Solutions Manual Applied Mathematics, 3rd Edition

1 Solutions ManualApplied mathematics , 3rd EditionJ. David LoganWilla Cather Professor of MathematicsUniversity of Nebraska LincolnNovember 8, 2010iiContentsPrefacev1 Scaling, Dimensional Analysis (Secs. 1 & 2) Dimensional Analysis .. Scaling .. 62 Perturbation Regular Perturbation .. Singular Perturbation .. Boundary Layer Analysis .. Initial Layers .. WKB Approximation .. Asymptotic Expansion of Integrals .. 373 Calculus of Variational Problems .. Necessary Conditions for Extrema .. The Simplest Problem .. Generalizations .. The Canonical Formalism .. Isoperimetric Problems .. 55iiiivContentsPrefaceThis Manual contains hints or full Solutions to many of the problems in Chapters1, 2, and 3 of the text: J.

2 David Logan, mathematics , 3rd ed.,Wiley Interscience, New would like to thank Glenn Ledder, my colleague at UNL, who has taughtthe course many times and who has been the source of many examples, exercises,and and corrections will be greatly David LoganWilla Cather ProfessorDepartment of MathematicsUniversity of Nebraska LincolnLincoln, NE 1 Scaling, DimensionalAnalysis (Secs. 1 & 2) Dimensional AnalysisExercises, page period cannot depend only on the length and mass; there is no waythat length and mass can be combined to yield a time dimension. If weassume there is a physical lawf(P, L, g) = 0, thenP=F(L, g). Theright side must be time dimensions, and the only way that we can gettime dimensions withgandLis to take L/g.

3 Thus,P=C L/gforsome (D, e) = 0 then we can solve and gete=F(D). Now,eis energy permass, or length-squared per time-squared. So the right hand side of theequation must be proportional toD2. Thene=cD2for some nonlinear regression, writer=bt2/5, b= (E/ )1/5,The sum of the squares of the errors isS=8 i=1(bt2/5i ri) the derivative with respect toband set it equal to zero to getb= the energy in kilotons isE= 1012 b5= 1. SCALING, DIMENSIONAL ANALYSIS (SECS. 1 & 2)Another method is to average. We haveE= each data point to getEi= , i= 1,2, .. , average theEiand divide by 1012to getE= variables aret, r, , e, P. We already know one dimensionless quantity 1= r5/et2. Try to find another that usesP, which is a pressure, or forceper unit area, that is, mass per length per time-squared.

4 By inspection, 2=P gris another dimensionless quantity. Thus we havef( r5/et2,P gr) = we cannot isolate therandtvariables in one dimensionless expres-sion. If we solve for the first dimensionless quantity we get r5/et2=F(P gr).Thenr=(et2 )1/5F(P gr).Because the second dimensionless variable containsrin some unknownmanner, we cannot conclude thatrvaries liket2/5. However, if one canargue that the ambient pressure is small and can be neglected, then wecan setP= 0 and obtain the resultr=(et2 )1/5F(0),which does imply thatrvaries liket2 , then =x/gt2is dimensionless and the physical law is =12. If we include mass, thenmmust be some function oft, x, g, whichis 12gt2+vt, then, by inspection,y=x/gt2ands=v/gtaredimensionl ess.

5 Dividing the equation bygt2gives the dimensionless formy= 1/2 + DIMENSIONAL ANALYSIS3 Exercises, page thatf(v, , g) = 0. If is dimensionless[ ] = [v 1 2g 3],= (LT 1) 1L 2(LT 2) we have the homogeneous system 1+ 2+ 3= 0, 1 2 3= rank of the coefficient matrix is one, so there is one dimensionlessvariable. Notice that ( 2,1,1) is a solution to the system, and thus = the Pi theorem,F( ) = 0 or g/v2= dimensionless variables are Vm,SV2 Vm=f(SV2/3). length, time and mass as fundamental and writex= 1x,t= 2t,m= writev= 1 12v, and so on for the other variables. Show thatv 29r2 g 1(1 l/ ) = 1 12(v 29r2 g 1(1 l/ ))So, by definition, the law is unit ,L, andT(mass, length, and time) as fundamental is only one dimensionless variable amongE,P, andA, namelyPA3/2/E.

