Ch_06 - PDE - Exam_Soln - Exam Problems of Partial...

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Exam Problems of Partial Differential Equations
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Partial Differential Equations Combination of Variables Combination of Variables 1. (’05) (’06) Solve the following PDE . 2 2 uu t x α = ( 0 > ) I.C. 0 (,0 ) ux u = B.C.’s 1 0 (0, ) (,) ut u = ∞= Solution: Use the combination of variables, let 0 Θ= − , then the given PDE becomes 2 2 t x Θ∂ Θ = ( 0 > ) I.C. ) 0 x Θ = B.C.’s 01 0 (0, ) 0 tu u t Θ =Θ = Θ∞ = And again let 0 ( ) φη ΘΘ == Θ− and 4 x t η = then, the I.C and B.C’s become 0 ( ) 0 ( ) 0 0 0 (0) 1 t x x φ ηφ = =∞ ∞ = = The partial derivative terms are converted as 0 22 00 1 2 4 4 d tt d dd xd t t ηα ∂Θ =− Θ ∂Θ ∂ Θ =⇒ = ∂∂ Then the given PDE is converted to an ODE as 2 0 0 2 1 24 td t d φφ Θ −Θ = or 2 2 2 0 d d += and ( ) 0 0 (0) 1 = ∞∞ = = =
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Partial Differential Equations Combination of Variables let d G d φ η = , then 2 0 dG G d += or dG G d =− 2 1 ln ln Gc + 2 1 d e d == Integrate above with B.C. 1 2 1 0 dc e d φη = ∫∫ 2 1 ced = Using B.C.2 2 0 1 1 = 2 1 0 1 c ed = Then 2 22 2 2 2 2 2 00 0 0 0 0 ( ) 2 ( ) ( ) erf erfc ηη π −− ∞∞ === = where the error function () erf is defined as 2 0 2 erf e d = and 2 0 2 1 erf e d Finally, 0 1 1 ( ) 4 x erf erf t α Θ = Θ 0 10 1 44 uu xx erf erfc tt αα =
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Partial Differential Equations Combination of Variables 2. Combination of Variables for 1-D Diffusion Equation with Convection An energy or mass balance on a fluid running down a wall ( z -distance down the wall, x - distance into the fluid from the wall) leads to the following equation for the temperature or concentration in the fluid. 2 2 uu x z x = D ( 0 z <∞ , 0 > D ) B.C’s: 0 ( ,0) 0 for all 0 (0, ) for all 0 (,) 0 ux z uz u x uz = > = > ∞= where D is the corresponding diffusivity. Find the general solution of this equation. z First Step: Combination of variables. z Second Step: Solve ODE. (1) First Step: Combination of Variables Let () 0 u f xg z u φη == and n x z η = The partial derivative terms are converted as 1 22 2 2 2 n n nnn un d nxz zz z d u z xx ud z x d φ ηφ ηη ∂∂ = ⎛⎞ = ⎜⎟ ∂∂∂ ⎝⎠ Then, the given partial differential equation becomes 2 1 nn d n d x d d + and 2 n z xd = DD 1 3 1 n dd d nz z z d φφ ++ Let’s take 1/3 n =− in order to make the z -term dimensionless, then 1/3 x z = and 2 0 3 d d += D with new B.C’s of 0 ( ) 0 ( ) 0 0 0 (0) 1 z x x = =∞ ∞ = =
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Partial Differential Equations Combination of Variables (2) Second Step: Solve ODE. Let / Gd d φ η = , then 2 0 3 dG G d += D or 2 3 dG G d =− D 2 3 dG d G D 3 1 ln ln 9 Gc + D 3 /9 1 d GC e d == D Integrate over to with () 0 = 3 1 () ( ) dC e d ηη φφ ηφ ∞∞ = = ∫∫ D Apply B.C. (0) 1 = : 3 0 1 (0) 1 Ce d φη D 33 1 0 0 11 C ed −− DD NOTE: Gamma function 1 0 sx x es d s Γ= and (1 ) ( ) x xx Γ +=Γ Let 3 s = D , then 3 9 s = D and 2/3 3 1 9 3 ds d s = D , then 3 3 1 00 9 9 (1/3) 9 (4/3) s Ce d e s d s = Γ = Γ D D Finally, 3 3 0 ( ) 9 (4/3) 9 (4/3) u u = ΓΓ D D ( 1/3 x z = )
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This note was uploaded on 04/27/2011 for the course CHEM 101 taught by Professor Suh during the Spring '11 term at University of Toronto- Toronto.

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Ch_06 - PDE - Exam_Soln - Exam Problems of Partial...

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