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[SOLVED] Variable transformation

dwsmith

Well-known member
Feb 1, 2012
1,673
Given the following ODE
\[
\left(\frac{du}{dx}\right)^2 = \mu u^2 - \frac{2\alpha}{\sigma + 2}u^{\sigma + 2} - \frac{\gamma}{\sigma + 1}u^{2(\sigma + 1)}
\]
How do I obtain
\[
u(x) = \left(\frac{A}{B + \cosh(Dx)}\right)^{1/\sigma}
\]
where
\(A = \frac{(2 + \sigma)B\mu}{\alpha}\), \(B = \text{sgn}(\alpha)\left[1 + \frac{(2 + \sigma)^2\gamma}{(1 + \sigma)\alpha^2}\mu\right]^{-1/2}\), and \(D = \sigma\sqrt{\mu}\) with the variable transformation \(y = u^{-\sigma}\)?
 
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Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
Is the power on the second term rhs of the equation correct?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Is the power on the second term rhs of the equation correct?
It should be \(\sigma + 2\).

I don't see how this variable change makes anything easier.
\(u = y^{-1/\sigma}\) so \(du = -\frac{1}{\sigma}y^{-(1 + 1/\sigma)}dy\) and \(u^{\sigma + 2} = y^{-(1+2/\sigma)}\) and \(u^{2(\sigma + 1)} = y^{-2(1+1/\sigma)}\).
 
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Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
You'll notice that each term has $u^{-2/\sigma}$ common so they cancel. So when you simplify your equation you should get

$\left(\dfrac{du}{dx}\right)^2 = au^2 + bu + c$

for some $a, b$ and $c$. This is solvable.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
You'll notice that each term has $u^{-2/\sigma}$ common so they cancel. So when you simplify your equation you should get

$\left(\dfrac{du}{dx}\right)^2 = au^2 + bu + c$

for some $a, b$ and $c$. This is solvable.
What method is used to solve an ODE of this form?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
\(\DeclareMathOperator{\arcsec}{arcsec}\)
\(\DeclareMathOperator{\sech}{sech}\)

How can I get to the final result from (see work below)?

For the moment, let \(\mu\sigma^2 = a\), \(\frac{2\alpha\sigma^2}{\sigma + 2} = b\), and \(\frac{\gamma\sigma^2}{\sigma + 1} = c\).
Then our ODE becomes
\[
\int\frac{du}{\sqrt{au^2 - bu - c}} = \pm x.
\]
For expression inside the square root, we will need to complete the square, make the subtition \(s = u\sqrt{a} - \frac{b}{2\sqrt{a}}\), \(s = i\sec(t)\sqrt{\frac{-b^2 - 4ac}{4a}}\), and \(t = \arcsec\Big(\frac{-2iu\sqrt{a}}{\sqrt{-b^2 - 4ac}}\Big)\).
After completing the square, we will have obtained
\[
\frac{1}{\sqrt{\frac{-b^2 - 4ac}{4a} +
\Big(u\sqrt{a} - \frac{b}{2\sqrt{a}}\Big)^2}}.
\]
Using our \(s\) substitution, \(ds = \sqrt{a}du\) so the integral becomes
\[
\int\frac{ds}{\sqrt{\frac{-b^2 - 4ac}{4a} + s^2}} = \pm x\sqrt{a}.
\]
Using our final substitution for \(s\), we have \(ds = i\tan(t)\sec(t)\sqrt{\frac{-b^2 - 4ac}{4a}}dt\), and then using our \(t\) substitution, we now have
\begin{align*}
\int\sec(t)dt &= \pm x\sqrt{a}\\
\ln(\tan(t) + \sec(t)) &= \pm x\sqrt{a}\\
\tan(t) + \sec(t) &= \big\{e^{x\sqrt{a}}, e^{-x\sqrt{a}}\big\}\\
2\sqrt{a}\sqrt{au^2 - bu - c} + 2au - b &=
\big\{\cosh\big(x\sqrt{a}\big), \sinh\big(x\sqrt{a}\big)\big\}
\end{align*}
Then when I keep the \(\cosh\) term and solve for \(u\), I have
\[
u(x) = \frac{2b + \cosh(Dx) + (b^2 + 4ac)\sech(Dx)}{4a}
\]
I have not idea how to back sub in the definitions of \(a\), \(b\), and \(c\) to get the correct form of \(u\).
 
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Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
You can do what you did but try differentiating. In doing so we obtain

$2 u' u'' = (2 a u +b) u' $

giving $u ' = 0$ or $u'' = au + b/a$. I think the latter is easier to solve than your approach.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
You can do what you did but try differentiating. In doing so we obtain

$2 u' u'' = (2 a u +b) u' $

giving $u ' = 0$ or $u'' = au + b/a$. I think the latter is easier to solve then your approach.
This should be $u'' = au + b/2$, correct?

