The reasoning behind the steps to form the partial different

The reasoning behind the steps to form the partial differential equation is important for me to understand what I\'m doing, my intuition leads me to believe i need to use some derivation of the heat equation/diffusion equation, any insight would be very much appreciated, thank you

Imagine a pipe sealed at one end with the other end going to infinity. The pipe is full of stagnant, not moving, water. We desire an equation of concentration of a chemical in the water. At the start there is constant concentralion of 1 Karl per million of e chemical in Lhe section between 1 meter and 2 meters from the sealed end with concentratian e to change aver time. Assume all concentrations in parts per million, all distances in meters, and all time in secands. The chemical diffuses in Lhe water with a diffusion coefficient of 0.5 a) Considering one spalial dimension, wha is the inilial boundary value problem represenling the concentration of the chemical in the water. What are the domains at the independent variables? Solve the initial problem showing and justifying the step by step process. How do the boundary conditions affect yo solution method? ur c) What does the concentration of the chemical look like as a tunction of time 10 meters tram the sealed end initial time until 500 seconds after

Solution

You are correct. The mass diffusion equation is the same as the thermal diffusion equation and derived similarly.

One can do this by using dimensional analysis, and this is found in seral text books.

A more ab initio approach is as follows:

Concentration C = M/dxdydx

C1-C2 is the difference between concentrations at two locationsx1 and x2 in a tube ( we drop the y and z later)

Let the probability that matter moves from left to right be kM1, and from right to left -kM2 , then qx= K( M1-M2)

dC/dx = (C1-C2)/(x1-x2) = (M1 -M2)/[dx(x1-x2)] where C1 = M1/dx

put dydz=1 for 1 D case.

Then (M1-M2) = dx(x1-x2) dC/dx = qx/k

for x1-x2 small we have qx = kdx²dC/dx

The RHS depends on k and x. Since q should be independent of the position, kdx² is constant caled the diffuion coefficient

This is the reasoning for the 1st law of mass diffusion (analogous to thermal diffusion)

Units of D =[L][.L]/[t]

For the dependence on time: dM/dt = m\' in -m\' out

qx,in = D dc/dx| ₁

qx,out = D dc/dx|₂

Total mass flux = D ( dc/dx|₁- dc/dx|₂)

By Taylors expansion dC/dx |₂ = dC/dx|₁ +d/dx( dC/dx|₁)dx + second order terms)

dM/dt|_x = D d²C/dx² dx ( putting dydz =1 as before)

since M = c.1 one obtains the time varying 2nd law dC/dt = D d²c/dx²

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Initial BVP, conditions

C=0 x=0 to 1, C=0 x=2 to infinity

C=10^-6 for x= 1 to 2 ( one can use a Heaviside function for this, superimposing two of them to give the pillar at 1 and 2)

Domains of the ind variables x and t: x from 0 to infinity, t from 0 to infinity

b) The solution needs a Laplace transform method to deal with the Heaviside function, or a Kernal integral approach.

An approximate method would be to look at two problems

one for the LHS from x =0 to 1, and one for the RHS from x=2 to infinity

You have to deal with the fact that the region with the tracer does not actual remain constant at a unit length.

I would consider this as a thrid probelm one where the unit length diffuses out into zero concentration with time.

For linear problems one can superpose, but this is not a Linear problem.

You may have to solve it on the computer. But have fun buddy.!