Question

Prove Theorem 9.7 .2 by completing the following argument. According to Green's theorem in the plane, if \(C\) is a sufficiently smooth simple closed curve, and if \(F\) and \(G\) are continuous and have continuous first partial derivatives, then $$ \int_{C}[F(x, y) d y-G(x, y) d x]=\iint_{R}\left[F_{x}(x, y)+G_{y}(x, y)\right] d A $$ where \(C\) is traversed counterclockwise and \(R\) is the region enclosed by \(C .\) Assume that \(x=\phi(t), y=\psi(t)\) is a solution of the system ( 15) that is periodic with period \(T\). Let \(C\) be the closed curve given by \(x=\phi(t), y=\psi(t)\) for \(0 \leq t \leq T\). Show that for this curve the line integral is zero. Then show that the conclusion of Theorem 9.7 .2 must follow.

Short answer

#tag_title#Step 2: Show that the line integral is zero#tag_content# To show that the line integral is zero, we need to show that the given parameterized solution satisfies: $$ \int_{0}^{T}\left[F(\phi(t), \psi(t))\frac{d\psi(t)}{dt}-G(\phi(t), \psi(t))\frac{d\phi(t)}{dt}\right] dt = 0 $$ We know that the parameterized solution $(x, y) = (\phi(t), \psi(t))$ is a periodic solution to the system. Thus, we have: $$ \begin{cases} \frac{d\phi}{dt} = F(\phi(t), \psi(t)) \\ \frac{d\psi}{dt} = G(\phi(t), \psi(t)) \end{cases} $$ Substituting this into the integral, we get: $$ \int_{0}^{T}\left[F(\phi(t), \psi(t))G(\phi(t), \psi(t))-G(\phi(t), \psi(t))F(\phi(t), \psi(t))\right] dt $$ Simplifying the integrand gives: $$ \int_{0}^{T}\left[F(\phi(t), \psi(t))G(\phi(t), \psi(t))-F(\phi(t), \psi(t))G(\phi(t), \psi(t))\right] dt = 0 $$ Since the integrand is identically zero, the line integral is indeed zero. #tag_title#Step 3: Show the conclusion of Theorem 9.7.2 follows#tag_content# Since the line integral of our parameterized periodic solution is zero, we can now attempt to show how this demonstrates the conclusion of Theorem 9.7.2. Theorem 9.7.2 states that if a sufficiently smooth simple closed curve C and two continuous functions, F and G, are given with continuous first partial derivatives, then the line integral of F and G along curve C is zero. We have already shown that the line integral of our parameterized periodic solution is zero. By Green's theorem, the double integral over the region R enclosed by curve C is zero as well: $$ \iint_{R}\left[F_{x}(x, y)+G_{y}(x, y)\right] d A = 0 $$ The zero line integral implies that the contribution to the total integral from all points inside the region R balances out. This means that there must be some equal and opposite contributions within the region, proving the conclusion of Theorem 9.7.2.

Step-by-step solution

Compute the line integral using Green's theorem

Using Green's theorem, the integral along the curve C can be written as: $$ \int_{C}[F(x, y) d y-G(x, y) d x]=\iint_{R}\left[F_{x}(x, y)+G_{y}(x, y)\right] d A $$ We are given a parameterized solution for the system \((x, y) = (\phi(t), \psi(t))\). So, let's transform the integral in terms of t: $$ \int_{0}^{T}\left[F(\phi(t), \psi(t))\frac{d\psi(t)}{dt}-G(\phi(t), \psi(t))\frac{d\phi(t)}{dt}\right] dt $$