Partial differential equation - change of variables.

[mathman44]

Homework Statement

Consider the PDE:

2dz/dx - dz/dy = 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 dz/dt = 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.

 

[mathman44]

Shameless bump. This is driving me nuts!

 

[HallsofIvy]

Homework Statement

Consider the PDE:

2dz/dx - dz/dy = 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: \partial f/\partial x= (df/du)(\partial u/\partial x)= f'(u)(1)= f'(u) and \partial f/\partial y= (df/du)(\partial u/\partial y)= f'(u)(2)= 2f'(u). Put those into the equation and see what happens.

Also, the change in variables t = x+2y, s=x reduces the above PDE to dz/dt = 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, \partial z/\partial x= (\partial z/\partial t)(\partial t/\partial t)+(\partial z/\partial s)(\partial s/\partial x)= 1(\partial z/\partial t)+ 1(\partial z/\partial s)= \partial z/\partial t+ \partial z/\partial s and \partial z/\partial y= (\partial z/\partial t)(\partial t)\partial y)+ (\partial z/\partial s)(\partial s/\partial y)= 2 \partial z/\partial t+ 0\partial z/\partial s= 2\partial z/\partial t.

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

The two "2\partial z/\partial t" terms cancel leaving 2\partial z/\partial s= 0 which is equivalent to \partial z/\partial s= 0 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.

 

[mathman44]

Thanks so much, that makes it seem very intuitive.