220-HW1s

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

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

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

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