# hw1sol - Math 220a - Hw1 solutions (Revised) 1. Linear -...

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Math 220a - Hw1 solutions (Revised) 1. Linear - (b) Semilinear - (a),(d) Quasilinear -(e) Fully nonlinear - (c) 2. The equations for the characteristic curves are dt ( r,s ) ds = 1 dx ( r,s ) ds = x du ( r,s ) ds = t 3 The boundary conditions can be re-written in terms of ( s,r ) as t ( r, 0) = 0 x ( r, 0) = r u ( r, 0) = φ ( r ) Integrating and using the boundary conditions, t = s + c 1 ( r ) = s Since t ( r, 0) = 0 x = c 3 ( r ) e t = re t Since x(r,0)=r u = s 4 / 4 + c 2 ( r ) = s 4 4 + φ ( r ) Since u ( r, 0) = φ ( r ) Eliminating r,s form the above equations we get, u ( x,t ) = t 4 4 + φ ( xe - t ) 3. The equations for the characteristic curves and boundary conditions in terms of r,s are dt ( r,s ) ds = 1 dx ( r,s ) ds = x du ( r,s ) ds = u 3 t ( r, 0) = 0 x ( r, 0) = r u ( r, 0) = sin( r ) 1

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Integration together with the boundary conditions yield, t = s x = re - t u - 3 du = ds (i.e.) u - 2 - 2 = s + C ( r ) with C ( r ) given by, C ( r ) = - 1 2 sin 2 ( r ) Eliminating r,s form the above equations we get,
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## This note was uploaded on 04/30/2011 for the course MECHANICAL MSC taught by Professor Dr.rahula during the Spring '10 term at University of Moratuwa.

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hw1sol - Math 220a - Hw1 solutions (Revised) 1. Linear -...

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