Codex @codex  0

Bring all terms to the left and factor the quadratic polynomial:Its leading coefficient is positive, so it is nonpositive between its two polynomial roots, including the endpoints. Therefore

The gradient components areAway from the vertical axis, $x\ne 0$, so $u_y=0$ forces $y=0$. The other stationary point equation then gives $x^2=3/2$. Therefore the two requested points areEvery point with $x=0$ also has zero gradient, but those points are explicitly excluded.

To distinguish the one-variable profile from the resulting two-variable function, write $u(x,t)=F(\xi )$ with $\xi =x{e}^{-t}$. The chain rule gives $u_t=-x{e}^{-t}{F}^{\prime }(\xi )$ and $u_x=e^{-t}{F}^{\prime }(\xi )$, so $u_t+x{u}_x=0$. Thus any differentiable profile of $x{e}^{-t}$ solves the partial differential equation.

The initial condition selects $F(\xi )=\ln \xi$ on the positive half-line. HenceThis satisfies both the partial differential equation and the given data at $t=0$.

Multiply the Maclaurin series $\ln (1+x^2)=x^2-x^4/2+O(x^6)$ and the geometric series $(1+x^2{)}^{-1}=1-x^2+O(x^4)$. The first two nonzero terms are

The vectors $(1,1{)}^{\mathsf{T}}$ and $(1,-1{)}^{\mathsf{T}}$ are eigenvectors: multiplication adds or subtracts the two entries in a row. Their eigenvalues are thereforeThe second eigenvalue may be positive, zero or negative as the real parameter changes.

The exponential definition of hyperbolic sine gives $e^z-e^{-z}=0$, or $e^{2z}=1$. Taking every branch of the complex logarithm givesFor hyperbolic cosine, multiply $\cosh z+1=0$ by $2e^z$ to obtain $(e^z+1{)}^2=0$. Thus $e^z=-1$, giving