R. von Mises - Mathematical theory of compressible fluid flow, страница 8

In contrast to the exact equations (1.1) and ( l . I I ) , thepresent equations are linear in the unknowns ρ and q, which makes themmore accessible to solution. On the other hand, the full implications ofthis linearization are not easily determined. See, however, a few commentsin Sec. 13.3.2. One-dimensional case. D'Alembert's solutionSome information on the behavior of a small perturbation may be ob­tained quickly by a consideration of the one-dimensional problem: it isassumed that all particles move in the same fixed direction, say that of x,while all unknowns are independent of y and z\ i.e.,qy= qz= 0and= |±= 0.Then only the x-component of (3) remains, and in (4) div is to be replacedby d/dx:dqx2po —=/ x·K(5)dtdpdx-αοdqPo —x—,dx=-dp—.dtOn multiplying the second equation by a , differentiating with respectto t in the first and χ in the second, and subtracting, we obtain0\f n(6a)dq22xW=a o2dqa*'2xwhile if the multiplication by a is omitted and / and χ are interchangedin the differentiation step, the result is20/«ι \d"p2(a? -6b)dp2α ολ?·I t can also be seen that any twice differentiable function u of the variablesq and ρ satisfies an equation of exactly the same form,x36I.INTRODUCTIONif terms of higher order in the derivatives of q and ρ are neglected.

In fact,xif u = u iq ,p),xthen222du _ du dqdu dpd u _ du d qdu d ρdxdp dx'dxdp dx 'xdq dxxxdq dx22x2where in the second equation terms involving (dq /dx)\etc., have beenxomitted. Similarlyd u _ du d q22dt2". du d p2xdq ~d¥dp dt '2xThese expressions, together with (6a) and ( 6 b ) , give ( 6 ) .Equation (6) is the (one-dimensional) wave equation. T h e general solu­tion of this equation was given by d'Alembert: if /i(z) and/ (z) denote ar­bitrary twice differentiable functions of the real variable z, then (6) issatisfied by21(7)u = fi(x+ ooO + h(x2— OoOiwhere ζ has been replaced by χ + aot in the first function and by χ — a tin the second; for, letting/( denote dfi(z)/dz, etc., we get022d u »//ι-//— = j \ - r h ,d u2//·/ /ι= o Ui"a2pff\Γ J2),from which (6) follows.T h e small perturbation under consideration is assumed to have beencaused by an initial disturbance imposed on a state of rest.

T h e disturbancemay be represented by giving the values of q and ρ — po at t = 0, so thatq (x, 0) and p(x, 0) are given functions of x. Knowing q at some time tmeans knowing also dp/dt at the same time, according to the second of Eqs.(5). Thus, if u stands for ρ — p , the initial conditions for u can be writtenin this form:xxx0(8)at t = 0,=u = f(x),g(x),dtwhere / and g are given (sufficiently differentiable) functions of x. liu standsfor q , we derive from the given p(x, 0) the value of dq /dt by means of thefirst Eq.

( 5 ) , so that again the initial conditions for u have the form (8).The problem before us is to find an integral of Eq. (6) satisfying the con­ditions ( 8 ) . W e are going to show that the solution readsxx0 = \ lf(x(9)+ aot) + fix-aot)]1+7Γ- [Gix + aot) ZaoG(x -aot)],4.2where G(x)D'ALEMBERT'S37SOLUTIONis a function whose derivative is g(x):(9')G(x)=fρ®dt ;the second part of the solution (9) can also be expressed by-fdt.g(00x—a tJ0Indeed, the solution (9) has the form (7) with /i = i f + G/2a andfi = \f — G/2a , so that (9) satisfies the differential equation ( 6 ) .

Further,it is seen immediately that u(x, 0) = f(x), while00ττ)[ ο / ' ( * + «οΟ -=dot)]a /'(aα0+J— [a>oG'(x + dot) + d G'(x2d002doG'(x)2d=— a t)]\0Jg(x),0as required by ( 8 ) .Equation (9) shows that the value of u for a given position χ at a giventime t depends only on the values of the initial disturbance u and du/dtin the interval (x + dot, χ — dot) and is independent of the values of theinitial disturbance at all other points. On the other hand, the initial dis­turbance on an interval (χι , x ) determines the perturbation u(x, t) onlyfor values (x, t) that lie in the triangle in the x, £-plane having base (x\ , x )and sides with slopes l/oo and — I/do, respectively.22T o discuss the solution ( 9 ) , we assume first that g is identically zero andt h a t / ( x ) vanishes outside some interval AB:( — c < χ < c).

