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Method of characteristics
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== Example == As an example, consider the [[advection equation]] (this example assumes familiarity with PDE notation, and solutions to basic ODEs). :<math>a \frac{\partial u}{\partial x} + \frac{\partial u}{\partial t} = 0</math> where <math>a</math> is constant and <math>u</math> is a function of <math>x</math> and <math>t</math>. We want to transform this linear first-order PDE into an ODE along the appropriate curve; i.e. something of the form :<math> \frac{d}{ds}u(x(s), t(s)) = F(u, x(s), t(s)) ,</math> where <math>(x(s),t(s))</math> is a characteristic line. First, we find :<math>\frac{d}{ds}u(x(s), t(s)) = \frac{\partial u}{\partial x} \frac{dx}{ds} + \frac{\partial u}{\partial t} \frac{dt}{ds}</math> by the chain rule. Now, if we set <math> \frac{dx}{ds} = a</math> and <math>\frac{dt}{ds} = 1</math> we get :<math> a \frac{\partial u}{\partial x} + \frac{\partial u}{\partial t} </math> which is the left hand side of the PDE we started with. Thus :<math>\frac{d}{ds}u = a \frac{\partial u}{\partial x} + \frac{\partial u}{\partial t} = 0.</math> So, along the characteristic line <math>(x(s), t(s))</math>, the original PDE becomes the ODE <math>u_s = F(u, x(s), t(s)) = 0</math>. That is to say that along the characteristics, the solution is constant. Thus, <math>u(x_s, t_s) = u(x_0, 0)</math> where <math>(x_s, t_s)\,</math> and <math>(x_0, 0)</math> lie on the same characteristic. Therefore, to determine the general solution, it is enough to find the characteristics by solving the characteristic system of ODEs: * <math>\frac{dt}{ds} = 1</math>, letting <math>t(0)=0</math> we know <math>t=s</math>, * <math>\frac{dx}{ds} = a</math>, letting <math>x(0)=x_0</math> we know <math>x=as+x_0=at+x_0</math>, * <math>\frac{du}{ds} = 0</math>, letting <math>u(0)=f(x_0)</math> we know <math>u(x(t), t)=f(x_0)=f(x-at)</math>. In this case, the characteristic lines are straight lines with slope <math>a</math>, and the value of <math>u</math> remains constant along any characteristic line.
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