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MATH 534HEditor : Byeongho BanProof. We only need to find specific function G from the general form in 4-(b).From the general form, observe that11 u(x, 0) G(x)e 2 x .11Thus, only possibility is when G(x) e 2 x . Therefore, we have G(x y) e 2 (x y) then our specific solution would be 111 x 2y1 x 2yu(x, y) e 2 (x y) e 2 (x y) e ex 2y e y e ex 2y .44And observe that1u(x, 0) e0 (ex ex ) 1.4Thus, it is correct one. Proof. 1.3.6Note that the Heat flow equation iscρwhere κ, c and ρ are constants.Then observe that u · (κ u) κ u, t u 1 u u1 u1 2u 2u,, r 2 2 2. r r θ zr r rr θ zSince the temperature u does not depend on z and θ, derivatives with respect to z and θ are 0. Therefore, observe that u u1 u1 u 2urcρut cρ κr κ r 2 κ urr . tr r rr r rrκSetting k cρ , we get ur ut k urr ras desired. u · u ·Page 10

MATH 534HEditor : Byeongho BanProof. 1.3.9Suppose that D the ball of radius a centered at the origin and F r2 x with x (x, y, z). We should verifyZZZZZ · Fdx F · ndS.DbdyD r2x2 , y r 2y 2 and z 2z 2 ,Let’s verify the LHS first. Observe that, noting thatZZZZZZ · Fdx · r2 x, r2 y, r2 z dxDZ Z ZD r2 2x2 r2 2y 2 r2 2z 2 dxZ Z ZD (3r2 2r2 )dxDZZZ 5r2 r2 sin φdrdφdθZ Da 4π5r4 dr r x0 4πa5 .On the other hands,ZZZZr2 (x, y, z) ·F · ndS bdyDbdyDZZ(x, y, z)dSrr(x2 y 2 z 2 )dS bdyDZZr3 dS ZbdyDπ Z 2π φ 05a3 a2 sin φdφdθθ 0 4πa .Therefore, the left and right hand sides are equal so divergence theorem is valid in this case. Proof. 1.3.10Let r 0 be given and let Dr be a ball of radius r. Then observe that, by the divergence theorem,ZZZZZf · ndS · f dx bdyDrDrZZ f dSbdyDrZZ1 dS3 1 x bdyDrZZ1 dS3 1rbdyDr 4πr2.1 r3Observe thatZZZ4πr2 0.r 1 r 3ZZZ · f dx limr all space · f dx limDrTherefore,ZZZZZZ · f dx 0 all space · f dx 0.all space D Page 11

MATH 534HEditor : Byeongho BanProof. 1.4.1Let u(x, t) x2 2t. And note thatut (2t x2 )t 2 (2t x2 )xx uxx .Thus, u solves the PDE. Also, observe thatu(x, 0) x2 2 · 0 x2 .Thus, u also satisfies the initial condition. Therefore, the u is the function we are finding. Page 12

MATH 534HEditor : Byeongho BanDue Date : February 28thByeongho BanAdditional Problem 1- (a)Find the general solution to Problem 8 in Section 1.2. Specify what method you are using and explain step by step yourwork. Show all your work.Problem 8 of Section 1.2Solve aux buy cu 0.Proof. Let’s use variation method. Let(ζ ax byη bx ay.Then observe thatux u u ζ u η auζ buη x ζ x η xuy u ζ u η u buζ auη . y ζ y η yBy plugging them into the PDE, we haveaux buy cu a(auζ buη ) b(buζ auη ) cu (a2 b2 )uζ cu 0.Then observe thatuζcc 2 ln u 2ζ f (η)ua b2a b2By letting G(η) ef (η) , we have c ζu(ζ, η) G(η)e a2 b2 .By changing the variable back, our solution is(a2 b2 )uζ cu u(x, y) G( bx ay)ec a2 b2 (ax by)for some differentiable function f .Observe thatc a2 b2 (ax by)aux buy cu abG0 e baG0 ec a2 b2 (ax by) a ba2ac c (ax by)Ge a2 b22 bbc c (ax by) c (ax by) cGe a2 b2Ge a2 b2a2 b2 0.Thus, our u solves the PDE. Additional Problem 1-(b)Choose a 2, b 5, and c 29 and find the solution u(x, y) to part (a) that also satisfies u(x, 0) e 3x .Proof. From part (a), we know that the solution for 2ux 5uy 29u 0 isu(x, y) G( 5x 2y)e 2229(2x 5y) 52 G( 5x 2y)e (2x 5y) ,where G(x) ef (x) for some differntiable function f . Then observe that, by given condition,e 3x u(x, 0) G( 5x)e 2x .Then we have, with α 5x,αe x G( 5x) G(α) e 5 .Therefore,u(x, y) G( 5x 2y)e (2x 5y) e 5x 2y523e (2x 5y) e 3x 5 y .Then note that2ux 5uy 29u 6u 23u 29u 0.Thus, our solution is23u(x, y) e 3x 5 y . Page 13

