This appendix addresses the issue of solving the PDE (2.5).
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Start by considering the homogeneous equation
to which the following ODE relating the independent variables t and x is associated
(A.2)
The first integral of (A.2) is a relation of the form
p(x, t) = C
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for C an arbitrary constant, satisfied by any solution x = x (t) of (A.2), where the function p is not identically constant for all the values of x and t. In other words, the function p is constant along each solution of (A.2), with the constant C depending on the solution. Since in the case of equation (a1e1) there are 2 independent variables, there is only one functionally independent integrals Ibragimov (1999). By integrating (A.2) with respect to time, its first integral is found to be
A function p(x, t) is a first integral of (A.2) if and only if it is a solution of the homogeneous PDE (A.1). Furthermore, or t > t0, the general solution of (A.1) is given by
T(x, t) = F(P(x, t)), (A.4)
where F is an arbitrary function Ibragimov (1999). The general solution of (A.1) is thus given by
t
T(x, t) = F(x – u(a)da), (A.5)
t0
where F(x) for x є [0, L] is the temperature distribution along the pipe at t = t0.
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Consider now the non-homogeneous Eq. (2.5) that, for the sake of simplifying the notation, is written as
9 9
T(z, t) = u T(z, t) + g(x, t, T), (A.6)
91 9 z
where
g(x, t, T) := aR(t) – у T(z, t). (A.7)
To solve (A.6), and according to a procedure known as Laplace’s method Ibragimov (1999), introduce the new set of coordinates (^, t) given by
f = ni(x, t) := p(x, t)
and
t = n2(x, t) := t
where p is the first integral (A.3). The chain rule for derivatives yields
|
9 T |
_ dpi |
9 T dp2 Ji + ~9x ‘ |
9 T |
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9 x |
9 x |
9t |
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9 T |
_ dpi |
9 T 9p2 ‘ + ~df |
9 T |
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~dt |
= a t |
9t |
Since the transformation of variables is defined by (A.8, A.9), these equations reduce to
dT _ dp 9T 9x 9x df ’
9T _ dp dT dT
at at af dr
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Therefore, in the new coordinates, and regrouping terms, the non-homogeneous equation (a1e7) is written
Since p satisfies the homogeneous equation (A.1) and r = t, this equation reduces to the ODE
dT
= g (A.15)
dt
where x is to be expressed in terms of t and f by solving (A.8) with respect to x. Since g is given by (A.7), the above ODE reads
d
— Tf t) = -YTf t) + aR(t). (A.16)
Equation(A.16) is a linear, scalar, first order ODE whose solution is
t
T(f’ t) = T(f’ t0)e-y(t-t0) + a j R(a)ey(a-t)da. (A.17)
t0
Inverting the change of variable (A.8) and using (A.3) yields (2.6).