In Fig. 7 thevalues of u = \\f{x + dot) + f(x — dot)] corresponding to certain valuesof t (t = 0, t = h , t = t ) are illustrated. For a fixed position a; to the rightof Α Β (χ > c) the perturbation u will be zero except for t such that χ — dot2• x=-c+ a tQ\x=-c-a t-^0\/χV*7777\γτ777)/\/ΑΟF i g . 7. Propagation with velocity a0Βof one-dimensional small disturbance.38I.INTRODUCTIONfalls between — c and c; in other words, u(x, t) ^ 0 only for (x — c)/a <t < (x + c)/oo. Within this time interval w takes successively one-halfof the values that /(x) has in the interval χ = — c to χ = c.

For a fixed t,the perturbation is zero in the three intervals0(10)χ ^ — c — dot;c — aot ^ χ ^ —c + aot;χ ^ c + Oo£.This whole situation can be described in the following terms: the initialdisturbance/(x) of width 2c splits into two equal parts, each again of width2c but each half as high as the original / ( x ) , one moving to the right, theother to the left, and each with velocity a . Thus, a may be called thevelocity of propagation of a small disturbance. This velocity depends only onthe nature of the fluid and the original state of rest; it is independent of theexact form of the disturbance.

Typical disturbances of air at rest are noiseor acoustic signals, so the shorter term sound velocity is used mostly. Ifthe given (p, p)-relation is E q . (1.5b): p/p = constant, characterizing polytropic flow, then a = dp/dp = κρ/ρ. For air, considered as a perfect gasand under standard conditions, i.e., p = 2,116 lb/ft , p = 0.002378 slug/ft ,and κ = 7 = 1.4, the value of ao is 1,116 ft/sec or 760.9 mph.00K2203022T h e results of the above discussion are not essentially changed if therestriction g(x) = 0 is omitted. Suppose that q and ρ — p at t = 0 bothvanish outside a common interval AB.

If u is identified with ρ — p , /(x)again vanishes outside AB, and g(x) = (dp/dt) =o, which by (5) is propor­tional to (dq /dx) =o,must also vanish outside AB. N e x t , q must vanishat A and B, since it vanishes outside this interval and is continuous byhypothesis. Thusx00txtxBut then also Jf 9f(x) dx = 0; and from G(x) = J* g(£) dl- it follows thatG(A)= G(B). For a suitable choice of the constant of integration, thiscommon value is 0. Then G ( x ) has the further property G(x) = 0 for all χoutside AB, which was the essential property of /(x) used in the previousdiscussion. Again it follows that the perturbation is zero whenever E q . (10)is satisfied, so that the disturbance is propagated to the right and left withvelocity oo exactly as before.

This time, however, the left and right halfwaves are not equal to each other.3. The wave equation in three dimensionsBefore returning to the general case covered by Eqs. (3) and ( 4 ) , twoformal relations should be recalled. First, the divergence of a gradient4.3WAVE EQUATIONIN THREE DIMENSIONS39equals the Laplacian operator, or, as applied to any scalar /,™>~l(IK(fK(I)(n)dxdy2dz22JSecondly, as shown in Sec.

2.6, the flux of a vector through a closed surfaceS is the integral of the divergence of that vector over the volume enclosedby S, or, as applied to some vector v,(12)[ divvdV=[v dS.nW i t h the use of (11), the equations (3) and (4) governing a small per­turbation of an inviscid elastic fluid originally at rest can be suitably trans­formed. Taking the divergence of both sides of ( 3 ) , applying (11), anddifferentiating (4) with respect towe obtainΙ·dqpo div — =dt2-a0AΔρ,dpo — div q = dtdp—.dr2Since d/dt and div are interchangeable, these give(13)| £ = α Δρ.02On the other hand, if we differentiate (3) with respect to t and take thegradient of ( 4 ) , there resultsΡ°τ7ϊ=dt1~ ° έ S d p,dtα2po grad (div q) =m-grad^.dtHere the right-hand members are equal, except for the factor a ; therefore0|^ = a022grad (div q ) .Taking the divergence of both sides of this equation, and applying (11)once more (with / = div q ) , we obtaind2(14)T h e equation— (divq) = a A(divq).0240I.INTRODUCTIONis called the (three-dimensional) wave equation.