# Homework Help: Change of Variables Question with chain rule

1. Apr 10, 2015

### RYANDTRAVERS

1. The problem statement, all variables and given/known data
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.

3. 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.

#### Attached Files:

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• ###### Scan4.pdf
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Last edited: Apr 10, 2015
2. Apr 10, 2015

### 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 ux = uf fs + ug gt
uf and ug are trivially 1 and I propose to carry it a little further:
ux = fs fx + gt gx
where fx and gx are trivially 1 also, leaving ux = fs + gt
(which incidentally may well be what you intended).

Similarly, uy = fs fy + gt gy = - fs + gt/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:

uxx is the derivative of fs + gt and that certainly is not the same as ux ux !

Simply repeat what you did with u to get ux:

uxx = ( fs + gt )x = ( fs )x + ( gt )x

analogously with uyy and uxy (which I think is (uy)x -- someone correct me if I am wrong here -- too long ago )

-  corrected myself -- see post #4

Last edited: Apr 11, 2015
3. Apr 10, 2015

### RYANDTRAVERS

I agree with you that uf and ug are trivially 1; however, with the uxx there is the derivative operator with respect to x acting on ux to give uxx. I wasn't multiplying ux by ux.

4. Apr 11, 2015

### 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.

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: $f_{xy} = (f_x)_y = \partial_{yx}f = {\partial^2f\over \partial y \partial x}={\partial\over \partial y}\left(\partial f\over \partial x\right )$

-

Last edited: Apr 11, 2015
5. Apr 11, 2015

### Staff: Mentor

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.