R. von Mises - Mathematical theory of compressible fluid flow (798534), страница 32
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7 ) .* B y setting Ω =(£ + η ) ω , we obtain1ίθωwhose general solution is given by Z\ in (39).θω\168III.ONE-DIMENSIONALFLOW5. Application of the speedgraph. Initial-value problemT h e two variables u,p or u,v are determined as functions of χ and t bythe pair of planar differential equations (18) or (20), respectively.
W e haveseen (Sec. 2) that the problem is hyperbolic in all cases; thus there are realcharacteristics and the theorems of Sec. 10.3 can be used. T h e first of thesestates that if both u and ν are given at all points of an arc A Β in the x,tplane nowhere tangent to a characteristic, then the solution is determined(at most) in the characteristic quadrangle having A and Β as opposite corners. A natural problem is to ask for the solution when the initial values,i.e., the values of u and υ at t = 0, are given on a certain interval of thear-axis, say from χ = 0 to χ = I.
In this case the arc A Β is a segment of thex-axis; now the characteristics have slopes u dz a as measured from theZ-axis, so that no characteristic can have the same direction as A Β as longas u and a have finite values.T h e numerical approximation method described in detail in Sec. 10.3consisted of a step-by-step construction of a network of characteristic arcstogether with the determination of the values of the unknowns at the nodalpoints of the network. Both parts of this program take a very simple formtF I G . 60.
Construction of the characteristic directions at a point Ρ of thex,i-plane,for ν = 5a. These directions are given by ΜΝι , MN* .12.5A P P L I C A T I O N OF T H E169SPEEDGRAPHin the present problem if the speedgraph is used. T h e pairs of values givenfor the points of the segment A Β map this segment into an arc A'B' in thew,?;-plane (Fig.
60). L e t P' with coordinates u v correspond to a pointΡ on AB. T h e characteristic directions at Ρ make the angles arc tan (u ± a)with the £-axis. Suppose that κ = 1.4, so that υ = 5a. Then lines from P'with slopeZ B 5 meet the w-axis in the pointsIVI and N with abscissas u — aand u + a respectively. If OM = 1, then the tangents of the angles betweenthe vertical direction and the lines MNi and MN are u — a and u + a, respectively. Therefore lines through Ρ parallel to MNand MNgive thecharacteristic directions at P.23y22X2Also we have seen (Sec.
3) that the (u dz a)-characteristics in the x,tplane correspond to the T 4 5 ° lines respectively in the ΐ/,ν-plane. Thus theimages of the characteristics are very easily constructed and form a rectangular network. Furthermore, the ^ - c o o r d i n a t e s of a point of intersection give the values of the unknowns at the corresponding intersectionin the χ,Ζ-plane.T o carry out the computation, we 1) choose a sequence of points Ρ onAB (see Fig.
61), 2) locate the corresponding points P' on A'B', 3) conttΑcρρρΒXν»ΥF I G . 61. Step-by-step construction of an approximate solution to an initial-valueproblem using the speedgraph.170III. ONE-DIMENSIONALFLOWstruct the characteristic directions at the points Ρ by means of the parallelmethod described above, leading to a series of intersections Q in the x,tplane, 4) find the image points Q'—and thus the values of the unknownsat Q—by drawing ± 4 5 ° lines through the points P'. T h e process may thenbe repeated, starting from the points Q and finding a new set of intersections R and corresponding points R' whose u,v-coordinates supply the valuesof the unknowns at the points R, until the complete mapping in the x,tplane of the network in A'B'Chas been determined.
Then the values of thetwo flow variables u,v will have been found in the interior of the characteristic triangle ABC, that is, in the whole domain (for t > 0) in which thesevalues are determined by the initial conditions. In addition, the directionof the particle line, dx/dt = u, at any point Ρ may be found by drawing aparallel to the line Μ Ν in the w,v-plane (Fig.
60), where Ν is the projection of the corresponding point P' onto the u-axis.This method works rapidly and with sufficient accuracy for all practicalcases. Three remarks may be added. First, it is not necessary to takeOM = 1 in Fig. 60, provided a suitable adjustment is made in the scale oft. For example, if it is convenient to choose OM = c ft/sec, then drawingparallels still gives the correct lines in the #,£-plane provided that 1 secon the /-axis has the same length as c feet on the #-axis.Second, in obtaining the solution for t > 0 only half the characteristicquadrangle is used, and it is necessary to choose the appropriate part of theimage rectangle in the w,v-plane.
