The Student Room Group

Characteristics

Solve: xux+yuy=0;u(1,y)=f(y) x \frac {\partial u}{\partial x} + y\frac {\partial u}{\partial y} = 0 ; u(1,y) =f(y)
Cauchy problem for characteristics:
xs=x,ys=y,us=0\frac {\partial x}{\partial s} = x, \frac {\partial y}{\partial s} = y, \frac {\partial u}{\partial s} =0
x(0;τ)=tor,y(0;τ)=τ,u(0;τ)=f(τ) x (0;\tau)=tor, y(0;\tau)=\tau, u(0;\tau)=f(\tau)
x=es,y=tes,u=f(τ) x = e^s, y=te^s, u =f(\tau)

I understand it up until getting x, y (and therefore u)
Where does the e^s come from in both?
(edited 12 years ago)
dqdr=qq=er\frac{dq}{dr} = q \rightarrow q = e^r (+ constant).
Reply 2
Original post by marcusmerehay
dqdr=qq=er\frac{dq}{dr} = q \rightarrow q = e^r (+ constant).


:banghead: that's come about from separation of variables hasn't it?!?

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