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Partial derivatives and change of variables

  • Thread starter miew
  • Start date
  • #1
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Homework Statement



Sorry I tried to use Latex but it didn't work out :/

Make the change of variables r = x + vt and s = x vt in the wave equation
partial^2y/partialx^2-(1/v^2)(partial^2y/partialt^2)=0

Homework Equations


partialy/patialx=(partialy/partialr)(partialr/partialx)+(partialy/partials)(partials/partialx)
partialy/partialy=(partialy/partialr)(partialr/partialt)+(partialy/partials)(partials/partialy)

partialr/partialx=1
partials/partialx=1
partialr/partialt=v
partials/partialt=-v


The Attempt at a Solution


I am a little bit confused about what to do. This is what I did:

Rearranging:
partial^2y/partialr^2(1-1/v)+partial^2y/partials^2(1+1/v)=0

I feel this is completely wrong, but I can't think of another way of doing it.
 

Answers and Replies

  • #2
hunt_mat
Homework Helper
1,739
18
Start from the basics:

[tex]
\begin{array}{ccc}
\frac{\partial}{\partial x} & = & \frac{\partial r}{\partial x} \frac{\partial}{\partial r}+\frac{\partial s}{\partial x} \frac{\partial}{\partial s} \\
\frac{\partial}{\partial t} & = & \frac{\partial r}{\partial t} \frac{\partial}{\partial r}+\frac{\partial s}{\partial t} \frac{\partial}{\partial s}
\end{array}
[/tex]

Computing these yields:

[tex]
\begin{array}{ccc}
\frac{\partial}{\partial x} & = & \frac{\partial}{\partial r}+\frac{\partial}{\partial s} \\
\frac{\partial}{\partial t} & = & v\frac{\partial}{\partial r}-v\frac{\partial}{\partial s}
\end{array}
[/tex]

Then just apply these again to get the second order derivatives.
 
  • #3
27
0
Yes, that is what i did and I got:

partial^2y/partialx^2=partial^2y/partialr^2+partial^2y/partials^2
partial^2y/partialt^2=vpartial^2y/partialr^2-v(partial^2y/partials^2)

Is that right? And then I just plugged that in the initial equation.
 
  • #4
hunt_mat
Homework Helper
1,739
18
Just use:
[tex]
\frac{\partial^{2}}{\partial x^{2}}=\left(\frac{\partial}{\partial r}+\frac{\partial }{\partial s}\right)\left(\frac{\partial}{\partial r}+\frac{\partial }{\partial s}\right)
[/tex]
 
  • #5
27
0
Could you explain that ? And why what I did is wrong ? Can't I just take the partial derivative of partialy/partialx ?
 
  • #6
hunt_mat
Homework Helper
1,739
18
Where dies y come into it, I thought the original variables were x and t which you were transforming into r and s?
Okay, you should know that:
[tex]
\frac{\partial}{\partial x}=\left(\frac{\partial}{\partial r}+\frac{\partial }{\partial s}\right)
[/tex]

So just apply this operator twice.
 
  • #7
27
0
Y comes from the original equation,

partial^2y/partialx^2-(1/v^2)(partial^2y/partialt^2)=0

I did what you said and I got

4partial^2y/(partialr * partials)

Is that right ?
 
  • #8
hunt_mat
Homework Helper
1,739
18
That looks very good. set that equal to zero and you have your original equation which you have now reduced to:
[tex]
\frac{\partial^{2}y}{\partial r\partial s}=0
[/tex]
 
  • #9
27
0
Thank you very much !

How can I solve that equation ?
 
  • #10
hunt_mat
Homework Helper
1,739
18
integrate once and what do you get?
 
  • #11
27
0
do i get y=rs? I am confused...
 
  • #12
hunt_mat
Homework Helper
1,739
18
Integrate once w.r.t. r say and you obtain:
[tex]
\frac{\partial y}{\partial s}=f(s)
[/tex]
Now what would you do?
 
  • #13
27
0
now Integrate with respect to s and I get: f(s)f(r)??
 
  • #14
hunt_mat
Homework Helper
1,739
18
No, you're forgetting your basic integration:
[tex]
y=\int f(s)ds+g(r)
[/tex]
The integral of f(s) w.r.t. s is just another function, call it h(s) amd trhat is your solution.
 
  • #15
27
0
Oh okay ! thank you so much ! you were really helpful :)
 

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