For t > 0 the (u + a)-characteristicthrough a point P(whose image is the —45° line through the correspondingP') must meet the (u — a)-characteristic (whose image is the + 4 5 ° line)through the next point to the right on AB within the half-quadrangle.Therefore, as one goes along A'B' in the sense corresponding to increasing x,the right side of A'B' is to be used as long as dv/du lies between —1 and 1,while the left side is chosen when dv/du lies in the interval from + 1 throughzb oo to —1 (as in Fig. 61).Finally, it can happen that some domain in the w,v-plane is covered morethan once. For example, suppose that the w,y-values given along the segment ABC (Fig.
62) lead to a curve A'B'Cin the speedgraph with dv/du =1 at B'. Then the right side is used from A' to B', giving the solution in thetriangle ABD corresponding to A'B'D',and the left side is used from B'to C", giving the solution in BCE. T h e solution in DBEF is determined, according to the second theorem of Sec. 10.3, from the (compatible) values ofu and ν along the two characteristics BD and BE. T h e speedgraph of DBEFis the rectangle D'B'E'F'.Thus the region A'B'D'is covered twice T ocomplete the picture in the #,2-plane, it is best to transfer the lines D'B'and ΒΈ', together with the points at which they are intersected by characteristics already plotted, to a new speedgraph figure.
T h e ± 4 5 ° lines may24t12.6VALUES GIVEN ON TWOCHARACTERISTICS171F I G . 62. Double covering of the speedgraph.(a) T h e characteristic net in the x,2-plane.( b ) Image regions in the speedgraph with Α'Β'Ώ'covered twice.( c ) Image region B'D'F'E'shown separately.De drawn in the rectangle, and the network mapped onto the £,2-plane asdescribed above. Each (u + a)-characteristic has an inflection point as itcrosses BD, corresponding to a reversal of direction along the image lineB'D' in the speedgraph plane.*6. Analytic solution: values given on two characteristicsA one-dimensional flow problem is completely solved when the values ofthe state variables u,p (or u,v) are known for each point x,t in the :r,/-plane.Our methods, however, which are based on the interchange of the originalindependent and dependent variables, will supply χ and t as functions ofu and v.
T o invert this solution, i.e., to solve for u,v in terms of x,t is notpossible, in general, by the use of known operators. When this situationoccurs, it is final: no other method of integration can supply u and ν asexplicit functions of χ and t.Giving χ and t as functions of u and v, however, does supply a parametric* See Sec. 19.6 for the general study of a "branch l i n e " , which appears here(namely BD) for the first time.172III.ONE-DIMENSIONALFLOWrepresentation for the characteristics in the :r,£-plane (by setting ν + u orν — u equal to a constant), and in most cases this is sufficient for practicalproblems.
T o find an analytic expression for the particle lines it is stillnecessary to integrate a first-order differential equation.In the remainder of this article we shall deal with the two importantboundary-value problems introduced in Art. 10, as they appear in onedimensional flow. W e shall consider here the linearized form of these problems where the speedgraph variables u,v are the independent variables andshall obtain complete and explicit solutions. In this investigation we shallwork with the function V(u,v) defined in (27) and the general solution forV given in (42).
From these equations we find, for the case κ = 1.4,,dV/' x — ut = —du =vf"9~ vgJ(47)at = -t5=-—dv3/ -3vf' +v*f"+ Sg - 3vg'+vg"where / is some function of ξ = ν + u and g is a function of η = ν — u.I t then remains to choose/(ξ) and g(y) so that given initial conditions aresatisfied. Of course, / and g are not uniquely determined; certain arbitraryexpressions in the one function may be compensated for by terms in theother.Consider now first the characteristic boundary-value problem, wherein weare given compatible values of u and ρ along two characteristics starting from26CA/F I G . 63.
Boundary-value problem with data on two characteristics PA, PB, showing a quadrilateral in which solution is determined in physical plane and in speedgraph.12.6VALUES G I V E N ON TWO173CHARACTERISTICSa common point Ρ in the χ,ί-plane (as in the second of the theorems of Sec.10.3). This problem corresponds to " w a v e penetration'' (see A r t . 13).Of course, u and ν and the shape of the curves cannot be given independently, as it is necessary that ν — u be constant along one curve, say(Fig. 63), and ν +u be constant along the other, PB.PALet us use the subscript 1 to denote the value of any variable at the point P.
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