Is there a method for taking an asymptotic solution of a PDE and using it to find a general solution to the equation?

Question

I have found a special solution to the Estevez-Mansfield-Clarkson equation.
$u(y,t) = \frac{y^2}{3 \beta t}$ solves $u_{yttt} + \beta u_yu_{yt} + \beta u_{yy} u_t + u_{tt} = 0$ This is the asymptotic solution, which is the general solution with all the constants set equal to zero. Is there any way to use this special solution to find a more general solution?

Answer 1

Yep, this is useful method. Helps to solve the Schrodinger equation in various scenarios.

$-\hbar^2/2m (d^2\psi/dx^2)+(1/2)kx^2\psi = E\psi$

$d^2\psi/dx^2+(2mE/\hbar^2-mkx^2/\hbar^2)\psi =0$

Let $x=\alpha u$

$\frac{1}{\alpha^2} \frac{d^2\psi}{du^2}+(2mE/\hbar^2-mkx^2/\hbar^2)\psi=0$

$ \frac{d^2\psi}{du^2}+(2mE\alpha^2/\hbar^2-mkx^2\alpha^2/\hbar^2)\psi=0$

$mkx^2\alpha^2/\hbar^2=mk\alpha^4u^2/\hbar^2$

We simplify the equation if $mk \alpha^4/\hbar^2=1\implies \alpha=(\hbar^2/mk)^{1/4}\implies \alpha ^2=\hbar/m\omega$

and $(1/2)m\omega^2x^2=(1/2)kx^2\implies \omega^2=k/m\implies m^2\omega^2=mk$

$\frac{d^2\phi}{du^2}+(2E/\omega \hbar-u^2)\phi=0$

Asymptotically, this becomes $\frac{d^2 \phi}{du^2}-u^2\phi=0$

$\psi = e^{bx^2}$

$\psi' =2bxe^{bx^2}$

$\psi''= 2be^{bx^2}+4b^2x^2e^{bx^2}$

$2e^{bx^2}+4b^2x^2e^{bx^2}-x^2e^{bx^2}=0 \implies b=\pm (1/2)$ where we take the minus sign to yield a normalizable function.

$\phi = He^{-u^2/2}$

$\phi ' = -uHe^{-u^2/2}+H'e^{-u^2/2}$

$\phi '' =-He^{-u^2/2}-uH'e^{-u^2/2}+u^2He^{-u^2/2}+H''e^{-u^2/2}-uH'e^{-u^2/2}$

$\phi'' = -He^{-u^2/2}-2uH'e^{-u^2/2}+u^2He^{-u^2/2}+H''e^{-u^2/2}$

$-He^{-u^2/2}-2uH'e^{-u^2/2}+u^2He^{-u^2/2}+H''e^{-u^2/2}+(2E/\hbar \omega)He^{-u^2/2} - u^2He^{-u^2/2}=0$

$\implies H''-2uH'+(2E/\hbar \omega -1)H=0$

So by stripping out the asymptotical part, the equation has been transformed into terms of H as a function of $x$.

This is similar to the metho of Frobenius in that effort is made to simplify the powers of the independent variable involved in the solution.