Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 17 / b

Yes. The differential operator $\mathcal{L}=2y{\partial }_x-x{\partial }_y$ is linear, even though its coefficients depend on position. For constants $A,B$,If each homogeneous right-hand side is zero, so is their combination. Hence the superposition principle holds.

The homogeneous equation is the dot product condition $(2y,-x)\cdot \nabla u=0$. A suitable vector field is thereforeWhere nonzero, it is tangent to the level curves of $u$ and perpendicular to its gradient. At the origin this vector vanishes, so no direction is defined there.

For $\varphi =x^2+2y^2$, the gradient is $\nabla \varphi =(2x,4y)$, andAway from the origin, both $\nabla \varphi$ and $\nabla u$ lie in the one-dimensional subspace perpendicular to $(2y,-x)$, so they are parallel. More precisely,for some scalar $\lambda$, including $\lambda =0$. If $\nabla u\ne 0$, this can also be inverted to express $\nabla \varphi$ as a multiple of $\nabla u$. At a stationary point of $u$ with $\nabla \varphi \ne 0$, that literal inverse wording is not valid; the geometric assertion is parallelism of the nonzero gradient directions.

Write $u(x,y)=F(\varphi (x,y))$ for any differentiable $F$. The multivariable chain rule gives $u_x=2x{F}^{\prime }(\varphi )$ and $u_y=4y{F}^{\prime }(\varphi )$. ConsequentlyThus $\boxed{u=F(x^2+2y^2)}$ solves the homogeneous equation. The method of characteristics also explains completeness on the full plane: the characteristic curves are the connected ellipses $x^2+2y^2=\text{constant}$, and $u$ is constant on each one. These are the elliptic characteristics of a planar transport equation.

For a monomial trial $u_p=A{x}^m{y}^n$, the operator givesChoosing $m=n=2$ matches the two powers in the forcing, and $A=1/2$ matches their coefficients. A particular solution is therefore $u_p=x^2{y}^2/2$. Adding the homogeneous solutions givesThe arbitrary function labels the values on the characteristic curves; it must have sufficient regularity for the desired classical solution.

Substitution of $y=1$ into the general solution givesThus $F(s)=(s-2)/2$ for $s\ge 2$. Choosing this same formula everywhere gives the polynomial solutionIt directly satisfies the differential equation and $u(x,1)=x^2$.

The boundary condition does not determine a unique solution on the entire plane. Every characteristic ellipse $x^2+2y^2<2$ misses the data line $y=1$. More generally, $F$ is arbitrary for $0\le s<2$, subject to matching the chosen regularity at $s=2$ and at the origin. The displayed polynomial is one global extension; it is forced on $x^2+2y^2\ge 2$. A smooth function supported strictly inside $0<s<2$ may be added to $F$ without changing the boundary data.