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Dickfore
Dickfore is offline
#14
Aug22-10, 12:42 PM
P: 3,015
Well, I can't seem to find an error in my derivation and I can't decipher your writing, so I will assume the derivation I posted is correct. As someone already mentioned in this thread, it seems impossible that the equation can be solved in a closed form. So, I'm afraid you are left with solving it numerically.

Last time we ended at this step:

Quote Quote by Dickfore View Post
[tex]
\left\{\begin{array}{rcl}
\frac{d v}{d x} & = & p \\

\frac{d p}{d x} & = & \frac{v}{a} \, \left[p + \frac{n}{2 \, b} \, v^{2} \, (v^{2} - c) \right]
\end{array}\right.
[/tex]

where we also made the transformation [itex]p/2 \rightarrow p[/itex] again.
The boundary conditions that you gave translate to the following:

[tex]
\begin{array}{l}
x \rightarrow -\infty \Rightarrow v = \sqrt{c} \\

x = 0 \Rightarrow v = \sqrt{b + c}
\end{array}
[/tex]
The connection with the old variables is given by:

[tex]
u(x) = [v(x)]^{2}
[/tex]

When working with numerics, it is best to get rid of as many parameters as possible. Let's scale everything:

[tex]
\begin{array}{l}
x = x_{0} \, \bar{x} \\

v = v_{0} \, \bar{v} \\

p = p_{0} \, \bar{p}
\end{array}
[/tex]

Then the equations become:

[tex]
\begin{array}{rcl}
\frac{v_{0}}{x_{0}} \, \frac{d \bar{v}}{d \bar{x}} & = & p_{0} \, \bar{p} \\

\frac{p_{0}}{x_{0}} \, \frac{d \bar{p}}{d \bar{x}} & = & \frac{v_{0}}{a} \, \left[ p_{0} \, \bar{p} + \frac{n}{2 b} \, v^{2}_{0} \, \bar{v}^{2} \left(v^{2}_{0} \, \bar{v}^{2} - c \right) \right]
\end{array}
[/tex]

I think it is most convenient to make this choice:

[tex]
v^{2}_{0} = c, \; p_{0} = \frac{n}{2 b} \,v^{4}_{0}. \; \frac{p_{0}}{x_{0}} = \frac{v_{0} \, p_{0}}{a}
[/tex]

[tex]
x_{0} = \frac{a}{\sqrt{c}}, p_{0} = \frac{n \, c^{2}}{2 b}, \; v_{0} = \sqrt{c} \Rightarrow \frac{x_{0} \, p_{0}}{v_{0}} = \frac{a n c}{2 b} \equiv k
[/tex]

Also, let's get rid of the bars above the symbols again:

[tex]
\begin{array}{rcl}
\frac{d v}{d x} & = & k \, p \\

\frac{d p}{d x} & = & v \, \left[ p + v^{2} \, (v^{2} - 1) \right]
\end{array}
[/tex]

With the boundary conditions being:

[tex]
\left\{\begin{array}{l}
\sqrt{c} \, v = \sqrt{c}, \; x \rightarrow -\infty \\

\sqrt{c} \, v = \sqrt{b + c}, \; x = 0
\end{array}\right. \Rightarrow \left\{\begin{array}{l}
v = 1, \; x \rightarrow -\infty \\

v = \sqrt{1 + \frac{b}{c}} = M, \; x = 0
\end{array}\right.
[/tex]

Instead of having this limit as [itex]x \rightarrow -\infty[/itex], let us make the simultaneous substitution [itex]x \rightarrow -x, p \rightarrow -p[/itex]. The equations become:

[tex]
\begin{array}{rcl}
\frac{d v}{d x} & = & k \, p \\

\frac{d p}{d x} & = & v \, \left[ p + v^{2} \, (v^{2} - 1) \right]
\end{array}
[/tex]

with the boundary conditions:

[tex]
\left\{\begin{array}{l}
v = \sqrt{1 + \frac{b}{c}} = M, \; x = 0 \\

v \rightarrow 1, \; x \rightarrow \infty \\
\end{array}\right.
[/tex]

Can you find the stationary points for this autonomous non-linear system. What is their type?