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Hydrology Program Quantitative Methods in Hydrology
Basic solution to Heat Diffusion In general, one-dimensional heat diffusion in a material is defined by the linear, parabolic PDE
2 0
2
∂
x
T
D
t
T
or Tt = D Txx (41)
where we assume that T is defined on the domain of interest. Substitution (your homework) will show that that
Dt
x t
A
T x t 4
( , ) exp
2 1 / 2 (42a)
is a general solution of this equation. (See p. 11-12 of Crank, 1975.)
What is the total heat content in the domain at any time t? The heat content at a point is given by the product of specific heat c [J kg-1 °C -1^ ], the material density ρ [kg m-3^ ], and the temperature T [°C], or ρcT. The total heat content depends on the size of the domain. If we assume that the domain is infinite (this is easier than dealing with a finite domain, for reasons that will become clearer later), then the total heat content will be given by the integral^16
H cTdx c Tdx cA D d cA D ρ cA π D
= (^) ∫ ρ =ρ ∫ =ρ ∫ −ζ ζ =ρ =
∞ −∞
∞ −∞
∞ −∞ 4 exp( )^44
where we’ve assumed that specific heat and the density are constants, and where we’ve used the
change of variables, ζ 2 = x^2 / 4 Dt. We can solve for A , which is a constant since heat is conserved
in this model (no source or sink in the heat diffusion equation).
D
H c A
Consequently, we can now write out the general solution as
Dt
x c Dt
H
T x t 4
exp 4
2
ρ (^) π
(42b)
But what are there boundary and initial conditions that correspond to this solution?
What happens at t =0? That is, what is the initial condition for the solution in (42) applied over an infinite domain?
{ { ( ) 4
exp 4
lim 4
exp
( , 0 ) lim
2
0
2 1 / 2 0
x c
H
Dt
x c Dt
H
Dt
x t
T x A t t
→ →
where we’ve introduced a new variable and concept, the Dirac delta function, δ.
The Dirac delta funciton is not the typical function you study in freshman calculus. It has strange properties. For example, it is not differentiable using the rules with which you are familiar. How is it defined? One way of defining it is simply the second limit in (45). We can augment this with the following additional properties.
δ ( x )= 0 for all x≠ 0 (46a)
(^16) One would integrate over finite or semi-infinite domains similarly, by changing the limits of integration.
Hydrology Program Quantitative Methods in Hydrology
∞ − ∞
δ ( x )= 1 (46b)
That is, the delta function integrates to one and is zero except at the location where the argument is zero. It’s a “spike” or “pulse.” δ ( x ) has units of [L -1^ ], that is, one over the units of the argument^17. The delta function is a very useful representation of a point (narrow) input pulse, in this case of heat.
Should the initial temperature not be zero other than at the delta function input, superposition can be used to represent that effect. For uniform initial temperature one would simply add (42) to that initial temperature.
What are the boundary conditions for this problem? Basically, you can see from the solution that
{ {lim {^ }^0
exp
( , ) lim
(^22) 1 / 2 = =
→±∞ →±∞
ζ ρ (^) π ζ
e c Dt
H
Dt
x t
T t A x
That is, the boundaries are at x = ±∞, and they are homogeneous Dirichlet boundaries. Should the Dirichlet boundaries be non-homogeneous, then superposition of solutions can be used to represent the sum of the delta input and non-zero Dirichlet boundaries.
Other example applications Another example application of the delta function would be a narrow pulse of solute mass in a fluid where the solute redistributes by diffusion. The PDE is Ct = D Cxx , with state variable concentration, C ( x,t ), and solute diffusion coefficient, D. Then A=M/√ 4 πD , M is the total solute mass, the initial condition is C ( x, 0) = M δ (x), the Dirichlet boundaries are C (±∞ ,t ) = 0, and the solution is C ( x,t ) = ( M/√ 4 πDt ) exp( -x^2 /4Dt ).
A third example application of the delta function would be a narrow pulse of water injected (recharged) into a “vertically integrated phreatic aquifer” where the water redistributes by so-called hydraulic diffusion following Darcy’s Law. The PDE is Ht = D Hxx , where the state variable is the rise of water table elevation, H ( x,t ), and D is the hydraulic diffusivity, D=T/Sy , with T = transmissivity and Sy = specific yield. Then A= V/ Sy√ 4 πD , V is the total water volume injected, the initial condition is H ( x, 0) = ( V/ Sy ) δ (x), the Dirichlet boundaries are H (±∞ ,t ) = 0, and the solution is H ( x,t ) = ( V/ S (^) y√ 4 πDt ) exp( -x^2 /4Dt ).
(^17) δ ( t ) has units of one over time, while the triple product δ ( x ) δ ( y ) δ ( t ), for a Dirac delta function in two space
dimensions and one time dimension, has SI dimensions s-1m-^.
x
δ(x)
Hydrology Program Quantitative Methods in Hydrology
This superposition concept can be extended to any number of spatially and time varying inputs.
