Recent content by Stalker_VT

isn't this just 0? f(x) = \frac{\pi}{3} \frac{df}{dx}= 0 \frac{dx}{df}= 0

Thanks for the reply, I got the first step of what you did, and got the equation \frac{\partial u}{\partial t} = \frac{\partial v}{\partial t} - c \frac{\partial v}{\partial \xi} And I realized (by trying do to do it) that when you do a similar thing for ux you get an equation very similar...

O I am sorry, I am new to these forums and was not sure which section to post in because it was "like" a homework question, but I really was just looking for the concept behind that one step. Sorry again...Didn't mean to waste your time reading two posts.

I think i found the solution to my problem but i was hoping to have someone check to make sure i did not make a mistake. \xi = x - ct...... (1) u(t,x) = v(t,\xi)......(2) Taking the derivative d[u(t,x) = v(t,\xi)] \frac{\partial u}{\partial t}dt + \frac{\partial u}{\partial x}dx =...

Homework Statement I am trying to solve the transport PDE using a change of variables and the chain rule, and my problem seems to be with the chain rule. The PDE is: \frac{\partial u}{\partial t}+c\frac{\partial u}{\partial x} = 0 ......(1) The change of variables (change of reference frame)...

But what if there is no equation describing the relation between x and t? As in they are completely independent. Does this not imply that for a differential change in t there would be NO change in x? i.e. \frac{dx}{dt}=0 Also, micromass and Curious3141, in reference to wheter one can take...

So then does \frac{dx}{dt} = 1? meaning that \frac{D f(x,t)}{D t} = \frac{\partial f(x,t)}{\partial x } + \frac{\partial f(x,t)}{\partial t} Thank you

That's what i thought. Thanks! Now what if you had an equation of multiple independent variables such as f(x,t) and you take the Total Derivative of f(x,t) with respect to t...using the chain rule \frac{D f(x,t)}{D t} = \frac{\partial f(x,t)}{\partial x }\frac{dx}{dt} + \frac{\partial...

how do you use the change of variable: \xi = x - ct on the PDE ut + c ux = 0 where u(t,x) = v(t,x-ct) = v(t,\xi) to get vt = 0 I am trying to follow the steps of Peter J. Olver in this document http://www.math.umn.edu/~olver/pd_/lnw.pdf starting on pg 4 but get stuck here...i...

If you are given an equation y(x), where y is the dependent variable, and x is the independent variable, can you take a derivative with respect to the dependent variable? ex. y = 5x dy/dx = 5 dx/dy = 1/5 Thank you for any help