt, b, x ≥bt,(22.31) which is graphed in Figure 22.17. In fact, it can be shown, [ 105], that this is the only solution that preserves the causality condition (22.28). These prototypical solutions epitomize the basic phenomen a modeled by the nonlinear transport equation: rarefaction waves , that emanate from regions where the initial data satisfies f′(x)>0, where the solution spreads out as time progresses, and compression waves, emanting from regions where f′(x)<0, that progressively steepen and eventually break into a shock discontinuity. Anyone caught in a traffic ja m recognizes the com- pression waves, where the vehicles are bunched together and almost stationary, while the interspersed rarefaction waves correspond to freely movin g traffic. (An intelligent driver will take advantage of the rarefaction waves moving through the jam to switch lanes!) The familiar, frustrating traffic jam phenomenon, even on acc ident- or construction-free stretches of highway, is an intrinsic effect of the nonlinear transport model that governs the traffic flow, [ 188]. Our derivation of the Rankine–Hugoniot condition (22.27)p rescribing the shock speed relies on the fact that we can write the original partial diffe rential equation in the form of a conservation law. But there are other ways to do this; for instance, multiplying the nonlinear transport equation (22.13) by uallows us write it in the alternative conservative form u∂u ∂t+u2∂u ∂x=∂ ∂t/parenleftbig1 2u2/parenrightbig +∂ ∂x/parenleftbig1 3u3/parenrightbig = 0. (22.32) Here, the conserved density is T=1 2u2, and the associated flux X=1 3u3. The integral form equation (22.25) of the conservation law is d dt/integraldisplayb a1 2u(t,x)2dx=1 3/bracketleftbig u(t,a)3−u(t,b)3/bracketrightbig . (22.33) In some physical models, the integral on the left hand side re presents the energy within the interval [ a,b], and the conservation law tells us that energy can only ente r the interval as a flux through its ends. If we assume that energy is conserved a t a shock, then, repeating 12/11/12 1187 c/circleco√yrt2012 Peter J. Olver our previous argument, we are led to the alternative conditi on ds dt=1 3(u−(t)3−u+(t)3) 1 2(u−(t)2−u+(t)2)=2 3u−(t)2+u−(t)u+(t)+u+(t)2 u−(t)+u+(t)(22.34) for the shock speed. Thus, a shock that conserves energy move s at a different speed than one that conserves mass! The evolution of a shock depends not just on the underlying differential equation, but also on the physical assumptions governing the selection of a suitable entropy condition. The mathematical property that characterizes the shock dyn amics is known as an entropy condition . Entropy conditions, such as the Rankine–Hugoniot Equal Ar ea Rule (22.27), or the alternative (22.34), allow us to follow the s olution beyond the formation of a simple shock. Once a shock forms, it cannot suddenly disa ppear — the discontinuity remains as the solution propagates. One consequence is the i rreversibility of the solutions to the nonlinear transport equation. One cannot simply run t ime backwards and expect shocks to spontaneously vanish. However, this irreversibi lity is of a different character than that of the ill-posedness in the backwards heat equatio n. The nonlinear transport equation can be solved for t <0, but this would result, typically, in the formation of a different collection of shocks, and would not be just the time reversal of the solution. Continuing past the initial shock formation, as other chara cteristic lines start to cross, additional shocks appear. The shocks themselves continue p ropagate, often at different velocities. When a fast moving shock catches up with a slow mo ving shock, one must then decide how to merge the shocks together to retain a physicall y consistent solution. The selected entropy condition continues to resolve the ambigu ities. However, at this point, the mathematical details have become too complicated for us to pursue in any more detail, and we refer the interested reader to Whitham’s book, [ 188], which includes a wide range of applications to equations of gas dynamics, flood waves in r ivers, motion of glaciers, chromotography, traffic flow, and many other physical systems . 