# Second derivative relationships of substituted varaibles

1. Aug 6, 2012

### PhilDSP

I'm using variable substitution to solve a problem. Finding the relationships between the first derivatives of both sets is straightforward using the chain rule, but I'm uncertain if the way I'm determining the second derivative relationships is correct.

Given a description of a problem expressed in the x and y variables, I make the substitution as follows

$r = 3x + y \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = x + \frac{1}{5}y$

The chain rule gives

$\frac{\partial}{\partial x} = 3 \frac{\partial}{\partial r} + \frac{\partial}{\partial s} \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\partial}{\partial y} = \frac{\partial}{\partial r} + \frac{1}{5} \frac{\partial}{\partial s}$

Solving for $\frac{\partial}{\partial r}$ and $\frac{\partial}{\partial s}$ I get

$\frac{\partial}{\partial r} = - \frac{1}{2} \frac{\partial}{\partial x} + \frac{5}{2} \frac{\partial}{\partial y} \ \ \ \ \ \ \ \ \frac{\partial}{\partial s} = \frac{5}{2} \frac{\partial}{\partial x} - \frac{15}{2} \frac{\partial}{\partial y}$

Are the second derivative relationships from the chain rule then ?

$\frac{\partial^2}{\partial x^2} = - \frac{1}{2} \frac{\partial^2}{\partial r^2} + \frac{5}{2} \frac{\partial^2}{\partial s^2} \ \ \ \ \ \ \ \ \frac{\partial^2}{\partial y^2} = \frac{5}{2} \frac{\partial^2}{\partial r^2} - \frac{15}{2} \frac{\partial^2}{\partial s^2}$

Maybe this a strange example as the $\frac{5}{2} \frac{\partial}{\partial y}$ and $\frac{5}{2} \frac{\partial}{\partial x}$ in the separate equations seem to make the relationships symmetrical.

2. Aug 6, 2012

### Millennial

What are you differentiating?

3. Aug 6, 2012

### PhilDSP

There will be a function f(x, y) which is undefined at the moment because this example is only a mock up. So the differential operators operate on f(). Then the differential operators for the substituted variables also operate on f() as f(r, s).

That is to say $\frac{\partial}{\partial x} \rightarrow \frac{\partial f}{\partial x}$ and $\frac{\partial}{\partial r} \rightarrow \frac{\partial f}{\partial r}$

Last edited: Aug 6, 2012
4. Aug 6, 2012

### Muphrid

No, you should have cross terms. Try taking $\partial/\partial x = 3 \partial/\partial r + \partial/\partial s$ and applying it again. You should see that you get a cross term equal to $4 \partial^2/\partial r \partial s$. Think about what happens when you multiply $(3r+s)(3r+s)$.

5. Aug 8, 2012

### PhilDSP

Excellent hints Muphrid, thanks. But this is a bit tricky and I don't seem to have arrived at a result that is in sync with yours. Here is the breakdown:

The first order chain rule is

$$\frac{\partial}{\partial x} = \frac{\partial r}{\partial x} \frac{\partial}{\partial r} + \frac{\partial s}{\partial x} \frac{\partial}{\partial s} \ \ \ \ \ \ \ \ \ \ \ \ \frac{\partial}{\partial y} = \frac{\partial r}{\partial y} \frac{\partial}{\partial r} + \frac{\partial s}{\partial y} \frac{\partial}{\partial s}$$

The second order chain rule for x should be

$$\frac{\partial^2}{\partial x^2} = \frac{\partial}{\partial x} (\frac{\partial r}{\partial x} \frac{\partial}{\partial r} + \frac{\partial s}{\partial x} \frac{\partial}{\partial s})$$
$$\ \ \ \ \ \ \ = \ \ \frac{\partial r}{\partial x} \cdot \frac{\partial}{\partial x} (\frac{\partial}{\partial r}) \ \ + \ \ \frac{\partial}{\partial x} (\frac{\partial r}{\partial x}) \frac{\partial}{\partial r} \ \ + \ \ \frac{\partial s}{\partial x} \cdot \frac{\partial}{\partial x} (\frac{\partial}{\partial s}) \ \ + \ \ \frac{\partial}{\partial x} (\frac{\partial s}{\partial x}) \frac{\partial}{\partial s}$$

Then the results would seem to be

$$\frac{\partial^2}{\partial x^2} = \ \ 3 \cdot [3 \frac{\partial}{\partial r} + \frac{\partial}{\partial s}] \frac{\partial}{\partial r} \ \ + \ \ [3 \frac{\partial}{\partial r} + \frac{\partial}{\partial s}] (3) \frac{\partial}{\partial r} \ \ + \ \ 1 \cdot [3 \frac{\partial}{\partial r} + \frac{\partial}{\partial s}] \frac{\partial}{\partial s} \ \ + \ \ [3 \frac{\partial}{\partial r} + \frac{\partial}{\partial s}] (1) \frac{\partial}{\partial s}$$
$$\ \ \ \ \ \ \ \ = \ \ 9 \frac{\partial^2}{\partial r^2} \ \ + \ \ 3 \frac{\partial^2}{\partial r \partial s} \ \ + \ \ 9 \frac{\partial^2}{\partial r^2} \ \ + \ \ 3 \frac{\partial^2}{\partial r \partial s} \ \ + \ \ 3 \frac{\partial^2}{\partial r \partial s} \ \ + \ \ \frac{\partial^2}{\partial s^2} \ \ + \ \ 3 \frac{\partial^2}{\partial r \partial s} \ \ + \ \ \frac{\partial^2}{\partial s^2}$$
$$\ \ \ \ \ \ \ \ = \ \ 18 \frac{\partial^2}{\partial r^2} \ \ + \ \ 12 \frac{\partial^2}{\partial r \partial s} \ \ + \ \ 2 \frac{\partial^2}{\partial s^2}$$

Last edited: Aug 8, 2012
6. Aug 8, 2012

### PhilDSP

But now I'm thinking that the
$$\frac{\partial}{\partial x} (\frac{\partial r}{\partial x}) \frac{\partial}{\partial r} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\partial}{\partial x} (\frac{\partial s}{\partial x}) \frac{\partial}{\partial s}$$

terms should be interpreted as $\frac{\partial}{\partial x}$ operating on $\frac{\partial r}{\partial x}$ and $\frac{\partial s}{\partial x}$ rather than their separate values being multiplied.

Then those terms would vanish and the end result would be

$$\frac{\partial^2}{\partial x^2} \ \ = \ \ 9 \frac{\partial^2}{\partial r^2} \ \ + \ \ 6 \frac{\partial^2}{\partial r \partial s} \ \ + \ \ \frac{\partial^2}{\partial s^2}$$

Last edited: Aug 8, 2012
7. Aug 8, 2012

### Muphrid

Yeah, you've got it right now. The way I would've done it is really as simple as

$$\frac{\partial^2}{\partial x^2} = \frac{\partial}{\partial x} \frac{\partial}{\partial x} = \left(3 \frac{\partial}{\partial r} + \frac{\partial}{\partial s} \right)\left(3 \frac{\partial}{\partial r} + \frac{\partial}{\partial s} \right)$$

Which gets the same result.