Change of variables theorem problem

Question

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.

Answer 1

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 $$\phi(x,t):=\tilde\phi(x+ct,x-ct)\ .$$

Answer 2

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