Write down the ode satisfied by a characteristic curve

[gtfitzpatrick]

Homework Statement

i)write down the general form of a semi lenear first order pde in the unknown u(x,y)
ii)write down the ode satisfied by a characteristic curve in the x-y plane for your pde
ii)give a careful derivation of the ode satisfied by u(x,y) along such a charcteristic curve.

Homework Equations

The Attempt at a Solution

i)a(x,y) \frac{\partial u}{\partial x} + b(x,y) \frac{\partial u}{\partial y} = g(x,y,u)

ii) the characteristic traces are given by \frac{dx}{dt} = a(x,y) and \frac{dy}{dt} = b(x,y) and \frac{du}{dt} = g(x,y,u) so is one of these what I'm looking for?

iii) since \frac{dx}{dt} = a(x,y) along our characteristic we get t in term of x, provided a(x,y) \neq 0. We can also express y in terms of x.

the chainrule gives\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt} and so \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{b(x,y)}{a(x,y)} along the projected trace

=> the projected trace satisfies the ODE \frac{dy}{dx} = \frac{b(x,y)}{a(x,y)}

 

[gtfitzpatrick]

Anyone know where i can find information on region of influence of initial conditions?
I can't get my head around it...

 

[gtfitzpatrick]

Homework Statement

i)write down the general form of a semi lenear first order pde in the unknown u(x,y)
ii)write down the ode satisfied by a characteristic curve in the x-y plane for your pde
ii)give a careful derivation of the ode satisfied by u(x,y) along such a charcteristic curve.

Homework Equations

The Attempt at a Solution

i)a(x,y) \frac{\partial u}{\partial x} + b(x,y) \frac{\partial u}{\partial y} = g(x,y,u)

ii) the characteristic traces are given by \frac{dx}{dt} = a(x,y) and \frac{dy}{dt} = b(x,y) and \frac{du}{dt} = g(x,y,u) so is one of these what I'm looking for?

iii) since \frac{dx}{dt} = a(x,y) along our characteristic we get t in term of x, provided a(x,y) \neq 0. We can also express y in terms of x.

the chainrule gives\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt} and so \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{b(x,y)}{a(x,y)} along the projected trace

=> the projected trace satisfies the ODE \frac{dy}{dx} = \frac{b(x,y)}{a(x,y)}

Can anyone confirm, is my theory right here?

 

[hunt_mat]

I would give the thumbs up with the above.