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I'm having trouble with the change of variables theorem in two variables.

The theorem says: $$\iint f(x,y)dxdy=\iint f(x(u,v),y(u,v))|J|dudv$$ Where J is the Jacobian.

If $f(x,y)=xy$ where $x,y \in \mathbb{R} $

$$\iint f(x,y)dxdy= \frac{(x·y)^2}{4}$$

if: $x=u+v; y=u-v$

$$\iint f(x,y)dxdy=\iint (u+v)(u-v)|J|dudv=2(\frac{u^3v}{3}-\frac{v^3u}{3})$$

Where |J|=2. If I undo the change I'd have to get the same result at the two cases but if I do I don't. What I am doing wrong?

Where my problem appeared

I was solving:

$$c^2 \phi_{xx}-\phi_{tt} =h(t,x)$$ with c constant and $$h(x,t)=tsin(x);x,t \in \mathbb{R}$$

I find the characteristics are: $$\xi=x+ct;\eta=x-ct$$ Then I make a change of variables and rewrite the PDE to: $$\phi_{\xi\eta}=\frac{1}{4c^2}H(\xi,\eta)$$ Then I have: $$\phi(\xi,\eta)=\frac{1}{4c^2}\int(\int (H(\xi,\eta)d\eta))d\xi+A(\xi)+B(\eta)$$ with $$H(\xi,\eta)=h(x=x(\xi,\eta),t=t(\xi,\eta)))=\frac{\xi-\eta}{2c}sin(\frac{\xi+\eta}{2}) $$

The problem

Solving this: $$\frac{1}{4c^2}\int(\int (H(\xi,\eta)d\eta))d\xi$$ If I undo with the change of variables theorem I just have: $$\frac{1}{4c^2} \iint 2c t sin(x) dxdt $$ Where $2c$ is the Jacobian. That integral is much easier than the one with $\eta$ and $\xi$ but I did both integrals and get to different results and I don't know what I am doing wrong.

Progress

As a commenter stated I could be mistaking the notion of double integral and iterative integral. I'm reviewing those concepts but I have solved nothing yet.

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    $\begingroup$ What set are you integrating the function over? $\endgroup$ – wj32 Feb 14 '13 at 12:55
  • $\begingroup$ I know the example I wrote is really easy but I just want to know what I am doing wrong if the theorem must be met. $\endgroup$ – Jorge Feb 14 '13 at 12:56
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    $\begingroup$ The notion of indefinite integral does not make sense in two (or more) dimensions! (Except in a more advanced sense involving differentials, but that is beside the point here.) $\endgroup$ – Harald Hanche-Olsen Feb 14 '13 at 12:57
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    $\begingroup$ @George: There is a very big difference between a double integral (integrating a function $f:\mathbb{R}^2\to\mathbb{R}$ on some set) and an iterated integral. You should review these concepts as well as the precise statement of Fubini's theorem. (This is why the notation $\iint_D f$ can be misleading. I like $\int_D f$ instead.) $\endgroup$ – wj32 Feb 14 '13 at 13:02
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    $\begingroup$ Note that a) you can get a proper double integral using \iint instead of \int\int, and b) equations with integrals and fractions look much nicer and are easier to read if you typeset them as displayed equations by enclosing them in double dollar signs. $\endgroup$ – joriki Feb 14 '13 at 13:27
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The purpose of the variable transformation $(x,t)\to(\xi,\eta)$ is to convert the differential operator $c^2D_{xx}-D_{tt}$ ("a sum of squares") into a product $D_\xi D_\eta\>$, in order to make the following procedure possible. If there would be some magic to perform a corresponding thing in $(x,t)$-space one never would have bothered with the transformation in the first place.

You have $$\tilde\phi_{\xi\eta}(\xi,\eta)={1\over 4 c^2} H(\xi, \eta)={\xi-\eta\over 8c^3}\sin{\xi+\eta\over2}$$ (if your computations are correct). It follows that $$\tilde\phi_\xi(\xi,\eta)={1\over 8c^3}\int (\xi-\eta)\sin{\xi+\eta\over2}\ d\eta=:\Phi(\xi,\eta)\ ,$$ where $\Phi(\xi,\eta)$ is an explicit expression in $\xi$ and $\eta$, and the implicit integration constant may depend on $\xi$ (it's the $A'(\xi)$ of your argument). Integrating once more we obtain $$\tilde\phi(\xi,\eta)=\int \Phi(\xi,\eta)\ d\eta=:\Psi(\xi, \eta)\ .$$ Here again $\Psi(\xi,\eta)$ is an explicit expression in $\xi$ and $\eta$, and we now have arrived at two arbitrary added functions $A(\xi)$ and $B(\eta)$.

Having $\tilde \phi(\xi,\eta)$ in our hands we now can go back to the original variables $x$ and $t$ by writing $$\phi(x,t):=\tilde\phi(x+ct,x-ct)\ .$$

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  • $\begingroup$ What are you writing is correct but that's not my issue. I can do the integral in $\xi$ and $\eta$ with any problem but I was wondering if I could undo the variable change first to make the integral easier. I did it by the two ways and obtaining different results. What I've found so far I've explained it at the my answer to the question. I'm not sure if it is correct but it has worked for me in an invented example. Check it out if you wish. Thanks for your time. $\endgroup$ – Jorge Feb 14 '13 at 17:22
  • $\begingroup$ @George: See my edit. $\endgroup$ – Christian Blatter Feb 15 '13 at 13:12
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It seems that to apply the change of variables theorem I must know the intervals first.

If I apply the change of variables of an indefinite multiple integral I am, in fact, integrating over an interval and that interval is not the same as the one I was integrating at the first indefinite multiple integral so I get different results.

Source: http://www.physicsforums.com/showthread.php?t=211258

In fact: lets try to integrate the area of a circle with radius 1.

In cartesian coordinates: $$\int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}dxdy=\pi$$

In polar coordinates: $$\int_{0}^{1}\int_{0}^{2\pi}rdrd\theta=\pi$$

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  • $\begingroup$ So in the end. At my case it's better to integrate at $\xi,\eta$ and then undo the variable change to avoid any problem. $\endgroup$ – Jorge Feb 15 '13 at 9:50

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