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Chapter 6 Partial Differential Equations

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24 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS<br />

P 0<br />

C 0<br />

C 1<br />

Example 6.2.16. Recall the Example 6.2.12<br />

Solution Surface for Each Curve C 1<br />

u t + cu x =0,<br />

u(x, 0) = F (x)<br />

with solution given by<br />

u(x, t) =F (x + ct).<br />

We note that the initial surface is t = 0 and the characteristics are determined by<br />

dx<br />

dτ = c, x(s, 0) = s, dt<br />

=1,t(s, 0)=0, ⇒ t(s, τ) =τ, x(s, τ) =cτ + s, ⇒ x − ct = s.<br />

dτ<br />

Thus we see that the solution is constant along characterisitics, i.e., (x, t) on a characteristic<br />

means that x − ct = s = constant and the solution is given by u(x, t) =F () and u is<br />

constant on a characteristic, i.e., a line x − ct = s for s ∈ R.<br />

A generalization of this problem is the case in which c = c(x, t). Then the characteristics<br />

are determined by<br />

and along this curve<br />

dx<br />

dτ<br />

= c(x, t), x(s,<br />

0) = s, dt<br />

dτ<br />

=1,t(s, 0)=0,<br />

du<br />

dt (x(t),t)=u dx<br />

x<br />

dt + u dt<br />

t<br />

dt = u xc(x(t),t)+u t ≡ 0.

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