Past exam of the differential equations and applications course of the University of Oxford / 2010 / B / 6

For $\eta =x/y^{\beta }$ the chain rule gives $u_y=-\beta \eta f^{\prime }/y$, $u_x=y^{-\beta }{f}^{\prime }$ and $(u{u}_x{)}_x=y^{-2\beta }[f{f}^{\prime \prime }+(f^{\prime }{)}^2]$. Multiplication of the equation by $y$ givesA nontrivial scale-invariant balance therefore selects $\boxed{\beta =1/2}$ and yieldsIts left side is the derivative of $f{f}^{\prime }+\eta f/2$. The boundary conditions give $f(0)=1$ and $f^{\prime }(0)=0$, so the integration constant is zero:On the positive branch before the first zero, divide by $f$ and integrate, obtaining $f=1-{\eta }^2/4$. Its first positive zero is ${\eta }_F=2$. HenceA direct check gives $u_y=x^2/(4y^2)$ and $(u{u}_x{)}_x+u/(2y)=x^2/(4y^2)$. The zero extension past this front is also a natural weak compact-support continuation: $u$ is continuous and the degenerate diffusive flux $u{u}_x$ tends to zero at the front. This is a source-driven porous-medium similarity front.

Assume a classical solution $u\in C^2(D)\cap C(\overline{D})$, as needed to interpret the boundary values. At an interior local minimum its Hessian is positive semidefinite, hence $\Delta u\ge 0$. But the equation gives $\Delta u=-e^{-x^2}<0$, a contradiction. Since $D$ is bounded and $u=0$ on its boundary, a nonpositive interior value would either furnish an interior negative minimum or itself be a zero minimum. Thus $\boxed{u>0\text{ in }D}$.

For the upper bound use the barrier $w(x,y)=(1-x^2-y^2)/4$, which satisfies $-\Delta w=1$. Set $z=w-u$. On the boundary $z=w\ge 0$, because the domain lies inside the unit disk, whileThe quadratic perturbation proof of the minimum principle gives $z\ge 0$. The strong minimum principle for elliptic operators makes this strict in the connected interior unless $z$ is constant. Constancy is impossible since $1-e^{-x^2}$ is not identically zero on any open domain. HenceThis last strong-principle step handles points with $x=0$, where the comparison forcing itself vanishes. Simple connectedness is not needed for the comparison argument.

With source point $\xi \in D$ and field point $X=(x,y)$, the Dirichlet Green function for the operator $-\Delta$ satisfiesClassically it is harmonic off $\xi$, with the two-dimensional normalization $G(X,\xi )=-(2\pi {)}^{-1}\log |X-\xi |+H(X,\xi )$, where $H$ is regular near the source. Equivalently $-{\lim }_{r\downarrow 0}{\int }_{|X-\xi |=r}{\partial }_{r}Gds=1$. This sign convention gives the representation $u(X)={\int }_{D}G(X,\xi )e^{-{\xi }_1^2}d\xi$.