m344lecture14

m344lecture14 - M344 - ADVANCED ENGINEERING MATHEMATICS...

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Lecture 14: Other Boundary Conditions for the Heat Equation We have seen that the one-dimensional heat equation can be used to model the temperature in a rod which is insulated around its sides. The first case we solved assumed that both ends of the rod were held at 0 0 for all time t > 0. Other boundary conditions can be modelled, and we will treat two different cases. Heat Equation with both ends insulated If the ends of the rod are insulated , the boundary conditions on the partial differential equation u t = α 2 u xx must be changed to u x (0 ,t ) = u x ( L,t ) = 0 for all t > 0. This implies that no heat can flow across either end of the rod. The Method of Separation of Variables proceeds exactly as before, except that the boundary conditions on the Sturm-Liouville problem X 00 + λX = 0 must be changed. With u ( x,t ) = X ( x ) T ( t ), the condition u x (0 ,t ) X 0 (0) T ( t ) = 0, for all t > 0 implies that X 0 (0) = 0, and similarly the condition u x ( L,t ) = 0 implies that X 0 ( L ) = 0. To solve the Sturm-Liouville problem X 00 + λX = 0 with X 0 (0) = X 0 ( L ) = 0, we again need to treat the three cases λ < 0 = 0 ,λ > 0. 1. If λ < 0, let λ = - K 2 . Then the general solution can be written as X ( x ) = A cosh( Kx ) + B sinh( Kx ) , X 0 ( x ) = AK sinh( Kx ) + BK cosh( Kx ) . Using the boundary conditions, X 0 (0) = AK sinh(0) + BK cosh(0) = 0 + BK = 0 B = 0 , and X 0 ( L ) = AK sinh( KL ) = 0 A = 0 , since neither K nor L can be zero. This means that X ( x ) 0 is the only solution, and therefore there are no negative eigenvalues. 2. If λ = 0, then the general solution is X ( x ) = C + Dx , and X 0 ( x ) = D . Both conditions X 0 (0) = X 0 ( L ) = 0 are satisfied if, and only if, the constant D = 0. This means that X ( x ) C , where C is a non-zero constant, is a non-zero solution corresponding to the eigenvalue λ 0 = 0. 1
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m344lecture14 - M344 - ADVANCED ENGINEERING MATHEMATICS...

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