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...
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