A First Course In Quasi-Linear Partial Di Erential Equations For .

16CONTENTSSolution.(a) This can be done by plugging in the equations.(b) Plug in.(c) We have sup{ un (x, 0) 1 : x R} n1 sup{ sin nx : x R} n1 .(d) We have sup{ un (x, t) 1 : x R} 2en t.n(e) We have limt sup{ un (x, t) 1 : x R, t 0} limt Hence, the solution is unstable and thus the problem is ill-posed2en tn .Problem 2.13Find the general solution of each of the following PDEs by means of directintegration.(a) ux 3x2 y 2 , u u(x, y).(b) uxy x2 y, u u(x, y).(c) uxtt e2x 3t , u u(x, t).Solution.(a) u(x, y) x3 xy 2 f (y), where f is an arbitrary differentiable function.R3 2(b) u(x, y) x 6y F (x) g(y), where F (x) f (x)dx and g(y) is anarbitrary differentiable function.RR1 2x 3te t f1 (x)dx f2 (x)dx g(t)(c) u(x, t) 18Problem 2.14Consider the second-order PDEuxx 4uxy 4uyy 0.(a) Use the change of variables v(x, y) y 2x and w(x, y) x to showthat uww 0.(b) Find the general solution to the given PDE.Solution.(a) Using the chain rule we finduxuxxuyuyyuxy 2uv uw 4uvv 4uvw uww uv uvv 2uvv uvw .

SOLUTIONS TO SECTION 217Substituting these into the given PDE we find uww 0.(b) Solving the equation uww 0 we find uw f (v) and u(v, w) wf (v) g(v). In terms of x and y the general solution is u(x, y) xf (y 2x) g(y 2x)Problem 2.15Derive the general solution to the PDEutt c2 uxxby using the change of variables v x ct and w x ct.Solution.We haveututtuxuxx cuv cuw c2 uvv 2c2 uwv c2 uww uv uw uvv 2uvw uwwSubstituting we find uvw 0 and solving Rthis equation we find uv f (v)and u(v, w) F (v) G(w) where F (v) f (v)dv.Finally, using the fact that v x ct and w x ct; we get d’Alembert’ssolution to the one-dimensional wave equation:u(x, t) F (x ct) G(x ct)where F and G are arbitrary differentiable functions

18CONTENTSSolutions to Section 3Problem 3.1Solve the IVP: y 0 2ty t, y(0) 0Solution.R2Since p(t) 2t, we find µ(t) e 2tdt et . Multiplying the given equation2by et to obtain 2 02et y tetIntegrating both sides with respect to t and using substitution on the righthand integral to obtain1 22et y et C22Dividing the last equation by et to obtain2y(t) Ce t 12Since y(0) 0, we find C 12 . Thus, the unique solution to the IVP isgiven by12y (1 e t )2Problem 3.2Find the general solution: y 0 3y t e 2tSolution.Since p(t) 3, the integrating factor is µ(t) e3t . Thus, the general solutionisRy(t) e 3t Re3t (t e 2t )dt Ce 3t3tt 3t e 3t 3t (te e )dt Ce e 3t e9 (3t 1) et Ce 3t3t 1 e 2t Ce 3t9Problem 3.3Find the general solution: y 0 1t y 3 cos t, t 0

SOLUTIONS TO SECTION 319Solution.R dtSince p(t) 1t , the integrating factor is µ(t) e t eln t t. Using themethod of integrating factor we findR1y(t) 3t cos tdt Ctt 3t (t sin t cos t) Ctt 3 sin t 3 cos CttProblem 3.4Find the general solution: y 0 2y cos (3t).Solution.We have p(t) 2 so that µ(t) e2t . Thus,Z 2ty(t) ee2t cos (3t)dt Ce 2tButZ2e2te cos (3t)dt sin (3t) e2t sin (3t)dt33Ze2t2 e2t2 sin (3t) ( cos (3t) e2t cos (3t)dt)3333Z13e2te2t cos (3t)dt (3 sin (3t) 2 cos (3t))99Ze2te2t cos (3t)dt (3 sin (3t) 2 cos (3t))13Z2tHence,y(t) 1(3 sin (3t) 2 cos (3t)) Ce 2t13Problem 3.5Find the general solution: y 0 (cos t)y 3 cos t.Solution.Since p(t) cos t we have µ(t) esin t . Thus,Z sin ty(t) eesin t ( 3 cos t)dt Ce sin t 3e sin t esin t Ce sin t Ce sin t 3

