Geology Reference
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Laplace transforms conveniently circumvent these steps, in the process
minimizing the possibility of algebraic error. We merely solve our ordinary
differential equation in turn with Laplace transforms. If we analogously define
L {U(x,s)} = U ( ,s) = e - x U( x , s ) dx , > 0
(1.159)
0
it is clear that s 2 U(x,s) - c 2 U"(x,s) = F(s) (x-x*) transforms into s 2 U ( ,s) -
c 2 { 2 U ( ,s) - U(0,s) U x (0,s)} = F(s) e - x (x-x*) dx = F(s)e - x* , where we
have again used the “sifting property” of delta functions.
So far, we have defined a Laplace transform with respect to time, the
conventional one used by electrical engineers, and then a second one with
respect to space. Thus, we obtain the double Laplace transform
U ( ,s) = F(s)e - x* /(s 2 - c 2
2 )
- c 2
U( 0 ,s ) / ( s 2 - c 2
2 ) - c 2 U x (0,s)/(s 2 - c 2
2 )
(1.160)
Now, U(0,s) vanishes by our assumption of a pinned left end (this is obtained by
transforming the boundary condition u(0,t) over s). The presence of U x (0 , s ),
also for x = 0, might suggest that we have incorrectly applied an initial value
problem solver, assuming a time-like variable x, to a boundary value problem.
This is not so. Although we assumed a pinned right end u(L,t) = 0, thus
formulating a two-point boundary value problem, it is perfectly valid to define a
Laplace transform over (0, ) even though our domain is 0 < x < L. This use is
less common, but the idea is simple: in applying spatial Laplace transforms to
finite domains, we in fact assume that a semi-infinite space holds, although we
ultimately confine our attention to the finite space between x = 0 and the
position x = L hosting the constraint u(L,t) = 0.
While U x (0,s) is not prescribed as part of the formulation, it can be
regarded as an unknown “constant” that is fully determined by the condition
u(L,t) = 0 at the opposite boundary. Continuing with our solution of Equation
1.160, the simplification U(0,s) = 0 leads to
U ( ,s) = U x (0,s)/( 2 - s 2 /c 2 ) + F(s)e - x* /(s 2 - c 2
2 )
(1.161)
for the double Laplace transform of the function u(x,t). We perform the
inversion with respect to the parameter . For the first term shown above, table
look-up readily shows that the function corresponding to 1/( 2 - s 2 /c 2 ) is just
(c/s) sinh {(s/c)x}. Laplace transform tables also show that the exact inverse of
e - x* /(s 2 - c 2 2 ) or (-1/c 2 )e - x* /( 2 -s 2 /c 2 ) is represented by the simple function
(-1/c 2 )(c/s) (x-x*) sinh {(s/c)(x-x*)}, where (x-x*) = 1 for x > x* and 0 for x
< x* (an “exponential shift” property has been exploited). Therefore, we have
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