Question

Partial Derivative

Answer 1

\[\psi=c \frac{\partial u}{\partial x} \]
find
\[\frac{\partial \psi}{\partial t} \]

Answer 2

c is constant? :o

Answer 3

yea

Answer 4

This is probably pretty simple.. But it has been so long since Ive delt with partial derivatives that I have forgotten

Answer 5

Assuming u is a function of both x and t,
and you're not taking a full derivative, then I think this is pretty straight forward:\[\Large\rm \psi=c u_x\]then,\[\Large\rm \psi_t=c u_{xt}\]yes? :o

Answer 6

If you prefer to write it out using the Leibniz notation, it would look like ummmmmmmmmmmmmmmmmm
Ahh I forget which one goes on the left,\[\Large\rm \frac{\partial \psi}{\partial t}=\frac{\partial^2 u}{\partial x \partial t}\]

Answer 7

I dunno, the question is confusing XD
I assume this has some larger context?

Answer 8

http://www.math.tau.ac.il/~turkel/notes/wave.pdf.. yea I am trying to characterize second order pde's. I was using this paper as a reference and just trying to figure out how the author came up with his substitution

Answer 9

I think what you did echoed what he did

Answer 10

So what are we trying to do? :O

I can vaguely remember Method 1,
the operators separate similar to the way we do the different of squares with real numbers.

Answer 11

essentially I am trying to reduce 2nd order PDE's into first order systems. Then with the system I can find the eigenvalues and classify the PDE. The problem I am running into is knowing which dang substitutions to make

Answer 12

Yah this v is getting me a little confusing >.<
Trying to follow the trail.

Answer 13

I think it is supposee to be psi.. that is the only logical explanation.

Answer 14

Yah that would make more sense.
\[\Large\rm (u_t)_t-c^2(u_x)_x=0\]Making the substitution:\[\Large\rm \phi = c u_x\]\[\Large\rm \psi=u_t\]Gives us:\[\Large\rm (\phi)_t-c(\psi)_x=0\]
Looks like we have to mess around with our substitution that we established.
Taking the derivative of \phi with respect to t,\[\Large\rm \phi_t = c u_{xt}\]There is probably some justification to show that this is equal to,\[\Large\rm \phi_t = c u_{tx}\]I can't remember how to do that though >.<
Anyway, when you get to that point,\[\Large\rm \phi_t = c (u_{t})_x\]\[\Large\rm \phi_t = c (\psi)_x\]Understand how they're establishing that `by definition` line using this approach?

Answer 15

Woops, we're trying to get the second part, aren't we? :)
lol my bad.

Answer 16

no, the definition is where I am confused. I get the second part

Answer 17

The definition is based on the substitution actually.
I didn't need to go through that stupid process.
After we plug in our sub:\[\Large\rm \phi_t-c\psi_x=0\]Add c\psi_x to each side.

Answer 18

I hate these stupid greek letters so much -_- lol

Answer 19

especially psi and phi together.. thanks for the work through.. makes since now.. I will never understand how people just come up with these substitutions on the fly

Answer 20

ya weird stuff >.<