Higher order differential equations and the chain rule (2 variables)

[simba_]

Homework Statement

The function F is defined by F (r, θ) = f (x(r, θ), y(r, θ)), where f is twice continuously
differentiable and
x(r, θ) = r cos θ, y(r, θ) = r sin θ.
Use the chain rule to find

d²F/dθ²

Homework Equations

The Attempt at a Solution

I know that dF/dθ = (df/dx)(dx/dθ) + (df/dy)(dy/dθ)
and i can solve this

I know the next step is (d/dθ)(dF/dθ) but this is were i get lost. I must of missed the class this was explained in :(

 

[HallsofIvy]

$$\frac{\partial F}{\partial \theta}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}$$

Since $x= r cos(\theta)$, $y= r sin(\theta)$, $x_\theta= -r sin(\theta)$ and $y_\theta= r cos(\theta)$

So
$$\frac{\partial F}{\partial \theta}= - r sin(\theta)\frac{\partial f}{\partial x}+ r cos(\theta) \frac{\partial f}{\partial y}$$
which, I presume, is what you got.

Now,
$$\frac{\partial^2 F}{\partial \theta^2}= \frac{\partial}{\partial \theta}\left(\frac{\partial F}{\partial \theta}\right)$$

But the formula
$$\frac{\partial \phi}{\partial \theta}= \frac{\partial \phi}{\partial x}\frac{\partial x}{\partial \theta}+ \frac{\partial \phi}{\partial y}\frac{\partial y}{\partial \theta}$$
is true for any function, $\phi$ and, in particular, for
$$\phi= \frac{\partial F}{\partial \theta}= - r sin(\theta)\frac{\partial f}{\partial x}+ r cos(\theta) \frac{\partial f}{\partial y}$$


That is,
$$\frac{\partial^2 F}{\partial \theta}= \frac{\partial}{\partial \theta}\left(-r sin(\theta)\frac{\partial f}{\partial x}+ r cos(\theta)\frac{\partial f}{\partial y}\right)$$
$$+ \frac{\partial}{\partial\theta}\left((-r sin(\theta)\frac{\partial f}{\partial x}+ r cos(\theta)\frac{\partial f}{\partial y}\right)$$
$$= - r cos(\theta)\frac{\partial f}{\partial x}- r sin(\theta)\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\right)$$
$$- r sin(\theta)\frac{\partial f}{\partial y}+ r cos(\theta)\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial y}\right)$$

This is getting awkward to write as a single formula so just note that you do
$$\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\right)$$
and
$$\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial y}\right)$$
using that same "chain rule formula":
$$\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\right)$$
$$= -r sin(\theta)\frac{\partial^2 f}{\partial x^2}+ r cos(\theta)\frac{\partial^2 f}{\partial x\partial y}$$
and
$$\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial y}\right)$$
$$= r cos(\theta)\frac{\partial f}{\partial x\partial y}- r sin(\theta)\frac{\partial^2 f}{\partial y^2}$$

 

[simba_]

Thanks for that, I know it must of taken a while to write out.
I had a good look at it there now but some parts still don't make sense to me. I'll have another look at it tomorrow with fresh eyes :)