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I am making the following variable change from $x_i,y_i\to u_i,v_i,w_i$ where we define

$$u_1=x_1+x_2+y_1,\quad u_2=x_1+x_2-y_1,\quad u_3=x_1-x_2+y_1,\\ v_1=x_3+x_4+y_2,\quad v_2=x_3+x_4-y_2,\quad v_3=x_3-x_4+y_2,\\ w_1=x_5+x_6+y_3,\quad w_2=x_5+x_6-y_3,\quad w_3=x_5-x_6+y_3, $$

which gives us that $$x_1=\frac{1}{2}(u_2+u_3),\quad x_2=\frac{1}{2}(u_1-u_3),\quad x_3=\frac{1}{2}(v_2+v_3),\quad x_4=\frac{1}{2}(v_1-v_3),\quad x_5=\frac{1}{2}(w_2+w_3),\quad x_6=\frac{1}{2}(w_1-w_3),\\ y_1=\frac{1}{2}(u_1-u_2),\quad y_2=\frac{1}{2}(v_1-v_2),\quad y_3=\frac{1}{2}(w_1-w_2).$$

The limits on all the original variables, $x_i,y_i$ is $$0\leq x_i<\infty,\,0\leq y_i<\infty,$$

and I'm trying to find the new integration limits. Analysing the above I get

$$x_1=0,\,\rightarrow u_2=-u_3,\quad u_1\in[0,\infty],\quad u_2\in[-\infty,\infty],\quad u_3\in[-\infty,\infty],\\ x_2=0,\,\rightarrow u_1=u_3,\quad u_1\in[0,\infty],\quad u_2\in[-\infty,\infty],\quad u_3\in[0,\infty],\\ x_3=0,\,\rightarrow v_2=-v_3,\quad v_1\in[0,\infty],\quad v_2\in[-\infty,\infty],\quad v_3\in[0,\infty],\\ x_4=0,\,\rightarrow v_1=v_3,\quad v_1\in[0,\infty],\quad v_2\in[-\infty,\infty],\quad v_3\in[0,\infty],\\ x_5=0,\,\rightarrow w_2=-w_3,\quad w_1\in[0,\infty],\quad w_2\in[-\infty,\infty],\quad w_3\in[-\infty,\infty],\\ x_6=0,\,\rightarrow w_1=w_3,\quad w_1\in[0,\infty],\quad w_2\in[-\infty,\infty],\quad w_3\in[0,\infty],\\ y_1=0,\,\rightarrow u_1=u_2,\quad u_1\in[0,\infty],\quad u_2\in[0,\infty],\quad u_3\in[-\infty,\infty],\\ y_2=0,\,\rightarrow v_1=v_2,\quad v_1\in[0,\infty],\quad v_2\in[0,\infty],\quad v_3\in[-\infty,\infty],\\ y_3=0,\,\rightarrow w_1=w_2,\quad w_1\in[0,\infty],\quad w_2\in[0,\infty],\quad w_3\in[-\infty,\infty]. $$

This led me to take the following limits for the new variables

$$\int_{-u_3}^{u_1} du_2\,\int_{-v_3}^{v_1} dv_2\,\int_{-w_3}^{w_1} dw_2\,\int_{0}^{\infty} du_1\,\int_{0}^{\infty} dv_1\,\int_{0}^{\infty} dw_1\,\int_{-u_2}^{u_1} du_3\,\int_{-v_2}^{v_1} dv_3\,\int_{-w_2}^{w_1} dw_3,$$

is this correct? Or can I use the condition $u_1=u_2$ etc. to write the limits independent of $u_1$? Thanks!

SOLUTION:

Following the answer given, if we just focus on the $u_i$ terms. Holding $u_1=x_1+x_2+y_1$ fixed,

$$u_2=x_1+x_2-y_1\\ \rightarrow u_2=u_1-2y_1\leq u_1\\ \rightarrow u_2=2(x_1+x_2)-u_1\geq -u_1$$

Now holding $u_2$ fixed

$$u_3=x_1-x_2+y_1\\ \rightarrow u_3=2x_1-u_2\geq -u_2\\ \rightarrow u_3=u_2-2(x_2-y_1)=u_1-2x_2\leq u_1 $$

So this implies to me the limits $$\int_0^{\infty}\,du_1\,\int_{-u_1}^{u_1}\,du_2\,\int_{-u_2}^{u_1}\,du_3$$

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You cannot directly make all the limits independent of each other. To show why using a simplified example, consider the case of

$$\int_0^\infty \int_0^\infty f(x, y) dy dx$$

under the transformation

$$\begin{eqnarray} u & = & x + y \\ v & = & x - y \end{eqnarray}$$

It's pretty clear that $u \in [0, \infty)$ because it's the sum of two positive numbers, but for any fixed value of $u$ notice that we can write

$$\begin{eqnarray} v & = & x - y\\ & = & u - 2y \leq u \\ & = & 2x - u \geq -u \end{eqnarray}$$

and those two inequalities are tight, meaning that $v \in [-u, u]$ and so after substitution our limits of integration become $\int_0^\infty \int_{-u}^u dv du$.

The fact that the limits of the inner integrals have become dependent on the outer variables is inherent to the chosen transformation. You can change the way the dependencies behave by changing the order the integrals are nested but in this example that just means you'll have an inner integral with respect to $u$ whose limits are functions of $v$ instead of the other way around.

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  • $\begingroup$ Thanks for your answer! In my case then does that mean that, choosing $u_1,v_1,w_1$ as fixed, that the $u_2,\,u_3,\,v_2,\,v_3,\,w_2,\,w_3,$ integrals will have limits from $[-u_1,u_1],\,[-v_1,v_1],\,[-w_1,w_1]$? $\endgroup$ Sep 21, 2023 at 8:36
  • $\begingroup$ Not quite. I recommend working very carefully through the limits for $u_1, u_2, u_3$ since the ones for the $v$s and $w$s will end up being very similar. Fix $u_1$, and work out the biggest and smallest values $u_2$ can take relative to it, then fix $u_2$ as well and work out what effect that has on the range of $u_3$. $\endgroup$
    – ConMan
    Sep 21, 2023 at 13:58
  • $\begingroup$ I have edited my question now with an attempt, is what I did correct? Also I accidentally edited your comment so sorry about that. $\endgroup$ Sep 21, 2023 at 15:33
  • $\begingroup$ Looks about right to me. $\endgroup$
    – ConMan
    Sep 21, 2023 at 23:50
  • $\begingroup$ Great, I will choose your answer now, thanks for your help! $\endgroup$ Sep 22, 2023 at 11:35

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