Partial differential equation - change of variables.

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Homework Statement



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?

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.
 

Answers and Replies

  • #2
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Shameless bump. This is driving me nuts!
 
  • #3
HallsofIvy
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Homework Statement



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)?
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.

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?
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.

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.
 
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  • #4
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Thanks so much, that makes it seem very intuitive.
 

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