R. von Mises - Mathematical theory of compressible fluid flow, страница 9

From (13) and (14) itfollows that this equation holds for u = ρ and for u = div q. A s in thepreceding section, it can be shown that any twice differentiable function ofρ and div q also satisfies Eq. (15), if terms of higher order in the perturbationvariables and their derivatives are neglected. In particular, the total energyper unit volume, p(q /2 + c T), is, at this level of approximation (neglectingq ), a function of ρ only and thus also satisfies (15). (T depends only onρ because, by the equation of state, Τ is a function of ρ and p, and hereρ is a function of p.) I t can further be seen, by differentiation of (15), thatif an arbitrary function u satisfies the wave equation, then so also do itsderivatives du/dx, du/dt, e.g., any component of grad p, etc.

T h e equationpreceding (14) can also be written2v2^ 5 = a (Aq + curl curl q).02( W e denote by curl q a vector with components dq /dy — dq /dz, dq /dz —dq /dx, and dq /dx — dq /dy, which will be considered in detail in A r t . 6.)Moreover from Eq. (3)zzyyxx- (curl q) = 0.otHence in any region where curl q = 0 initially, it is always zero, andconsequently q, as well as any component of q, satisfies the wave equation.4.

Poisson's solution23Functions u(x, y, z, t) satisfying the w a v e equation may be found asfollows. Let f(x, y, z) be an arbitrary and sufficiently differentiable func­tion of the space coordinates, Ρ a point with coordinates x, y, z, and Sthe spherical surface with center at Ρ and radius R = a t. For any position(x, y, z) and time t, we define f(x, y, z, t) by0/ ( * , y, z, t) = J ^ -(16)2J/(ξ, υ, f ) dS = i - | / da,where the first integral is to be taken over the total surface S. Since the areaof S is 47τβ , / represents the mean or average value of / on S, the variable tentering only in the radius R = ad of the sphere.

T h e third term in (16)is obtained from the second by writing da = dS/R , where da is the solidangle under which d§> is viewed from P. Then the function of four variables22(17)u(x, y, z, t) = t](x, y, z, t)satisfies Eq. (15).T o prove this, we note that the operations of averaging over the sphere4.4POISSON'S41SOLUTIONand of differentiating with respect to x, y, or ζ may be interchanged, sothat(18)Au =t A f ^2J A f d ^ ^ j A fd, .As to du/dt, it is seen that a change dt in t causes a change dr = ao dt inthe radius of S, and therefore a change (df/dr) dr in the value of / corre­sponding to a da on the sphere, where df/dr is the directional derivative of/ along the radius of the sphere, or radial derivative.

ThusdrTherefored*dt\dt2)Jdtdt2t dt\dt)'which may be written, using R = aot and (19), asu _ι a(r2< a< \ α2(20)02a/\ιθ<// , f a/a4ττα ί θ< \i47τα ^ d£ V^00\arr//Using Gauss' Theorem (12) with ν = grad/, and observing that the normaland radial directions are the same when S is a sphere, we obtainf ^d§> = J div (grad /) dV = jΔ f dV,where the volume integral is extended over the interior of S.

T h e time rateof change of this volume integral depends only on the change of the domainof integration, since the integrand is independent of t. Increasing t by anamount dt adds a spherical shell of thickness a dt to the volume; for this0shell dV=a dt dS, so that0/(21),rχa dt0J Δ / d§>dth{ifr V=dThen, substituting this in (20) and using (18), we getd\_1Wiwt[Δ/ASand the proposition is proved.=ao2j Afd§> =α Δκ,0242I.INTRODUCTIONAs in the one-dimensional case, the main problem is to adapt the solution(17) to given initial conditions. Suppose that(22)at t = 0,du— = g(z, y, z ) ,u = /(x, y,z),dtwhere / and g are differentiable.

Consider(23)u(x, y, z, t)-(tj)d+ tg,dtwhere / and g are functions of x, y, z, and t derived from / and g by theaveraging process (16). This expression certainly satisfies the wave equa­tion (15), since it is the sum of two solutions: the second term is exactlyof the form (17), and the first is the derivative of (17), and we have al­ready seen that derivatives of a solution of the wave equation are alsosolutions. I t remains to be shown that the initial conditions are fulfilled.Since d(tj)/dt = / + t(dj/dt), the right-hand member of (23) reducesto / for t = 0; but /(x, y, z, 0) is precisely /(x, y, z).

Indeed, /(x, y, z, 0)represents the limit, as R —» 0, of the mean value of /(x, y, z) on a sphereof radius R about (x, y z), and this limit must b e / ( x , y> z ) , since / is con­tinuous. T h e first condition (22) is therefore satisfied.

Computing du/dtfrom Eq. (23), one obtainsydt ~2dt+9+\dtt2+dt)'For t = 0, the last term is 0, and the second term gives g(x, y, z, 0) =g{x, y, z) as above for/. T h e term dj/dt may be evaluated from (19). Ast —> 0, so also does the radius of the sphere on which the values of df/drare to be taken. In the limit, the contributions from diametrically oppositepoints on the sphere cancel, since the values of df/dr are the directionalderivatives of / at (x, y, z) for exactly opposite directions. Thus for t = 0,du/dt reduces to #(x, y, z ) , as required.T h e solution (23) of the initial value problem for small perturbations isknown as Poisson's formula.5.DiscussionConsider a perturbation that originates within a bounded region of threedimensional space; i.e., / and g vanish identically outside a finite domain D(see Fig.