22.2. Nonlinear Diffusion. First order partial differential equations, beginning with elementary scalar transport equations, and progressing on to the equations of gas dynami cs, the full-blown Euler equa- tions of fluid mechanics, and yet more complicated systems fo r plasmas and other compli- cated physical processes, are used to model conservative wa ve motion. Such systems fail to account for frictional and/or viscouseffects, whichare typ icallymodeled by a parabolicdif- fusion equation such as the heat equation. In this section we investigate the consequences of combining nonlinear wave motion with linear diffusion by a nalyzing the simplest such model. As we will see, the viscous term helps smooth out abrup t shock discontinuities, and the result is a well-determined and smooth dynamical proces s. Moreover, in the inviscid limit, as the diffusion term becomes vanishingly small, the s mooth viscous solutions con- verge non-uniformly to the appropriate discontinuous shoc k wave, leading to an alternative mechanism for analyzing conservative nonlinnear dynamica l processes. 12/11/12 1188 c/circleco√yrt2012 Peter J. Olver Burgers’ Equation The simplest nonlinear diffusion equation is known as†Burgers’ equation ut+uux=γuxx, (22.35) and is obtained by appending a linear diffusion term to the non linear transport equation (22.13). In fluids and gases, one can interpret the right hand side as modeling the effect of viscosity, and so Burgers’ equation represents a very simpl ified version of the equations of viscous fluid mechanics, [ 188]. As with the heat equation, the diffusion coefficient γ >0 must be positive in order that initial value problem be well- posed in forwards time. Since Burgers’ equation is first order in t, we expect that its solutions are uniquely prescribed by their initial values, say, u(0,x) =f(x),−∞< x <∞. (22.36) (For simplicity, we will ignore boundary effects here.) Smal l, slowly varying solutions — more specifically, those for which both |u(t,x)|and|ux(t,x)|are small — tend to act like solutions to the heat equation, smoothing out and decaying t o 0 as time progresses. On the other hand, when the solution is large or rapidly varying , the nonlinear term tends to play the dominant role, and we might expect the solution to behave like the nonlinear waves that we analyzed in Section 22.1, perhaps steepening i nto some sort of shock. But, as we will see, the smoothing effect of the diffusion term, no ma tter how small, ultimately prevents the appearance of a discontinuous shock. Indeed, i t can be proved that, under rather mildassumptions onthe initialdata, the solutionto theinitialvalueproblem (22.35, 36) remains smooth and well-defined for all subsequent times , [188]. The simplest explicit solutions are the traveling waves , for which u(t,x) =v(ξ) =v(x−ct),where ξ=x−ct, indicates a fixed profile, moving to the right with constant sp eedc. By the chain rule, ∂u ∂t=−cv′(ξ),∂u ∂x=v′(ξ),∂2u ∂x2=v′′(ξ). Substituting these expressions into Burgers’ equation (22 .35), we conclude that v(ξ) must satisfy the nonlinear second order ordinary differential eq uation −cv′+vv′=γv′′. This equation can be solved by first integrating both sides wi th respect to ξ, and so γv′=k−cv+1 2v2, †The equation is named after the applied mathematician J.M. Burgers, [ 36], and so the apostrophe goes after the “s”. Burgers’ equation was apparently first studi ed as a physical model by Bateman, [ 14], although its solution already appears as an exercise in a nineteenth ce ntury ordinary differential equations text, [ 72; vol. 6, p. 102]. 