Chapter_4_Solutions

Chapter_4_Solutions - CHAPTER 1 Partial Differential...

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Unformatted text preview: CHAPTER 1 Partial Differential Equations on Bounded Domains 1. Separation of Variables Exercise 1. The solution is u ( x,t ) = summationdisplay n =1 a n e n 2 t sin nx where a n = 2 integraldisplay / 2 sin nxdx = 2 n (( 1) n cos( n/ 2)) Thus u ( x,t ) = 2 e t sin x 2 e 4 t sin 2 x + 2 3 e 9 t sin3 x + 2 5 e 25 t sin 5 x + Exercise 2. The solution is given by formula (4.14) in the text, where the coeffi- cients are given by (4.15) and (4.16). Since G ( x ) = 0 we have c n = 0. Then d n = 2 integraldisplay / 2 x sin nx dx + 2 integraldisplay / 2 ( x )sin nx dx Using the antiderivative formula integraltext x sin nx dx = (1 /n 2 )sin nx ( x/n )cos nx we integrate to get d n = 4 n 2 sin n 2 Exercise 3. Substituting u ( x,y ) = ( x ) ( y ) we obtain the Sturm-Liouville prob- lem = , x (0 ,l ); (0) = ( l ) = 0 and the differential equation = 0 The SLP has eigenvalues and eigenfunctions n = n 2 2 /l 2 , n ( x ) = sin( nx/l ) and the solution to the equation is n ( y ) = a n cosh( ny/l ) + b n sinh( ny/l ) 1 2 1. PARTIAL DIFFERENTIAL EQUATIONS ON BOUNDED DOMAINS Therefore u ( x,y ) = summationdisplay n =1 ( a n cosh( ny/l ) + b n sinh( ny/l ))sin( nx/l ) Now we apply the boundary conditions: u ( x, 0) = F ( x ) = summationdisplay n =1 a n sin( nx/l ) and u ( x, 1) = G ( x ) = summationdisplay n =1 ( a n cosh( n/l ) + b n sinh( n/l ))sin( nx/l ) Thus a n = 2 l integraldisplay F ( x )sin( nx/l ) dx and a n cosh( n/l ) + b n sinh( n/l ) = 2 l integraldisplay G ( x )sin( nx/l ) dx which gives the coefficients a n and b n . Exercise 4. Substituting u = y ( x ) g ( t ) into the PDE and boundary conditions gives the SLP y = y, y (0) = y (1) = 0 and, for g , the equation g + kg + c 2 g = 0 The SLP has eigenvalues and eigenfunctions n = n 2 2 , y n ( x ) = sin nx, n = 1 , 2 ,... The g equation is a linear equation with constant coefficients; the characteristic equations is m 2 + km + c 2 = 0 which has roots m = 1 2 ( k radicalbig k 2 4 c 2 n 2 2 ) By assumption k < 2 c , and therefore the roots are complex for all n . Thus the solution to the equation is (see the Appendix in the text on ordinary differential equations) g n ( t ) = e kt ( a n cos( m n t ) + b n sin( m n t )) where m n = 1 2 radicalbig 4 c 2 n 2 2 k 2 ) Then we form the linear combination u ( x,t ) = summationdisplay n =1 e kt ( a n cos( m n t ) + b n sin( m n t ))sin( nx ) Now apply the initial conditions. We have u ( x, 0) = f ( x ) = summationdisplay n =1 a n sin( nx ) 2. FLUX AND RADIATION CONDITIONS 3 and thus a n = integraldisplay 1 f ( x )sin nx dx The initial condition u t = 0 at t = 0 yields u t ( x, 0) = 0 = summationdisplay n =1 ( b n m n ka n )sin( nx ) Therefore b n m n ka n = 0 or...
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Chapter_4_Solutions - CHAPTER 1 Partial Differential...

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