Chain Rule for Secon Partial Derivatives

helenli89

New member
Joined
Oct 1, 2009
Messages
48
Q: Let f,g: R->R where f and g are twice differentiable. Show that u(c,t)=f(x-at)+g(x+at) is a solution of the wave equation: utt = a^2*uxx, where utt is the derivative interms of t twice and same to x.

My attempt of solving this question is to first find the first derivative of u interms of t, as follows:
Let w=x-at and v=x+at.
ut (x,t) = (df/dw)(?w/?t) + (dg/dv)(?v/?t) = f'(w)(-a) + g'(v)(a).
utt (x,t)= (?ut/?w)(?w/?t) + (?ut/?v)(?v/?t) = [f''(w)(-a)+g'(v)(a)](-a) + [f'(w)(-a) + g''(v)(a)](a) = a^2 * f''(w) - a^2 g'(v) - a^2 * f'(w) + a^2 * g''(v) = a^2 * [f''(w)-g'(v)-f'(w)+g''(v)]
ux (x,t) = (df/dw)(?w/?x) + (dg/dv)(?v/?x) = f'(w)(1) + g'(v)(1)
uxx (x,t) = (?ux/?w)(?w/?x) + (?ux/?v)(?v/?x) = [f''(w) + g'(v)](1) + [f'(w) + g''(v)](1) = f''(w)+g'(v)+f'(w)+g''(v)

But f''(w)-g'(v)-f'(w)+g''(v) ? f''(w)+g'(v)+f'(w)+g''(v), how do go from here to utt = a^2*uxx.

Thank you.
 
helenli89 said:
Q: Let f,g: R->R where f and g are twice differentiable. Show that u(c,t)=f(x-at)+g(x+at) is a solution of the wave equation: utt = a^2*uxx, where utt is the derivative interms of t twice and same to x.

My attempt of solving this question is to first find the first derivative of u interms of t, as follows:
Let w=x-at and v=x+at.
ut (x,t) = (df/dw)(?w/?t) + (dg/dv)(?v/?t) = f'(w)(-a) + g'(v)(a).
utt (x,t)= (?ut/?w)(?w/?t) + (?ut/?v)(?v/?t) = [f''(w)(-a)+g'(v)(a)](-a) + [f'(w)(-a) + g''(v)(a)](a) = a^2 * f''(w) - a^2 g'(v) - a^2 * f'(w) + a^2 * g''(v) = a^2 * [f''(w)-g'(v)-f'(w)+g''(v)]
ux (x,t) = (df/dw)(?w/?x) + (dg/dv)(?v/?x) = f'(w)(1) + g'(v)(1)
uxx (x,t) = (?ux/?w)(?w/?x) + (?ux/?v)(?v/?x) = [f''(w) + g'(v)](1) + [f'(w) + g''(v)](1) = f''(w)+g'(v)+f'(w)+g''(v)

But f''(w)-g'(v)-f'(w)+g''(v) ? f''(w)+g'(v)+f'(w)+g''(v), how do go from here to utt = a^2*uxx.

Thank you.

u[sub:3e3y3mns]x[/sub:3e3y3mns] = f'(x-at) + g'(x+at)

u[sub:3e3y3mns]xx[/sub:3e3y3mns] = f''(x-at) + g''(x+at)

u[sub:3e3y3mns]t[/sub:3e3y3mns] = a[-f'(x-at) + g'(x+at)]

u[sub:3e3y3mns]tt[/sub:3e3y3mns] = a[sup:3e3y3mns]2[/sup:3e3y3mns][f''(x-at) + g''(x+at)] = a[sup:3e3y3mns]2[/sup:3e3y3mns] * u[sub:3e3y3mns]xx[/sub:3e3y3mns]

I don't quite know what you are doing.....
 
Don't warry I already solved this problem and the assignment is dued long time ago. But, hey, thanks for the reply anyway.
it should be f''(x) + g''(x) = a^2(f''(x) + g''(x)).
 
Top