Past exam of the differential equations and applications course of the University of Oxford / 2008 / B / 8

At a positive interior maximum, $u_x=0$, $u_{xx}\le 0$, and $u_t\ge 0$ if the point is compared with its past. The PDE instead gives $u_t\le -u^3<0$, a contradiction. Thus a positive global/past-strip maximum must come from the spatial boundary or initial-time limit. The far value is positive, so any attained global maximum is positive; the argument does not prohibit a negative local maximum in isolation.

For uniqueness put $w=u_1-u_2$. It satisfiesThe parabolic comparison principle with zero difference data gives $w\le 0$ and then $-w\le 0$, hence $w=0$. As usual on this noncompact singular-time domain, this uses an admissible comparison class in which the initial and far-field limits justify truncation; for example bounded/subquadratic differences on finite past strips with the corresponding uniform zero traces suffice. A barrier $\delta (1+x^2+2t)$ controls spatial infinity, since ${\partial }_t-{\partial }_x^2+[x/(2t)]{\partial }_x+a$ applied to it is nonnegative. The printed fixed-point limits alone should not silently be substituted for all uniform/growth hypotheses of this comparison argument.

Use drift cancellation in a parabolic similarity reduction. For the similarity ansatz, derivative scaling givesThus $\boxed{\beta =1/2}$, the drift terms cancel, and $f^{\prime \prime }=f^3-f/2$ with $f(\infty )=1/\sqrt{2}$. Use the positive-equilibrium separatrix for a cubic profile equation. Multiply by $f^{\prime }$ and use the limiting equilibrium to determine the integration constant:The branch approaching the positive far equilibrium satisfies $f^{\prime }=(1/2-f^2)/\sqrt{2}$; the constant equilibrium is included as well. Separation givesFor a finite boundary value at $x=0$ and no interior pole, take $A>-1$. Values $-1<A<0$ describe the corresponding coth-type approach from above; $A=0$ is constant, and positive $A$ gives the tanh-type branch. The other first-integral sign cannot approach the positive equilibrium nontrivially as $\eta \to \infty$.

The requested deficit integral is, after $x=\sqrt{t}\eta$,Therefore $\boxed{A=1}$, and the selected solution is $\boxed{u(x,t)=(2t{)}^{-1/2}\tanh [x/(2\sqrt{t})]}$.