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I've decided to finish my education through completing my last exam (I've been working for 5 years). The exam is in multivariable calculus and I took the classes 6 years ago so I am very rusty. Will ask a bunch of questions over the following weeks and I love you all for helping me.

Q: Suppose that $f(x,y)$ fulfills the Laplace equation

$$\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2} = 0$$

Show that $g(x,y) = f(2x+y,x-2y)$ also fulfills the equation.

I understand everything about the teachers answer except one early part.

A: $$u=2x+y\\\\ v=x-2y$$ The chain rule give: $$\frac{\partial g}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x} = 2\frac{\partial f}{\partial u}+\frac{\partial f}{\partial v}$$ $$\frac{\partial g}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y} = \frac{\partial f}{\partial u}-2\frac{\partial f}{\partial v}$$

...

My question is:

How is he simplifying that last step, where he get 2.. + .. and .. - 2 ..?

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    $\begingroup$ He just plugs the values of $\frac{\partial u}{\partial x}$ and $\frac{\partial v}{\partial y}$. He differentiates the formulas $u = 2x+y$ and $v = x- 2y$. $\endgroup$
    – Evgeny
    Commented Aug 8, 2013 at 15:06

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You have

$$ u(x,y) = 2x + y $$

and

$$ v(x,y) = x-2y $$

So, you get :

$$ \frac{\partial u}{\partial x} = 2 $$

and

$$ \frac{\partial v}{\partial y} = -2 $$

which you can plug in the equations you have in your post.

Anyway, be careful to your notations. Use $\partial$ instead of $\mathrm{d}$.

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  • $\begingroup$ Thank you so much. I can't remember when to use the one and the other. Is it when it's partial derivatives? $\endgroup$
    – Mattis
    Commented Aug 8, 2013 at 15:37
  • $\begingroup$ It is. Thank you for your answer! $\endgroup$
    – Mattis
    Commented Aug 8, 2013 at 15:38
  • $\begingroup$ $\frac{\mathrm{d}}{\mathrm{dx}}$ is usually used for functions of one variable (for example : $\mathbb{R} \, \rightarrow \, \mathbb{R}$ or $\mathbb{R} \, \rightarrow \, \mathbb{R}^{n}$). $\frac{\partial}{\partial x}$ is used when you have more than one variable (for example $\mathbb{R}^{p} \, \rightarrow \, \mathbb{R}^{n}$, $p \in \mathbb{N}, \, p \geq 2$) $\endgroup$
    – pitchounet
    Commented Aug 8, 2013 at 15:42
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    $\begingroup$ $\frac{\text{d}}{\text{d}x}$ is also used for total derivative of multivariable functions. $\endgroup$
    – Ruslan
    Commented Aug 8, 2013 at 18:47
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Besides to correct points in @jibount's answer, you can use the chain rule to see the final results: $$g_x=f_u\cdot u_x+f_v\cdot v_x\to g_{xx}=(f_u\cdot u_x+f_v\cdot v_x)'=(f_u\cdot u_x)'+(f_v\cdot v_x)'$$ $$=(f_u)_x\cdot u_x+f_u\cdot u_{xx}+(f_v)_x\cdot v_x+f_v\cdot v_{xx}=...$$

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  • $\begingroup$ $\ddot\smile +1$ $\endgroup$
    – amWhy
    Commented Aug 9, 2013 at 0:07

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