20CONTENTSProblem 3.6Given that the solution to the IVP ty 0 4y αt2 , y(1) 13 exists on theinterval t . What is the value of the constant α?Solution.Solving this equation with the integrating factor method with p(t) find µ(t) t4 . Thus,Z1Cy 4 t4 (αt)dt 4ttα 2 C t 46t4tweSince the solution is assumed to be defined for all t, we must have C 0.On the other hand, since y(1) 13 we find α 2Problem 3.7Suppose that y(t) Ce 2t t 1 is the general solution to the equationy 0 p(t)y g(t). Determine the functions p(t) and g(t).Solution.RThe integrating factor is µ(t) e2t . Thus, p(t)dt 2t and this implies thatp(t) 2. On the other hand, the function t 1 is the particular solutionto the nonhomogeneous equation so that (t 1)0 2(t 1) g(t). Hence,g(t) 2t 3Problem 3.8Suppose that y(t) 2e t et sin t is the unique solution to the IVPy 0 y g(t), y(0) y0 . Determine the constant y0 and the function g(t).Solution.First, we find y0 : y0 y(0) 2 1 0 1. Next, we find g(t) : g(t) y 0 y ( 2e t et sin t)0 ( 2e t et sin t) 2e t et cos t 2e t et sin t 2et cos t sin tProblem 3.9Find the value (if any) of the unique solution to the IVP y 0 (1 cos t)y 1 cos t, y(0) 3 in the long run?

SOLUTIONS TO SECTION 321Solution.RThe integrating factor is µ(t) e (1 cos t)dt et sin t . Thus, the general solution isZ (t sin t)y(t) eet sin t (1 cos t)dt Ce (t sin t) 1 Ce (t sin t)Since y(0) 3, we find C 2 and therefore y(t) 1 2e (t sin t) . Finally,lim y(t) lim (1 2e sin t e t ) 1t t Problem 3.10Solve the initial value problem ty 0 y t, y(1) 7Solution.Rewriting the equation in the form1y0 y 1twe find p(t) 1t and µ(t) 1t . Thus, the general solution is given byy(t) t ln t CtBut y(1) 7 so that C 7. Hence,y(t) t ln t 7tProblem 3.11Show that if a and λ are positive constants, and b is any real number, thenevery solution of the equationy 0 ay be λthas the property that y 0 as t . Hint: Consider the cases a λ anda 6 λ separately.Solution.Since p(t) a we find µ(t) eat . Suppose first that a λ. Theny 0 ay be at

22CONTENTSand the corresponding general solution isy(t) bte at Ce atThus,limt y(t) limt ( ebtat eCat ) limt aebat 0Now, suppose that a 6 λ theny(t) b λte Ce ata λThus,lim y(t) 0t Problem 3.12Solve the initial-value problem y 0 y et y 2 , y(0) 1 using the substitution1u(t) y(t)Solution.Substituting into the equation we findu0 u et , u(0) 1Solving this equation by the method of integrating factor with µ(t) e t wefindu(t) tet CetSince u(0) 1, C 1 and therefore u(t) tet et . Finally, we havey(t) ( tet et ) 1Problem 3.13Solve the initial-value problem ty 0 2y t2 t 1, y(1) Solution.Rewriting the equation in the form12y0 y t 1 tt12

SOLUTIONS TO SECTION 3Since p(t) 2t23we find µ(t) t2 . The general solution is then given byy(t) Since y(1) 12we find C 1.12y(t) t2t 1 C 243 2 tHence,t2t 11 43 2 12t2Problem 3.14Solve y 0 1t y sin t, y(1) 3. Express your answer in terms of the sineRtintegral, Si(t) 0 sins s ds.Solution.Since p(t) 1t we find µ(t) 1t . Thus, R 0t sin s 0 s1y(t) Si(t) Cty(t) tSi(t) Ct 01ytSince y(1) 3, C 3 Si(1). Hence, y(t) tSi(t) (3 Si(1))t

