Partial derivatives and change of variables

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Homework Help Overview

The discussion revolves around the application of partial derivatives and change of variables in the context of the wave equation. The original poster attempts to make a change of variables from \(x\) and \(t\) to \(r\) and \(s\) in the equation \(\partial^2y/\partial x^2 - (1/v^2)(\partial^2y/\partial t^2) = 0\).

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the computation of first and second-order partial derivatives after the change of variables. Some express confusion about the correctness of their attempts and seek clarification on the application of the chain rule. Others suggest specific forms for the derivatives and inquire about the role of the original variables in the transformation.

Discussion Status

There is an ongoing exploration of the correct application of derivatives and the implications of the change of variables. Some participants have provided guidance on how to approach the second-order derivatives, while others are questioning their understanding and the steps taken so far. The discussion reflects a mix of interpretations and attempts without a clear consensus on the final approach.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can reference or the methods they can employ. There is also a noted confusion regarding the integration steps and the relationship between the variables involved.

miew
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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.
 
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Start from the basics:

<br /> \begin{array}{ccc}<br /> \frac{\partial}{\partial x} &amp; = &amp; \frac{\partial r}{\partial x} \frac{\partial}{\partial r}+\frac{\partial s}{\partial x} \frac{\partial}{\partial s} \\<br /> \frac{\partial}{\partial t} &amp; = &amp; \frac{\partial r}{\partial t} \frac{\partial}{\partial r}+\frac{\partial s}{\partial t} \frac{\partial}{\partial s}<br /> \end{array}<br />

Computing these yields:

<br /> \begin{array}{ccc}<br /> \frac{\partial}{\partial x} &amp; = &amp; \frac{\partial}{\partial r}+\frac{\partial}{\partial s} \\<br /> \frac{\partial}{\partial t} &amp; = &amp; v\frac{\partial}{\partial r}-v\frac{\partial}{\partial s}<br /> \end{array}<br />

Then just apply these again to get the second order derivatives.
 
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.
 
Just use:
<br /> \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)<br />
 
Could you explain that ? And why what I did is wrong ? Can't I just take the partial derivative of partialy/partialx ?
 
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:
<br /> \frac{\partial}{\partial x}=\left(\frac{\partial}{\partial r}+\frac{\partial }{\partial s}\right)<br />

So just apply this operator twice.
 
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 ?
 
That looks very good. set that equal to zero and you have your original equation which you have now reduced to:
<br /> \frac{\partial^{2}y}{\partial r\partial s}=0<br />
 
Thank you very much !

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

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