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Partial differential equation - change of variables.

  1. Nov 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider the PDE:

    [tex]2dz/dx[/tex] - [tex]dz/dy[/tex] = 0

    How can I show that if f(u) is differentiable function of one variable, then the PDE above is satisfied by z = f(x +2y)?

    Also, the change in variables t = x+2y, s=x reduces the above PDE to [tex]dz/dt[/tex] = 0. But how can I show this?

    3. The attempt at a solution

    I simply don't understand, these are my notes from class today and I want to understand while it's early... All I know is that this involves the chain rule.
     
  2. jcsd
  3. Nov 5, 2009 #2
    Shameless bump. This is driving me nuts!
     
  4. Nov 5, 2009 #3

    HallsofIvy

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    Let u= x+ 2y and use the chain rule: [itex]\partial f/\partial x= (df/du)(\partial u/\partial x)[/itex]= f'(u)(1)= f'(u) and [itex]\partial f/\partial y= (df/du)(\partial u/\partial y)= f'(u)(2)= 2f'(u)[/itex]. Put those into the equation and see what happens.

    If t= x+ 2y and s= x, then x= s and t= s+ 2y so 2y= t- s and y= (t-s)/2. Now, [itex]\partial z/\partial x= (\partial z/\partial t)(\partial t/\partial t)+[/itex][itex] (\partial z/\partial s)(\partial s/\partial x)[/itex][itex]= 1(\partial z/\partial t)+ 1(\partial z/\partial s)= \partial z/\partial t+ \partial z/\partial s[/itex] and [itex]\partial z/\partial y= (\partial z/\partial t)(\partial t)\partial y)+ (\partial z/\partial s)(\partial s/\partial y)[/itex][itex]= 2 \partial z/\partial t+ 0\partial z/\partial s= 2\partial z/\partial t[/itex].

    Putting those into [itex]2\partial z/\partial x-\partial z/\partial y= 0[/itex] and it becomes [itex]2(\partial z/\partial t+ \partial z/\partial s)- 2\partial z/\partial t= 0[/itex].

    The two "[itex]2\partial z/\partial t[/itex]" terms cancel leaving [itex]2\partial z/\partial s= 0[/itex] which is equivalent to [itex]\partial z/\partial s= 0[/itex] which says that z does not depend upon s at all. Since any dependence on t alone satisfies that, z= f(t), for t any differentiable function, satisfies that equation.

     
    Last edited: Nov 5, 2009
  5. Nov 5, 2009 #4
    Thanks so much, that makes it seem very intuitive.
     
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