MATH 534HEditor : Byeongho BanAdditional Problem 1-(c)Check that 61 (ex y e3x y ) is a particular solution to the inhomogeneous equation2ux 5uy 29u (6ex y 5e3x y ).Proof. Observe that 1112ux 5y 29u 2 ex y 3e3x y 5 ex y e3x y 29 ex y e3x y666 2 5 29 x y5 29 3x y ee 1 6 6666 6ex y 5e3x y .Thus, 16 (ex y e3x y ) solves the inhomogeneous equation so it is a particular solution. Additional Problem 1-(d)Use part (a) a 2, b 5, and c 29 together with part (c) to find the general solution to the inhomogeneous equation2ux 5uy 29u (6ex y 5e3x y ).Proof. LetL 2 5 29. x yThen letuh (x, y) G( 5x 2y)e (2x 5y)for any differentiable function G andui (x, y) 1 x y(e e3x y ).6Then observe thatL(uh ui ) Luh Lui 0 (6ex y 5e3x y )since it is clear that L is linear.Therefore, for any differentiable function G, our general solution to the PDE is1G( 5x 2y)e (2x 5y) (ex y e3x y ).6 Proof. 1.5.1The solution to the PDE is not unique depending on the value of L. Consider the function, when L nπ for any n Z,u(x) eix e ix .Observe thatd2 u u i2 eix ( i)2 e ix eix e ix eix e ix eix e ix 0dx2and thatu(0) ei0 e i0 1 1 0 and u(L) u(nπ) enπi e nπi cos(nπ) cos ( nπ) 0.Thus, u solves the ODE.Note that the general solution for the ODE isψ(x) Aeix Be ix .And observe thatψ(0) 0 Ae0 Be0 0 A Bso thatψ(L) 0 ψ(0) A(eiL e iL ) 0 cos(2L) i sin(2L) e2iL 1 2L 2πk L kπ k Z.Thus, in order for any nonzero function to solve the given ODE, L should be integer multiple of π. Therefore, the solutionof the ODE is unique if and only if L is not an integer multiple of π. Page 14

MATH 534HEditor : Byeongho BanProof. 1.5.4(a)Let u be the solution to the given Neumann problem. Then, for arbitrary constant C, observe that (u C) (u) (C) f 0 f in Dand u C(u C) 0 0 on D. n n nThus, as long as C is constant, we can add C to the solution of the problem to make another different solution. Therefore,the solution to the PDE is not unique so this problem is ill-posed.(b)Suppose that the problem has solution, u. Then, by the divergence theorem, observe thatZZZZZZf (x, y, z)dxdydz udxdydzDZ ZD ( u) · n dS DZZ udS n D 0.Therefore,ZZZf (x, y, z)dxdydz 0Dis a necessary condition for the Neumann problem to have a solution.(c)Note that, for heat flow or diffusion,f (x, y, x) k u tfor some constant k 0.The physical interpretation from (a) is that the diffusion or heat flow only depends on the difference of concentration ortemperature level between two regions. Thus, when we increase a certain amount of heat or concentration equally, evenif the increased level is very high, the diffusion and flow would be same as before the increasing.Note from (b) thatZZZkD u dxdydz k t tZZZudxdydz k M 0 tD uwhere M is the total mass over all D. Note that this hold when n 0 on D which means there is no lose of mass in D.Thus, physical interpretation from (b) is that, when there is no mass going out from D, the total mass is preserved overtime. Page 15

