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Write down the ode satisfied by a characteristic curve

  • #1

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)[itex] a(x,y) \frac{\partial u}{\partial x} + b(x,y) \frac{\partial u}{\partial y} = g(x,y,u) [/itex]

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

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

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

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

Answers and Replies

  • #2


Anyone know where i can find information on region of influence of initial conditions?
I cant get my head around it...
 
  • #3


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)[itex] a(x,y) \frac{\partial u}{\partial x} + b(x,y) \frac{\partial u}{\partial y} = g(x,y,u) [/itex]

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

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

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

=> the projected trace satisfies the ODE [itex]\frac{dy}{dx} = \frac{b(x,y)}{a(x,y)}[/itex]
Can anyone confirm, is my theory right here?
 
  • #4
hunt_mat
Homework Helper
1,741
25


I would give the thumbs up with the above.
 

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