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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 π integraldisplay π π/ 2 sin nxdx = − 2 nπ (( − 1) n − cos( nπ/ 2)) Thus u ( x,t ) = 2 π e − t sin x − 2 π e − 4 t sin 2 x + 2 3 π e − 9 t sin3 x + 2 5 π 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 π integraldisplay π/ 2 x sin nx dx + 2 π integraldisplay π π/ 2 ( π − 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 πn 2 sin nπ 2 Exercise 3. Substituting u ( x,y ) = φ ( x ) ψ ( y ) we obtain the Sturm-Liouville prob- lem − φ ′′ = λφ, x ∈ (0 ,l ); φ (0) = φ ( l ) = 0 and the differential equation ψ ′′ − λψ = 0 The SLP has eigenvalues and eigenfunctions λ n = n 2 π 2 /l 2 , φ n ( x ) = sin( nπx/l ) and the solution to the ψ –equation is ψ n ( y ) = a n cosh( nπy/l ) + b n sinh( nπy/l ) 1 2 1. PARTIAL DIFFERENTIAL EQUATIONS ON BOUNDED DOMAINS Therefore u ( x,y ) = ∞ summationdisplay n =1 ( a n cosh( nπy/l ) + b n sinh( nπy/l ))sin( nπx/l ) Now we apply the boundary conditions: u ( x, 0) = F ( x ) = ∞ summationdisplay n =1 a n sin( nπx/l ) and u ( x, 1) = G ( x ) = ∞ summationdisplay n =1 ( a n cosh( nπ/l ) + b n sinh( nπ/l ))sin( nπx/l ) Thus a n = 2 l integraldisplay π F ( x )sin( nπx/l ) dx and a n cosh( nπ/l ) + b n sinh( nπ/l ) = 2 l integraldisplay π G ( x )sin( nπx/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 ′′ = λy, y (0) = y (1) = 0 and, for g , the equation g ′′ + kg ′ + c 2 λg = 0 The SLP has eigenvalues and eigenfunctions λ n = n 2 π 2 , y n ( x ) = sin nπx, n = 1 , 2 ,... The g equation is a linear equation with constant coefficients; the characteristic equations is m 2 + km + c 2 λ = 0 which has roots m = 1 2 ( − k ± radicalbig k 2 − 4 c 2 n 2 π 2 ) By assumption k < 2 πc , 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 π 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( nπx ) Now apply the initial conditions. We have u ( x, 0) = f ( x ) = ∞ summationdisplay n =1 a n sin( nπx ) 2. FLUX AND RADIATION CONDITIONS 3 and thus a n = integraldisplay 1 f ( x )sin nπx 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( nπx ) Therefore b n m n − ka n = 0 or...

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