Question: Consider a distribution that at the initial time t = 0 has the form of a Dirac delta function (x). A delta function can he

Consider a distribution that at the initial time t = 0 has the form of a Dirac delta function δ(x). A delta function can he represented by a Fourier integral:

0(x,0) = dkexp(ikx) 8(x) %3D 2n

At later times the pulse becomes

0(x,0) = dkexp(ikx) 8(x) %3D 2n

or by use of(10),

Evaluate the integral to obtain the result (14). The method can be extended to describe the time development of any distribution given at t = 0. If the distribution is f(x, 0), then by the definition of the delta function


f(x, 0) = dx’ f(x’, 0)δ(x – x’).


The time development of δ(x – x’)is

by (14). Thus at time t the distribution f(x, 0) has evolved to

This is a powerful general solution.

0(x,0) = dkexp(ikx) 8(x) %3D 2n

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