Can you explain why $du/\sqrt{c-2u} = d\varepsilon$ using hyperbolic substitution integral becomes $A\operatorname{sech}^2(t)$?

Question

Here I have a 3rd order ODE wave equation $$ -cu' + 6uu' + u''' = 0 $$ where $u(\varepsilon) = u$ and $\varepsilon = x - ct$; a wave (assume $\varepsilon$ is one variable; hence $u' = du/d\varepsilon$).

I've integrated w.r.t $d\varepsilon$ and came up with $$ -cu + 3u^2 + u'' = \alpha $$ and again $$ \tfrac12 (u')^2 + u^3 - \tfrac12 c u^2 = \alpha u + \beta $$ Hence $$ \frac{du}{u\sqrt{c-2u}} = d\varepsilon $$

I've been told that the solution is: $$ u = 2 k^2 \operatorname{sech}^2 \bigl( k(\varepsilon - x_0) \bigr) $$ where $x_0$ is a constant and $2k^2$ is the amplitude of the wave.

Can you please explain WHY and HOW $$ u = 2k^2 \operatorname{sech}^2 \bigl( k(\varepsilon-x_0) \bigr) $$ using hyperbolic substitution integral?

Thanks.

Answer 1

It's hard to read but you seem to have the tnegral $$\int\frac{du}{u\sqrt{c-2u}}$$ where $c$ seems to be a constant. If I have misinterpreted your question, please disregard the rest of this post. Tf my interpretation is correct, the obvious substitution to make is $v=\sqrt{c-2u}$ Then $$u=\frac{1}{2}(c-v^2)$$ and the integral becomes $$-2\int \frac{vdv}{(c-v^2)v}=2\int\frac{dv}{v^2-c}$$ If $c>0$, the integral is $$\frac{1}{\sqrt c}\int(\frac{1}{v-\sqrt c}-\frac{1}{v+\sqrt c})dv$$ $$=\frac{1}{\sqrt c}(\ln(\vert (v-\sqrt c\vert)-\ln(\vert v+\sqrt c\vert))+K$$ If $c<0$, the integral is $$\frac{2}{\sqrt{\vert c\vert}}\arctan\frac{v}{\sqrt{\vert c\vert}}+K$$

Answer 2

I'll take it from $-c u + 3u^2 + u'' = \alpha$, see note below. Upon multiplication by $u'$ and integration, we end up with $$ -\tfrac12 c u^2 + u^3 + \tfrac12(u')^2 = \alpha u + \beta $$ where $\beta$ is a constant of integration (as done in OP). Setting $\alpha=0$ and $\beta=0$, we get $$ (u')^2 = u^2 \left(c - 2 u\right) . $$ For the case $u' = u \sqrt{c - 2 u}$ with $c > 0$, we get the separable equation in OP: $$\int \frac{\text d u}{u \sqrt{c - 2 u}} = \int \text{d}\varepsilon ,$$ where \begin{aligned} \int \frac{\text d u}{u \sqrt{c - 2 u}} &= \frac1{\sqrt c} \int \frac{\text d v}{v \sqrt{1 - v}}, \qquad {\scriptstyle [v=2u/c]}\\ &= -\frac2{\sqrt c} \int \frac{\text d w}{1-w^2}, \qquad {\scriptstyle [w=\sqrt{1-v}]} \\ &= -\frac2{\sqrt c} \operatorname{artanh} w . \end{aligned} Here, we have assumed that the integration constant vanishes. Thus, from $\int \text{d}\varepsilon = \varepsilon - x_0$, we get $$ \sqrt{1-2u/c} = -\tanh\left(\tfrac12\sqrt{c} (\varepsilon - x_0)\right) , $$ $$ \text{i.e.,}\qquad u = \frac{c}2\, \operatorname{sech}^2\left(\tfrac12 \sqrt{c}\; (\varepsilon-x_0)\right) . $$ Finally, letting $c = 4k^2$ with $k>0$ provides the desired solution.

Note: The initial differential equations follows from the Korteweg–De Vries (KdV) - type equation $u_t + 6uu_x + u_{xxx} = 0$, for which the traveling wave Ansatz $u = u(\varepsilon)$ with $\epsilon = x-ct$ produces $$ -cu'+6uu'+u''' = 0. $$ Integrating once (as done in OP), we get $-c u + 3u^2 + u'' = \alpha$ where $\alpha$ is a constant of integration.