6 Thus,PA3/2/E= two dimensionless variables areat L,bt .4 CHAPTER 1. SCALING, DIMENSIONAL ANALYSIS (SECS. 1 & 2) quantities are e, , by the pi theorem,v= Ef( e). ,L, andT(mass, length, and time) as fundamental there is a physical lawf(T, V, C, Y, r) = 0. We have =T 1V 2C 3Y 4r 5and so1 =T 1(L3) 2(ML 3) 3(MT 1) 4(MT 1V 3) the powers ofT,L, andMequal to zero and solving gives 1= 4+ 5, 2= 4, 3= 4 leads to two dimensionless quantitiesTYV C, ,L, andT(mass, length, and time) as fundamental ,L, andT(mass, length, and time) as fundamental lengthL, timeT, and massMas fundamental dimensions. Thenthe dimension matrix has rank three and there are 5 3 = 2 dimension-less variables; they are given by 1= and 2=R l/ P.

7 Thusf( 1, 2) = 0 implies =R 1 P/ lG( )for some dimensions are[E] =energymass,[T] = temp,[k] =energymass is clear there is only one dimensionless variable, =E/kT. ThusE/kT= DIMENSIONAL have dimensions[F] =MLT 1,[V] =LT 1,[C] =L3T 1,[K] =ML 1T a physical lawf(F, V, C, K) = 0. If is dimensionless, then =F 1V 2C 3K gives1 = (MLT 1) 1(LT 1) 2(L3T 1) 3(ML 1T 2) leads to the system of equations 1+ 2+ 3 3 4= 0, 2 1 2 3 2 4= 0, 1+ 4= system has rank 3 and so there is one solution, ( 1, 1,1,1), whichgives the dimensionless variableCK/FV= have dimensions[w] =L,[C0] = [C 1] =ML 3,[d] =L2T 1,[ ] =ML 2T a physical lawf(w, C0, C1, d, ) = 0. If is dimensionless, then =w 1C 20C 31d 4 gives1 =L 1(ML 3) 2(ML 3) 3(L2T 1) 4(ML 2T 1) leads to the system of equations 1 3 2 3 3+ 2 4 2 5= 0, 2+ 3+ 5= 0, 4+ 5= system has rank 3 and so there are two independent Solutions (0, 1,1,0,0)and (1, 1,0, 1,1).

8 This gives dimensionless variablesC0C1,w dC0=G(C0C1),which gives the form of the flux , =dC0wG(C0C1).6 CHAPTER 1. SCALING, DIMENSIONAL ANALYSIS (SECS. 1 & 2) ScalingExercises, page (a) we haveu=Asin tand sou = Acos t. ThenM=Aandmax|u |= A. Then we havetc= 1/ . In (b) we haveu=Ae tandu = Ae t. Thentc= max|u|/max|u |= 1/ . In part (c) wehaveu=Ate tandu = (1 t)Ae t. The maximum ofuoccurs att= 1/ and isM=A/ e. To find the maximum ofu we calculate thesecond derivative to getu =A ( t 2)e t. So the maximum derivativeoccurs att= 2/ or at an endpoint. It is easily checked that the maximumderivative occurs att= 0 and has value max|u |=Aon the given (A/ e)/A= 1/ 1+exp( t/ ) andu = exp( t/ )/.

9 Thentc= max|u|/max|u |=2/ 1= 2 . The time scale is very small, indicating rapid change in a smallinterval. But a graph shows that that this rapid decrease occurs only in asmall interval neart= 0; in most of the interval the changes occur two time scales are suggested, one near the origin and one out in theinterval wheretis order havem =ax2 bx3, m= ( x3) = 3x2 x =ax2 bx3,givingx =a3 b3 haveagiven in mass per time per length-squared andbin mass pertime per volume. Scaling time by /band length bya/bleads to thedimensionless modely =13 (0) = 0 theny(0) = 0 and the solution to the dimensionless model isy( ) = 1 e , this is a reasonable model. The organism grows exponentially towarda limiting value.

10 This is, in fact, observed with most constants in the problem,V,k, andahave dimensions[V] =LT,[k] =MT2,[a] = time scale is m/kwhich is based on damping. Another is m/aV,which is based on the restoring force. To rescale, letTandLbe scales SCALING7be chosen and lety=x/Land =t/T. Then the model becomesy = aTLmy|y | T2kmy, y(0) = 0, y (0) =V want the restoring force to be small and have the small takeT= m/aV. Then we gety = aL m/aVmy|y | kaVy, y(0) = 0, y (0) =V m/aVLNow choose the length scaleLso that the coefficient ofy|y |is one. Thedifferential equation then becomesy = y|y | y, y(0) = 0, y (0) = 1, , the small coefficient is in front of the small damping dimensions of the constants are[I] =MLT,[a] =MT,[k] = (I/a) and =t/T, whereTis yet to be determined, wegetmaTu = u kTI2a3u, u(0) = 0,maTu (0) = the mass is small, we want to chooseTso that the coefficient of theu term is small.