Change of Variables Question with chain rule

[RYANDTRAVERS]

Homework Statement

Consider the function of two variables:
u(x,y) = f(x-y) + g(x+(1/3)y)
where f(s) and g(t) are each arbitrary functions of a single variable.

Using the change of variables:
s = x-y
t = x-(1/3)y

use the chain rule to determine the first and second derivatives of u with respect to x and y in terms of derivatives of f and g.

Hence, show that the second derivatives satisfy
u_xx = 2u_xy + 3u_yy

where u_xx is the second derivative of u with respect to x, etc.

The Attempt at a Solution

My attempt, along with the original question paper, is attached as a PDF. It looks very fiddly but I have attempted the question a few times and still can’t satisfy the last equation.

 

[BvU]

Dear RY&T,

It's not all that difficult to use the subscript button on the green menu bar. makes life a lot easier for you and us...

Anyway, I'm not completely happy with u_x = u_f f_s + u_g g_t
u_f and u_g are trivially 1 and I propose to carry it a little further:
u_x = f_s f_x + g_t g_x
where f_x and g_x are trivially 1 also, leaving u_x = f_s + g_t
(which incidentally may well be what you intended).

Similarly, u_y = f_s f_y + g_t g_y = - f_s + g_t/3

The next step (on scan2 -- by the way: PF frowns on posting pictures when you can also type it out using the buttons -- or better: LaTeX) is a derailment:

u_xx is the derivative of f_s + g_t and that certainly is not the same as u_x u_x !

Simply repeat what you did with u to get u_x:

u_xx = ( f_s + g_t )_x = ( f_s )_x + ( g_t )_x

analogously with u_yy and u_xy (which I think is (u_y)_x -- someone correct me if I am wrong here -- too long ago )

- [edit] corrected myself -- see post #4

 

[RYANDTRAVERS]

I agree with you that u_f and u_g are trivially 1; however, with the u_xx there is the derivative operator with respect to x acting on u_x to give u_xx. I wasn't multiplying u_x by u_x.

 

[BvU]

Hi,

I see what you mean. It just looked a lot like multiplying. Still a derailment to interpret $\partial\over \partial x$ as $(f_s {\partial\over \partial f }+ g_t {\partial\over \partial g})$.

The incomplete way to write $u_x$ avenges itself somewhat.

Instead you want to differentiate $u_x = u_f f_s s_x + u_g g_t t_x = f_s+g_t$ as $(f_s)_x + (g_t)_x$. Write it out.

[edit]Or, if you want to write the whole thing, e.g. for the first term:

${\partial\over \partial x} u_f f_s s_x = u_{fx} f_s s_x+u_f f_{sx} s_x+u_f f_s s_{xx}$​

And treat $f_s$ in $(f_s)_x$ the same way as you treated $f$ in $(f)_x$ I have now also corrected myself in post #2. Really had to look it up: $f_{xy} = (f_x)_y = \partial_{yx}f = {\partial^2f\over \partial y \partial x}={\partial\over \partial y}\left(\partial f\over \partial x\right )$

-

 

[Mark44]

I have now also corrected myself in post #2. Really had to look it up: $f_{xy} = (f_x)_y = \partial_{yx}f = {\partial^2f\over \partial y \partial x}={\partial\over \partial y}\left(\partial f\over \partial x\right )$

Yes, the notation is counterintuitive, with $f_{xy}$ being read one way and $\frac{\partial}{\partial y}\left( \frac{\partial f}{\partial x} \right)$ being read the other way. IOW, with $f_{xy}$, you read this left to right, meaning that you differentiate wrt x first, then with respect to y.