12/11/12 1189 c/circleco√yrt2012 Peter J. Olver γ=.25 γ=.1 γ=.025 Figure 22.18. Traveling Wave Solutions to Burgers’ Equation. wherekis a constant of integration. As in Section 20.1, the non-con stant solutions to such an autonomous first order ordinary differential equation ten d to either ±∞or to one of the equilibrium points, i.e., the roots of the right hand sid e, ast→ ±∞. Thus, to obtain a bounded traveling wave solution v(ξ), the quadratic polynomial on the right hand side must have two real roots, which requires k <1 2c2. Assuming this holds, we rewrite the equation in the form 2γdv dξ= (v−a)(v−b),where c=1 2(a+b). (22.37) To obtain the bounded solutions, we concentrate on the case w hena < v < b . Integrating (22.37) by the usual method, we find /integraldisplay2γdv (v−a)(v−b)=2γ b−alog/parenleftbiggb−v v−a/parenrightbigg =ξ−δ, forδa constant of integration, and hence v(ξ) =ae(b−a)(ξ−δ)/(2γ)+b e(b−a)(ξ−δ)/(2γ)+1. Thus, the bounded traveling wave solutions all have the expl icit form u(t,x) =ae(b−a)(x−ct−δ)/(2γ)+b e(b−a)(x−ct−δ)/(2γ)+1. Observe that lim x→−∞u(t,x) =b, lim x→∞u(t,x) =a, and hence our solution is a monotonically decreasing functi on going from btoa. The wave travels to the right, unchanged in form, with speed equal to t he average of its asymptotic values. In Figure 22.18 we graph sample profiles correspondi ng toa=.1,b= 1 for three different values of the diffusion coefficient. Note that the sma llerγis, the sharper the transition layer between the two asymptotic values of the so lution. In the inviscid limit γ→0, the solutions converge to the step shock wave wave solutio n (22.30)to the nonlinear transport equation, which, asa result, is often referred to astheinviscid Burgers’ equation . Indeed, the profound fact is that, in the inviscid limit as the diffusion becomes van- ishingly small, γ→0, the solutions to Burgers’ equation (22.35) converge to th e shock 12/11/12 1190 c/circleco√yrt2012 Peter J. Olver wave solution to (22.13) constructed by the Equal Area Rule. This observation is in accor- dance with our physical intuition, that all physical system s retain a very small dissipative component, that serves to smooth out discontinuities that m ight appear in a theoretical model that fails to take the dissipation/viscosity/dampin g/etc. into account. In modern theory, this so-called viscosity solution method has been successfully used to characterize the discontinuous solutions to a broad range inviscid nonli near wave equations is as the limit, as the viscosity goes to zero, of classical solutions to a diffusive version. Thus, the viscosity solutions to the nonlinear transport equation re sulting from Burgers’ equation are consistent with the Equal Area Rule for drawing the shock dis continuities. More generally, this method allows one to monitor the solutions as they evolv e into regimes where multiple shocks merge and interact. We refer the interested reader to [119,188]. The Hopf–Cole Transformation By a remarkable stroke of luck, the nonlinear Burgers’ equat ion can be converted into the linear heat equation and thereby explicitly solved . Thelinearization of Burgers’ equation first appeared in an obscure exercise in a nineteent h century differential equations textbook, [ 72; vol.6, p. 102]. Its modern rediscovery by Eberhard Hopf, [ 104], and Julian Cole, [43], was a milestone in the modern era of nonlinear partial diffe rential equations, and is named the Hopf–Cole transformation in their honor. Finding a way to covert a nonlinear differential equation int o a linear equation is extremely challenging, and, in, most instances, impossibl e. On the other hand, the re- verse process — “nonlinearizing” a linear equation — is triv ial: any nonlinear changes of variables will do the trick! However, the resulting nonli near equation, while evidently linearizable through the inverse change of variables, is ra rely of any independent inter- est. Sometimes there is a lucky accident, and such “accident al” linearizations can have a profound impact on our understanding of more complicated no nlinear systems. In the present context, our starting point is the linear heat equation vt=γvxx. (22.38) Among all possible nonlinear changes of dependent variable , one of the simplest that might spring to mind is an exponential function. Let us, therefore , investigate the effect of an exponential change of variables v(t,x) =eαϕ(t,x),so ϕ(t,x) =1 αlogv(t,x), (22.39) whereαis a nonzero constant. The function ϕ(t,x) is real provided v(t,x)>0 is apositive solution to the heat equation. Fortunately, this is not hard to arrange: if the initial data v(0,x)>0 is strictly positive, then the resulting solution v(t,x) is positive for all t >0. This follows from the Maximum Principle for the heat equatio n, cf. Theorem 14.3. To determine the differential