 Solution

Introduce the Euler differential operator $D=x{\partial }_x$. The fourth-order spatial expression is $D^{4}u$. The moving boundary has similarity coordinate $\xi =1$, so its two conditions giveHence $\alpha =0$ and $v^{\prime }(1)=1$. With $D_{\xi }=\xi {\partial }_{\xi }$, substitution in the equation yieldsThis is a Cauchy-Euler equation; putting $r=\log \xi$ turns it into the constant-coefficient equation $V^{(4)}-\beta V^{\prime }=0$. Its characteristic polynomial is $n(n^3-\beta )$.

If $\beta >0$, write $a={\beta }^{1/3}$. The roots are $0,a,-a/2\pm i\sqrt{3}a/2$. Decay excludes the constant and growing modes, leavingThe boundary conditions force $A=0$ and $B=2/(\sqrt{3}a)$. If $\beta <0$, writing $\beta =-a^3$ leaves only one decaying mode, $C{\xi }^{-a}$; $v(1)=0$ then forces the zero function, which fails the derivative condition. If $\beta =0$, the solutions are polynomials of degree at most three in $\log \xi$, and none can both decay and satisfy $v^{\prime }(1)=1$.

Consequently all real self-similar solutions are the one-parameter familyThe decaying profile for a fourth-order dilation equation oscillates in logarithmic distance. No positivity condition on $u$ was imposed, so these sign changes are permitted.