Hence neither side can depend upon either x or t in

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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.

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