Past exam of the differential equations and applications course of the University of Oxford / 2009 / B / 6

At such an interior backward-in-time maximum, spatial differentiation gives $u_x=0$ and $u_{xx}\le 0$. Comparison with earlier times at the same spatial point gives the one-sided time derivative $u_y\ge 0$; differentiability makes this the usual derivative. Consequently $x{u}_{xx}-u_y\le 0$, contradicting the strictly positive source $x{y}^{-2}{e}^{-x/y}$. Therefore no such maximum can occur. The temporal one-sided argument matters because the condition concerns only the past of the candidate point.

The stated pointwise boundary limits do not imply uniqueness. A counterexample to homogeneous uniqueness isIt is smooth for $x,y>0$. Writing $\eta =x/y$, differentiation givesFor fixed positive $y$, $w\to 0$ as $x\to 0$ and as $x\to \infty$; for fixed positive $x$, it also tends to zero as $y\downarrow 0$. Thus if $u_0$ is one solution, $u_0+Cw$ satisfies the same equation and every printed boundary limit for any constant $C$. Part (d) supplies such a particular solution, so this is a genuine family of distinct solutions, not merely a formal obstruction.

This is corner-source nonuniqueness for a degenerate parabolic problem. The peak $w(y,y)=1/(ey)$ diverges towards the missing corner $(0,0)$; pointwise initial limits at each fixed $x>0$ do not exclude it. The preceding maximum argument fails to force uniqueness because a supremum over an entire past strip need not be attained or finite.

A customary bounded-solution condition restores the intended result. If the difference $z$ of two solutions is bounded on every finite time strip, subtract $\delta (1+x+y)$. Its operator $x{\partial }_x^2-{\partial }_y$ equals $\delta >0$, excluding a positive past maximum; the linear barrier controls spatial infinity, and the homogeneous initial/boundary values control the parabolic boundary. Letting $\delta \downarrow 0$ gives $z\le 0$; applying the same argument to $-z$ gives $z=0$. This proves uniqueness in that additional bounded class, which the printed pointwise limits alone do not specify.

Choose the positive causal sign convention $\mathcal{L}={\partial }_y-x{\partial }_x^2$, so the forced equation is $\mathcal{L}u=-x{y}^{-2}{e}^{-x/y}$. With source $(\xi ,\tau )$, a forward Green's function satisfiesIt vanishes at $x=0$, decays at spatial infinity in the chosen bounded evolution class, and has the temporal jump $G(x,{\tau }^{+};\xi ,\tau )=\delta (x-\xi )$. Away from the source it solves the homogeneous equation. Equivalently one may reverse both the Green-function operator and its source sign to use the original printed operator.

To select the physical zero-initial-data solution one also excludes extra corner-source terms of the kind found in (b). With this causal bounded convention the solution representation isA mere pointwise statement of the three boundary limits does not by itself provide that uniqueness convention.

The forcing contains $x/y$, so matching it with a single similarity variable requires $\beta =1$. For $u=y^{\alpha }f(\eta )$, $\eta =x/y$, substitution givesMatching the source $y^{-1}\eta e^{-\eta }$ selects $\boxed{\alpha =0,\beta =1}$ and gives $f^{\prime \prime }+f^{\prime }=e^{-\eta }$. Multiplying the first-order equation for $f^{\prime }$ by $e^{\eta }$ yields $f^{\prime }=(\eta +C)e^{-\eta }$, so $f=-(\eta +1+C)e^{-\eta }+D$. The limits at zero and infinity give $D=0$, $C=-1$. HenceFor fixed positive $x$, the initial limit is also zero because the exponential dominates $x/y$. The function $-\eta e^{-\eta }$ decreases to $-1/e$ at $\eta =1$ and increases thereafter. ThereforeThe axis limits are zero, but there is no joint limit at $(0,0)$: approaching along $x=y$ gives $-1/e$. This does not violate any of the printed fixed-variable limits. It is this bounded similarity solution whose minimum was calculated; the unbounded corner-source additions in (b) need not have the same minimum.