III. ESTRUCTURA PROGRAMÁTICA
2. PRESUPUESTO SOCIAL PARA EL BIENESTAR
The boundary conditions in the plane wall problem are now modified. As before, the wall is initially at T = T1, yet the surface at x = 0 is maintained at T = T1 and the surface at x = L is instantaneously brought to T = T2 at t = 0. By defining the non–dimensional temperature as T = (T − T1)/(T2− T1), the problem becomes
∂T
∂t = ∂2T
∂x2 (3.66)
T (x = 0, t) = 0 T (x = 1, t) = 1
T (x, t = 0) = 0 (3.67)
The overbar notation will now be dropped from the variables. For this problem the BC at x = 1 is now inhomogeneous, and the IC is homogeneous. Because of this, SOV cannot be directly applied to the problem as stated. This is because the time ‘direction’ of the problem (which has the homogeneous condition) cannot be put in the Sturm–Liouville form, i.e. time–dependent eigenfunctions would not result from the separated problem.
In a subsequent chapter a more robust method of solving PDEs (know as variation of parame-ters) will be introduced, which allows one to get around this problem. For now, though, a solution can be obtained by (again) appealing to physics. It is not hard to see that this particular problem will attain a steady–state condition for t → ∞. Denoting the steady–state solution as s(x), this behavior would be
T (x, t → ∞) = s(x) = x (3.68)
That is, the steady state temperature profile in the wall is simply the linear profile x. Continuing with this concept, the function defined by w(x, t) = T (x, t) − s(x) would go to zero for t → ∞.
Consequently, the solution for the temperature can be split into the two parts:
T (x, t) = w(x, t) + s(x) (3.69)
in which w(x, t) represents the transient portion (that goes to zero for large t) and s(x) is the steady–state portion. The next step is to determine the PDE, BCs, and IC for w. By replacing the above equation into the PDE for T , one obtains
∂w
∂t + ∂s
|{z}∂t
= 0
= ∂2w
∂x2 + d2s dx2
|{z}= 0
(3.70)
The second derivative of s is zero in the above because s satisfies the steady–state conduction equation. The homogeneous 1–D and transient conduction PDE is therefore obtained for w. The boundary conditions for w are found in a similar manner, in that T = w + s is substituted into the BCs and IC;
T (x = 0, t) = w(x = 0, t) + s(x = 0)
| {z }
= 0
= 0 −→ w(x = 0, t) = 0 (3.71)
T (x = 1, t) = w(x = 1, t) + s(x = 1)
| {z }
= 1
= 1 −→ w(x = 0, t) = 0 (3.72)
T (x, t = 0) = w(x, t = 0) + s(x) = 0 −→ w(x = 0, t) = −s(x) = −x (3.73) The problem for w is now has homogeneous BCs and an inhomogeneous IC – which is the form that can be tackled by SOV. Note that the IC for w is simply the negative of the steady–state temperature profile – which makes physical sense.
Following the separation of variables procedure, we set w = u · v as before, and find that v = exp(−λ2t), as before. The solution for u will be
u = A cos(λx) + B sin(λx)
The result A = 0 will satisfy the BC at x = 0, and the BC at x = 1 gives the eigencondition:
sin(λn) = 0 −→ λn= nπ, n = 1, 2, . . .
The specific case of n = 0 can be eliminated here, because it does not give the correct time–
asymptotic behavior for w. The general solution for w is then
w = X∞ n=1
Ansin(λnx)e−λ2nt
Application of the initial condition w = −x gives
− x = X∞ n=1
Ansin(λnx)
By use of the orthogonality properties of the eigenfunctions, the expansion coefficients are found as
An= − Z 1
0 x sin(nπx) dx ·
Z 1 0
sin2(nπx) dx
−1
= 2 cos(nπ)
nπ = 2(−1)n
nπ (3.74)
And the complete solution is T = w + s, or
T = x + 2 π
X∞ n=1
(−1)nsin(nπx)
n e−(nπ)2t (3.75)
The method of splitting the solution for T into two parts, i.e., T = w +s, is sometimes known as
‘partial solutions’. The goal in using this method is to transform a problem that cannot, directly, be solved with SOV into a problem (or problems) that can. The partial solutions technique is one example of a general method known as superposition, in which two or more solutions to a modified problem are superimposed (or, equivalently, added) to form a solution to the whole problem (DE, BCs, and IC) under consideration. The feature of the DE and BCs that allows for this method is linearity – for which a sum of independent solutions to the DE will also be a solution.
Inhomogeneous DE
Consider now an example in which heat generation occurs in the wall. Say the surface at x = 0 is adiabatic and the surface at x = 1 is maintained at T = T1. Initially the wall is at a uniform temperature of T1. At time t = 0 uniform heat generation occurs in the wall, of strength q′′′0.
The dimensionless temperature in this case will be T → (T − T1)k/q0′′′L2, and the dimensionless problem is
Both the BCs and the IC are homogeneous, yet the DE is inhomogeneous. Again, SOV cannot be directly applied to this problem, because the resulting ODE for u(x) will not be in the Sturm-Liouville form. However, the method of partial solutions can be applied because this particular situation will have a steady–state. As before, let
T (x, t) = w(x, t) + s(x) (3.80)
in which s(x) is the steady state temperature distribution in the wall. The problem presented by s is
s′′+ 1 = 0, s′(0) = 0, s(1) = 0 (3.81) which has the solution
s = 1
So again, a homogeneous DE and homogeneous BCs are obtained for w, and the initial condition for w is equal to the negative of the steady state solution.
The general solution for w is
w = X∞ n=1
Ancos(λnx)e−λ2nt (3.87)
in which the eigenvalues are
λn= (2n − 1)π Using the orthogonality of the eigenfunctions gives
An= −
At this point the formula for s could be inserted and the formula integrated (which would be easy with Mathematica), yet it is instructive to demonstrate how this integral can be easily evaluated by making use of the DE and BCs for s. Let φn denote the eigenfunction cos(λnx). The previous integral becomes
Integration by parts was used in evaluating the integrals. The boundary terms are zero by virtue of the BC’s on φn and s. Also, s′′ was eliminated using the DE for s, i.e., s′′ = −1. The formula for the Anbecomes
An= 16(−1)n
(2n − 1)3π3 (3.88)
which, together with Eqs. (3.87) and (3.82), gives the complete solution for the temperature distri-bution in the wall.