equation satisfied by the funct ionϕ, we invoke the chain rule to differentiate (22.39): vt=αϕteαϕ, vx=αϕxeαϕ, vxx=/parenleftbig αϕxx+α2ϕ2 x/parenrightbig eαϕ. 12/11/12 1191 c/circleco√yrt2012 Peter J. Olver Substituting the first and last formulae into the heat equati on (22.38)and canceling a com- mon exponential factor, we conclude that ϕ(t,x) satisfies the nonlinear partial differential equation ϕt=γϕxx+γαϕ2 x, (22.40) known as the potential Burgers’ equation , for reasons that will soon become apparent. The second step in the process is to differentiate the potenti al Burgers’ equation with respect to x; the result is ϕtx=γϕxxx+2γαϕxϕxx. (22.41) If we now set∂ϕ ∂x=u, (22.42) so thatϕhas the status of a potential function , then the resulting partial differential equation ut=γuxx+2γαuux coincides with Burgers’ equation (22.35) with α=−1/(2γ). In this manner, we have arrived at the famous Hopf–Cole transformation . Theorem 22.7. Ifv(t,x)>0is any positive solution to the linear heat equation vt=γvxx, then u(t,x) =∂ ∂x/parenleftbig −2γlogv(t,x)/parenrightbig =−2γvx v(22.43) solves Burgers’ equation ut+uux=γuxx. Do all solutions to Burgers’ equation arise in this way? In or der to decide, we run the argument in reverse. First, choose a potential function /tildewideϕ(t,x) that satisfies (22.42); for example /tildewideϕ(t,x) =/integraldisplayx 0u(t,y)dy. Ifu(t,x) is any solution to Burgers’ equation, then /tildewideϕ(t,x) satisfies (22.41). Integrating both sides of the latter equation with respect to x, we conclude that /tildewideϕt=γ/tildewideϕxx+γα/tildewideϕ2 x+h(t), for some integration “constant” h(t). Thus, unless h(t)≡0, our potential function /tildewideϕ doesn’t satisfy the potential Burgers’ equation (22.40), b ut that’s because we chose the “wrong” potential. Indeed, if we define ϕ(t,x) =/tildewideϕ(t,x)−η(t),where η′(t) =h(t), then ϕt=/tildewideϕt−h(t) =γ/tildewideϕxx+γα/tildewideϕ2 x=γϕxx+γαϕ2 x, and hence the modified potential ϕ(t,x)isa solution to the potential Burgers’ equation (22.40). From this it easily follows that v(t,x) =e−ϕ(t,x)/(2γ)(22.44) 12/11/12 1192 c/circleco√yrt2012 Peter J. Olver -7.5 -5-2.5 2.5 57.5 -2-1.5-1-0.50.511.52 -7.5 -5-2.5 2.5 57.5 -2-1.5-1-0.50.511.52 -7.5 -5-2.5 2.5 57.5 -2-1.5-1-0.50.511.52 -7.5 -5-2.5 2.5 57.5 -2-1.5-1-0.50.511.52 Figure 22.19. A Solution to Burgers’ Equation. is a positive solution to the heat equation, from which u(t,x) can be recovered via the Hopf –Cole transformation (22.43). Thus, we have proved tha t every solution to Burgers’ equation comes from a positive solution to the heat equation via the Hopf–Cole transfor- mation. Example 22.8. As a simple example, the separable solution v(t,x) =a+be−γω2tcosωx to the heat equation leads to the solution u(t,x) =2γbωe−γω2tsinωx a+be−γω2tcosωx to Burgers’ equation; a typical example is plotted in Figure 22.19. We should require that a >|b|in order that v(t,x)>0 be a positive solution to the heat equation for t≥0; otherwise the resulting solution to Burgers’ equation will have singularities at the roots of u— see the first graph in Figure 22.19. This particular solutio n primarily feels the effects of the diffusivity, and rapidly goes to zero. To solve the initial value problem (22.35–36) for Burgers’ e quation, we note that, under the Hopf–Cole transformation, v(0,x) =h(x) = exp/parenleftbigg −ϕ(0,x) 2γ/parenrightbigg = exp/parenleftbigg −1 2γ/integraldisplayx 0f(y)dy/parenrightbigg , (22.45) Remark: The lower limit of the integral can be changed from 0 to any ot her convenient value without affecting the final form of u(t,x) in (22.43). The only effect is to multiply v(t,x) by an overall constant, which does not change u(t,x). 