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u t = K 2 u x 2 , t > 0 , x > 0 u ( x , 0 ) = u I C u ( 0 , t ) = u B C

The PDE, IC and BC are made dimensionless with respect to reference quantities.

u * = u - u I C u o t * = t t o x * = x x o u * t * = K t o x o 2 2 u * x * 2 u * ( x * , 0 ) = 0 u * ( 0 , t * ) = u B C - u I C u o = 1 , u o = u B C - u I C

There are three unspecified reference quantities and two dimensionless groups. The BC can be specified to equal unity. However, the system does not have a characteristic time or length scales to specify the dimensionless group in the PDE. This suggests that the system is over specified and the independent variables can be combined to specify the dimensionless group in the PDE to equal 1/4.

K t o x o 2 = 1 4 η = x 4 K t is dimensionless u ( x , t ) = u ( η )

The partial derivatives will now be expressed as a function of the derivatives of the transformed similarity variable.

η x = 1 4 K t η t = - η 2 t u t = d u d η η t = - η 2 t d u d η u x = 1 4 K t d u d η 2 u x 2 = 1 4 K t d 2 u d η 2

The PDE is now transformed into an ODE with two boundary conditions.

d 2 u * d η 2 + 2 η d u * d η = 0 u * ( η = 0 ) = 1 u * ( η ) = 0
Let v = d u * d η d v d η + 2 η v = 0 d v v = d ln v = - 2 η d η v = C 1 e - η 2 u * = C 1 0 η e - η 2 d η + C 2 u * ( η = 0 ) = 1 C 2 = 1 u * ( η ) = 0 = C 1 0 e - η 2 d η + 1 C 1 = - 1 0 e - η 2 d η u * ( η ) = - 0 η e - η 2 d η 0 e - η 2 d η + 1 = erfc ( η ) u * x , t = erfc x 4 K t

Therefore, we have a solution in terms of the combined similarity variable that is a solution of the PDE, BC, and IC.

Complex potential for irrotational flow

Incompressible, irrotational flows in two dimensions can be easily solved in two dimensions by the process of conformal mapping in the complex plane. First we will review the kinematic conditions that lead to the PDE and boundary conditions. Because the flow is irrotational, the velocity is the gradient of a velocity potential. Because the flow is solenoidal, the velocity is also the curl of a vector potential. Because the flow is two dimensional, the vector potential has only one non-zero component that is identified as the stream function. The kinematic condition at solid boundaries is that the normal component of velocity is zero. No condition is placed on the tangential component of velocity at solid surfaces because the fluid must be inviscid in order to be irrotational.

v = ϕ = ϕ x , ϕ y , 0 = v x , v y , v z v = 0 = 2 ϕ v = × A = ( A 3 y , - A 3 x , 0 ) = ψ y , - ψ x , 0 = v x , v y , v z × v = w = 0 = 2 ψ v = ϕ = × A ϕ x , ϕ y , 0 = ψ y , - ψ x , 0 or ϕ x = ψ y and ϕ y = - ψ x

Both functions are a solution of the Laplace equation, i.e., they are harmonic and the last pair of equations corresponds to the Cauchy-Riemann conditions if ϕ and ψ are the real and imaginary conjugate parts of a complex function, w ( z ) .

w ( z ) = ϕ ( z ) + i ψ ( z ) z = x + i y = r e i θ = r cos θ + i sin θ , r = z = x 2 + y 2 1 / 2 , θ = arctan y / x or ϕ ( z ) = real w ( z ) ψ ( z ) = imaginary w ( z )

The Cauchy-Riemann conditions are the necessary and sufficient condition for the derivative of a complex function to exist at a point z o , i.e., for it to be analytical . The necessary condition can be illustrated by equating the derivative of w ( z ) taken along the real and imaginary axis.

d w ( z ) d z = Re w ' ( z ) + i Im w ' ( z ) = lim δ ϕ + i δ ψ δ x + i 0 = ϕ x + i ψ x = lim δ ϕ + i δ ψ 0 + i δ y i i = ψ y - i ϕ y ϕ x = ψ y and ψ x = - ϕ y , Q . E . D .

Also, if the functions have second derivatives, the Cauchy-Riemann conditions imply that each function satisfies the Laplace equation.

Questions & Answers

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Source:  OpenStax, Transport phenomena. OpenStax CNX. May 24, 2010 Download for free at http://cnx.org/content/col11205/1.1
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