24CONTENTSSolutions to Section 4Problem 4.1Solve the (separable) differential equation2 ln y 2y 0 tet.Solution.At first, this equation may not appear separable, so we must simplify theright hand side until it is clear what to do.2 ln y 2y 0 tett2 ln te · e12 tet · 2yt 2 2 et .y1y2 Separating the variables and solving the equation we find2y 2 y 0 tetZZ123 0(y ) dt tet dt31 3 1 t2y e C323 23y et C2Problem 4.2Solve the (separable) differential equationy0 t2 y 4y.t 2

SOLUTIONS TO SECTION 425Solution.Separating the variables and solving we findy 0 t2 4 t 2yt 2ZZ0(ln y ) dt (t 2)dtln y t2 2t C2t2y(t) Ce 2 2tProblem 4.3Solve the (separable) differential equationty 0 2(y 4).Solution.Separating the variables and solving we findy02 y 4 tZZ20(ln y 4 ) dt dttln y 4 ln t2 Cy 4ln 2 Cty(t) Ct2 4Problem 4.4Solve the (separable) differential equationy 0 2y(2 y).Solution.Separating the variables and solving (using partial fractions in the process)

26CONTENTSwe findy0 2y(2 y)y0y0 22y 2(2 y)ZZZ1100(ln y ) dt (ln 2 y ) dt 2dt22yln 4t C2 yy Ce4t2 y2Ce4ty(t) 1 Ce4tProblem 4.5Solve the IVPy0 4 sin (2t), y(0) 1.ySolution.Separating the variables and solving we findyy 0 4 sin (2t)(y 2 )0 8 sin (2t)ZZ2 0(y ) dt 8 sin (2t)dty 2 4 cos (2t) Cpy(t) C 4 cos (2t).Since y(0) 1, we find C 5 and hencepy(t) 5 4 cos (2t)Problem 4.6Solve the IVP:πyy 0 sin t, y( ) 2.2

SOLUTIONS TO SECTION 427Solution.Separating the variables and solving we findZ y22 0Zdt sin tdty2 cos t C2y 2 2 cos t C.pSince y( π2 ) 2, we find C 4. Thus, y(t) ( 2 cos t 4). Fromy( π2 ) 2, we havepy(t) ( 2 cos t 4)Problem 4.7Solve the IVP:y 0 y 1 0, y(1) 0.Solution.Separating the variables and solving we find(ln (y 1))0 1ln (y 1) t Cy 1 Ce ty(t) Ce t 1.Since y(1) 0, we find C e. Thus, y(t) e1 t 1Problem 4.8Solve the IVP:y 0 ty 3 0, y(0) 2.

28CONTENTSSolution.Separating the variables and solving we findZ0 3Zy y dt tdtZ 2 0yt2dt C 221t2 2 C2y21y2 2. t CSince y(0) 2, we find C have y(t) 1.4Thus, y(t) q4. 4t2 1Since y(0) 2, we 2 4t2 1Problem 4.9Solve the IVP:πy 0 1 y 2 , y( ) 1.4Solution.Separating the variables and solving we findy0 11 y2arctan y t Cy(t) tan (t C).Since y( π4 ) 1, we find C π2 . Hence, y(t) tan (t π2 ) cot tProblem 4.10Solve the IVP:1y 0 t ty 2 , y(0) .2

SOLUTIONS TO SECTION 429Solution.Separating the variables and solving we findy0y2 10yy0 y 1 y 1y 1lny 1y 1y 1 t 2t t2 C2 Ce t21 Ce ty(t) .1 Ce t2Since y(0) 12 , we find C 31 . Thus,23 e ty(t) 3 e t2Problem 4.11Solve the equation 3uy uxy 0 by using the substitution v uy .Solution.Letting v uy we obtain 3v vx 0. Solving this ODE by the method ofseparation of variables we findvx 3vln v(x, y) 3x f (y)v(x, y) f (y)e 3x .Hence, u(x, y) Rf (y)e 3x dy F (y)e 3x G(x) where F (y) Problem 4.12Solve the IVP(2y sin y)y 0 sin t t, y(0) 0.Rf (y)dy

