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I am trying to solve this problem here:

Consider the function of two variables:
$ u(x,y) = f(x−y) + g (x+ \frac{1}{3}y) $ , where $f(s)$ and $g(t)$ are each arbitrary functions of a single variable. Using the change of variables $s=x−y$, $t = x+\frac{1}{3}y$ 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 $u_{xx} = \frac{∂^2u}{∂x^2}$ etc.

I have tried this using the chain rules for partial derivatives ($\frac{\partial u}{\partial x} = \frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial t} \frac{\partial t}{\partial x}$)

So like this:

$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial s} + \frac{\partial u}{\partial t} \Rightarrow \frac{\partial^2u}{\partial x^2} = \frac{\partial(\frac{\partial u}{\partial s} + \frac{\partial u}{\partial t})}{\partial s} + \frac{\partial(\frac{\partial u}{\partial s} + \frac{\partial u}{\partial t})}{\partial t} = \frac{\partial^2u}{\partial s^2} + 2\frac{\partial^2u}{\partial s \partial t} + \frac{\partial^2u}{\partial t^2} $

I then did this for each double differential, however in doing this I get that:

$ 2u_{xy} + 3u_{yy} = \frac{\partial^2u}{\partial s^2} - \frac{10}{3}\frac{\partial^2u}{\partial s \partial t} + \frac{\partial^2u}{\partial t^2} $

Which is quite clearly wrong, is the way I am going about the question wrong? Or am I just making a silly mistake somewhere within my working?

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For instance, $\frac{\partial}{\partial x}u = f'(x-y)+g'(x+\frac{1}{3}y)$. This is just the chain rule for functions of one variable. Can you go on? – Siminore Apr 23 '12 at 14:38
How do you find $\frac{\partial}{\partial y} u $ Using that method? or $ \frac{\partial^2}{\partial x \partial y} u $ ? – simonthumper Apr 23 '12 at 14:43
Well, $\frac{\partial}{\partial y}u=-f'(x-y)+\frac{1}{3}g'(x+\frac{1}{3}y)$, for example. The point is that $f$ and $g$ are functions of a single variable, so that you just need to differentiate as usual. – Siminore Apr 23 '12 at 15:01
Ahhhhh right I've got you! Didn't quite click in my mind what you meant until you showed $ \frac{\partial}{\partial y} $ Thanks, I'll have another go at it, then mark you correct :D – simonthumper Apr 23 '12 at 15:07
The only problem I have still is with your $ ' $ notation? What does that mean? is $f'(x-y) = \frac{ \partial f(x-y)}{\partial y} $ or $ \frac{ \partial f(x-y)}{\partial x} $? – simonthumper Apr 23 '12 at 15:27

You are told that $ u(x,y) = f(x−y) + g (x+ \frac{1}{3}y) $. You are also told that $s=x-y$ and $t=x+\frac{1}{3} y$. So : $$u(x,y) = f(x−y) + g (x+ \frac{1}{3}y)=u(x,y) = f(s) + g (t)$$ As you pointed out the chain rule for the derivative of $u(x,y)$ with respect to $x$ is: $$u_x=\frac{\partial u}{\partial x} = \frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial t} \frac{\partial t}{\partial x}$$

and with respect to $y$: $$u_y=\frac{\partial u}{\partial y} = \frac{\partial u}{\partial s}\frac{\partial s}{\partial y}+\frac{\partial u}{\partial t} \frac{\partial t}{\partial y}$$

You have the formula for the partial derivatives of $u(x,y)$ above. So: $$u(x,y) = f(s) + g (t)$$ $$\frac{\partial u}{\partial s}=\frac{\partial f}{\partial s}=f_s,\;\frac{\partial u}{\partial t}=\frac{\partial g}{\partial t}=g_t$$ $$s=x-y,\; \frac{\partial s}{\partial x}=\frac{\partial }{\partial x}[x-y]=1,\;\frac{\partial s}{\partial y}=\frac{\partial }{\partial y}[x-y]=-1$$ $$t=x+\frac{1}{3} y,\;\frac{\partial t}{\partial x}=1,\;\frac{\partial t}{\partial y}=\frac{1}{3}$$

Using the above stuff you will see that:

$$u_x=\frac{\partial f}{\partial s}\frac{\partial }{\partial x}[x-y]+ \frac{\partial g}{\partial t}\frac{\partial }{\partial x}[x+\frac{1}{3} y]=\frac{\partial f}{\partial s}+\frac{\partial g}{\partial t}=f_s+g_t$$

$$u_y=\frac{\partial f}{\partial s}\frac{\partial }{\partial y}[x-y]+\frac{\partial g}{\partial t}\frac{\partial }{\partial y}[x+\frac{1}{3} y]=-\frac{\partial f}{\partial s}+\frac{\partial g}{3\partial t}=-f_s+\frac{1}{3} g_t$$

Now it is easy to compute the second partial derivatives to verify $u_{xx} = 2u_{xy} + 3u_{yy}$. Just use the chain rule again: $$u_{xx}=\frac{\partial^2 u}{\partial x^2}=\frac{\partial u_x}{\partial x} = \frac{\partial u_x}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u_x}{\partial t} \frac{\partial t}{\partial x}= \frac{\partial }{\partial s}[f_s+g_t]\frac{\partial s}{\partial x}+\frac{\partial }{\partial t}[f_s+g_t] \frac{\partial t}{\partial x}=f_{ss}+g_{tt}$$ $$u_{xy}=\frac{\partial u_x}{\partial y}=\frac{\partial u_x}{\partial s}\frac{\partial s}{\partial y}+\frac{\partial u_x}{\partial t} \frac{\partial t}{\partial y}=\frac{\partial }{\partial s}[f_s+g_t]\frac{\partial s}{\partial y}+\frac{\partial }{\partial t}[f_s+g_t] \frac{\partial t}{\partial y}=-f_{ss}+\frac{1}{3} g_{tt}$$ $$u_{yy}=\frac{\partial u_y}{\partial y}=\frac{\partial u_y}{\partial s}\frac{\partial s}{\partial y}+\frac{\partial u_y}{\partial t} \frac{\partial t}{\partial y}=\frac{\partial }{\partial s}[-f_s+\frac{1}{3} g_t]\frac{\partial s}{\partial y}+\frac{\partial }{\partial t}[-f_s+\frac{1}{3} g_t] \frac{\partial t}{\partial y}=-f_{ss}+\frac{1}{9} g_{tt}$$

Note that $\frac{\partial }{\partial s}[f_s+g_t]=f_{ss}$ and$\frac{\partial }{\partial t}[f_s+g_t]=g_{tt}$, i.e, we take the derivative of a derivative of f(s) and g(t) with respect to $s$ and $t$ respectively. So here you go. Now you can verify $u_{xx} = 2u_{xy} + 3u_{yy}$ just by plugging in the derivatives of $u$ in terms of $f_{ss}$ and $g_{tt}$.

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