Shearer And Levy: Partial Di Erential Equations { Solutions

Shearer and Levy: Partial Differential Equations – SolutionsHunter Stufflebeam2017ForwardThe following is a collection of my solutions for Michael Shearer and Rachel Levy’s text Partial DifferentialEquations: An Introduction to Theory and Applications. These solutions were worked out over the summerof 2017, and will almost certainly contain errors. If you happen to find any, or have suggestions for moreelegant/interesting/general approaches to problems, please drop me a line at hstufflebeam@utexas.edu. Thefigures and diagrams were made with Mathematica.As an update, I have become to privy to the existence of an errata sheet for the text, which explainssome of the funkiness of certain questions. I have adjusted the problems here to represent what is writtenon the errata sheet, when necessary.Contents1 Introduction1.1 . . . . . .1.2 . . . . . .1.3 . . . . . .1.4 . . . . . .1.5 . . . . . .1.6 . . . . . .1.7 . . . . . .1.8 . . . . . .1.9 . . . . . .1.10 . . . . . .223333445662 Beginnings2.1 . . . . .2.2 . . . . .2.3 . . . . .2.4 . . . . .2.5 . . . . .2.6 . . . . .8810101112143 First Order PDE3.1 . . . . . . . . .3.2 . . . . . . . . .3.3 . . . . . . . . .3.4 . . . . . . . . .3.5 . . . . . . . . .3.6 . . . . . . . . .3.7 . . . . . . . . .3.8 . . . . . . . . .3.9 . . . . . . . . .15151516161718182021.1

3.103.113.123.133.143.154 4Wave Equation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2424252526272830303131325 The Heat Equation336 Separation of Variables and Fourier Series337 Eigenfunctions and Convergence of Fourier Series338 Laplace’s Equation and Poisson’s Equation339 Green’s Functions and Distributions3310 Function Spaces3311 Elliptic Theory and Sobolev Spaces3312 Traveling Wave Solutions of PDE3313 Scalar Conservation Laws3314 Systems of First Order Hyperbolic PDE3315 The Equations of Fluid Mechanics331Introduction1.1Show that the traveling wave u(x, t) f (x 3t) satisfies the linear transport equation ut 3ux 0 for anydifferentiable function f : R R.Solution. This is obvious: if f is differentiable then ut 3f 0 (x 3t) and ux f 0 (x 3t). Hence ut 3ux 3f 0 (x 3t) 3f 0 (x 3t) 0.2

1.2Find an equation relating the parameters k, m, n so that the function u(x, t) emt sin(nx) satisfies the heatequation ut kuxx .Solution. We have ut mu(x, t) and uxx n2 u(x, t). Hence, u will satisfy the heat equation aboveprovided that m kn2 .1.3Find an equation relating the parameters c, m, n so that the function u(x, t) sin(mt) sin(nx) satisfies thewave equation utt c2 uxx .Solution. We have utt m2 u(x, t) and uxx n2 u(x, t). Hence, u will satisfy the wave equation aboveprovided that m2 c2 n2 .1.4Find all functions a, b, c : R R such that u(x, t) a(t)e2x b(t)ex c(t) satisfies the heat equation ut uxxfor all x, t.Solution. We have ut a0 (t)e2x b0 (t)ex c0 (t) and uxx 4a(t)e2x b(t)ex . If u satisfies the heat equationabove, then 4a(t) a0 (t), b(t) b0 (t), and c0 (t) 0. As such, we conclude that a, b, c in general have theforms a(t) λ1 e4t , b(t) λ2 ex , and c(t) λ3 for λi R.1.5For m 1, define the conductivity k k(u) so that the porous medium equation ut 2 (um ) can bewritten as the quasilinear heat equation ut · (k(u) u)Solution. We first recall that, in general, if f, g : Rm R are differentiable, then · (g f ) h g, f i g 2 f.Define k k(u) mum 1 . Then we havennXXmut (u ) (u )xi xi (mum 1 uxi )xi2mi 1i 1 nX{m(m 1)um 2 u2xi mum 1 uxi xi }i 1 m(m 1)um 2 k uk2 mum 1 2 uOn the other hand, by the first remark · (k(u) u) h k(u), ui k(u) 2 uand sinceh k(u), ui h mum 1 , ui m(m 1)um 2 h u, ui m(m 1)um 2 k uk2we see that indeed ut · (k(u) u) with the conductivity k(u) mum 1 .3

