# How does this equation hold? $u_x=u_{x'}\frac {\partial x’}{\partial x}+u_{y'}\frac {\partial y’}{\partial x}$

Let $x'=x+a, \quad y'=y+b$

How are these two equations hold?

$u_x=u_{x'}\frac {\partial x’}{\partial x}+u_{y'}\frac {\partial y’}{\partial x} (*)$

$u_{xx}=u_{x'x'}\frac {\partial x’}{\partial x}+u_{x'y'}\frac {\partial y’}{\partial x}$

In $(*)$ isn't RHS equal to $2u_x$ whereas LHS is $u_x$ ? And also how do they get the second equation?

Also there is another one:

Here

$x=r\cos \theta ,\quad y=r\sin \theta ,\quad r=\sqrt{x^2+y^2},\quad \theta=\arctan(\frac y x )$

$u_x=u_{r}\frac {\partial r}{\partial x}+u_{\theta}\frac {\partial \theta}{\partial x}$

$u_{xx}=[u_{rr}\frac {\partial r}{\partial x}+u_{r\theta}\frac {\partial \theta}{\partial x}]\frac {\partial r}{\partial x}+u_{r}\frac {\partial^2 r}{\partial x^2}+[u_{\theta r}\frac {\partial r}{\partial x}+u_{\theta \theta}\frac {\partial \theta}{\partial x}]\frac {\partial \theta}{\partial x}+u_{\theta}\frac {\partial^2 \theta}{\partial x^2}$

• By chain rule for the first it should be $u_x=u_{x'}$ – user Dec 16 '17 at 23:21
• For the second by chain rule again:$u_{xx}=u_{x'x'}$ – user Dec 16 '17 at 23:24
• Infact $\frac {\partial x’}{\partial x}=1$ and $\frac {\partial y’}{\partial x}=0$ – user Dec 16 '17 at 23:26
• the new system make sense and seems a correct application of chain rule – user Dec 17 '17 at 0:01
• have you solved your doubts? – user Dec 17 '17 at 8:07

Of course, your confusion in this case has been caused by the lack of a clear picture about the several-variable chain rule. So, I'm going to give you an intuition about it at the end of the post, which is going to be pretty long and boring.

I'm just going to show you how $(*)$ is obtained and leave the rest to you. Consider $u$ as a function of two variables $x'$ and $y'$ where $x'$ and $y'$ themselves are functions of $x$ and $y$. By the chain rule, we have:

$$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial x'}\cdot \frac{\partial x'}{\partial x} + \frac{\partial u}{\partial y'}\cdot \frac{\partial y'}{\partial x}$$ This is the same as $(*)$ written in an expanded notation. This is also equal to:

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

Because $\frac{\partial x'}{\partial x}=1$ and $\frac{\partial y'}{\partial x}=0$.

Now I'll leave it to you to use the chain rule on this again. Can you do the rest now?

Why does chain rule work the way it does?

Suppose that $f: \mathbb{R}^2 \to \mathbb{R}$ is a scalar function of two real variables. We want to find a way to understand the change in the value of $f$ when we move a small step in our domain. In other words, we're looking for a way to define the derivative of $f$ at the point $(a,b)$ when we have traveled a small distance $\Delta =(\Delta x, \Delta y)$ in some direction. In other words, we want to study the value of

$$\Delta f(a,b) = f(a+\Delta x, b+\Delta y)-f(a,b), \hspace{10px} |\Delta| \to 0$$

I will define two families of auxiliary functions $g_b(x): \mathbb{R} \to \mathbb{R}$ and $h_a(y): \mathbb{R} \to \mathbb{R}$ given by $g_b(x)=f(x,b)$ and $h_a(y)=f(a,y)$. We are going to assume continuity and differentiability without mentioning them directly now to avoid confusion and too much rigor.

