I was wondering on how can we transform a pde to an ode using
$$\partial_{tt}u+ \partial_{xx}u^2+\partial_{xxxx} u= 0$$
$$u=f(w)=f(x-vt)$$
how can I transform it into the ode
$$f'' +v^2f+f^2=aw+b$$
where the derivative here refer to $d/{dw}$?
My approach
Since $ u(x-vt)$
$u_x = f'(x-vt)$
$u_{xx} = f''(x-vt)$
and
$ u_t= -vf'(x-vt)$
$u_{tt} =v^2f''(x-vt)$
Then I integrate it to time with respect to $w$ and got the equation, My issue is that my derivative $u_x$ and $u_{xx}$ and $u_t$ and $u_{tt}$ aren't in respect to $w$.
For Part 2, we need to show that if
$\displaystyle \lim_{|w|\to\infty} f(w)=K $
and if
$\displaystyle \lim_{|w|\to\infty} \frac{df}{dw}=0$
and if
$\displaystyle \lim_{|w|\to\infty} \frac{d^2f}{dw^2}=0$
then $ K_+={(-v^2+ \sqrt{v^4 +4b})}/2$ , and $ K_-={(-v^2- \sqrt{v^4 +4b})}/2$.
I'm not sure how would I approach the second part but I know for sure that I will have something of the form $f^2+v^2f-b=0 $, which I fail to show using the conditions given by the question.
If $u=f(w)$ with $w = x-vt$, then you correctly showed that all its partial derivatives are also functions of $w$. Substituting the partial derivatives in the PDE, we have $$ v^2 f'' + (f^2)'' + f^{(4)} = 0 $$ which gives the second-order differential equation $$ f'' + v^2 f + f^2 = aw+b $$ after double integration w.r.t. $w$, where $a$, $b$ are constants of integration. The fact that $f\to K$ and that the derivatives of $f$ vanish at infinity yields $$ a=0 \qquad\text{and}\qquad v^2 K + K^2 = b \, . $$ In other words, we assume that the steady states of the differential equation are reached by $u = f(w)$ at infinity. Solving this quadratic equation gives the result.