How to transform diffusion PDE into an ODE via Laplace transform. Two examples (arrangements of boundary conditions) are supplied.Descripción completa...
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Diffusion in Solid Metals and Alloys
Descripción: examen de instalaciones eléctricas
Descrição: Transformador Relação de Transformação Auto transformador funcionamento modelo elétrico impedância
2 2 −1 − pt −1 (∇ C − D Cɺ )e dt = 0 = ∇ L{C} − D −1 − pt ∞ Cɶ − D ( Ce |
0
)
− pL{−C} = ∇
2
∫
∞
0
ɺ Ce
− pt
dt
(1.2)
2 ɶ −1 ɶ ( x, p) Cɶ − D pCɶ → D∇ C ( x, p) = pC
The laplace transform ate up the time-derivative. Now we have an ODE. co ndition C (0, t ) = C 0 , and a Example: find the concentration profile in three dimensions for boundary condition constant-diffusivity-coefficient. Laplace transform the boundary condition as, L{C (0, t )} = L{C0 } → Cɶ (0) =
∫
∞
0
C0 e − pt dt = − (C0 / p )e− pt |∞t =0 = C0 / p → Cɶ (0) = C0 / p
(1.3)
Example: consider going to 1D, and consider having the same boundary condition C (0, t ) = C 0 , whose Laplace
transform was (1.3). Then: pCɶ ( x, p ) of (1.2) constitutes a homogeneous ODE with constant coefficients, the 1
ɶ (0) , solution to which is a dying-exponential with amplitude A± = C0 / p = C ∇
2
Cɶ =
→ C ( x, t ) =
d 2 dx2
+ Cɶ = ( p / D)Cɶ → Cɶ ( x, p) = A+ e
C 0 p
−1
L {e
± p / Dx
} = C0 erfc(
p / Dx
−
+ A− e
p / Dx
Cɶ ( 0 ) = C0 / p
=
x
) ; erfc z ≡ 1 − erf z = 1 − 4 Dt
The inverse laplace transform that was needed is L−1{ p
− ( 12 n +1)
e
−a
p
}≡
±
p / Dx
A± e
2 π
z
∫
0
2
− z′
e
(1.4) dz′;
4 n t 3n ⋅ erfc( 12 a / t ) , which requires
positive-frequency a ≥ 0 and n ≥ 1 . Plotting, CHx,tL 2.0
1.5
t
=
0.0001
t
=
0.01
t
=
1
1.0
0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
x; D
1 C0
=
=
(1.5)
1
Inhomogeneities Inhomogeneities are amenable to solution-methods described in EM 04 - 035 - greens theorem and BVPs in electrostatics, an (elementary) example of which is found in EM 04 - 094 - pr 35 - green function for potential of spherical conductor held at V0.