The solution to this ODE after making the substitution \(u^{-\sigma} = y\) is
\[
u(x) = \left(\frac{1}{\frac{b}{2a} + c_1\cosh(Dx)}\right)^{1/\sigma} = \left(\frac{1}{\frac{\alpha}{(\sigma + 2)\mu} + c_1\cosh(Dx)}\right)^{1/\sigma}
\]
as \(x\to\infty\), \(u\to 0\), and \(u'\to 0\).
\[
u'(x) = -\frac{c_1 D \sinh (D x) \left(\frac{1}{\frac{b}{2 a}+c \cosh (D x)}\right)^{\frac{1}{\sigma }+1}}{\sigma }
\]
I don't see a meaningful way to find \(c_1\). If I assume \(c_1 = 1\), then I can't produce the desired result either. So I think \(c_1\) isn't 1 since then I could factor out something to potential get \(A\) and \(B\).
 
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dwsmith

Well-known member
Feb 1, 2012
1,673
I have determined that \(c_1 = \frac{1}{A} = \frac{\alpha}{(2 + \sigma)\mu B}\).
How do I get this value though?

I just worked through the problem to see what I needed for \(c_1\) without using the BC at infinity.
\[
u(x) = \left(\frac{A}{B + Ac_1\cosh(Dx)}\right)^{1/\sigma}
\]
 
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Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
Yes, that was a typo. It should be $b/2$. As for finding the constant. Go back to the original equation after the substitution, namely

$ y'^2 = \mu \sigma^2 y^2 - \dfrac{2 \alpha \sigma^2}{\sigma+2} y - \dfrac{ \gamma \sigma^2}{\sigma+1}$

and substitute your solution.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Yes, that was a typo. It should be $b/2$. As for finding the constant. Go back to the original equation after the substitution, namely

$ y'^2 = \mu \sigma^2 y^2 - \dfrac{2 \alpha \sigma^2}{\sigma+2} y - \dfrac{ \gamma \sigma^2}{\sigma+1}$

and substitute your solution.
With \(y(x) = \frac{b}{2a} + c_1\cosh(Dx)\), I get
\[
\left\{c\to -\frac{i \sqrt{\alpha ^2 (\sigma +1)+\gamma \mu (\sigma +2)^2}}{\sqrt{\mu } \sqrt{\sigma +1} (\sigma +2) \sqrt{\mu \left(\mu \sigma ^2 \sinh ^2\left(\mu \sigma ^2 x\right)-\cosh ^2\left(\mu \sigma ^2 x\right)\right)}}\right\},\left\{c\to \frac{i \sqrt{\alpha ^2 (\sigma +1)+\gamma \mu (\sigma +2)^2}}{\sqrt{\mu } \sqrt{\sigma +1} (\sigma +2) \sqrt{\mu \left(\mu \sigma ^2 \sinh ^2\left(\mu \sigma ^2 x\right)-\cosh ^2\left(\mu \sigma ^2 x\right)\right)}}\right\}
\]
which isn't \(\frac{1}{A}\) and the solution is complex. I used Mathematica to solve for the coefficient so I doubt there is a mistake.
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
First, is it $\dfrac{b}{2a}$ or $- \dfrac{b}{2a}$. It really makes a difference. Second, you can use the hyperbolic sinh-cosh identity. Try that.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
\(\DeclareMathOperator{\sgn}{sgn}\)
First, is it $\dfrac{b}{2a}$ or $- \dfrac{b}{2a}$. It really makes a difference. Second, you can use the hyperbolic sinh-cosh identity. Try that.
If I use \(-\frac{b}{2a}\) and simplify, I get:
\[
\text{c1}\to \pm\frac{\sqrt{\alpha ^2 (\sigma +1)-\gamma \mu (\sigma +2)^2}}{\sqrt{\mu ^2 (\sigma +1) (\sigma +2)^2}}
\]
When I do some algebra, I get
\[
\frac{\alpha\sqrt{1 - \frac{\gamma\mu(2 + \sigma)^2}{\alpha^2(1 + \sigma)}}}{\mu(2 + \sigma)}
\]
but \(\frac{1}{A}\) is
\[
\frac{\alpha\sqrt{1 + \frac{\gamma\mu(2 + \sigma)^2}{\alpha^2(1 + \sigma)}}}{\mu(2 + \sigma)\sgn(\alpha)}
\]
So I have a negative when I need a positive and how do I pick up the \(\sgn(\alpha)\)?

Since \(\alpha^2\) is being factor out through the radical, \(\alpha\) can be both + or - so that is where the \(\sgn(\alpha)\) is picked up, correct? However, I still have a negative issue in the radical.
 
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