12/11/12 1193 c/circleco√yrt2012 Peter J. Olver -2 -1 1 20.20.40.60.81 -2 -1 1 20.20.40.60.81 -2 -1 1 20.20.40.60.81 Figure 22.20. Shock Wave Solution to Burgers’ Equation. According to formula (14.61) (adapted to general diffusivit y, as in Exercise ), the solution to the initial value problem (22.38,45) for the hea t equation can be expressed as a convolution integral with the fundamental solution: v(t,x) =1 2√πγt/integraldisplay∞ −∞e−(x−y)2/(4γt)h(y)dy. Therefore, the solution to the Burgers’ initial value probl em (22.35,36) is u(t,x) =/integraldisplay∞ −∞x−y teF(t,x,y)dy /integraldisplay∞ −∞eF(t,x,y)dywhere F(t,x,y) =−1 2γ/integraldisplayy 0f(z)dz−(x−y)2 4γt. (22.46) Example 22.9. To demonstrate the smoothing effect of the diffusion terms, le t us see what happens to the initial data u(0,x) =/braceleftbigga, x <0, b, x >0,(22.47) in the form of a step function. We assume that a > b, which would correspond to a shock wave in the inviscid limit γ= 0. (In Exercise , the reader is asked to analyze the case a < bwhich corresponds to a rarefaction wave.) In this case, F(t,x,y) =−(x−y)2 4γt−  −ay 2γ, y < 0, −by 2γ, y > 0. After some algebraic manipulations, the solution is found t o have the explicit form u(t,x) =a+b−a 1+h(t,x) expb−a 2γ(x−ct)(22.48) where c=a+b 2, h (t,x) =1−erf/parenleftbiggx−bt√4γt/parenrightbigg 1−erf/parenleftbiggx−at√4γt/parenrightbigg, (22.49) 12/11/12 1194 c/circleco√yrt2012 Peter J. Olver -2 2 4 6 810 120.20.40.60.811.2 -2 2 4 6 810 120.20.40.60.811.2 -2 2 4 6 810 120.20.40.60.811.2 -2 2 4 6 810 120.20.40.60.811.2 Figure 22.21. Triangular Wave Solution to Burgers’ Equation. and erfzdenotes the error function (14.63). The solution, with a= 1,b=.1 andγ=.03 is plotted in Figure 22.20 at times t=.01,1.0,2.0. Note that the sharp transition region for the shock is immediately smoothed, and the solution rapi dly settles into the form of a continuously varyingtransitionlayer between the two step heights. Thelarger thediffusion coefficient in relation to the initial solution heights a,b, the more significant the smoothing effect. Observe that, as γ→0, the function h(t,x)→1, and hence the solution converges to the shock wave solution (22.30) to the transport equation , in which the speed of the shock is the average of the two initial values. Example 22.10. Consider the case when the initial data u(0,x) =δ(x) is a concen- trated delta function impulse at the origin. In the solution formula (22.46), starting the integral for F(t,x,y) at 0 is problematic, but as noted earlier, we are free to sele ct any other starting point, e.g., −∞. Thus, we take F(t,x,y) =−1 2γ/integraldisplayy −∞δ(z)dz−(x−y)2 4γt=  −(x−y)2 4γt, y < 0, −1 2γ−(x−y)2 4γt, y > 0. Substituting this into (22.46), we can evaluate the upper in tegral in elementary terms, while the lower integral involves the error function (14.63 ); after a little algebra, we find u(t,x) =/radicalbigg 4γ πte−x2/(4γt) coth/parenleftbigg1 4γ/parenrightbigg −erf/parenleftbiggx√4γt/parenrightbigg, (22.50) where cothz=coshz sinhz=ez+e−z ez−e−z=e2z+1 e2z−1 12/11/12 1195 c/circleco√yrt2012 Peter J. Olver is the hyperbolic cotangent function. A graph of this soluti on when γ=.02 anda= 1, at times t= 1,5,10,50, appears in Figure 22.21. As you can see, the initial conce ntration diffuses out, but, unlike the heat equation, the wave does not remain symmetric owing to the advection terms in the equation. The effect is to steepen i n front as it propagates. Eventually the triangular wave spreads out as the diffusion p rogresses. 22.3. Dispersion and Solitons. Finally, we study a remarkable third order evolution equati on that originally arose in the modeling of surface water waves, that serves to introd uce yet further phenomena, both linear and nonlinear. The third order derivative model s dispersion, in which waves of different frequencies move at different speeds. Coupled wi th the same nonlinearity as in the inviscid and viscous Burgers’ (22.13,35), the result is one of the most remarkable equations in all of mathematics, with far-reaching implica tions, not only in fluid mechanics and applications, but even in complex function theory, phys ics, etc., etc. Linear Dispersion So far, in our study of partial differential equations, we hav e not ventured beyond second order. Higher order equations do occur in applicatio ns, particularly in models for wave motion. The simplest linear partial differential equat ion of a type that we have not yet considered is the third order equation ut+uxxx= 0 (22 .51) It is the third in a hierarchy of simple evolution equations t hat starts with the simple ordinary differential equation ut=u, then proceeds to the transport equation ut=ux, and then the heat