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.

  • $\begingroup$ 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 $\mathbb R^2\to\mathbb R$. $\endgroup$ Dec 15, 2012 at 13:48
  • $\begingroup$ @HenningMakholm sorry, you are right, they are $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ $\endgroup$
    – Sunny88
    Dec 15, 2012 at 13:50
  • $\begingroup$ Also, are all of $f_1$, $f_2$, $u_1$, $u_2$ given? $\endgroup$ Dec 15, 2012 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, 2012 at 13:53
  • $\begingroup$ I mean, do you know what they are? $\endgroup$ Dec 15, 2012 at 13:53

1 Answer 1


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.

  • $\begingroup$ 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, 2012 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. $\endgroup$ Dec 15, 2012 at 16:20

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