Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 58 / 3

Set $f(A)={\mu }_{0}A+\alpha A^3-A^5$ and $V(A)={\mu }_0{A}^2/2+\alpha A^4/4-A^6/6$. Multiplication of the stationary reaction-diffusion equation by $A_x$ gives the first integral $A_x^2/2+V(A)=E$. A regular front with finite endpoint limits has $A_x\to 0$ there, and its nonzero endpoint must satisfy $f(C)=0$. Since $V(0)=0$, the first integral gives $E=0$ and $V(C)=0$. Write $q=C^2>0$. The two endpoint conditions areEliminating ${\mu }_0$ gives the Maxwell balance for a scalar reaction-diffusion front:This is a necessary condition for a stationary front, because the two spatially homogeneous states must have equal potential. It is also sufficient here. With these values, $-2V(A)=A^2(A^2-q{)}^2/3$. For the increasing positive front chooseIntegrating this logistic equation and writing $\beta =2q/\sqrt{3}=\sqrt{3}\alpha /2$ givesThe positive constant multiplying the exponential is arbitrary and can be absorbed into $X_0$; choosing it to be $q$ gives the printed normalization exactly. The profile tends to zero and $q$ at the two ends and its first-order equation implies $A_0^{\prime \prime }+f(A_0)=0$. Its negative is the front to $-\sqrt{q}$, with the same squared profile.

Differentiation of the stationary equation produces the translation modeThe operator $\mathcal{L}$ is self-adjoint. Twice applying integration by parts gives, for $R$ with vanishing boundary Wronskian,Bounded $R,R^{\prime }$ suffice, because the derivatives of the front decay exponentially. This proves the identity as a Fredholm solvability condition associated with translation invariance.

Use $\xi =x-\hat{X}(T)$, $T=\epsilon t$. At order $\epsilon$, the moving-profile expansion givesThe slow time derivative of $\epsilon R$ is of order ${\epsilon }^2$. Multiplication by $A_0^{\prime }$ and integration eliminate $\mathcal{L}R$ by the Fredholm solvability condition. Thus the leading velocity isFor $\nu (x,T)=K\delta (x+cT)$ the Dirac delta samples $\xi =-\hat{X}-cT$. From the first-order profile equation,ConsequentlyThe prefactor $q$ in the printed profile is important for the position shift in $Z$. This is a leading-order velocity in slow time; the physical velocity is $d\hat{X}/dt=\epsilon {\hat{X}}_T+O({\epsilon }^2)$.

The relative position obeys the autonomous moving-defect locking of a cubic-quintic front equation $Z_T=c-K{\mathrm{sech}}^2(\beta Z/2)$. For $K\ne 0$, a finite constant-velocity front must have constant $Z$, and hence must travel with the defect at ${\hat{X}}_T=-c$. Such locked fronts exist exactly whenFor the strict inequality, their positions areLinearizing the relative-position equation gives ${\eta }_T=K\beta {\mathrm{sech}}^2(\beta Z_{\ast }/2)\tanh (\beta Z_{\ast }/2)\eta$. Therefore the root with $K{Z}_{\ast }<0$ is stable and the other root unstable. At $c=K$, they coalesce at $Z=0$ in a saddle-node bifurcation. Since $Z_T=K{\beta }^2{Z}^2/4+O(Z^4)$ there, the double root attracts from only one side and is not two-sided asymptotically stable. Locking within its existence range still requires an initial relative position in the stable root's basin; beyond the unstable root the front escapes instead.

These positional stability conclusions also describe the leading weakly perturbed front. Indeed $v=A_0^{\prime }>0$ has no zeros, and $\mathcal{L}=({\partial }_{\xi }+v^{\prime }/v)({\partial }_{\xi }-v^{\prime }/v)$ is nonpositive in the $L^2$ inner product. The limiting coefficients $f^{\prime }(0)=-3{\alpha }^2/16$ and $f^{\prime }(\sqrt{q})=-3{\alpha }^2/4$ are negative. Thus the unperturbed shape modes are damped, with the neutral translation mode singled out by the solvability condition; the defect determines its stability at first order.

If $c\ne 0$ and the locking inequality fails, $Z_T$ has no finite zero and has the sign of $c$. The separation grows without bound, the localized forcing of the front decays exponentially, and ${\hat{X}}_T\to 0$: the defect passes the front, which approaches a stationary position in laboratory coordinates. The same escaping behavior occurs outside the locking basin when roots exist. If $c=0$ and $K\ne 0$, the stationary defect drives the front away from it with ever decreasing speed; $Z$ grows in magnitude logarithmically rather than tending to a finite locked position. Finally, when $K=0$, the front remains stationary for every $c$; when also $c=0$, every front position is a neutral equilibrium. These degenerate cases must not be included by dividing by $K$.