# Write down the ode satisfied by a characteristic curve

## 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.

## 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 $\frac{dy}{dx} = \frac{b(x,y)}{a(x,y)}$

## Answers and Replies

Related Calculus and Beyond Homework Help News on Phys.org

Anyone know where i can find information on region of influence of initial conditions?
I cant get my head around it...

## 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.

## 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
Homework Helper

I would give the thumbs up with the above.