30CONTENTSSolution.Separating the variables and solving we findZZ 3x0f (y)e(2y sin y)y dt (sin t t)dtf (y)e 3x y 2 cos y cos t t2 C.2(4.1)(4.2)Since y(0) 0, we find C 2. Thus,y 2 cos y cos t t2 22Problem 4.13State an initial value problem, with initial condition imposed at t0 2,having implicit solution y 3 t2 sin y 4.Solution.Differentiating both sides of the given equation we find3y 2 y 0 cos y 2t 0, y(2) 0Problem 4.14Can the differential equationdy x2 xydxbe solved by the method of separation of variables? Explain.Solution.If we try to factor the right side of the ODE, we getdy x(x y).dxThe second factor is a function of both x and y. The ODE is not separable

SOLUTIONS TO SECTION 531Solutions to Section 5Problem 5.1Classify each of the following PDE as linear, quasi-linear, semi-linear, or nonlinear.(a) xux yuy sin (xy).(b) ut uux 0(c) u2x u3 u4y 0.(d) (x 3)ux xy 2 uy u3Solution.(a) Linear (b) Quasi-linear, non-linear (c) Non-linear (d) Semi-linear, nonlinearProblem 5.2Show that u(x, y) ex f (2x y), where f is a differentiable function of onevariable, is a solution to the equationux 2uy u 0.Solution.Let w 2x y. Then ux 2uy u ex f (w) 2ex fw (w) 2ex fw (w) ex f (w) 0Problem 5.3 Show that u(x, y) x xy satisfies the equationxux yuy usubject to the constraintu(y, y) y 2 , y 0.Solution. 1 1 3 1 3 2 2We have xux yuy x 2 x y y 21 x 2 y 2 x xy u. Also, u(y, y) y2Problem 5.4Show that u(x, y) cos (x2 y 2 ) satisfies the equation yux xuy 0subject to the constraintu(0, y) cos y 2 .

32CONTENTSSolution.We have yux xuy 2xy sin (x2 y 2 ) 2xy sin (x2 y 2 ) 0. Moreover,u(0, y) cos y 2Problem 5.5Show that u(x, y) y 12 (x2 y 2 ) satisfies the equation111ux uy xyysubject to u(x, 1) 21 (3 x2 ).Solution.We have x1 ux y1 uy x1 ( x) y1 (1 y) y1 . Moreover, u(x, 1) 12 (3 x2 )Problem 5.6Find a relationship between a and b if u(x, y) f (ax by) is a solution to theequation 3ux 7uy 0 for any differentiable function f such that f 0 (x) 6 0for all x.Solution.Let v ax by. We haved(ax by) afv (v)dxd(ax by)uy fv (v) bfv (v).dyux fv (v)Hence, by substitution we find 3a 7b 0Problem 5.7Reduce the partial differential equationaux buy cu 0to a first order ODE by introducing the change of variables s ax by andt bx ay.

SOLUTIONS TO SECTION 533Solution.By the chain rule we findux us sx ut tx aus butuy us sy ut ty bus aut .Thus,0 aux buy cu (a2 b2 )us cuorus a2cu 0. b2This is a first order linear ODE that can be solved using the method ofseparation of variablesProblem 5.8Solve the partial differential equationux uy 1by introducing the change of variables s x y and t x y.Solution.Using the chain rule we findux us sx ut tx us utuy us sy ut ty us utSubstituting these into the PDE to obtain us 21 . Solving this ODE wefind u(s, t) 12 s f (t) where f is an arbitrary differentiable function in onevariable. Now substituting for s and t we find u(x, y) 21 (x y) f (x y)Problem 5.9Show that u(x, y) e 4x f (2x 3y) is a solution to the first-order PDE3ux 2uy 12u 0.

34CONTENTSSolution.We haveux 4e 4x f (2x 3y) 2e 4x f 0 (2x 3y)uy 3e 4x f 0 (2x 3y)Thus,3ux 2uy 12u 12e 4x f (2x 3y) 6e 4x f 0 (2x 3y) 6e 4x f 0 (2x 3y) 12e 4x f (2x 3y) 0Problem 5.10Derive the general solution of the PDEaut bux u, a, b 6 0by using the change of variables v ax bt and w a1 t.Solution.We have ut buv a1 uw and ux auv . Substituting we find uw uand solving this equation we find u(v, w) f (v)ew where f is an arbitrarytdifferentiable function in one variable. Thus, u(x, t) f (ax bt)e aProblem 5.11Derive the general solution of the PDEaux buy 0, a, b 6 0by using the change of variables s(x, y) ax by and t(x, y) bx ay.Assume a2 b2 0.Solution.According to the chain rule for the derivative of a composite function, wehaveux us sx ut tx aus butuy us sy ut ty bus autSubstituting these into the given equation to obtaina2 us abut b2 us abut 0

