Past exam of the differential equations 1 course of the University of Oxford / 2023 / 3

Let $\mathcal{P}u=u_t-\Delta u+a{u}_x+b{u}_y$ on the unit disk over $0<t\le 1$. Smoothness is understood to include continuity on the closed cylinder. If $u$ exceeded its nonnegative parabolic boundary maximum $A$, it would attain a positive global maximum $M>A$ at some spatial interior point at time $t_0>0$. There,At $t_0=1$ the last inequality uses the derivative from the past; at an interior time it is equality. These signs imply $\mathcal{P}u+u^3\ge M^3>0$, contradicting the given differential inequality. ThusIf $A>0$, equality at any point of $Q$ gives the same contradiction with $M=A$, so $u<A$ throughout $Q$. If $A=0$, equality is possible: $u\equiv 0$ is an example. More precisely, write the inequality as $\mathcal{P}u+cu\le 0$ with $c=u^2\ge 0$. The strong parabolic maximum principle shows that an interior zero at time $t_0>0$ forces $u\equiv 0$ on $B\times [0,t_0]$.

That conclusion need not extend into the future for a subsolution. For $t_0<1$, take a spatially constant function which is zero for $t\le t_0$ and equals $-\exp (-1/(t-t_0))$ for $t>t_0$. It is smooth, nonincreasing and nonpositive, so $\mathcal{P}u+u^3\le 0$, but it becomes negative after $t_0$.

With zero drift, choose the smooth functionIt equals $-1$ on the entire parabolic boundary, so $A=-1$. On the closed cylinder, $-1\le u\le -7/8$, whileConsequentlyNevertheless the maximum at $(0,0,1)$ isThe negative cubic term permits an interior increase while the boundary stays fixed, explaining why the sign condition in the preceding part matters.

The parabolic boundary values are zero. An exact solution satisfies the preceding differential inequality, so its maximum estimate with $A=0$ gives $u\le 0$. The function $v=-u$ satisfies the same equation because both the linear differential operator and the cubic term change sign. Applying the same maximum principle to $v$ gives $-u\le 0$. Hence the complete solution set isSubstitution verifies existence, and the two inequalities prove uniqueness among the assumed smooth solutions for every continuous choice of the drift coefficients.

Define the squared L2 norm $E(t)={\int }_0^{1}u(x,t{)}^{2}dx$. Multiplication of the semilinear partial differential equation by $u$, followed by integration by parts, givesAt $x=1$ the Dirichlet boundary condition makes the boundary term zero. At $x=0$ the Robin boundary condition is $u_x(0,t)=\alpha u(0,t)$, so subtraction of the lower endpoint gives $-\alpha u(0,t{)}^2$. In particular the sign is dissipative. Thus the exact energy identity iswhich implies

For every real $r$ and integer $n\ge 1$, $|\sin r|\le \min (1,|r|)$ givesThis estimate covers both even and odd $n$, without any assumption on the sign of $u$. Since $a(t)\ge 0$, the energy estimate from part (i) becomesEquivalently, an integrating factor givesThe bracket is nonnegative and starts at zero, so it vanishes identically. This is also the zero-data case of the Gronwall inequality. The vanishing L2 norm and continuity of $u$ imply