By Ioannis P. Stavroulakis, Stepan A. Tersian
Read or Download Partial Differential Equations: An Introduction With Mathematica and Maple, 2nd Edition PDF
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Additional resources for Partial Differential Equations: An Introduction With Mathematica and Maple, 2nd Edition
Example text
11) at a point P (x,y) E R is: ( i ) hyperbolic, i f A (x,y) > 0, (ii) parabolic, i f A (x,y) = 0 , ( i i i ) elliptic, i f A (x,y) < 0. The equation is hyperbolic (parabolic, elliptic) in a subset G c R i f it i s hyperbolic (parabolic, elliptic) at every point of G. < Next we will show that we can find new coordinates and 7 so that in terms of the new coordinates the form of Eq. 11) is such that its principal part is particularly simple. Then we say that the equation is in canonical fomn. 2. Assume that Eq.
ProzyI? e. As before, the proofs are based on the existence and uniqueness theorem for ODES and IMT. 8. Find the solution of the eikonal equation u2, + u; = 1 through the initial curve I? : x = coss, y = sins, u = 1, 0 5 s 5 27r. Solution. Functions po ( s ) and qo (s) such that Po" (4+ ao" (4= 1, duo- 0 = P O (s) (-sins) + qo (s) (coss) ds are po (s) = cos s and qo ( s ) = sin s. 39) is fulfilled First-order Partial Differential Equations Integrating the system 35 i x = 2p y = 24 u=2 p=0 q=0 with initial conditions x ( s , O ) = cos s , Y (0) = sin s, 1, u(s,O) = P(S,O) = cos s, q(s,O) = sins, we get + x = (2t 1)COSS y = (2t+I)sins u = (2t 1).
Exercises 1. ) which means that at each point P of I? the normal vectors n i i and n i i are linearly independent, then I' is a characteristic curve. 2. Solve the following initial value problems: (a) u, yu, = 2u, ~ ( 1 , s=) s. (b) U , + uY = u2, U ( S , 0) = s2. (c) xu, + (y + 2 ) U Y = u , U ( 2 ) s) = s - 4. + + 3. Show that the solution of the quasi-linear PDE uy a(u)u, = 0 with the initial condition u ( s , 0) = h ( s ) is given implicitly by u = h (z - a (u) y) . Show that the solution becomes singular for some positive y unless a ( h ( s ) )is a nondecreasing function.