Partial derivative proof

  • Thread starter The1TL
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  • #1
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Homework Statement


Suppose f: R^2 --> R is differentiable and (df/dt) = c(df/dx) for some nonzero constant c.
Prove that f(x, t) = h(x + ct) for some function h.


Homework Equations



hint: use (u, v) = (x, x+ct)

The Attempt at a Solution



df/dt = limk-->0 (f(x, x+ct+k) - f(x, x+ct))/k
multiplying this by c gives:
limk-->0 (f(x, x+ct+ck) - f(x, x+ct))/k

I'm not sure where to go from here
 

Answers and Replies

  • #2
HallsofIvy
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Are you required to use the basic definition of the derivative? You were given a hint which works nicely with the chain rule but you have ignored it. If u= x, v= x+ ct, then [itex]\partial f/\partial x= (\partial f/\partial u)(\partial u/\partial x)+ (\partial f/\partial v)(\partial v/\partial x)[/itex][itex]= \partial f/\partial u+ \partial f/\partial v[/itex]. Similarly, [itex]\partial f/\partial t= (\partial f/\partial u)(\partial u/\partial t)+ (\partial f/\partial v)(\partial v/\partial t)[/itex][itex]= c\partial f/\partial v[/itex].

Now, the original differential equation, in terms of u and v, becomes
[itex]c\partial f/\partial v= c(\partial f/\partial u+ \partial f/\partial v)[/itex].

We can now subtract [itex]c\partial f/\partial v[/itex] from both sides leaving [itex]c\partial f/\partial u= 0[/itex] which says that f does not depend upon u at all, it is simply a function of v.
 
  • #3
I like Serena
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Hi The1TL! :smile:

Can you calculate ##\frac d {du}f(x(u,v), t(u,v))## with partial derivatives?
 

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