# [SOLVED]Variable transformation

#### dwsmith

##### Well-known member
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 $$y = u^{-\sigma}$$?

Last edited:

#### Jester

##### Well-known member
MHB Math Helper
Is the power on the second term rhs of the equation correct?

#### dwsmith

##### Well-known member
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 $$u^{2(\sigma + 1)} = y^{-2(1+1/\sigma)}$$.

Last edited:

#### Jester

##### Well-known member
MHB Math Helper
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
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
$$\DeclareMathOperator{\arcsec}{arcsec}$$
$$\DeclareMathOperator{\sech}{sech}$$

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 $$u$$.

Last edited:

#### Jester

##### Well-known member
MHB Math Helper
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
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 $$B$$.

Last edited:

#### dwsmith

##### Well-known member
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.
$u(x) = \left(\frac{A}{B + Ac_1\cosh(Dx)}\right)^{1/\sigma}$

Last edited:

#### Jester

##### Well-known member
MHB Math Helper
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

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

#### dwsmith

##### Well-known member
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

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

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
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
$$\DeclareMathOperator{\sgn}{sgn}$$
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.

Last edited: