Partial derivatives and change of variables

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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