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Chain rule proof

Yankel

Active member
Jan 27, 2012
398
Hello,

I need to do this proof here:

Capture.PNG

I tried but didn't get what I wanted, so I was re-thinking the whole thing.

If I say u=y+ax and v=y-ax, should I do something like (dz/df)*(df/du)*(du/dx)+....?

Because I tried just with u and v (without f and g), and I got almost what I wanted, with a little minus away from proofing, but I think I got it wrong from the start...
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hello,

I need to do this proof here:

View attachment 2008

I tried but didn't get what I wanted, so I was re-thinking the whole thing.

If I say u=y+ax and v=y-ax, should I do something like (dz/df)*(df/du)*(du/dx)+....?

Because I tried just with u and v (without f and g), and I got almost what I wanted, with a little minus away from proofing, but I think I got it wrong from the start...
Soppose to have to solve the second order PDE...

$\displaystyle z_{x x} = \frac{1}{a^{2}}\ z_{y y}\ (1)$

Substituting the x and y variable $u = x - a\ y$, $v=x + a\ y$ and applying the chain rule You arrive to write from (1)...


$\displaystyle z_{x x} = z_{u u} + 2\ z_{u v} + z_{v v}$

$\displaystyle a^{2}\ z_{y y} = z_{u u} - 2\ z_{u v} + z_{v v}\ (2)$

... and the means that...

$\displaystyle z_{u,v} = 0\ (3)$

It is easy to see that the solution of (3) is...

$\displaystyle z (u,v) = f(u) + g(v) = f( x - a\ y) + g(x + a\ y)\ (3)$

Kind regards

$\chi$ $\sigma$
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The chain rule tells us that:

(1) \(\displaystyle \frac{\partial z}{\partial x}=\frac{\partial z}{\partial f}\frac{\partial f}{\partial x}+\frac{\partial z}{\partial g}\frac{\partial g}{\partial x}\)

Let $t$ be the variable by which $f$ and $g$ are defined.

The chain rule also tells us:

(2) \(\displaystyle \frac{\partial f}{\partial x}=af'(t)\)

(3) \(\displaystyle \frac{\partial g}{\partial x}=-ag'(t)\)

And so we have:

\(\displaystyle \frac{\partial z}{\partial x}=a\left(f'(t)\frac{\partial z}{\partial f}-g'(t)\frac{\partial z}{\partial g} \right)\)

Now, differentiating this with respect to $x$, we find:

\(\displaystyle \frac{\partial^2 z}{\partial x^2}=a\left(f'(t)\frac{\partial^2 z}{\partial f^2}\frac{\partial f}{\partial x}+af''(t)\frac{\partial z}{\partial f}-g'(t)\frac{\partial^2 z}{\partial g^2}\frac{\partial g}{\partial x}+ag''(t)\frac{\partial z}{\partial g} \right)\)

Using (2) and (3), we find:

\(\displaystyle \frac{\partial^2 z}{\partial x^2}=a^2\left(\left(f'(t) \right)^2\frac{\partial^2 z}{\partial f^2}+f''(t)\frac{\partial z}{\partial f}+\left(g'(t) \right)^2\frac{\partial^2 z}{\partial g^2}+g''(t)\frac{\partial z}{\partial g} \right)\)

Now you may proceed similarly to show that:

\(\displaystyle \frac{\partial^2 z}{\partial y^2}=\left(f'(t) \right)^2\frac{\partial^2 z}{\partial f^2}+f''(t)\frac{\partial z}{\partial f}+\left(g'(t) \right)^2\frac{\partial^2 z}{\partial g^2}+g''(t)\frac{\partial z}{\partial g}\)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
I find this easier conceptually without Liebniz:

$z_x = f'(y + ax)(a) + g'(y - ax)(-a)$

$z_{xx} = (a)f''(y + ax)(a) - (a)g''(y - ax)(-a) = a^2(f''(y + ax) + g''(y - ax))$

$z_y = f'(y + ax)(1) + g'(y - ax)(1)$

$z_{yy} = f''(y + ax) + g''(y - ax) = \dfrac{1}{a^2}z_{xx}$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I find this easier conceptually without Liebniz...
I did too, but felt the OP was likely supposed to invoke ol' Liebniz. I'm glad you posted it though, as an easier alternate method. (Yes)