Question

pde example problem

Answer 1

so basically I am stuck at this one part of example 1. I've been trying to redo it step by step but I really want to know why the r_x on one half of the equation becomes (r_x)^2 and when I am taking the second derivative, there is a r_x hanging around. Why is that happening?

Answer 2

like at the r_xx part ... I know that I have to take the derivative again, but what is that r_x doing there?

Answer 3

at
u_xx=f''(r)(r_x)^2+f'(r)r_xx
why is it (r_x)^2?

Answer 4

u_x = f'(r)r_x

if I take the derivative to bring u_xx
I will have f'(r)+r_xx+ r_xf'(r) but that isn't the case since there is another r_x lurking which makes it f'(r)+r_xx+(r_x)^2f'(r) ummm what?! why?

Answer 5

\[\LARGE u_x=f'(r)r_x\] Yeah this has messed me up before, so many little details. So now when we take the derivative wrt x we just have to be careful when applying the chain and rules.\[\LARGE u_{xx}= \frac{\partial}{ \partial x}[f'(r)r_x]\]
\[\LARGE \frac{\partial}{ \partial x}[f'(r)]r_x+ f'(r)\frac{\partial}{ \partial x}(r_x)\]

\[\LARGE [f''(r)r_x]r_x+ f'(r)[r_{xx}]\]

Answer 6

I know it's like ODE remix... like I've been reading the book and I get all the ode stuff and then there's extras and I'm like O________O

Answer 7

so there is an extra r_x due to the chain rule?

Answer 8

because I understand the f'' part... not the why is there an extra r_x what :O

Answer 9

Well remember what the prime really means? \[\LARGE f'(r)=\frac{df}{dr}\] so if we take the derivative with respect to x, then we have: \[\LARGE \frac{\partial}{\partial x}[f'(r)]=\frac{\partial}{\partial x}[\frac{df}{dr}]\] But what you're doing as your next step is really just: \[\LARGE \frac{d}{dr}[f'(r)]=f''(r)\] So you're forgetting to account for the fact that r is a function of x and you're taking the derivative with respect to x, not r. I hope that makes it clearer maybe?

Answer 10

can you give me an example?

Answer 11

using that book example ^^

Answer 12

ok so how to I take the derivative with respect to r?

Answer 13

Well as it stands, remember \[\LARGE r(x,y,z)=\sqrt{x^2+y^2+z^2}\] So if you have a function of r, like f(r) you really have this, and your confusion comes from it being shorthand from lazy authors who don't want to write it all out: \[\LARGE f(r)=f(r(x,y,z))\] So if you take the derivative of f with respect to x you have from the chain rule: \[\LARGE \frac{\partial}{\partial x}[f(r(x,y,z))]=\frac{\partial f}{\partial x}\] But of course df/dx is not something we can easily calculate unless we just plug in the function r(x,y,z) everywhere in f(r) where r occurs. So instead what we do is expand it in terms of functions we have already so that we can more easily play with it. \[\LARGE \frac{\partial f}{\partial x}=\frac{d f}{dr}\frac{\partial r}{\partial x}\]

Answer 14

ahhh so that's where it came from . it's like chain rule within a chain rule.

Answer 15

Yeah that's all, nothing too fancy really, just sort of like getting to the point in math where there are all sorts of techniques for making what you write smaller and easier to write, at first its confusing then it makes sense because, yeah, screw writing out all those little d's.

Answer 16

the book was written by 2 retired professors from my university. So is that where the confusion comes from?

Answer 17

I doubt it, this is pretty standard stuff, they're not doing anything new here that you won't see anywhere else.

Answer 18

hmmmmmmmmmm, but some books have errors or leave something out right?

Answer 19

No, all books are perfect and never leave anything out ever.

Answer 20

not differential equations by baer. crazy ode text

Answer 21

lol