Simplifying the Chain Rule for Partial Derivatives in PDEs

[x-is-y]

If $$z = f(x,y)$$ and $$x = r \cos{v}$$, $$y = r\sin{v}$$ the object is to show that $$d = \partial$$ since it's easier to do on computer

Show that:
$$\frac{d^2 z}{dr^2} + \frac{1}{r} \frac{dz}{dr} + \frac{1}{r^2} \frac{d^2 z}{dv^2} = \frac{d^2 z}{dx^2} + \frac{d^2 z}{dy^2}$$

It's from Adams calculus, will show where I get lost.

We have from the chain rule
$$\frac{dz}{dr} = \frac{dz}{dx} \frac{dx}{dr} + \frac{dz}{dy} \frac{dy}{dr}$$

But $$\frac{dx}{dr} = \cos{v}$$ and $$\frac{dy}{dr} = \sin{v}$$, so gives:

$$\frac{dz}{dr} = \frac{dz}{dx} \cos{v} + \frac{dz}{dy} \sin{v}$$

Now we need $$\frac{d^2 z}{dr^2}$$, where I end up in trouble. We have

$$\frac{d^2 z}{dr^2} = \frac{d}{dr} \cos{v} \frac{dz}{dx} + \frac{d}{dr} \sin{v} \frac{dz}{dy}$$ which can be written as

$$\frac{d^2 z}{dr^2} = \cos{v} \frac{d}{dr} \frac{dz}{dx} + \sin{v} \frac{d}{dr} \frac{dz}{dy}$$

Now my books says that

$$\frac{d}{dr} \frac{dz}{dx} = \cos{v} \frac{d^2 z}{dx^2} + \sin{v} \frac{d^2 z}{dy dx}$$ if I understand correct. But I don't see how you get to this ...

 

[elibj123]

Since z is a function of (x,y), the derivative of z in respect to x (or y), is another function of (x,y). Therefore the operator d/dr acts on the derivative of z the same way it acts on z.

Look at this:
$$\frac{d}{dr}(z)=\frac{dx}{dr}\frac{d}{dx}(z)+\frac{dy}{dr}\frac{d}{dy}(z)=$$

Substitude z by dz/dx and you get the final result which you have written.

 

[x-is-y]

Substitude z by dz/dx and you get the final result which you have written.

Thank you very much, the equation for d^2 z/dv^2 got even messier. But I think I understand, I solved it and I'm going to try other examples on this.