MATH 534HEditor : Byeongho BanProof. 1.5.5Note that0 ux yuy (1, y) · (ux , uy ) (1, y) · u.Then solve the ODEyy0dy y 1 ln y x C y Aex A ye xdx1yfor some constant C and A eC . Thus, the characteristic curve isA ye x .Then the solution to the PDE isu(x, y) f (ye x ) for any differentiable function f.(a)Suppose that, when φ(x) x, the solution exists.Then f (0) u(x, 0) φ(x) x. Note f (0) should be constant and x is not constant. Thus, it is a contradiction.Therefore, there does not exist such f .(b)Suppose that φ(x) 1. Then observe thatf (0) u(x, 0) φ(x) 1.Note that ga (x) e , for any constant a, satisfies g(0) 1 which says for any constant a, f can be ga . Therefore, forinfinitely many a, u(x, y) ga (ye x ) is a solution. So there are infinitely many solution to the problem. axProblem 6(modified)Solve the equation ux 2xy 2 uy 0 and find a solution that satisfies the auxiliary condition u(0, y) y.Proof. Note thatux 2xy 2 uy (1, 2xy 2 ) · (ux , uy ) (1, 2xy 2 ) · u 0.Thus, the tangent of characteristic curve isdy2xy 2 2xy 2 .dx1Then observe that1y0 2x x2 C2yywhere C is some constant. Then our characteristic curve is1x2 C.yThus, our solution is 12u(x, y) f x for some differentiable function f .yBy the given auxiliary condition, we get 1.y u(0, y) fyThen we have1f (x) .xTherefore, our final solution is 11y2u(x, y) f x 2 1 2.yyx 1x yAdditionally, note that, if y were 0, then our PDE isux 0which implies that u(x, y) g(y) for some differentiable function g. Then, by the auxiliary condition,g(y) u(0, y) y 0.Thus, u(x, y) y 0 which is consistent with our result. Page 16

MATH 534HEditor : Byeongho BanProof. 1.6.1(a)Note that we can reduce the PDE asuxx uxy 2uy uyy 3uyx 4u 0 uxx 4uxy uyy 2uy 4u 0.Then observe thata11 a22 a212 1 · 1 ( 2)2 1 4 3 0.Thus, the equation is Hyperbolic.(b)Observe thata11 a22 a212 9 · 1 (3)2 9 9 0.Thus, the equation is Parabolic. Proof. 1.6.2Observe thata11 a22 a212 (1 x)( y 2 ) (xy)2 y 2 xy 2 x2 y 2 y 2 (1 x x2 ).LetE {(x, y) R2 : y 2 xy 2 x2 y 2 0} {(x, y) R2 : 1 x x2 0 & y 6 0}H {(x, y) R2 : y 2 xy 2 x2 y 2 0} {(x, y) R2 : 1 x x2 0 & y 6 0}P {(x, y) R2 : y 2 xy 2 x2 y 2 0} {(x, y) R2 : 1 x x2 0 or y 0}Then the equation is Elliptic in E, Hyperbolic in H and Parabolic in P .Further note that, since 2 312 0 x R,x x 1 x 24we have E {(x, y) R2 : y 6 0}, H , and P {(x, y) R2 : y 0}. Therefore, the equation is Elliptic in R2 exceptx axis and is Parabolic in x axis. And the equation is never hyperbolic. Page 17

MATH 534HEditor : Byeongho BanProof. 1.6.4Suppose the general form of second order PDEa11 uxx 2a12 uxy a22 uyy a1 ux a2 uy a0 u 0From given PDE, note thata11 a22 a212 1 · 4 ( 2)2 4 4 0.Thus, the equation is Parabolic.Let f and g be arbitrary differentiable function and letu(x, y) f (y 2x) xg(y 2x).Then observe thatuxx 4uxy 4uyy (4f 00 2g 0 2g 0 4xg 00 ) 4(2f 00 g 0 2xg 00 ) 4(f 00 xg 00 ) (4 8 4)f 00 (2 2 4)g 0 (4x 8x 4x)g 00 0.Thus, u(x, y) f (y 2x) xg(y 2x) for arbitrary functions f and g is a solution to the PDE. Additional Problem 2Find the regions in R2 where x2 uxx 4uxy y 2 uyy 0 is respectively elliptic, parabolic, hyperbolic. Plot these regions.Proof. Let the general form of second order PDE isa11 uxx a12 uxy a22 uyy (low order terms) 0.2Then, in our case, a11 x , a12 2, and a22 y 2 . Then note thata11 a22 a212 x2 y 2 4.Then letE {(x, y) R2 : x2 y 2 4 0}H {(x, y) R2 : x2 y 2 4 0}P {(x, y) R2 : x2 y 2 4 0}.Then the PDE is Elliptic in E, Hyperbolic in H and Parabolic in P . The graph of x2 y 2 4 and the region of E, H andP are drawn indicated at below. (P is just the union of the lines.) Page 18

