Author Topic: HA10-P3  (Read 596 times)

Victor Ivrii

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Jeremy Li 2

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Re: HA10-P3
« Reply #1 on: November 30, 2015, 12:55:39 AM »
Part A

In this problem, the Lagrangian is
\begin{equation}
L=\frac{\sqrt{1+u'^2}}{c(x,u(x))}
\end{equation}

We need to calculate the Euler-Lagrange equation:
\begin{equation}
\frac{\partial L}{\partial u} - \frac{\partial}{\partial x} \frac{\partial L}{\partial u'} = 0
\end{equation}

Calculating each term
\begin{equation}
\frac{\partial L}{\partial u} = -\frac{\sqrt{1+u'^2}}{c^2\left(x,u(x)\right)} \frac{\partial c}{\partial y}
\end{equation}

\begin{equation}
\frac{\partial L}{\partial u'} = \frac{1}{c\left(x,u(x)\right)} \frac{u'}{\sqrt{1+u'^2}}\\
\frac{\partial}{\partial x} \frac{\partial L}{\partial u'} =
-\frac{1}{c^2\left(x,u(x)\right)} \frac{u'}{\sqrt{1+u'^2}} \left(\frac{\partial c}{\partial x}+\frac{\partial c}{\partial y}u'\right)
+\frac{1}{c\left(x,u(x)\right)}\frac{1}{\left(1+u'^2\right)^{3/2}} u''
\end{equation}

Plugging into $(2)$:
\begin{equation}
-\frac{\sqrt{1+u'^2}}{c^2\left(x,u(x)\right)} \frac{\partial c}{\partial y}
+\frac{1}{c^2\left(x,u(x)\right)} \frac{u'}{\sqrt{1+u'^2}} \left(\frac{\partial c}{\partial x}+\frac{\partial c}{\partial y}u'\right)
-\frac{1}{c\left(x,u(x)\right)}\frac{1}{\left(1+u'^2\right)^{3/2}}u''
=0
\end{equation}

We can simplify a little bit
\begin{equation}
\frac{1}{c\left(x,u(x)\right)^2\sqrt{1+u'^2}}\left(u'\frac{\partial c}{\partial x} - \frac{\partial c}{\partial y}\right)
-\frac{1}{c\left(x,u(x)\right)}\frac{1}{\left(1+u'^2\right)^{3/2}} u''
=0
\end{equation}
\begin{equation}
\frac{1}{\sqrt{1+u'^2}}\left(u'\frac{\partial c}{\partial x} - \frac{\partial c}{\partial y}\right)
=\frac{c\left(x,u(x)\right)}{\left(1+u'^2\right)^{3/2}} u''
\end{equation}

Will type up B later, Or if someone else wants to volunteer below. Since in part (b) the Lagrangian doesn't depend on $x$ explicitly, we can calculate the Hamiltonian ($H=u'L_{u'}-L$) and set it to a constant.
« Last Edit: November 30, 2015, 02:53:55 AM by Jeremy Li 2 »

Victor Ivrii

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Re: HA10-P3
« Reply #2 on: November 30, 2015, 09:34:06 AM »
Part (b) is simpler via Hamiltonian $H$

Jeremy Li 2

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Re: HA10-P3
« Reply #3 on: November 30, 2015, 04:31:16 PM »
Since $L$ does not depend on x explicitly, we have
\begin{equation}
H=u'L_{u'}=\frac{1}{c\left(u(x)\right)} \left(\frac{u'^2}{\sqrt{1+u'^2}} - \sqrt{1+u'^2}\right) = C
\end{equation}

Which simplifies to
\begin{equation}
H=-\frac{1}{c\left(u(x)\right)} \frac{1}{\sqrt{1+u'^2}}=C
\end{equation}

Letting $C=-1/A$,

\begin{equation}
-\frac{1}{c\left(u(x)\right)} \frac{1}{\sqrt{1+u'^2}}=-\frac{1}{A}\\
\frac{A^2}{c^2\left(u(x)\right)}=u'^2+1\\
\frac{du}{dx}=\sqrt{\frac{A^2}{c^2\left(u(x)\right)}-1}
\end{equation}

We get
\begin{equation}
x=\int\frac{1}{\sqrt{\frac{A^2}{c^2\left(u(x)\right)}-1}}du
\end{equation}

I don't think there's much more to be done without a formula for $c(y)$.