Chem Differential Eq HW Solutions Fall 2011 43

Chem Differential Eq HW Solutions Fall 2011 43 - k > 0....

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Section 3.6 Heat Conduction in Bars: Varying the Boundary Conditions 43 Solutions to Exercises 3.6 1. Since the bar is insulated and the temperature inside is constant, there is no exchange of heat, and so the temperature remains constant for all t> 0. Thus u ( x, t ) = 100 for all t> 0. This is also a consequence of (2), since in this case all the a n ’s are 0 except a 0 = 100. 5. Apply the separation of variables method as in Example 1; you will arrive at the following equations in X and T : X ±± - kX =0 ,X (0) = 0 ,X ± ( L )=0 T ± - kc 2 T =0 . We now show that the separation constant k has to be negative by ruling out the possibilities k = 0 and k> 0. If k = 0 then X ±± =0 X = ax + b . Use the initial conditions X (0) = 0 implies that b =0 , X ± ( L ) = 0 implies that a = 0. So X =0if k =0 . If k> 0, say k = μ 2 , where μ> 0, then X ±± - μ 2 X =0 X = c 1 cosh μx + c 2 sinh μx ; X (0) = 0 0= c 1 ; X = c 2 sinh μx ; X ± ( L )=0 0= c 2 μ cosh( μL ) c 2 =0 , because μ ± = 0 and cosh( μL ) ± = 0. So X
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Unformatted text preview: k > 0. This leaves the case k =-μ 2 , where μ > 0. In this case X ±± + μ 2 X = 0 ⇒ X = c 1 cos μx + c 2 sin μx ; X (0) = 0 ⇒ 0 = c 1 ; X = c 2 sin μx ; X ± ( L ) = 0 ⇒ 0 = c 2 μ cos( μL ) ⇒ c 2 = 0 or cos( μL ) = 0 . To avoid the trivial solution, we set cos( μL ) = 0, which implies that μ = (2 k + 1) π 2 L , k = 0 , 1 , . . . . Plugging this value of k in the equation for T , we Fnd T ± + μ 2 c 2 T = 0 ⇒ T ( t ) = B k e-μ 2 c 2 t = B k e-( (2 k +1) π 2 L ) 2 c 2 t . ±orming the product solutions and superposing them, we Fnd that u ( x, t ) = ∞ ± k =0 B k e-μ 2 c 2 t = ∞ ± k =0 B k e-( (2 k +1) π 2 L ) 2 c 2 t sin ² (2 k + 1) π 2 L x ³ . To determine the coefficients B k , we use the initial condition and proceed...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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