MATH 534HEditor : Byeongho BanProof. 2.1.1Note that ex C 2 and sin x C 1 . Consider the functionZ x ct 11 x ct11x ctsin sds [ex ct ex ct ] [cos(x ct) cos(x ct)].u(x, t) e e 22c x ct22cObserve thatu(x, 0) 1 x1[e ex ] [cos(x) cos(x)] ex22cand that1 x ct1[ce cex ct ] [ c sin(x ct) c sin(x ct)]22cso11ut (x, 0) [cex cex ] [ c sin(x) c sin(x)] sin x.22cThus, it satisfies the auxiliary condition. Also, observe that11utt (ut )t [c2 ex ct c2 ex ct ] [ c2 cos(x ct) c2 cos(x ct)]22cand that11ux [ex ct ex ct ] [ sin(x ct) sin(x ct)]22cso1 x ct1uxx [e ex ct ] [ cos(x ct) cos(x ct)].22cTherefore, utt c2 uxx so u solves the PDE. Therefore, u(x, y) is the solution to the IVP.ut Page 19

MATH 534HEditor : Byeongho BanProof. 2.1.2Note that log(1 x2 ) C 2 and 4 x C 1 . Then consider the functionu(x, t) Z x ct11(4 s)ds[log(1 (x ct)2 ) log(1 (x ct)2 )] 22c x ct 111[log(1 (x ct)2 ) log(1 (x ct)2 )] 8ct [(x ct)2 (x ct)2 ]22c2 111[log(1 (x ct)2 ) log(1 (x ct)2 )] 8ct [4xct]22c21[log(1 (x ct)2 ) log(1 (x ct)2 )] 4t xt21[log(1 x2 c2 t2 2xct) log(1 x2 c2 t2 2xct)] 4t xt.2Note that1[log(1 (x)2 ) log(1 (x)2 )] 0 log(1 x2 ).2And note that 12c2 t 2xc2c2 t 2xcut 4 x2 1 x2 c2 t2 2xct 1 x2 c2 t2 2xctand that 2xc 2xc1 4 x 4 x.ut (x, 0) 2 1 x21 x2Thus, u satisfies the initial conditions. Furthermore, observe thatu(x, 0) 12c2 t 2xc2c2 t 2xc 0 t 2 1 x2 c2 t2 2xct 1 x2 c2 t2 2xct 1 2c2 (1 x2 c2 t2 2xct) (2c2 t 2xc)22c2 (1 x2 c2 t2 2xct) (2c2 t 2xc)2 2(1 x2 c2 t2 2xct)2(1 x2 c2 t2 2xct)2 c2 2(1 x2 c2 t2 2xct) (2x 2ct)22(1 x2 c2 t2 2xct) (2x 2ct)2 2(1 x2 c2 t2 2xct)2(1 x2 c2 t2 2xct)2utt and thatuxx 2x 2ct2x 2ct 1 t x 2 1 x2 c2 t2 2xct 1 x2 c2 t2 2xct 1 2(1 x2 c2 t2 2xct) (2x 2ct)22(1 x2 c2 t2 2xct) (2x 2ct)2 .2(1 x2 c2 t2 2xct)2(1 x2 c2 t2 2xct)2Thus, c2 uxx utt so u satisfies the PDE. Therefore, u(x, t) is the solution to the IVP. Page 20

MATH 534HEditor : Byeongho BanProof. 2.1.8 (Hint : Recall a function f on R is odd if f (x) f ( x).)(a)Let v ru, then observe that 2 v vttv222 cutt c urr ur rrr rr r r r vtt2 rvr vrvr v2 crr2rr2r 2 vtt2 rvr v(r )(rvrr vr vr ) 2r(rvr v)2 crr4rr2 vtt2rvr 2v(r)(rvrr ) (2rvr 2v) c23rrr3 v vttrr c2rr vtt c2 vrr .(b)We use general solution of the wave equationv(r, t) f (r ct) g(r ct)for arbitrary functions f and g. Observe thatvtt c2 f 00 c2 g 00 c2 (f 00 g 00 ) c2 vxx .Thus, v solves the PDE. Then note thatu Then observe thatutt v1 (f (r ct) g(r ct)).rr v r tt 1 2 00c f (r ct) c2 g 00 (r ct)rand thatur rf 0 (r ct) rg 0 (r ct) f (r ct) g(r ct)r2 ur rr2 (f 0 (r ct) g 0 (r ct) rf 00 (r ct) rg 00 (r ct) f 0 (r ct) g 0 (r ct)) r400 2r(rf (r ct) rg (r ct) f (r ct) g(r ct)) r4(r3 )f 00 (r ct) (r2 2r2 r2 )f 0 (r ct) 2rf (r ct) r43 0022(r )g (r ct) (r r 2r2 )g 0 (r ct) 2rg(r ct) r4 0000(f (r ct) g (r ct)) 2 2rf 0 (r ct) 2rg 0 (r ct) 2(f (r ct) g(r ct)) rrr2 00utt2r(f (r ct) g (r ct)) (f (r ct) g(r ct)) 2 crr2utt2 2 ur .crurr Therefore, 2utt c2 urr ur .rThus,u vrsolves our wave equation. Page 21

