7
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To get details see: equations 29,30,31,34,44,50,51

We have known some solitary wave solutions, given by(equations 1 to 5) $$ \phi_1=p_1\cos \tau \tag{1}$$ $$\phi_2=\frac16 g_2p_1^2\left(\cos(2\tau)-3\right)\tag{2}$$ $$\phi_3=p_3\cos \tau+\frac{1}{72}(4g_2^2-3\lambda)p_1^3\cos(3\tau)\tag{3}$$ $$\phi_4= \frac{1}{360}p_1^4\left(3g_4-5g_2\lambda+5g_2^3\right)\cos(4\tau) -\frac{1}{72}\left(8g_2(\nabla p_1)^2-12g_4p_1^4+16g_2^3p_1^4 -24g_2p_1p_3-23g_2\lambda p_1^4-8g_2p_1^2\right)\cos(2\tau) -g_2p_1^2-g_2p_1p_3+\frac{1}{6}g_2\lambda p_1^4-g_2(\nabla p_1)^2 +\frac{31}{72}g_2^3p_1^4-\frac{3}{8}g_4p_1^4 \tag{4}$$ \begin{equation} p_5=\frac{\sqrt 2}{9\sqrt 3}\left( Y-\frac{1235}{32}S^2Z+\frac{1503}{16}Z-24S-\frac{17}{3}S^3 +\frac{11525}{384}S^5\right) \end{equation} Now we will set $\tau=0$ in these above equation. Now consider some conditions:

  1. $$S=p_1\sqrt{\lambda}$$
  2. \begin{equation} p_3=\frac{\sqrt{2}}{3\sqrt{3}}\left( \frac{65}{8}Z-\frac{8}{3}S-\frac{19}{12}S^3 \right)\,. \end{equation} 3.value of the some constants $g_2=-\frac32$, $g_3=\frac12$ and $g_i=0$ for $i\geq 4$ ,$ \lambda= \frac{3}{2}$

and equations of spherical symmetry

  1. \begin{equation} \frac{d^2S}{d\rho^2}+\frac{D-1}{\rho}\,\frac{dS}{d\rho} -S+S^3=0 \end{equation}

Now I need write the equations (4)and (5) as $$\phi_4^{(\tau=0)}=\frac{1}{9}\left[ \frac{65}{4}SZ+10\left(\frac{dS}{d\rho}\right)^2 +\frac{8}{3}S^2-\frac{125}{12}S^4 \right]$$ $$\phi_5^{(\tau=0)}=\frac{1}{9}\sqrt{\frac{2}{3}}\Biggl[ Y-\frac{2275}{64}S^2Z+\frac{1503}{16}Z -\frac{15}{32}S\left(\frac{dS}{d\rho}\right)^2 -24S-\frac{595}{96}S^3+\frac{11285}{384}S^5 \biggr]$$ My Problem is how the factor $\frac{dS}{d\rho}$ arises in the above two equations? For further details: equations 29,30,31,34,44,50,51 If you have problem to understand the questions then ask me please. Thanks in advance

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    $\begingroup$ $S = p_1 \sqrt{\lambda}$, assuming $\lambda$ is a constant we have $\nabla p_1 = \frac{1}{\sqrt{\lambda}} \nabla S$. Since $S$ is spherically symmetric $\nabla S = (\partial_\rho S) \partial_\rho$ where $\rho$ is the radial coordinate. $\endgroup$ – Willie Wong May 26 '13 at 15:33
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    $\begingroup$ The question is widely applicable to a large audience... It would help if you explained why you think so. $\endgroup$ – Did Jun 1 '13 at 9:59
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It's because the gradient $\nabla S$ is just $\partial S/\partial \rho$ (due to spherical symmetry), and $S$ is just a constant times $p_1$. Since everything else except $p_3$ is a constant, you can just plug this all in and get the answer.

For the fifth equation $\phi_5$, take equation 34 and plug in 47, and use the spherical gradient as in the previous case.

After multiplying out equation 34 and substituting $\tau=0$, the only term with the gradient is $\frac{12S(\nabla S)^2}{384\sqrt{\lambda}}$. But using the spherical gradient and formula for $\lambda$, this is $\sqrt{\frac{2}{3}}\frac{3S(\partial S/\partial \rho)^2}{96}$. But in the last equation of your post, the term with the derivative simplifies to $\sqrt{\frac{2}{3}}\frac{5S(\partial S/\partial \rho)^2}{96}$. Because they are so close, there is probably just a typo; equation 34 should probably say 15.

Take their equation 34 and set $\tau=0$ and you will get $p_5+\frac{S^5}{1152\sqrt{\frac{3}{2}}}(\frac{4g_5}{3}+2)-\frac{S}{384\sqrt{\frac{3}{2}}}[(\frac{40g_5}{3}+2)SZ+12S^2-12(\partial S/\partial \rho)^2-(\frac{20g_5}{3}-2)S^4]$. Plugging in for $p_5$, we get $\frac{\sqrt{2}}{9\sqrt{3}}(Y-\frac{1235}{32}S^2Z+\frac{1503}{16}Z-24S-\frac{17}{3}S^3+\frac{1125}{384}S^5)+\frac{S^5}{1152\sqrt{\frac{3}{2}}}(\frac{4g_5}{3}+2)-\frac{S}{384\sqrt{\frac{3}{2}}}[(\frac{40g_5}{3}+2)SZ+12S^2-12(\partial S/\partial \rho)^2-(\frac{20g_5}{3}-2)S^4]$.

Collecting like terms, we get

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    $\begingroup$ From en.wikipedia.org/wiki/… using the fact it's spherically symmetric. $\endgroup$ – Brian Rushton Jun 4 '13 at 15:51
  • $\begingroup$ Can you do it for $\phi_5$ please? $\endgroup$ – Complex Guy Jun 4 '13 at 15:55
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    $\begingroup$ Never mind, I found it. $\endgroup$ – Brian Rushton Jun 4 '13 at 16:10
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    $\begingroup$ Don't forget it is all multiplied by S/384. Doing it my way gives you 5/96 as a coefficient, but they have 3/96, so they may have a typo, because there is no other factor that has $\nabla$ in it. $\endgroup$ – Brian Rushton Jun 4 '13 at 17:34
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    $\begingroup$ That makes sense; looking through, they never state the full equations. I believe that they simply derived them outside of the paper and wrote them down. But you can get a lot of the terms correctly from 34 and 47, so I would use that as a guide. But I get the feeling that they derived the equations through a great deal of effort and just wrote it down without showing the difficult derivation. $\endgroup$ – Brian Rushton Jun 4 '13 at 19:12

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