equation ut=uxxmodeling basic diffusion processes. The third order case (22.51) is a simple model for linear dispersive waves. To avoid additional complications caused by boundary condi tions, we shall only look at the equation on the entire line, so x∈RThe solution to the equation is uniquely specified by initial data u(0,x) =f(x), −∞< x <∞. (22.52) See [1] for a proof. Let us apply the Fourier transform to solve the initial value problem. Let /hatwideu(t,k) =1√ 2π/integraldisplay∞ −∞u(t,x)e−ikxdx be the spatial Fourier transform of the solution, which is as sumed to remain in L2at allt, a fact that can be justified a posteriori. In view of the effect o f the Fourier transform on derivatives — see Corollary 13.22 — the Fourier transform co nverts the partial differential equation (22.51) into a first order, linear ordinary differen tial equation /hatwideut−ik3/hatwideu= 0, (22.53) 12/11/12 1196 c/circleco√yrt2012 Peter J. Olver parametrized by k, with initial conditions /hatwideu(0,k) =/hatwidef(k) =1√ 2π/integraldisplay∞ −∞f(x)e−ikxdx (22.54) given by the Fourier transform of (22.52). Solving the initi al value problem (22.53–54) by the usual technique, we find /hatwideu(t,k) =eik3t/hatwidef(k). Inverting the Fourier transform yields the explicit formul a for the solution u(t,x) =1√ 2π/integraldisplay∞ −∞eik3t+ikx/hatwidef(k)dk (22.55) to the initial value problem for the dispersive wave equatio n (22.51–52). Actually, to find the solutions to the differential equation, one does not need the full power of the Fourier transform. Note that (22.55) represent s a linear superposition of elementary exponential functions. Let us substitute an exp onential ansatz u(t,x) =eiωt+ikx(22.56) representing acomplexoscillatorywaveof frequency ω, whichindicatesthetimevibrations, andwave number k, which indicates the corresponding oscillations in space. Since ∂u ∂t= iωeiωt+ikx,∂3u ∂x3=−ik3eiωt+ikx, (22.56) satisfies the partial differential equation (22.51) if and only if its frequency and wave number are related by ω=k3. (22.57) The result is known as the dispersion relation for the partial differential equation. In general, any linear constant coefficient dynamical partial d ifferential equation admits a dispersion relation of the form ω=ω(k) which is straightforwardly found by substituting the exponential ansatz (22.56) and canceling the common exp onential factors in the re- sulting equation. In our particular case, the exponential s olution of wave number khas the form uk(t,x) =eik3t+ikx. Linear superposition permits us to combine them in integral form, and so, for any (rea- sonable) function a(k) depending on the wave number, u(t,x) =1√ 2π/integraldisplay∞ −∞eik3t+ikxa(k)dk is easily seen to be a solution to the partial differential equ ation. The Fourier transform solution (22.55) has this form. 12/11/12 1197 c/circleco√yrt2012 Peter J. Olver Example 22.11. Thefundamental solution corresponds to a concentrated initial disturbance u(0,x) =δ(x). since the Fourier transform of the delta function is just /hatwideδ(k) = 1/√ 2π, the resulting solution (22.55) is u(t,x) =1 2π/integraldisplay∞ −∞eik3t+ikxdk=1 π/integraldisplay∞ 0cos(k3t+kx)dk, since the solution is real (or, equivalently, the imaginary part of the integrand is odd) while the real part of the integrand is even. The second integ ral can be converted into that defining the Airy function, Ai(z) =1 π/integraldisplay∞ 0cos/parenleftbig sz+1 3s3/parenrightbig ds, as in (C.40), by the change of variables s=k3√ 3t, z =x 3√ 3t, and we conclude that the fundamental solution to the dispers ive wave equation (22.51) can be written in terms of the Airy function: u(t,x) =1 3√ 3tAi/parenleftbiggx 3√ 3t/parenrightbigg . SeeFigureee3 foragraph. Furthermore, writingthegeneralinitialdataa sasuperposition of delta functions f(x) =/integraldisplay∞ −∞f(ξ)dξδ(x−ξ), we conclude that the solution has the form u(t,x) =1 3√ 3t/integraldisplay∞ −∞f(ξ) Ai/parenleftbiggx−ξ 3√ 3t/parenrightbigg dξ. (22.58) Although energy is conserved, unlike the heat and diffusion e quations, the dispersion of waves means that the solution dies out. Group velocity and wave velocity. The Korteweg–deVries Equation The simplest wave equation that combines dispersion with no nlinearity is the cele- bratedKorteweg–deVries equation ut+uxxx+uux= 0. (22.59) The equation was first derived by the French applied mathemat ician Boussinesq, [ 24; eq.