MATH 534HEditor : Byeongho BanProof. 2.1.8(c)Note that vtt c2 vrr and that v(r, 0) ru(r, 0) rφ(r) and vt (r, 0) rut (r, 0) rψ(r). ThenZ r ct11sψ(s)dsv(r, t) [(r ct)φ(r ct) (r ct)φ(r ct)] 22c r ctis the solution to the PDE vtt c2 vrr . Then u(r, t) vrsolves the PDE 22utt c urr urrby the part (b).Now note thatu(r, t) Then1v(r, t) r2 1 ctr Z r ctct1sψ(s)ds.φ(r ct) 1 φ(r ct) r2cr r ct11u(r, 0) [(1 0) φ(r 0) (1 0) φ(r 0)] 22crZrsψ(s)ds φ(r) 0 φ(r).rAlso, note thato1 nc1ut (r, t) (φ(r ct) φ(r ct)) c(φ0 (r ct) φ0 (r ct)) (c(r ct)ψ(r ct) c(r ct)ψ(r ct)) .2 r2crTheno1 nc1ut (r, 0) (0) c(0) (crψ(r) crψ(r)) ψ(r).2 r2crThus, our u(r, t) satisfies the given initial condition. Therefore, u(r, t) solves the PDE 22utt c urr ur .r Page 22

MATH 534HEditor : Byeongho BanDue date: March 7thByeongHo BanAdditional Problem 1First find the solution to the linear homogeneous wave equation with wave speed 1 with initial condition u(x, 0) sin x,ut (x, 0) 0. Then calculate ut (0, t).Proof. We must solve the following IVP: utt uxx 0u(x, 0) sin x ut (x, 0) 0.The solution for IVP of wave equation is given byZ1 x t1ut (s, 0)dsu(x, t) [u(x t, 0) u(x t, 0)] 22 x tZ11 x t [sin(x t) sin(x t)] 0ds22 x t1 [sin(x t) sin(x t)]2Let’s double check:Note that, clearly,utt 11 sin(x t) ( 1)2 sin(x t) [ sin(x t) sin(x t)] uxx .22Now, observe thatu(x, 0) 11[sin(x 0) sin(x 0)] [2 sin x] sin x22and thatut (x, t) 11[cos(x t) cos(x t)] ut (x, 0) [cos x cos x] 0.22Thus, our solution solves the IVP.Lastly, note thatut (0, t) 1[cos t cos( t)] 02since cos is even function. Page 23

MATH 534HEditor : Byeongho BanAdditional Problem 2Find the solution to the wave equation utt 4uxx 0 and with initial conditions u(x, 0) sin x, ut (x, 0) 10. Calculatethen ut (0, t).Proof. We must solve the following IVP: utt 4uxx 0u(x, 0) sin x ut (x, 0) 10.The solution for IVP of wave equation is given byZ11 x 2tut (s, 0)ds[u(x 2t, 0) u(x 2t, 0)] 24 x 2tZ11 x 2t [sin(x 2t) sin(x 2t)] 10ds24 x 2t11 [sin(x 2t) sin(x 2t)] (4t)(10)241 [sin(x 2t) sin(x 2t)] 10t.2u(x, t) Let’s double check:Note that, clearly,utt 41 (2)2 sin(x 2t) ( 2)2 sin(x 2t) [ sin(x t) sin(x t)] 4uxx .22Now, observe thatu(x, 0) 11[sin(x 0) sin(x 0)] 0 [2 sin x] sin x22and that11[2 cos(x 2t) 2 cos(x 2t)] 10 ut (x, 0) [2 cos x 2 cos x] 10 10.22Thus, our solution solves the IVP.ut (x, t) Lastly, note thatut (0, t) 1[2 cos 2t 2 cos( 2t)] 10 102since cos is even function. Page 24

MATH 534HEditor : Byeongho Ban2.1.9We must solve following IVP: uxx 3uxt 4utt 0u(x, 0) x2 ut (x, 0) ex .Observe thatuxx 3uxt 4utt ( x t ) ( x 4 t ) u.Let v ( x 4 t ) u. Then since the solution of the transport equation vx vt 0 is v(x, t) f (x t) for any differentiablefunction f : R R, we have a following PDE( x 4 t ) u f (x t). Note that the solution of

Jan 31, 2018 · Show that the di erence of two solutions of an inhomogeneous linear equation Lu gwith the same gis a solution of the homogeneous equation Lu 0. Proof. . Suppose that wand vare given solution of the inhomogeneous equation so that Lw gand Lv g. Then observe that