1.6Solve the initial value problem x , t 0,ut 4ux 1,2 1u(x, 0) (1 x ) x .,Solution. The general form of the solution is u(x, t) f (x 4t) for some differentiable f : R R. By the1initial data, we have f (x) 1 x2 for all x R. Hence, the particular solution of the IVP isu(x, t) 1.1 (x 4t)2Figure 1: u(x, t)1.7Solve the initial boundary value problemut 4ux 0,u(x, 0) 0, tu(0, t) te ,0 x , t 0,0 x ,t 0.Why is there no solution if the PDE is changed to ut 4ux 0?Solution. The general solution is again of the form u(x, t) f (x 4t) for some differentiable functionf : R R. By the first boundary condition we have that f (x) 0 whenever t 0 and x 0. By the secondboundary condition we have f ( 4t) te t whenever x 0 and t 0. When x 4t 0, both imply that fvanishes. From the above, we gather that(0ξ 0f (ξ) ξ ξ4 4eξ 0which implies that the solution u of the BVP is(0u(x, t) x 4t4 x 4t4 e4x 4yx 4t

Figure 2: u(x, t) Note that Mathematica is having some trouble on the line x 4tSuppose, on the other hand, that the PDE were changed to ut 4ux 0. Then the general form ofthe solution would be u(x, t) f (x 4t), and the boundary conditions would give contradictory data, e.g.u(x, 0) f (x) 0 for all x 0 at the same time that u(0, t) f (4t) te t 0 when t 0.1.8Consider the linear transport equation ut cux 0 with initial and boundary conditionsu(x, 0) φ(x),if x 0,u(0, t) ψ(t),if t 0.where φ, ψ : [0, ) are differentiable.(a) Suppose the data φ, ψ are differentiable functions. Show that the function u : Q1 R given by(φ(x ct) x ct,u(x, t) ψ(t xc ) x 6 ctsatisfies the PDE away from the line x ct, the boundary condition, and the initial condition.(b) In the solution above, the line x ct which emerges from the origin x t 0 separates the quadrantQ1 into two regions. On the line, the solution has one-sided limits given by φ, ψ. Consequently, thesolution will in general have singularities on the line.(i) Find conditions on the data φ, ψ so that the solution is continuous across the line x ct.(ii) Find conditions on the data φ, ψ so that the solution is differentiable across the line x ct.Solution. (a) We have( cφ(x ct) x ctut ψ(t xc )x 6 ctand(ux henceφ(x ct)x ct1x c ψ(t c ) x 6 ct( cφ(x ct) cφ(x ct) 0 x ctut cux ψ(t xc ) ψ(t xc ) 0x ctprovided that we are also away from the boundary of Q1 .5

(b)(i) This seems to always be the case, since by the boundary conditions we get that φ(0) u(0, 0) ψ(0)(and finite!), which is exactly what we need to know to conclude that u is continuous across the linex ct. Indeed, if ((xk , tk )) is any sequence in Q1 which tends to a point (ξ, τ ) on the line x ct,then we can say the following: u takes the value u0 φ(0) ψ(0) along the line x ct, so it sufficesto show that for any ε 0 we can find a K Z 1 such that u(xk , tk ) B(u0 ; ε) for all k K.This is true, however, given the fact that both φ(x) and ψ(t) are continuous. We can simply chooseK so large such that all points of the sequence after time K lie in a ball about (ξ, τ ) small enoughso that φ(xk ctk ) B(u0 φ(0); ε) (in the case when xk ctk ) and ψ(tk xck ) B(u0 ψ(0); ε)(in the case when x 6 ctk ) for all k K. Hence, u(xk , tk ) u0 , so u is continuous across the linex ct.(ii) If we want u to be differentiable across the line x ct, then we must have cφ0 (0) ψ 0 (0). Indeed,this comes from the results above, since if cφ0 (0) ψ 0 (0), we can unambiguously define ut andux on the line x ct. If you look closely at the piecewise definitions of ut and ux above, you willsee that we actually get this condition for free as well, so as stated the solution is differentiableover the line x ct.As an aside, it may be possible that the intended problem involved NOT having data about how φ andψ behave at 0. If this is the case, then the conditions φ(0) ψ(0) and cφ0 (0) ψ 0 (0) are not had for free,and must be imposed to ensure continuity and differentiability, respectively, over the line x ct.1.9Let f : R R be differentiable. Verify that if u(x, t) is differentiable and satisfies u f (x ut), then u(x, t)is a solution of the initial value problemut uux 0,u(x, 0) f (x), x , t 0 x .Solution. It is immediately evident that u(x, 0) f (x) for all x. We then calculate ut (u ut t)f 0 (x ut)and ux (1 ux t)f 0 (x ut), from which we gatherut uux (u ut t)f 0 (x ut) u(1 ux t)f 0 (x ut) ut tf 0 (x ut) uux tf 0 (x ut) (ut uux )tf 0 (x ut)which gives us the relation(ut uux )(1 tf 0 (x ut)) 0.Since R is a field one of these factors must be zero; we show that in fact ut uux always vanishes, whichproves that u is a solution of the IVP. By inspection, the only thing we need to be worried about is thepossibility of having f 0 (x ut) 1t for some positive time. Recalling that ux (1 ux t)f 0 (x ut), if forsome positive time t it were true that f 0 (x ut ) t1 , then we would have ux t t t1 ux t t whichyields the absurdity 0 t1 . Thus, f 0 (x ut) 6 1t for all positive time, and we conclude that ut uux 0for all time t 0. Hence u as described is a solution of the IVP.1.10Let(1 x2u0 06 1 6 x 6 1otherwise.

