1
$\begingroup$

I need to find/define a function $G(x_1(t),x_2(t)) : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ such that the following holds: $$ \frac{\partial G(x_1(t),x_2(t))}{\partial x_1(t)} =u_1(t)$$ and $$ \frac{\partial G(x_1(t),x_2(t))}{\partial x_2(t)} =u_2(t)$$ where $x_1(t) = f_1(u_1(t),u_2(t))$ and $x_2(t) = f_2(u_1(t),u_2(t))$ $$ f_1,f_2: \mathbb{R} \times \mathbb{R} \to \mathbb{R} $$ How can I go about doing this?

Edit: $f_1$ and $f_2$ are bounded and continuously differentiable functions.

$\endgroup$
  • $\begingroup$ You write $f_1,f_2:\mathbb R\to\mathbb R$, but in the immediately preceding equations you're using them as if they are $\mathbb R^2\to\mathbb R$. $\endgroup$ – hmakholm left over Monica Dec 15 '12 at 13:48
  • $\begingroup$ @HenningMakholm sorry, you are right, they are $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ $\endgroup$ – Sunny88 Dec 15 '12 at 13:50
  • $\begingroup$ Also, are all of $f_1$, $f_2$, $u_1$, $u_2$ given? $\endgroup$ – hmakholm left over Monica Dec 15 '12 at 13:50
  • $\begingroup$ @HenningMakholm I am not sure what does "given" mean, but you can define G in terms of them. $\endgroup$ – Sunny88 Dec 15 '12 at 13:53
  • $\begingroup$ I mean, do you know what they are? $\endgroup$ – hmakholm left over Monica Dec 15 '12 at 13:53
1
$\begingroup$

You can't find $G$ uniquely, since your equation tells you only about its behavior near points that are hit by $(x_1,x_2)$.

What we can do is compute $\frac{d}{dt}G(x_1,x_2)$ by the chain rule:

$$ \frac{d}{dt}G(x_1,x_2) = u_1(t)\frac{d}{dt}x_1(t) + u_2(t)\frac{d}{dt}x_2(t) $$ where everything on the right is known. Arbitrarily setting $G(x_1(0),x_2(0))=0$ (any constant term can be added of course) we get $$ G(x_1(s),x_2(s)) = \int_0^s \left[u_1(t)\frac{d}{dt}x_1(t)+u_2(t)\frac{d}{dt}x_2(t) \right] \, dt $$

This gives you some values of $G$, and you'd better hope that if $(x_1(t),x_2(t))=(x_1(s),x_2(s))$ for any $t,s$, the computed values of $G$ match -- otherwise there's no solution. Similarly, you must have $u(t)=u(s)$, since otherwise you'll have conflicting demands on the partial derivatives of $G$ at that point.

Afterwards you need to choose neighboring values of $G$ such that the partial derivatives are right. This can be done if the curve described by $(x_1,x_2)$ is smooth enough, but not at all uniquely, of course. The simplest solution may be to use make $G(p)$ vary linearly on a short perpendicular to the main curve at $(x_1(t),x_2(t))$, with a slope chosen to make the partial derivatives come out right.

$\endgroup$
  • $\begingroup$ I didn't quite understand this part: "This gives you some values of G, and you'd better hope that if $(x_1(t),x_2(t))=(x_1(s),x_2(s))$ for any t,s, the computed values of G match -- otherwise there's no solution." So how do I know whether my $G()$ is correct or not? $\endgroup$ – Sunny88 Dec 15 '12 at 16:08
  • $\begingroup$ @Sunny88: The integral tells you what $G(a,b)$ must be for every $(a,b)$ that can be written as $(x_1(t),x_2(t))$. If there's an $(a,b)$ that can be written like that in two different ways, then the values prescribed by the integral had better be the same; otherwise your equation cannot be solved. $\endgroup$ – hmakholm left over Monica Dec 15 '12 at 16:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.