Let's suppose that we have: $$g_b(a+\Delta x)-g_b(a)=g'_b(a) \Delta x+o(|\Delta x|^2)$$ $$h_a(b+\Delta y)-h_a(b)=h'_a(b) \Delta y+o(|\Delta y|^2)$$

Now let's study $\Delta f(a,b)$: $$\Delta f(a,b) = f(a+\Delta x, b+\Delta y)-f(a+\Delta x, b)+f(a+\Delta x, b)-f(a,b)$$ $$\Delta f(a,b) = \left(f(a+\Delta x, b+\Delta y)-f(a+\Delta x, b)\right)+\left(f(a+\Delta x, b)-f(a,b)\right)$$ $$\Delta f(a,b) = \left(h_{a+\Delta x}(b+\Delta y)-h_{a + \Delta x}(b)\right)+\left(g_b(a+\Delta x)-g_b(a)\right)$$

Therefore,

$$\lim_{|\Delta| \to 0}\Delta f(a,b) = \lim_{|\Delta y| \to 0}\left(h_{a+\Delta x}(b+\Delta y)-h_{a + \Delta x}(b)\right)+\lim_{|\Delta x|\to 0}\left(g_b(a+\Delta x)-g_b(a)\right)$$

$$\lim_{|\Delta| \to 0}\Delta f(a,b) = \lim_{|\Delta y| \to 0}\left(h'_{a+\Delta x}(b) \Delta y+o(|\Delta y|^2)\right)+\lim_{|\Delta x|\to 0}\left(g'_b(a) \Delta x+o(|\Delta x|^2)\right)$$

Ignoring terms of higher order, it gives us:

$$\lim_{|\Delta| \to 0}\Delta f(a,b) = \lim_{|\Delta y| \to 0}\left(h'_{a+\Delta x}(b) \Delta y\right)+\lim_{|\Delta x|\to 0}\left(g'_b(a) \Delta x\right)$$

By continuity $\lim_{\Delta x \to 0}h'_{a+\Delta x}(b)=h'_{a}(b)$. Therefore,

$$\lim_{|\Delta| \to 0}\Delta f(a,b) = \lim_{|\Delta y| \to 0}\left(h'_{a}(b) \Delta y\right)+\lim_{|\Delta x|\to 0}\left(g'_b(a) \Delta x\right)$$

A little change of notation, $h'_{a}(b)=\frac{\partial f}{\partial y}(a,b)$ and $g'_{b}(a)=\frac{\partial f}{\partial x}(a,b)$, gives us the celebrated chain rule in multi-variable calculus for a scalar function of two variables:

$$df(a,b)=\lim_{|\Delta| \to 0}\Delta f(a,b) = \frac{\partial f}{\partial y}(a,b) \Delta y+\frac{\partial f}{\partial x}(a,b) \Delta x$$

If you take $f(x,y)=x$ and $f(x,y)=y$ separately, you'll see that $dx = \Delta x$ and $dy = \Delta y$.

Your equation is an extended form of chain rule. Suppose that your variables $x$ and $y$ are themselves functions of a third variable $t$. You can now see that:

$$\frac{df(a,b)}{dt}=\lim_{\Delta t \to 0} \frac{\Delta f(a,b)}{\Delta t} = \frac{\partial f}{\partial y}(a,b) \lim_{\Delta t \to 0} \frac{\Delta y}{\Delta t}+\frac{\partial f}{\partial x}(a,b) \lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t}$$

$$\frac{df}{dt} \mid_{t=t_0}=\frac{\partial f}{\partial y}\mid_{(a(t_0),b(t_0))} \frac{d y}{dt} \mid_{t=t_0}+\frac{\partial f}{\partial x} \mid_{(a(t_0),b(t_0))} \frac{dx}{dt} \mid_{t=t_0}$$

You have a common confusion with the chain rule. If $u=u\left(x,y\right)$ and $x^{\prime}=x^{\prime}\left(x,y\right)$, $y^{\prime}=y^{\prime}\left(x,y\right)$ are new variables, then

$$\frac{\partial u\left(x^{\prime},y^{\prime}\right)}{\partial x}=\frac{\partial u}{\partial x^{\prime}}\frac{\partial x^{\prime}}{\partial x}+\frac{\partial u}{\partial y^{\prime}}\frac{\partial y^{\prime}}{\partial x}\neq 2\frac{\partial u}{\partial x}$$

Given:

$$x'=x+a, \quad y'=y+b$$

It results that:

$$\frac {\partial x’}{\partial x}=1, \quad\frac {\partial y’}{\partial x}=0$$

Thus by chain rule:

$$u_x=u_{x'}\frac {\partial x’}{\partial x}+u_{y'}\frac {\partial y’}{\partial x}=u_{x'}$$

$$u_{xx}=u_{x'x'}\frac {\partial x’}{\partial x}+u_{x'y'}\frac {\partial y’}{\partial x}=u_{x'x'}$$

• Then do you agree with my interpretation? Why did you expect on RHS $2u_x$ of (*)? – user Dec 16 '17 at 23:40