Partial Differential Equations: Graduate Level Problems and ...

1 Partial Differential Equations: Graduate Level Problemsand SolutionsIgor Yanovsky1 Partial Differential EquationsIgor Yanovsky, 20052 Disclaimer:This handbook is intended to assist Graduate students with qualifyingexamination preparation. Please be aware, however, that the handbook might contain,and almost certainly contains, typos as well as incorrect or inaccurate solutions. I cannot be made responsible for any inaccuracies contained in this Differential EquationsIgor Yanovsky, 20053 Contents1 Trigonometric Identities62 Simple Eigenvalue Problem83 Separation of Variables:Quick Guide94 Eigenvalues of the Laplacian: Quick Quasilinear Equations.

2 Weak Solutions for Quasilinear Equations .. Conservation Laws and Jump Conditions .. GeneralNonlinearEquations .. ThreeSpatialDimensions .. 136 Second-Order CanonicalFormsandGeneralSolutions .. Well-Posedness .. Initial/Boundary Value Problem .. Duhamel sPrinciple .. HigherDimensions .. Spherical Means .. ApplicationtotheCauchyProblem .. Huygen EnergyMethods .. ContractionMappingPrinciple .. 308 Laplace Green PolarCoordinates.

3 Polar Laplacian inR2forRadialFunctions .. Spherical Laplacian Cylindrical Laplacian inR3forRadialFunctions .. MaximumPrinciple .. The Fundamental Solution .. 42 Partial Differential EquationsIgor Yanovsky, FourierTransform .. Multi-IndexNotation .. Solution of the Pure Initial Value Problem .. NonhomogeneousEquation .. Nonhomogeneous Equation with Nonhomogeneous Initial The Fundamental Solution .. 5010 Schr odinger Equation5211 Problems : Quasilinear Equations5412 Problems : Shocks7513 Problems : General Nonlinear.

4 9314 Problems : First-Order Systems10215 Problems : Gas Dynamics .. 13616 Problems : Wave Initial/Boundary Value Problem .. 18717 Problems : Laplace The Fundamental Solution .. Spherical Means .. Harmonic Extensions, Subharmonic Functions .. 249 Partial Differential EquationsIgor Yanovsky, 2005518 Problems : Heat .. 26419 Contraction Mapping and Uniqueness - Wave27120 Contraction Mapping and Uniqueness - Heat27321 Problems : Maximum Principle - Laplace and .. 28122 Problems : Separation of Variables - Laplace Equation28223 Problems : Separation of Variables - Poisson Equation30224 Problems : Separation of Variables - Wave Equation30525 Problems : Separation of Variables - Heat Equation30926 Problems : Eigenvalues of the Laplacian - Laplace32327 Problems : Eigenvalues of the Laplacian - Poisson33328 Problems : Eigenvalues of the Laplacian - Wave33829 Problems : Eigenvalues of the Laplacian - Heat Equation with Periodic Boundary Conditions in 2D(withextraterms).

5 36030 Problems : Fourier Transform36531 Laplace Transform38532 Linear Functional .. H .. 394 Partial Differential EquationsIgor Yanovsky, 200561 Trigonometric Identitiescos(a+b)=cosacosb sinasinbcos(a b)=cosacosb+sinasinbsin(a+b)=sinacosb+co sasinbsin(a b)=sinacosb cosasinbcosacosb=cos(a+b)+cos(a b)2sinacosb=sin(a+b)+sin(a b)2sinasinb=cos(a b) cos(a+b)2cos 2t=cos2t sin2tsin 2t=2sintcostcos212t=1+cost2sin212t=1 cost21+tan2t=sec2tcot2t+1 = csc2tcosx=eix+e ix2sinx=eix e ix2icoshx=ex+e x2sinhx=ex e x2ddxcoshx= sinh(x)ddxsinhx=cosh(x)

6 Cosh2x sinh2x=1 dua2+u2=1atan 1ua+C du a2 u2=sin 1ua+C L Lcosn xLcosm xLdx= 0n =mLn=m L Lsinn xLsinm xLdx= 0n =mLn=m L Lsinn xLcosm xLdx=0 L0cosn xLcosm xLdx= 0n =mL2n=m L0sinn xLsinm xLdx= 0n =mL2n=m L0einxeimxdx= 0n =mLn=m L0einxdx= 0n =0Ln=0 sin2xdx=x2 sinxcosx2 cos2xdx=x2+sinxcosx2 tan2xdx=tanx x sinxcosxdx= cos2x2ln(xy)=ln(x)+ln(y)lnxy=ln(x) ln(y)lnxr=rlnx lnxdx=xlnx x xlnxdx=x22lnx x24 Re z2dz= Re z22dz= 2 Partial Differential EquationsIgor Yanovsky, 20057A= abcd ,A 1=1det(A) d b ca Partial Differential EquationsIgor Yanovsky, 200582 Simple Eigenvalue ProblemX + X=0 Boundary conditionsEigenvalues nEigenfunctionsXnX(0) =X(L)=0 n L 2sinn Lxn=1,2.

