220-HW1s - MATH 220 PROBLEM SET 1 SOLUTIONS DUE THURSDAY...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 220: PROBLEM SET 1, SOLUTIONS DUE THURSDAY, OCTOBER 1, 2009 Problem 1. Classify the following PDEs by degree of non-linearity (linear, semi- linear, quasilinear, fully nonlinear): (1) (cos x ) u x + u y = u 2 . (2) u u tt = u xx . (3) u x − e x u y = cos x. (4) u tt − u xx + e u u x = 0 . Solution. They are: (1) semilinear, (2) quasilinear, (3) linear, (4) semilinear. Problem 2. (1) Solve u x + (sin x ) u y = y, u (0 , y ) = 0 . (2) Sketch the projected characteristic curves for this PDE. Solution. The characteristic ODEs are dx ds = 1 , dy ds = sin x, dz ds = y. We first solve the x ODE, substitute the solution into the y ODE, and then sub- stitute the solution into the z ODE. So: x ( r, s ) = s + c 1 ( r ) dy ds = sin( s + c 1 ( r )) ⇒ y ( r, s ) = − cos( s + c 1 ( r )) + c 2 ( r ) dz ds = − cos ( s + c 1 ( r )) + c 2 ( r ) ⇒ z ( r, s ) = − sin( s + c 1 ( r )) + c 2 ( r ) s + c 3 ( r ) . The initial condition is that the characteristic curves go through { (0 , r, 0) : r arbitrary } at s = 0, i.e. that (0 , r, 0) = ( c 1 ( r ) , − cos( c 1 ( r )) + c 2 ( r ) , − sin( c 1 ( r )) + c 3 ( r )) . Thus, c 1 ( r ) = 0, − 1 + c 2 ( r ) = r . i.e. c 2 ( r ) = r + 1, and c 3 ( r ) = 0, so the solution of the characteristic ODEs satisfying the initial conditions is ( x, y, z ) = ( s, − cos s + r + 1 , − sin s + ( r + 1) s ) . We need to invert the map ( r, s ) mapsto→ ( x ( r, s ) , y ( r, s )), i.e. express ( r, s ) in terms of ( x, y ). This gives s = x , and r = y + cos s − 1 = y + cos x − 1. The solution of the PDE is thus u ( x, y ) = z ( r ( x, y ) , s ( x, y )) = − sin x + ( y + cos x ) x. The projected characteristic curves are the curves along which r is constant, i.e. they are y = − cos x + C , C a constant (namely r + 1). 1 2 MATH 220: PROBLEM SET 1, SOLUTIONS DUE THURSDAY, OCTOBER 1, 2009 Problem 3. (1) Solve yu x + xu y = 0 , u (0 , y ) = e − y 2 . (2) In which region is u uniquely determined? Solution. This is a homogeneous linear PDE with no first order term, so its so- lutions are functions which are constant along the projected characteristic curves, i.e. the integral curves of the vector field V ( x, y ) = ( y, x ). Note also that the initial curve, the y-axis, is characteristic at exactly one point, namely the origin, where V vanishes. Elsewhere along the y axis V (0 , y ) = ( y, 0) which is not tangent to the...
View Full Document

This note was uploaded on 06/17/2010 for the course MATH 220 at Stanford.

Page1 / 5

220-HW1s - MATH 220 PROBLEM SET 1 SOLUTIONS DUE THURSDAY...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online