(a) Use the equation u u0 (x ut) to find a formula for the solution u u(x, t) of the inviscid Burgersequation ut uux 0 with 1 x 1 and 0 t 12 .(b) Verify that u(1, t) 0 for all 0 t 12(c) Differentiate the formula to find ux (1 , t), and deduce that ux (1 , t) as t 1 2 .Note that ux (x, t) is discontinuous at x 1.Solution. (a)(1 (x ut)2u u0 (x ut) 0 1 6 x ut 6 1otherwiseProvided that x ut [ 1, 1], then, we can write u 1 (x ut)2 1 x2 2uxt u2 t2 . Rearranging,we find thatt2 u2 (1 2xt)u x2 1 0.By the quadratic formula,u(x, t) (2xt 1) p(1 2xt)2 4t2 (x2 1).2t2where we must take the conjugate root solution, since the conjugate solution does not satisfy 1 6 x ut 6 1 over the region ( 1, 1) (0, 12 ).Figure 3: u(x, t)Figure 4: Graph of x ut corresponding to the conjugate solution for u(x, t)7

Figure 5: Graph of x ut corresponding to the conjugate solution for u(x, t) 2t 1 (1 2t)2(b) Note first that 1 2t 0 for 0 t 12 . Then u(1, t) 2t 1 (1 2t) 0 for all2t22t210 t 2.(c) By direct computation,2t 1 ux (x, t) 2t2soux (1 , t) which clearly tends to as t 2t(1 2xt)2 4t2 (x2 1)2t 1 2t1 2t2t2 2t 11 22tt(1 2t)1 2 .To see what this actually means, consider Figure 3. ux (1 , t) is the slope of the wave front at time t,and as time goes to 12 , this slope becomes infinitely steep, resulting in the shock.2Beginnings2.1(a) Determine the type of the equation uxx uxy ux 0(b) Determine the type of the equation uxx uxy αuyy ux u 0 for each real value of the parameterα.(c) Determine the type of the equation utt 2uxt uxx 0. Verify that there are solutions u(x, t) f (x t) tg(x t) for any twice differentiable functions f, g.(d) The equation (1 y)uxx x2 uxy xuyy 0 is hyperbolic, elliptic, or parabolic depending on the locationof (x, y) in the plane. Find a formula to determine where in the x y plane the equation is hyperbolic.Sketch the x y plane and label where the equation is hyperbolic, where it is elliptic, and where it isparabolic.Solution. (a) The principal symbol for the equation is L(p) [ξ1 , ξ2 ] ξ12 ξ1 ξ2 , which informs us that b2 ac,since 1 0. Hence, the PDE is hyperbolic.8

(b) The principal symbol for the equation is L(p) [ξ1 , ξ2 ] ξ12 ξ1 ξ2 αξ22 , which informs us that 0 α ( , 1)b2 ac 1 α is 0 α 1. 0 α (1, )Hence, the PDE is hyperbolic when α ( , 1), parabolic when α 1, and elliptic when α (1, ).(c) The principal symbol for the equation is L(p) [ξ1 , ξ2 ] ξ12 2ξ1 ξ2 ξ22 , which informs us that b2 ac,since 4 1. Hence, the PDE is hyperbolic. Suppose that f, g C 2 , and that u(x, t) f (x t) tg(x t).Thenutt f 00 (x t) 2g 0 (x t) tg 00 (x t)uxt f 00 (x t) g 0 (x t) tg 00 (x t)uxx f 00 (x t) tg 00 (x t)hence utt 2uxt uxx 0.(d) The principal symbol for the equation is L(p) [ξ1 , ξ2 ] (1 y)ξ12 x2 ξ1 ξ2 xξ22 , which says that b2 ac x4 x(1 y).Figure 6: Graph of b2 ac x4 x(1 y)The main item of interest is the region of the plane where this function is positive, since that is wherethe PDE is hyperbolic:Figure 7: Graph of b2 ac x4 x(1 y) 0. The function is positive of the shaded region, negativeon the light region, and zero on the boundary.9