Semi-infinite domain Suppose the injection of heat were to occur at the origin of a semi-infinite domain, x ≥ 0, such that the heat has to diffuse only toward positive values of x. Then we can assume the solution above for negative x is simply reflected about the origin and superposed, doubling the temperature for positive x from that give above (see Crank, 1975, p. 13). Then the general solution (42) becomes
Dt
x t
A
T x t 4
exp
2 1 / 2 (51)
or, for heat diffusion,
Dt
x c Dt
H
T x t 4
exp
2
The boundary condition at x= ∞ remains a homogeneous Dirichlet condition T (∞ ,t ) = 0, while due to the reflection at the origin the boundary condition at x =0 is a Neumann zero gradient condition,
x
T
at x =0, t ≥ 0 (53)
As ∂T/∂x = 0 at x =0 in the original solution (42), it “is clearly still zero after reflection and superposition.”
Integral of basic solution: step change in initial temperature Suppose that the initial condition were a step function, of the form
T=T 0 for x < 0 and T =0 for x > 0 (54)
An example, would be bring two metal bars of similar composition, each at a different temperature, into contact at their ends. We can use superposition of the infinite domain general solution to represent this process. Consider the “extended distribution to be composed to an infinite number of line sources” and superpose “the corresponding infinite number of elementary solutions” (Crank, 1975, p. 13-14). The heat source in some infinitesimal increment dx at location x 0 then has strength H=ρcT 0 dx. Then the temperature at some point x , a distance | x-x 0 | away, due to this increment of heat is
Dt
x x Dt
T dx 4
exp 4
2 0 0
The complete solution due to the initial condition in (54) is then determined by summing over the increments dx , or
Hydrology Program Quantitative Methods in Hydrology
= (^) ∫ ∫
∞ ∞ (^) −
Dt
x T erfc Dt
x erfc
T
e d Dt
T Dt dx Dt
x x Dt
T
T x t x x Dt
exp 4
0
0
/ 4
0
2 0 0 2
ζ
(56)
where ζ 2 =( x − x 0 )^2 / 4 Dt and we’ve used the definition of the error function and its complement,
∫ x = xe − d 0
erf ζ
ζ (^) z e d z z^1 erf
erfc
2 = (^) ∫ = −
∞ (^) −
ζ
erf (- z ) = -erf z , erf(0) = 0, erf(∞) = 1
What boundary conditions does this satisfy? From the properties of the error function we find that
T=T 0 as x → - ∞ and T =0 as x → +∞ (57)
Thus (56) is a solution of the heat diffusion equation with boundary and initial conditions that lead to its common application in hydrology and the geosciences.
Another example applications of the step change. Another example application would be a step change in initial solute concentration where the solute redistributes by diffusion. The PDE is Ct = D Cxx , with state variable concentration, C ( x,t ), and solute diffusion coefficient, D. The initial condition is C ( x <0 , 0) = C 0 and C ( x >0 , 0) = 0, the Dirichlet boundaries are C (-∞ ,t ) = C 0 and C (+∞ ,t ) = 0, and the solution is C ( x,t ) = 0.5 C 0 erfc(x/ √ 4 πD ).
Influence of mean flow. Suppose this heat or solute diffusion is taking place in the presence of a mean flow with velocity v. The PDE for heat advection-diffusion becomes
2
∂
x
T
D
x
T
v t
T
or Tt +vTx = D Txx (58)
In an infinite domain, with boundary and initial conditions as described above for the pulse input or step change, we can solve this equation using a transformed version of the solutions already produced. The basic idea is to transform (58) so that it looks like the diffusion equation (41). We use the characteristic of the advection equation, ∂ T ∂ t + v ∂ T ∂ x = 0 , (i.e., the secondary of (58)
and a chain rule operation. If v is constant then the characteristic is ( x − x 0 )− v ( t − t 0 )= 0 , or
x − vt =constant, suggesting a change of variables, to a moving coordinate system, x* , that follows a characteristic, or x*^ = x – vt (59)
Then we apply chain rules to implement this new coordinate system as follows,
Hydrology Program Quantitative Methods in Hydrology
Note that for the step change problem T (0, t ) = 0.5 T 0 , for t > 0. The step smears over time and, unlike the diffusion problem, the concentration at the origin changes. It is not a boundary condition.
Transform Solutions to Heat Diffusion (see Crank, §2.2) It can be shown by dimensional analysis that solutions to (30) are often of the form
Dt
x t
A
T x t 4
2 1 / 2 (68)
where Φ is a function to be determined 19. In particular, we’ve already found Gaussian and error function solutions. Notice that the function is self-similar. That is, the function is stretched one way or another, but the shape remains unchanged over time-space. You’ve seen such similarity solutions before, in the Theis Well function, which has similarity variable u=r^2 S/ 4 Tt , where the parameters take on the normal well-hydraulics definitions, and hydraulic diffusivity is D=T/S.
It is not unusual to assume a solution of the form of (68) and then to seek the coefficient A , that matches the boundary and initial conditions.
(^19) Typically its some combination of exponential or error functions.