8 ) . T h e functions / and g will certainly vanish whenever the sur­face of the sphere about Ρ does not intersect D. For a fixed point P(x\ y, z)outside Z>, there is a minimum radius R\ and a maximum radius R forspheres about Ρ which do intersect D\ for Ρ within D, Ri = 0. Then for24.543DISCUSSIONχF I G . 8. Perturbation due to disturbance in D perceptible at point Ρ only for<i(P) < t <U{P).R < Ri or R >Rl/θθ = h ,R,2R /(Iq =2(24)the spheres do not intersect D\ consequently, withUu(x, y, z, t) = 0for t < h or t >t,2whereand t vary with P . This means that the perturbation that origi­nated in D is perceptible at Ρ only within the time interval h to t .

On theother hand, for a fixed value of t, the perturbation is felt only at points Ρwhose distance from some point of D is exactly dot: the perturbation canexist only in the domain D consisting of the surface points of all spheres ofradius dot with center in D (see Fig. 9 ) .22tIf D shrinks to an infinitesimal neighborhood of a single point 0 , theDtF I G . 9. Perturbation due to disturbance in D perceptible at time / in domainonly.44I. I N T R O D U C T I O N/αF i g . 10.

Computation of a solid angle.perturbation is perceptible at Ρ only at the moment t = OP/a . On the0other hand, at a given time t, only points at a distance ad from Ο are af­fected, that is, D becomes the surface of a sphere of radius aot and center 0.tBriefly, a small disturbance in an inviscid elastic fluid originallypropagated in all directions with constant velocity a0=at rest is\/dp/dp.T h e manner in which the magnitude of the perturbation changes withtime can be seen from a simple example. L e t D be the interior of thesphere So : x + y + z = c , andtake/(rc, y> z) = U, where U is a positive221constant, for points in So, while / = 0 elsewhere, and g = 0.

Since the prob­lem is symmetrical, the perturbation u(x, y, z, t) must be a function onlyof t and of r = {x2+ y + z )\ L e t Ρ be a point outside S , so that τ > c220for P . From (16) it follows that the value of / at Ρ for time t is givenby Ϊ7/47Γ times the solid angle σ under which the intersection of S and S0is viewed from P , where S is the sphere of radius R = a t and center P .0T h e area on S cut out by a cone with vertex at Ρ and vertex angle a is27rP (l — cos a) = σ Ρ ; thus (see Fig. 10)22Then, substituting in the formula (23) for u, with g =have(25)u(x, y, z,t)= ~ (tf)dt=~T~ ~7 [c4ra dt-(r -0, R=aot,aot) ]= ~ (12 \-a - ) .r)(r +c)/ao = £2 since P i20This formula holds for t\ =(r — c)/a0^t ^we0and P2 of (24) are obviously r — c and r + c in this case. For all othervalues of £ we have u = 0 at P .

A t a given moment t, u(x, y, z, t) = 0except at points for which a t — c ^ r ^ aot + c; these points fill a spherical0shell of thickness 2c. Within the shell the value of u increases steadily fromthe value — Uc/2(aot — c) at the inner surface to Uc/2(aot + c) at the outersurface, becoming zero at r = aot. I t can be seen that, although the values4.6T W O - D I M E N S I O N A L CASE45of u within the shell decrease as the shell expands with t, the integral of uover this volume is constant:J=(26)Thus the perturbation is dying out in time, while it spreads from thesource. This behavior is different from that in the one-dimensional case.For points within S one has h = 0 as mentioned in Sec. 4, the perturba­0tion lasting until t = (r + c)/a .

A t the center Ο the perturbation dies out02at t =c/ao . B y time t =2c/a , it will have died out everywhere in S .00Thus, if a second disturbance is initiated in S after time 2c/a ,00this per­turbation will be propagated with the same velocity as the first wave andwill reach each point with a constant time lag after the first wave.

If cis very small, one can consider a succession of values of U, positive andnegative, and this succession will be perceptible, on a reduced scale, at anydistance r. This is the situation in the case of an acoustic signal. But thistype of problem, usually studied in the theory of acoustics, will not be pur­sued here. In any case it follows from (25) that the intensity of a dis­turbance originating in a very small neighborhood of 0 is the same at allpoints of the spherical surface S around Ο reached at any time t.6.

Two-dimensional case24If the motion is restricted to two dimensions, that is, if q = 0 and if allzderivatives with respect to ζ vanish, then the wave equation (15) reducestoIn this case the behavior of a small perturbation can be inferred from thesolution of the three-dimensional problem.In order that the initial conditions be those of a two-dimensional flow,we assume that at the beginning the disturbance is confined to some infinitecylinder in the 2-direction, with finite cross section Co in the £,7/-plane,and that the given functions / and g in (22) are independent of z, thatis,/(x, y, z) = f(x, y), etc.