Unformatted text preview: his eigenvalue problem.
b. Find the general solution to the heat equation
∂u
∂2u
=
∂t
∂x2
subject to the mixed boundary condition (4.15).
c. Find the function u(x, t), deﬁned for 0 ≤ x ≤ π and t ≥ 0, such that:
∂u
∂2u
,
=
∂t
∂x2 4.3 u(0, t) = ∂u
(π, t) = 0,
∂x u(x, 0) = 4 sin(x/2) + 12 sin(3x/2). Numerical solutions to the heat equation There is another method which is sometimes used to treat the initial value
problem described in the preceding section, a numerical method based upon
“ﬁnite diﬀerences.” Although it yields only approximate solutions, it can be
applied in some cases with variable coeﬃcients when it would be impossible to
apply Fourier analysis in terms of sines and cosines. However, for simplicity,
we will describe only the case where ρ and k are constant, and in fact we will
assume that c2 = L = 1.
Thus we seek the function u(x, t), deﬁned for 0 ≤ x ≤ 1 and t ≥ 0, which
solves the heat equation
∂u
∂2u
=
∂t
∂x2
subject to the boundary conditions u(0, t) = u(1, t) = 0 and the initial con...
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This document was uploaded on 01/12/2014.
 Winter '14
 Equations

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