LESSONS OF EASTER ISLAND
March 17th, 2016
This appendix addresses the issue of solving the PDE (2.5).
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
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 nonhomogeneous 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 
9 x 
9 x 
9t 

9 T 
_ dpi 
9 T 9p2 ‘ + ~df 
9 T 
~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
Therefore, in the new coordinates, and regrouping terms, the nonhomogeneous 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)ey(tt0) + a j R(a)ey(at)da. (A.17)
t0
Inverting the change of variable (A.8) and using (A.3) yields (2.6).
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