SOLUTIONS TO SECTION 535or(a2 b2 )us 0and since a2 b2 0 we obtainus 0.Solving this equation, we findu(s, t) f (t)where f is an arbitrary differentiable function of one variable. Now, in termsof x and y we findu(x, y) f (bx ay)Problem 5.12Write the equationut cux λu f (x, t)in the coordinates v x ct, w t.Solution.Using the chain rule, we findut uv vt uw wt cuv uwux uv vx uw wx uvSubstituting these into the original equation we obtain the equationuw λu f (v cw, w)Problem 5.13Suppose that u(x, t) w(x ct) is a solution to the PDExux tut Auwhere A and c are constants. Let v x ct. Write the differential equationwith unknown function w(v).

36CONTENTSSolution.Using the chain rule we findut cwvandux w v .Substititution into the original PDE givesvwv (v) Aw(v)

SOLUTIONS TO SECTION 737Solutions to Section 7Problem 7.1Solve ux yuy y 2 with the initial condition u(0, y) siny.Solution.We have a 1, b y, and f y 2 . Solvingk2dydx y we find y k1 ex . Solving y 2 k12 e2x we find u 21 e2x f (k1 ) 21 y 2 f (k1 ) 21 y 2 f (ye x ).Using the initial condition u(0, y) sin y we find sin y 21 y 2 f (y). Hence,u(x, y) 21 y 2 21 y 2 e 2x e2x sin (ye x )dudxProblem 7.2Solve ux yuy u2 with the initial condition u(0, y) sin y.Solution.dyWe have a 1, b y, and f u2 . Solving dx y we find y k1 ex . Solvingdu1 u2 we find x u1 k2 . Thus, u(x, y) f (ye x. Using the initialdx) xcondition u(0, y) sin y we find f (y) csc y. Hence, u(x, y) csc (ye1 x ) xProblem 7.3Find the general solution of yux xuy 2xyu.Solution.The system of ODEs isx dudy , 2xu.dxy dxSolving the first equation, we find x2 y 2 k1 . Solving the second equation,22we find u k2 ex . Hence, u(x, y) ex f (x2 y 2 ) where f is an arbitrarydifferentiable function in one variableProblem 7.4Find the integral surface of the IVP: xux yuy u, u(x, 1) 2 e x .Solution.The system of ODEs isy duudy , .dxx dxx

38CONTENTSSolving the first equation, we find y k1 x. Solving the second equation, weyfind u k2 x. Hence, u(x, y) xf x where f is an arbitrary differentiablefunction in one variable. From the initial condition u(x, 1) 2 e x we1find f (x) x(2 e x . Hence, the integral surface isu(x, y) y(2 e y )xProblem 7.5Find the unique solution to 4ux uy u2 , u(x, 0) 1.1 x2Solution.The system of ODEs can be written asdxdydu 2.41u dywe find x 4y k1 . Solving the equationSolving the equation dx41dydu11 u2 we find u(x, y) f (x 4y) y . Using the initial condition u(x, 0) 1 x2112we find f (x) 1 x . Hence, u(x, y) (x 4y)2 1 yProblem 7.62Find the unique solution to e2y ux xuy xu2 , u(x, 0) ex .Solution.The system of ODEs can be written asdydudx 2.2yexxuThus, xdx e2y dy which implies x2 e2y k1 . Solving the equationwe find y u1 k2 f (x2 e2y ). Hence,u(x, y) duu21.f (x2 e2y ) y2Using the initial condition u(x, 0) ex we find f (x) e (x 1) . Hence,u(x, y) 1e x2 e2y 1 y dy

SOLUTIONS TO SECTION 739Problem 7.

For each of the following PDEs, state its order and whether it is linear or non-linear. If it is linear, also state whether it is homogeneous or nonhomo-geneous: (a) uu x x2u yyy sinx 0: (b) u x ex 2u y 0: (c) u tt (siny)u yy etcosy 0: Solution. (a) Order 3, non-linear. (b) Order 1, linear, homogeneous. (c) Order 2, linear, non .