Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am given $u=\frac{x+y}{\sqrt 2}$ and $v=\frac{x-y}{\sqrt2}$, how would I find $\frac{d^2}{du^2},\frac{d^2}{dv^2}$?

I rearranged $u$ and $v$ in terms of $x$ and $y$, and I get $x = \frac{u + v}{\sqrt{2}}$ and $y = \frac{u - v}{\sqrt{2}}$. But how do I find $\frac{d^2}{du^2}$ and $\frac{d^2}{dv^2}$?

share|improve this question
    
Thanks, so is the answer for both, $0$? –  Ray Jan 16 '12 at 19:51
1  
What are you differentiating in this expression: $\frac{d^2}{dv^2}$? –  Emmad Kareem Jan 16 '12 at 19:52
    
@Emmad: It doesnt state anything in the problem, although you can take a look at the original question below (note the $v$ changes to $w$ in the original question). math.stackexchange.com/questions/99608/… –  Ray Jan 16 '12 at 19:55
    
sorry this is not too advanced for me :) –  Emmad Kareem Jan 16 '12 at 20:05
    
haha no problem, its ok. –  Ray Jan 16 '12 at 20:13

2 Answers 2

if $f$ is a function of $x,y$ then $$ \frac{\partial f}{\partial u}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u} $$ and $$ \frac{\partial^2 f}{\partial u^2}=\frac{\partial}{\partial u}\Big(\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}\Big) $$ $$ =\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial u^2}+\frac{\partial x}{\partial u}\Big(\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial u} +\frac{\partial^2 f}{\partial y\partial x}\frac{\partial y}{\partial u}\Big) +\frac{\partial f}{\partial y}\frac{\partial^2 y}{\partial u^2}+\frac{\partial y}{\partial u}\Big(\frac{\partial^2 f}{\partial x\partial y}\frac{\partial x}{\partial u}+\frac{\partial^2 f}{\partial y^2}\frac{\partial y}{\partial u}\Big) $$ $$ =\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial u^2}+\frac{\partial f}{\partial y}\frac{\partial^2 y}{\partial u^2}+2\frac{\partial^2 f}{\partial x\partial y}\frac{\partial x}{\partial u}\frac{\partial y}{\partial u}+\Big(\frac{\partial x}{\partial u}\Big)^2\frac{\partial^2 f}{\partial x^2}+\Big(\frac{\partial y}{\partial u}\Big)^2\frac{\partial^2 f}{\partial y^2} $$ note $$x=(u+v)\sqrt{2}, y=(u-v)\sqrt{2},\frac{\partial x}{\partial u}=\frac{\partial y}{\partial u}=\sqrt{2}$$ so the above simplifies to $$ 4\frac{\partial^2 f}{\partial x\partial y}+2\Big(\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}\Big) $$ you can do the same thing with respect to $v$ (i assumed equality of mixed partials above)

share|improve this answer
1  
Thanks, Im just wondering, how did you get $x=(u+v)\sqrt{2}$ and $y=(u-v)\sqrt{2}$? –  Ray Jan 16 '12 at 21:08

When you apply $\partial_u$ you take the directional derivative with respect to the vector $\frac 1{\sqrt 2}(1,1)$, so $\partial u =\frac 1{\sqrt 2}(\partial x+\partial y)$. Now take the directional derivative with respect to the vector $\frac 1{\sqrt 2}(1,1)$ of $\partial u$. We get $\partial^2_u=\frac 12(\partial^2_x+\partial_x\partial_y+\partial_y\partial_x+\partial_y^2)$. Now, do the same for $\partial_v$ and $\partial_v^2$.

share|improve this answer
    
Thanks, I understood what you did, but for $\partial^2u=\frac 12(\partial^2_x+\partial_x\partial y+\partial_y\partial_x+\partial_y^2)$, what exactly do I differentiate? What would $\partial^2_x$ be for example? –  Ray Jan 16 '12 at 21:12
    
$\partial_x^2$ apply to a function $f$ is only the the two times derivative with respect to $x$. –  Davide Giraudo Jan 16 '12 at 21:14
    
Thanks, but what would the $f$ be? $\frac{x+y}{\sqrt 2}$? –  Ray Jan 16 '12 at 21:22
    
Not necessary, here $f$ is only a function on which you apply the operator $\partial_u$. –  Davide Giraudo Jan 16 '12 at 21:27
    
Thanks, Im sorry but im really stuck on this question. How would I find $\partial^2_x,\partial_x\partial_y,\partial_y\partial_x,\partial_y^2$ etc.? As in, what do I differentiate when I am trying to find $\partial_x^2$? –  Ray Jan 16 '12 at 21:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.