MIT OpenCourseWarehttp://ocw.mit.edu 18.306 Advanced Partial Differential Equations with Applications Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Problem Set Number 02 18.306 — MIT (Fall 2009) Rodolfo R. Rosales MIT, Math. Dept., Cambridge, MA 02139 October 10, 2009 Due: Friday October 16. Contents 1.1 Statement: Linear 1st order PDE (problem 09) . . . . . . . . . . . . . . . . . . . . 1 1.2 Statement: Quasi-Linear 1st order PDE (problem 02) xtra Q’s . . . . . . . . . . . . 2 1.3 Statement: Linear 1st order PDE (problem 10) . . . . . . . . . . . . . . . . . . . . 2 1.4 Statement: Quasi-Linear 1st order PDE (problem 03) . . . . . . . . . . . . . . . . . 3 1.5 Statement: PDE numerical integration (problem 01) . . . . . . . . . . . . . . . . . . 4 1.1 Statement: Linear 1st order PDE (problem 09). Surface Evolution. The evolution of a material surface can (sometimes) be modeled by a pde. In evaporation dynamics, where the material evaporates into the surrounding environment, consider a surface described in terms of its “height” h = h(x, y, t) relative to the (x, y)-plane of reference. Under appropriate conditions, a rather complicated pde can be written1 for h. Here we consider a (drastically) simplified version of the problem, where the governing equation is A � ht = hr, for r = x2 + y2 > 0 and t > 0, where A > 0 is a constant. (1.1) r Axial symmetry is assumed, so that h = h(r, t). Obviously, h should be an even function of r. This is both evident from the symmetry, and necessary in the equation to avoid singular behavior 1From mass conservation, with the details of the physics going into modeling the flux and sink/source terms. 1�at the origin. Assume now h(r, 0) = H(r 2), (1.2) where H is a smooth function describing a localized bump. Specifically: (i) H(0) > 0, (ii) H is monotone decreasing. (iii) H 0 as r → ∞. Note that h(r, 0) is an even function of r.→ 1. Using the theory of characteristics, write an explicit formula for the solution of (1.1 – 1.2). 2. Do a sketch of the characteristics in space time — i.e.: r > 0 and t > 0. 3. What happens with the characteristic starting at r = ζ > 0 and t = 0 when t = ζ2/√2 A? 4. Show that the resulting solution is an even function of r for all times. 5. Show that, as t → ∞, the bump shrinks and vanishes. 1.2 Statement: Quasi-Linear 1st order PDE (problem 02) xtra Q’s. Do the ”Tasks left to the reader” at the end of the answer to the last problem in problem set # 1 — i.e.: The answer to the Quasi-Linear 1st order PDE (problem 02). 1.3 Statement: Linear 1st order PDE (problem 10). Integrating factors. Show that the pde (a(x, y) µ)y = (b(x, y) µ)x (1.3) is a necessary and sufficient condition guaranteeing that µ = µ(x, y) = 0 is an integrating factor for the ode a(x, y) dx + b(x, y) dy = 0, (1.4) in any open subset of the plane without holes. Part II. Assume that a = 3 x y + 2 y 2 and b = 3 x y + 2 x 2 . Find an integrating factor for (1.4) — i.e.: obtain a nontrivial solution of (1.3). Use it to integrate (1.4), and write (1.4) in the form Φ(x, y) = constant, for some function Φ. (1.5) Hint 1.1 Solving by characteristics (1.3) leads to (1.4), or equivalent, as part of the process — check this! To get out of this circular situation, note that: for a and b as above, µ = F (x, y) solves 2(1.3) iff µ = F (y, x) does. This suggests that you should look for solutions2 invariant under this symmetry; namely: µ(x, y) = µ(y, x). Hence write µ = µ(u, v), with u = x + y and v = x y, since solutions that satisfy µ(x, y) = µ(y, x) must have this form — see remark 1.1. Remark 1.1 The transformation (x, y) (u, v) is not one to one: it maps the whole xy-plane→ � � into the region v ≤ 1 u 2 of the uv-plane, with double valued inverse x =1 u ±√u2 − 4 v and 4 2 1 2y =2 � u �√u − 4 v � . Furthermore: (a) The two inverses are related by the x ↔ y switch. (b) The singular line u 2 = 4 v corresponds to the line x = y. (c) The map is a bijection between the regions x ≤ y and 4 v ≤ u 2 . (d) The map is a bijection between the regions x ≥ y and 4 v ≤ u 2 . From (c - d) we see that: for any µ = µ(x, y), µ = f(u, v) for x ≤ y and µ = g(u, v) for x ≥ y, for some f and g. Then, if µ(x, y) = µ(y, x), f = g, so that µ has the form µ = µ(u, v). Part III. Why is it that this approach CANNOT be generalized to three variables? That is, to find integrating factors for equations of the form a(x, y, z) dx + b(x, y, z) dy + c(x, y, z) dz = 0. (1.6) Note: this problem is based on Levine’s problem 11 in chapter 9. 1.4 Statement: Quasi-Linear 1st order PDE (problem 03). Simple waves in Gas Dynamics. Under the isentropic flow assumption, the Euler equations of Gas Dynamics in one dimension can be written in the form ρt + (ρ u)x = 0, conservation of mass, (1.7) (ρ u)t + (ρ u2 + p)x = 0, conservation of momentum, (1.8) where ρ is the mass density, u is the flow velocity, and p is the pressure. In addition, an equation of state must be provided: dP 2 p = P (ρ), satisfying = c > 0, (1.9)dρ where c = c(ρ) > 0 has the dimensions of a velocity (it is the sound speed). 2Note that you only need one nontrivial solution. 3This is a system of two equations for two unknowns. Interestingly, the system has solutions that depend on a single unknown function. Namely, solutions of the form ρ = R(ψ) and u = U(ψ), (1.10) where R and U are functions of the single argument ψ, and ψ = ψ(x, t) satisfies a scalar quasi-linear equation of the form: ψt + λ(ψ) ψx = 0. (1.11) Your TASK is to find these solutions; that is: find R, U, and λ. Hints. (i) Characterize the functions R and U as solutions of an ode — do not seek explicit expressions. (ii) After you substitute (1.10) into the equations, cast the system into …