MATH 212 Homework 1 Chapter 1, Fall 2014

MATH 212 HOMEWORK (# 1) Due Monday September 22, in class Mohammad El Smaily American University of Beirut September 14, 2014 Problem 1: PDEs solved via ODEs a) Show that φ(s) := tanh(s) := es−e−s es+e−s is a solution of the ordinary differential equation − φ00(s) + 2 � φ(s)2 − 1 � φ(s) = 0, s 2 R, (1) such that lim s!−1 φ(s) = −1 and lim s!+1 φ(s) = 1. For simplicity of the presentation presentation, you may omit the s variable in (1) and just write −φ00 + 2 � φ2 − 1 � φ = 0 in R. b) Consider the partial differential equation − ∆u + 2(u2 − 1)u = 0 for (x, y) 2 R2, (2) where u is a function of 2 variables (u = u(x, y)) and ∆u stands for uxx +uyy. Let e := (cos ↵, sin ↵) be a unit vector (|e| = 1) for some ↵ 2 (0, ⇡/2). Consider the change of variables u(x, y) = φ(x cos ↵+y sin ↵). 1) Find the differential equation that the function φ satisfies when- ever u(x, y) satisfies (2). 2) Use Part a) above to find a solution u for PDE (2) which satisfies lim (x,y)·e!−1 u(x, y) = −1 and lim (x,y)·e!+1 u(x, y) = 1 (Here, (x, y) · e stands for the dot product between the vectors (x, y) and e.) 1 Problem 2: Order 1 PDEs 1) Find a bunch of functions (3 or more) f(x, y) whose level curves are circles centred at the origin. Explain how you design these functions. 2) What are the components of V (x, y) which is the tangent vector, at a point (x, y), to a circle centred at the origin. 3) In this part, we intend to use the previous collected information in solving a PDE. Consider the partial differential equation − y@xu + x@yu = 0 for all (x, y) 2 R2. (3) Use a geometric method to find the general solution of the PDE (3) above. (Give all necessary explanations.) 4) Can you design a PDE whose solution u(x, y) admits elliptical level curves? (Hint: use the algorithm sketched in parts 1,2,3 above in the case of circles...) 5) Consider the parabolas (Pa)a2R : y = x2 + a (a is a random constant). Find a vector V (x, y) which is tangent to Pa at point (x, y). 6) Use the previous part to construct a PDE whose solution u(x, y) will have parabolic level curves of the form (Pa)a2R. Problem 3: Change of Coordinates Method Use a suitable change of coordinates to solve (i.e find a general solution u(t, x)) the PDE 9ut − 2014ux = 0. (4) 2