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Flows of the system $\dot{x}=y$ and $\dot{y}=x$ are area-preserving. Show by computation that the area of the unit disk is the same after time is moved forward 2 units by the flow.

My solution to the system is $x=\frac{1}{2}(e^{t}+e^{-t})$ and $y=\frac{1}{2}(e^{t}-e^{-t})$. I can easily check that these are solutions. But I don't know whether they are flows, because when t=0, both solutions equal constants.

More importantly, I don't understand why they make a disk. $x^{2}+y^{2}=(e^{2t}+e^{-2t})/2$. So a circle means $(e^{2t}+e^{-2t})/2\leq1$?

Since the original area is 1, I guess I need to show that the new integral is equal to 1, too. After the time moves forward, $x=\frac{1}{2}(e^{t+2}+e^{-t+2})$ and $y=\frac{1}{2}(e^{t+2}-e^{-t+2})$. I notice that we can let $u=x+y=e^{t+2}$ and $v=x-y=e^{-t+2}$. But how to set up the integration? Thanks.

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Your formulas only give the solution for the initial values $x(0)=1$ and $y(0)=0$. You need to find the solution for arbitrary initial conditions $x(0)=x_0$ and $y(0)=y_0$. –  Hans Lundmark Sep 20 '12 at 7:36

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This is a homogeneous linear system of two ordinary first-order differential equations, so its solution set forms a two-dimensional vector space. You've found one particular solution. To find all of them systematically, write

$$ \pmatrix{\dot x\\\dot y}=\pmatrix{0&1\\1&0}\pmatrix{x\\y} $$

and note that the eigenvalues of the matrix are $1$ and $-1$ with eigenvectors $\pmatrix{1\\1}$ and $\pmatrix{1\\-1}$, respectively, so the general solution is

$$ \pmatrix{x\\y}=c_1\pmatrix{1\\1}\mathrm e^t+c_2\pmatrix{1\\-1}\mathrm e^{-t}\;. $$

Your solution has $c_1=c_2=\frac12$.

The point $(x_0,y_0):=(x(0),y(0))$ flows to the point $(x(t),y(t))$ in time $t$. Since

$$ \pmatrix{x_0\\y_0}=c_1\pmatrix{1\\1}+c_2\pmatrix{1\\-1}\;, $$

$c_1$ and $c_2$ correspond to orthogonal components of $\pmatrix{x_0\\y_0}$ along the two orthogonal eigenvectors. The flow scales one component by $\mathrm e^t$, the other by $\mathrm e^{-t}$. Thus, the unit disk is deformed to an ellipse with semi-axes $\mathrm e^t$ and $\mathrm e^{-t}$ in the directions of the eigenvectors, and the area of such an ellipse is $\pi\mathrm e^t\mathrm e^{-t}=\pi$ as required.

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I understand the general solution part, but how to show that "the unit disk is deformed to an ellipse with semi-axes"? Thanks. –  user38335 Sep 20 '12 at 22:14
    
Do you understand the statements preceding that one? –  joriki Sep 20 '12 at 22:16

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