Can f(x, t) be expressed as a function of x + ct?

[The1TL]

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

I'm not sure where to go from here

 

[HallsofIvy]

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 $\partial f/\partial x= (\partial f/\partial u)(\partial u/\partial x)+ (\partial f/\partial v)(\partial v/\partial x)$$= \partial f/\partial u+ \partial f/\partial v$. Similarly, $\partial f/\partial t= (\partial f/\partial u)(\partial u/\partial t)+ (\partial f/\partial v)(\partial v/\partial t)$$= c\partial f/\partial v$.

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

We can now subtract $c\partial f/\partial v$ from both sides leaving $c\partial f/\partial u= 0$ which says that f does not depend upon u at all, it is simply a function of v.

 

[I like Serena]

Hi The1TL!

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