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Unformatted text preview: ACM 100c  Methods of Applied Mathematics Week 3 April 11, 2009 1 Examples of the method of finite transforms As an example of the finite transform and poison tooth approach approaches described last week above consider the diffusion equation φ t = φ xx + 1 < x < L with boundary conditions φ (0 , t ) = 0 φ ( L, t ) = 1 and initial condition φ ( x, 0) = 0 < x < L We’ll solve this problem using both the approaches discussed above. First we’ll use the “poison tooth” approach. We let φ ( x, t ) = ψ ( x, t ) + x/L This gives us a heat equation for ψ of the form ψ t = ψ xx + 1 The boundary conditions for ψ are homogeneous but the initial condition for ψ is now ψ ( x, 0) = − x/L. We can now use the formula derived earlier for the solution of such problems ψ ( x, t ) = ∞ summationdisplay n =1 α n ( x, t ) sin( nπx/L ) where the coefficients α n ( t ) are given by α n ( t ) = C n exp parenleftbigg − n 2 π 2 t L 2 parenrightbigg + exp parenleftbigg − n 2 π 2 t L 2 parenrightbiggintegraldisplay t exp parenleftbigg n 2 π 2 τ L 2 parenrightbigg w n ( τ ) dτ 1 The coefficients C n are gotten from C n = 2 L integraldisplay L ( − x/L ) sin( xπx/L ) dx and the w n were computed in a previous section. We therefore have α n ( t ) = C n exp parenleftbigg − n 2 π 2 t L 2 parenrightbigg + 2 L 2 n 3 π 3 (1 − cos( nπ )) bracketleftbigg 1 − exp parenleftbigg − n 2 π 2 t L 2 parenrightbiggbracketrightbigg Finally, we use this to get the answer for φ φ ( x, t ) = x/L + ψ ( x, t ) We note here that we could have gone a bit further by using the equilibrium solution for this problem. Recall the equilibrium solution is given by 0 = ∂ 2 φ Eq ∂x 2 + 1 with boundary conditions of φ Eq (0) = 0 , φ Eq ( L ) = 1 We can solve this boundary value problem directly to get φ Eq ( x ) = − x 2 / 2 + (2 + L 2 ) 2 L x If we use this equilibrium solution and let φ = ψ + φ Eq , then we see that the equation for ψ is now the heat equation with no source term and homogeneous boundary conditions but with the initial condition ψ ( x, 0) = − φ Eq . The solution here is simply going to be of the form ψ ( x, t ) = ∞ summationdisplay n =1 c n exp bracketleftbigg − n 2 π 2 t L 2 bracketrightbigg sin( nπx/L ) , where the c n get determined from the initial condition. The total solution is now φ ( x, t ) = φ Eq ( x ) + ∞ summationdisplay n =1 c n exp bracketleftbigg − n 2 π 2 t L 2 bracketrightbigg sin( nπx/L ) and we see that in the long time limit the solution approached the equilibrium solution as we suspected earlier but have now verified. 2 To use the method of finite transforms we transform our heat equation by multiplying all terms by (2 /L ) sin( nπx/L ) and integrating from to L : 2 /L integraldisplay L φ t sin( nπx/L ) dx = 2 /L integraldisplay L φ xx sin( nπx/L ) dx + 2 /L integraldisplay L sin( nπx/L ) dx We next integrate the second integral twice by parts, and do the third integral explicitly. WeWe next integrate the second integral twice by parts, and do the third integral explicitly....
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 Spring '09
 NilesA.Pierce

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