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Unformatted text preview: tegrands on the two
sides must be equal:
∂F
∂u
= −ρ(x)σ (x) .
∂x
∂t
It follows from (4.2) that
∂
∂x −κ(x) ∂u
∂x = −ρ(x)σ (x) ∂u
.
∂t In this way, we obtain the heat equation
ρ(x)σ (x) ∂u
∂
=
∂t
∂x κ(x) ∂u
∂x . (4.4) In the more general case in which heat is being created at the rate µ(x)u(x, t) +
ν (x) per unit length, one could show that heat ﬂow is modeled by the equation
ρ(x)σ (x) ∂u
∂
=
∂t
∂x κ(x) ∂u
∂x + µ(x)u + ν (x). (4.5) In the special case where the bar is homogeneous , i.e. its properties are the
same at every point, ρ(x), σ (x) and κ(x) are constants, say σ and κ respectively,
and (4.4) becomes
∂u
κ ∂2u
.
=
∂t
ρσ ∂x2 (4.6) This is our simplest example of a linear partial diﬀerential equation. Although
its most basic application concerns diﬀusion of heat, it arises in many other
contexts as well. For example, a slight modiﬁcation of the heat equation was
used by Black and Scholes to price derivatives in ﬁnancial markets.2
Exercises:
4.1.1. S...
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This document was uploaded on 01/12/2014.
 Winter '14
 Equations

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