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Unformatted text preview: . Hint: Let I denote the integral on the left hand side and note that
I2 = 1
4πt ∞ ∞ −∞ −∞ e−(x 2 +y 2 )/4t dxdy. Then transform this last integral to polar coordinates.
d. Use Mathematica to sketch u0 (x, t) for various values of t. What can you
say about the behaviour of the function u0 (x, t) as t → 0?
e. By diﬀerentiating under the integral sign show that if h : R → R is any
smooth function which vanishes outside a ﬁnite interval [−L, L], then
∞ u(x, t) =
−∞ ua (x, t)h(a)da (4.7) is a solution to the heat equation.
REMARK: In more advanced courses it is shown that (4.7) approaches h(x) as
t → 0. In fact, (4.7) gives a formula (for values of t which are greater than zero)
for the unique solution to the heat equation on the inﬁnite line which satisﬁes
the initial condition u(x, 0) = h(x). In the next section, we will see how to solve
the initial value problem for a rod of ﬁnite length. 4.2 The initial value problem for the heat equation We will now describe how to use the Fourier sine series to ﬁnd the solution to
an initial value problem for the heat equation in a rod of length L which is
insulated along the sides, whose ends...
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This document was uploaded on 01/12/2014.
- Winter '14