PDE Characteristic Method?

  • Thread starter The Head
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  • #1
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Homework Statement:
u_x + x*u_y = (y-1/2x^2)^2, u(0,y) = e^y
Relevant Equations:
du/dt = du/dx*dx/dt + du/dy*dy/dt
dx/dt =1, x(0,s)=0, dy/dt=x, y(0,s) = s, du/dt=(y-1/2x^2)^2, u(0,s)=e^s

I did well at the beginning to get x(t,s) =t and y(t,s)=1/2t^2 + s, but got stuck with the du/dt part.

You can sub in x=t and y=1/2t^2 +s for x and y to get du/dt = s^2. But that's still three variables, and I can't see any obvious substitutions.

This is a regular diffy Q course, so I don't have a lot of knowledge of advanced techniques, and this was just thrown in, so I'm a little unsure how to proceed.
 

Answers and Replies

  • #2
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Also, I tried u=e^s => s=ln(u), so du/dt = (ln(u))^2, but that gives me an Ei function, so I feel like that can't be correct.
 
  • #3
Orodruin
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You can sub in x=t and y=1/2t^2 +s for x and y to get du/dt = s^2.

The expression
$$
\frac{\partial u}{\partial t} = s^2
$$
can be directly integrated, ##s## is the other coordinate - not an unknown to be solved for.

Edit: Missed the square in the original PDE.
 
Last edited:
  • #4
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The expression
$$
\frac{\partial u}{\partial t} = s^2
$$
can be directly integrated, ##s## is the other coordinate - not an unknown to be solved for.

Edit: Missed the square in the original PDE.

Ohh, OK gotcha. So it can just be s^2 * t = u(s,t)? Thanks so much for your reply.
 
  • #5
Orodruin
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Ohh, OK gotcha. So it can just be s^2 * t = u(s,t)? Thanks so much for your reply.
No, but almost. Your function does not satisfy the initial condition.
 
  • #6
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No, but almost. Your function does not satisfy the initial condition.
Right, of course! Thank you.
 

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