7 X(0) =X (L)=0 (n 12) L 2sin(n 12) Lxn=1,2,..X (0) =X(L)=0 (n 12) L 2cos(n 12) Lxn=1,2,..X (0) =X (L)=0 n L 2cosn Lxn=0,1,2,..X(0) =X(L),X (0) =X (L) 2n L 2sin2n Lxn=1,2,..cos2n Lxn=0,1,2,..X( L)=X(L),X ( L)=X (L) n L 2sinn Lxn=1,2,..cosn Lxn=0,1,2,..X X=0 Boundary conditionsEigenvalues nEigenfunctionsXnX(0) =X(L)=0,X (0) =X (L)=0 n L 4sinn Lxn=1,2,..X (0) =X (L)=0,X (0) =X (L)=0 n L 4cosn Lxn=0,1,2,.. Partial Differential EquationsIgor Yanovsky, 200593 Separation of Variables:Quick GuideLaplace Equation: u= (x)X(x)= Y (y)Y(y)=.

8 X + X= (t)X(t)= Y ( )Y( )= .Y ( )+ Y( )= Equation:utt uxx= (x)X(x)=T (t)T(t)= .X + X= +3ut+u= T+3T T+1 =X X= .X + X= uxx+u= T+1 =X X= .X + X= + ut=c2uxx+ uxxt,( >0)X X= ,1c2T T+ c2T T= 1+ c2T T X Order:utt= kuxxxx. X X=1kT T= .X X= Equation:ut= T=kX X= .X + kX= Order:ut= T= X X= .X X= Eigenvalues of the Lapla-cian: Quick GuideLaplace Equation:uxx+uyy+ u= X+Y Y+ =0.( = 2+ 2)X + 2X=0,Y + 2Y= +uyy+k2u=0. X X=Y Y+k2= +c2X=0,Y +(k2 c2)Y= +uyy+k2u=0. Y Y=X X+k2= +c2Y=0,X +(k2 c2)X= Differential EquationsIgor Yanovsky, Quasilinear EquationsConsider the Cauchy problem for the quasilinear equation in two variablesa(x, y, u)ux+b(x, y, u)uy=c(x, y, u),with parameterized by (f(s),g(s),h(s)).

9 The characteristic equations aredxdt=a(x, y, z),dydt=b(x, y, z),dzdt=c(x, y, z),with initial conditionsx(s,0) =f(s),y(s,0) =g(s),z(s,0) =h(s).In a quasilinear case, the characteristic equations fordxdtanddydtneed not decouple fromthedzdtequation; this means that we must take thezvalues into account even to findthe projected characteristic curves in thexy-plane. In particular, this allows for thepossibility that the projected characteristics may cross each condition for solving forsandtin terms ofxandyrequires that the Jacobianmatrix be nonsingular:J xsysxtyt =xsyt ysxt = particular, att= 0 we obtain the conditionf (s) b(f(s),g(s),h(s)) g (s) a(f(s),g(s),h(s)) = s the Cauchy problem ut+uux=0,u(x,0) =h(x).

10 ( )The characteristic equations aredxdt=z,dydt=1,dzdt=0,and may be parametrized by (s,0,h(s)).x=h(s)t+s, y=t, z=h(s).u(x, y)=h(x uy)( )The characteristic projection in thext-plane1passing through the point (s,0) is thelinex=h(s)t+salong whichuhas the constant valueu=h(s). Two characteristicsx=h(s1)t+s1andx=h(s2)t+s2 intersect at a point (x, t)witht= s2 s1h(s2) h(s1).1yandtare interchanged herePartial Differential EquationsIgor Yanovsky, 200511 From ( ), we haveux=h (s)(1 uxt) ux=h (s)1+h (s)tHence forh (s)<0,uxbecomes infinite at the positive timet= 1h (s).