(30)], [ 25; eqs.(283,291)], in 1872 as a model for surface water waves. It was redis- covered by the Dutch mathematician Korteweg and his student de Vries, [ 120], over two 12/11/12 1198 c/circleco√yrt2012 Peter J. Olver decades later, and, despite Boussinesq’s priority, is name d after them. In the early 1960’s, the American mathematical physicists Martin Kruskal and No rman Zabusky, [ 195], red- erived it as a continuum limit of a model of nonlinear mass-sp ring chains studied by Fermi, Pasta and Ulam, [ 67]. Understanding the puzzling behavior of both systems comi ng from numerical experiments was the catalyst of one of the most rem arkable and far-ranging discoveries of modern mathematics: integrable nonlinear p artial differential equations. The most important special solutions to the Korteweg–deVri es equation are the trav- eling waves . We assume that the solution u=v(ξ) =v(x−ct),where ξ=x−ct, is a wave of permanent form, translating to the right with spe edc. By the chain rule, ∂u ∂t=−cv′(ξ),∂u ∂x=v′(ξ),∂3u ∂x3=v′′′(ξ). Substituting these expressions into the Korteweg–deVries equation (22.59), we conclude thatv(ξ) must satisfy the nonlinear third order ordinary differenti al equation v′′′+vv′−cv′= 0. (22.60) Let us further assume that the traveling wave is localized, meaning that the solution and its derivatives are small at large distances: lim x→±∞u(t,x) = lim x→±∞∂u ∂x(t,x) = lim x→±∞∂2u ∂x2(t,x) = 0. To this end, we impose the boundary conditions lim ξ→±∞v(ξ) = lim ξ→±∞v′(ξ) = lim ξ→±∞v′′(ξ) = 0. (22.61) (See Exercise for an analysis of the non-localized traveling wave solutio ns.) The ordinary differential equation (22.60) can, in fact be so lved in closed form. First, note that d dξ/bracketleftbig v′′+1 2v2−cv/bracketrightbig = 0,and hence v′′+1 2v2−cv=k, is a first integral, with kindicating the constant of integration. However, the local izing boundary conditions (22.61) imply that k= 0. Multiplying the latter equation by v′allows us to integrate a second time d dξ/bracketleftbig1 2(v′)2+1 6v3−1 2cv2/bracketrightbig =v′/bracketleftbig v′′+1 2v2−cv/bracketrightbig = 0. Thus, 1 2(v′)2+1 6v3−1 2cv2=ℓ, whereℓis a second constant of integration, which, again by the boun dary conditions (22.61), is also ℓ= 0. We conclude that v(ξ) satisfies the first order autonomous ordinary differential equation dv dξ=v/radicalBig c−1 3v . 12/11/12 1199 c/circleco√yrt2012 Peter J. Olver Figure 22.22. Solitary Wave. We integrate by the usual method, cf. (20.7): /integraldisplaydv v/radicalBig c−1 3v=ξ+δ. Using a table of integrals, and then solving for v, we conclude that the solution has the form v(ξ) = 3csech2/bracketleftbig1 2√cξ+δ/bracketrightbig , where sechy=1 coshy=2 ey+e−y, is thehyperbolic secant function . The solution has the form graphed in Figure 22.22; it has a global maximum at 3 csech0 = 3 caty= 0, and is an even function, exponentially decay to 0 as|ξ| → ∞. The resulting localized traveling wave solutions to the Ko rteweg–deVries equation are u(t,x) = 3csech2/bracketleftbig1 2√c(x−ct)+δ/bracketrightbig , (22.62) wherec >0 andδare arbitrary constants. The parameter cequals the speed of the wave. It is also equal to one third its amplitude, since the maximum value ofu(t,x) is 3cat the pointsx=ct, as well as the width, which is on the order of√c. The taller and wider the solitary wave, the faster it moves. The solution (22.62) is known as a solitary wave solution since it represents a localized wave that travels unchanged in shape. Such waves were first ob served by the British engineer J. Scott Russell, [ 165], who recounts how such a wave was generated by the sudden motion of a barge along an Edinburgh canal and then cha sing it on horseback for several miles. Russell’s observations were dismissed by hi s contemporary, the prominent mathematician George Airy, who claimed that such localized disturbances could not exist, basing his analysisupon a linearized theory. Much later, Bo ussinesq established the proper