= Solution
Use the characteristic parameter $t$ and write
$$
\dot x=3y-4u,\qquad\dot y=-3x+5u,\qquad\dot u=4x-5y.
$$
Direct <differentiation> gives
$$
\frac d{dt}(5x+4y+3u)=0,\qquad\frac d{dt}(x^2+y^2+u^2)=0.
$$
A characteristic beginning at $(s,0,0)$ therefore satisfies $5x+4y+3u=5s$ and $x^2+y^2+u^2=s^2$. Eliminating $s$ gives
$$
16u^2-(30x+24y)u+9y^2-40xy=0.
$$
Its discriminant as a quadratic in $u$ is $100(9x^2+40xy)$. Requiring the root to be zero when $y=0$ selects
$$
\boxed{u(x,y)=\begin{cases}\dfrac{15x+12y-5\sqrt{9x^2+40xy}}{16},&x>0,\\[5pt]\dfrac{15x+12y+5\sqrt{9x^2+40xy}}{16},&x<0.\end{cases}}
$$
The selected branch is differentiable in each of the two regions where the radical is strictly positive, and substitution into the differentiated quadratic identity verifies the <partial differential equation> there. On the initial line away from the origin the characteristic vector has nonzero $y$ component $-3s$, so these are the locally unique branches determined by that line.
The maximal real-valued continuation of these branches has the printed domain
$$
\boxed{D=\{9x^2+40xy\ge0\}=\{x\ge0,\ y\ge-9x/40\}\cup\{x\le0,\ y\le-9x/40\}.}
$$
At $x=0$ the <continuous> limiting value is $u(0,y)=3y/4$. The two bounding lines are $x=0$ and $y=-9x/40$, and the included sectors are the one above the sloping line on the right and the one below it on the left.
As in the <fold boundary of a characteristic graph>, the closed-domain statement concerns the formula's real values. A <classical solution> requires first derivatives on an <open set>. Along the nonzero boundary points $9x^2+40xy=0$, the radical's transverse <derivative> is singular, so the formula is generally not a <classical solution> there; its two classical domains are the interiors of the shaded sectors. The origin is characteristic for the initial data and does not supply a differentiable neighborhood solution. This qualification preserves the source's stated closed set while distinguishing it from the domain of a <classical solution>.
\Image[/past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2008-characteristic-domain.png]
{title=Real domain and classical interiors of the characteristic solution}
Back to article page
