I have a question on a project that I am working on. I have included a large amount of the background information so that all relevant information is included, however the question is as follows (it is also #12 below):
Using a simplified version of $H$ given as $H(p,x)=\frac{p^2}{2}+\frac{x^2}{2}$, consider a particle with initial position $a$ and no initial momentum. Solve Hamilton's equations for this particle. In other words, find the curve $\gamma(t)=(x(t),p(t))$ that solves Hamilton's equations with the initial conditions $x(0)=a$ and $p(0)=0$.
EDIT This is the answer that I have given so far:
Using the example of the Harmonic Oscillator, with potential energy is $V(x) = \displaystyle\frac{kx^2}{2}$ for some constant $k$, and allowing $k$ and $m$ to be equal to $1$ for simplicity, the Hamiltonian is given by: $$H(p,x) = \frac{p^2}{2} + \frac{x^2}{2}.$$ We can solve Hamilton's equations for a particle with initial position $a$ and no initial momentum, to find closed curve $\gamma(t)=(x(t),p(t))$, with $x(0)=a$ and $p(0)=0$. In actuality, we are finding $\dot{x}$ and $\dot{p}$. Starting with the former, $\dot{x}$:
$$ \frac{\partial x}{\partial t} =\dot{x} =\frac{\partial H}{\partial p} =\frac{\partial}{\partial p}\left(\frac{p^2}{2} + \frac{x^2}{2}\right) =\frac{\partial}{\partial p}\left(\frac{x^2}{2}\right) + \frac{\partial}{\partial p}\left(\frac{p^2}{2}\right) =0+\frac{2}{2}p =p.$$
We now integrate $\displaystyle\frac{\partial x}{\partial t}=p$ to get $x(t)$: $$x(t)=\int{\frac{\partial x}{\partial t} dt} = \int{p dt} = pt + c_1.$$ Since we know $x(0)=a$, $a$ is the initial condition of $x(t)$, which implies $c_1=a$, giving $x(t)=pt+a$. Similarly, we can find $p(t)$:
$$ \frac{\partial p}{\partial t} =\dot{p} =-\frac{\partial H}{\partial x} =-\frac{\partial}{\partial x}\left(\frac{p^2}{2} +\frac{x^2}{2 }\right) =-\frac{\partial}{\partial x}\left(\frac{x^2}{2}\right) + \frac{\partial}{\partial x}\left(\frac{p^2}{2}\right) =-\frac{2}{2}x+0=-x. $$ We now integrate $\frac{\partial p}{\partial t}=-x$ to get $p(t)$:
$$p(t)=\int{\frac{\partial p}{\partial t} dt} = \int{-x dt} = -xt + c_2.$$
Since we know $p(0)=0$, $0$ is the initial condition of $p(t)$, which implies $c_2=0$, giving $p(t)=-xt$. Therefore, $\gamma(t)=(x(t),p(t))$ is given by the closed curve solving Hamilton's equations $\gamma(t)=\big(pt+a,-xt\big)$.
Here is the full information:
With
$H = \dfrac{1}{2}(p^2 + x^2), \tag{1}$
we have
$\dfrac{\partial H}{\partial p} = p \tag{2}$
and
$\dfrac{\partial H }{\partial x} = x, \tag{3}$
so that Hamilton's equations become
$\dot x = \dfrac{\partial H}{\partial p} = p \tag{4}$
and
$\dot p = -\dfrac{\partial H}{\partial x} = -x; \tag{5}$
we thus have
$\ddot x = \dot p = -x, \tag{6}$
or
$\ddot x + x = 0. \tag{7}$
The solutions of (7) take the general form
$x = A\sin t + B \cos t \tag{8}$
as is well-known. From (8),
$p = \dot x = A\cos t - B\sin t, \tag{9}$
and we can find $A$ and $B$ by plugging in the initial conditions
$a = x(0) = A \sin 0 + B \cos 0 = B, \tag{10}$
$0 = p(0) = A \cos 0 - B \sin 0 = A. \tag{11}$
Thus
$x(t) = a\cos t, \tag{12}$
$p(t) = -a \sin t; \tag{13}$
the solution curve is
$\gamma(t) = (a\cos t, -a \sin t). \tag{14}$
Hope this helps. Cheers,
and as always,
Fiat Lux!!!