nonlinearsurfacewavemodel(22.59),validforlongwavesi nshallowwater,andalsoderived the solitary wave solution (22.62), thereby fully exonerat ing Scott Russell’s insight. These nonlinear traveling wave solutions were discovered b y Kruskal and Zabusky, [195], to have remarkable properties. For this reason they have b een given a special new name — soliton. Ordinarily, combining two solutions to a nonlinear equati on can be quite 12/11/12 1200 c/circleco√yrt2012 Peter J. Olver Figure 22.23. Interaction of Two Solitons. unpredictable, and one might expect any number of scenarios to occur. If you start with initial conditions representing a taller wave to the left of a shorter wave, the solution of the Korteweg–deVries equation runs as follows. The talle r wave moves faster, and so catches up the shorter wave. They then have a very complicate d nonlinear interaction, as expected. But, remarkably, after a while they emerge from th e interaction unscathed. The smaller wave is now in back and the larger one in front. After t his, they proceed along their way, with the smaller one lagging behind the high speed tall wave. the only effect 12/11/12 1201 c/circleco√yrt2012 Peter J. Olver of their encounter is a phase shift, meaning a change in the va lue of the phase parameter δin each wave. See Figure 22.23. After the interaction, the po sition of the soliton if it had traveled unhindered by the other is shown in a dotted line . Thus, they behave like colliding particles, which is the genesis of the word “solit on”. A similar phenomenon holds for several such soliton solutio ns. After some time where the various waves interact, they finally emerge with the larg est soliton in front, and then in order to the smallest one in back, all progressing at their own speed, and so gradually drawing apart. Remark: In the Korteweg–deVries equation model, one can find arbitr arily tall soliton solutions. In physical water waves, if the wave is too tall it will break. Indeed, it can be rigorously proved that the full water wave equations admit s olitary wave solutions, but there is a wave of greatest height, beyond which a wave will te nd to break. The solitary water waves are not genuine solitons, since there is a small, but measurable, effect when two waves collide. Moreover, it can be proved that, starting with an arbitrary initial disturbance u(0,x) =f(x) that decays sufficiently rapidly as |x| → ∞, after a sufficiently long time, the resulting solution u(t,x) disintegrates into a finite number of solitons of different h eights, moving off at their respective speeds to the right, and so are arrange d in order from smallest to largest, plus a small dispersive tail moving to the left that rapidly disappears. Proving this remarkable result is beyond the scope of this book. It relies on the method of inverse scat- tering, that effectively linearizes the Korteweg–deVries equatio n with a linear eigenvalue problem of fundamental importance in one-dimensional quan tum mechanics. The solitons correspond to the bound states of a quantum potential. We ref er the interested reader to the introductory text [ 61] and the more advanced monograph [ 1] for details. Like Burgers’ equation, the Korteweg–deVries equation can be linearized, but the linearization is considerably more subtle. It relies on the introduction of an auxiliary linear eigenvalue problem. There is a remarkable transformation, known as the inverse s cattering transform, which is a form of nonlinear Fourier transform, that can be us ed to solve the Korteweg– deVriesequation. Itsfascinating properties continue tob e ofgreat current research interest to this day. 22.4. Conclusion and Bon Voyage. These are your first wee steps in a vast new realm. We are unable to discuss nonlinear partial differential equations arising in fluid mechanics, i n elasticity, in relativity, in dif- ferential geometry, in computer vision, in mathematical bi ology. Chaos and integrability are the two great themes in modern nonlinear applied mathema tics, and the student is well-advised to pursue both. We bid you, dear reader, a fond adieu and wish you unparallele d success in your mathematical endeavors. 12/11/12 1202 c/circleco√yrt2012 Peter J. Olver