Solution of the Diffusion Equation

SOLUTION OF THE DIFFUSION EQUATION
Question 1
∂C
∂ t = D ∂2 C
∂ x2
The mathematical problem being solved concerns the distribution of the pollutant in the pipe as a
function of time and space with the initial conditions such as the concentration of the pollutant
being specified.
Solution of C(x,t)
Let C(x,t) = M(x)N(t) , differentiating C(x,t) with respect to t and x respectively we get,
∂C
∂ t = M dN
dt and ∂2 C
∂ x2 = M d2 M
d x2
We have M dN
dt = D d2 M
d x2 N hence dN
dt /DN = d2 M
d x2 /M = k
We obtain two ordinary differential equations from above,
d2 M
d x2 – kM = 0 and dN
dt – kDN = 0
Next we solve for M,
With k>0, the general solution is M(x) = Ceμx+ Deμx
Applying the boundary conditions, C+D = 0, Aeμx + Beμx =0 and therefore A = B = 0
With k = -p2
M(x) = C cos(px) + D sin(px) and the conditions at the boundary give,
M(0) = A = 0, M(L) = Dsin(pL) = 0 hence sin(pL) = 0 therefore Mn(x) = sin( nπ
L )x
Next we solve for N,
We have p = nπ
L and define λn = n Dπ
L , hence dN
dt + ¿λn2N = 0
The general solution is then given by: Nn (t) = Bne− λn 2
The solution to the one dimensional diffusion equation is thus: C(x,t) = ∑
n =1
∞
Cn ( x , t )= ∑
n =1
∞
Bn e−λ n 2
sin( nπ
L )x

The initial condition is C(x,0) ¿ ∑
n=1
∞
Bn sin( nπ
L )x = C(x), Bn = 2
L∫
0
L
C ( x ) sin( nπ
L )x dx
Question 2
From the equation, ∂C
∂ t = D ∂2 C
∂ x2 with C(x,0) = 0, C(0,t) = Co
Converting the diffusion equation into frequency domain yields;
∂ F (x , s)
∂ x = D ∂2 F ( x , s)
∂ x2 and F(0,s) = C o
s
F(x,s) = C o
s exp (-√s / Dx)
From inverse Laplace transform tables, L[ erfc ( a
2 √ t ) ] = exp ¿ ¿
Therefore, C(x,t) = Co erfc ( x
2 √Dt )
Question 3

Figure 1: A plot of C(x,t) as a function of x for t = 10,20,30,40,50,60,70,80,90,100.
The value of the diffusivity coefficient, D affects the speed of the concentration change.
Changing the value of D has a similar effect as changing time t. Thus changing D affects the rate
at which the concentration falls as distance from the initial location increase. If D increases,
initially the concentration reduces rapidly and the gradient of the curves is steep. On the other
hand, a low value of D means a low rate of diffusion hence the concentration changes slowly.
Question 4
The mathematical problem to be solved for C(x,t) revolves around the fact that the original
equation of diffusion was derived for a semi-infinite pipe which is an experiment that cannot be
carried out practically. The original derivation assumes that the concentration only varies in the
x-direction. Practically, this is not possible. This means that the results obtained vary from the
ones predicted by the ideal set up. Furthermore, the diffusivity constant is not really constant but
depends on the type of solute. D is only constant for dilute solutions.