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A travelling wave $u(x,t)=U(x-vt)$ obeys $m{U}^{\prime \prime }+v{U}^{\prime }+f(U)=0$, so the mechanical potential in the requested form is $V=F$, up to an additive constant. HereIts stationary points are maxima at $0,1$ and a minimum at $r$. Since $F(0)=0$ and $F(1)=(1-2r)/12$, the two maxima have equal height at $r=1/2$; the maximum at $1$ is lower for $r>1/2$ and higher for $r<1/2$.

Multiply the wave equation by $U^{\prime }$ and integrate along the heteroclinic orbit. The endpoint derivatives vanish, and $U(-\infty )=1$, $U(\infty )=0$, soConsequentlyThe integral is positive for a nonconstant front. In fact substitution of the same logistic shape from part (b) gives the exact velocity $v=\sqrt{m/2}(1-2r)$. Thus the $u=1$ phase invades rightwards when $r<1/2$, the zero phase invades leftwards when $r>1/2$, and the balanced front is stationary.

For a nonconstant invasion travelling wave, the relevant end states are $w(-\infty )=k$ and $w(+\infty )=0$. These fronts do not exist at every nonnegative speed: the phase plane calculation gives the threshold $2\sqrt{k}$. The constant solutions $w\equiv 0$ and $w\equiv k$ are, of course, nonnegative travelling waves for every speed, including zero.

Put $z=x-ct$ and $p=w^{\prime }$. The travelling-wave reduction of a reaction-diffusion system givesThe equilibrium points in the phase plane are $(k,0)$ and $(0,0)$. At $(k,0)$ the characteristic equation is $r^2+cr-k=0$, so there is one positive and one negative eigenvalue. Thus it is a saddle equilibrium. Its descending unstable manifold has $w<k$, $p<0$ and locallyAt $(0,0)$ the eigenvalues areFor $c>2\sqrt{k}$ it is a stable node; for $0<c<2\sqrt{k}$ it is a stable spiral. The repeated-root case $c=2\sqrt{k}$ must be included in the existence argument rather than inferred from distinct roots.

Suppose $c\ge 2\sqrt{k}$. Choose $a>0$ with $a(c-a)\ge k$; for example $a=(c-\sqrt{c^2-4k})/2$. Consider the closed triangular regionOn the horizontal side $p=0$, the vector field has $p^{\prime }=-w(k-w)\le 0$, pointing into $\mathcal{T}$. On the vertical side $w=k$, it has $w^{\prime }=p\le 0$, again pointing inward. For the sloping side use $h=p+aw$, so that on $h=0$,Thus trajectories cannot cross into $h<0$. The remaining boundary point $(0,0)$ is an equilibrium point and cannot be crossed in finite time by a nonconstant trajectory, by uniqueness theorem for ordinary differential equations. This proves forward invariance of $\mathcal{T}$. The argument works unchanged at $c=2\sqrt{k}$ with $a=\sqrt{k}$.

The descending unstable manifold of $(k,0)$ initially enters $\mathcal{T}$. Its forward trajectory remains in this compact region, so it exists for all positive $z$. Within the interior, $p<0$: a first contact with $p=0$ at $0<w<k$ would have $p^{\prime }<0$, contrary to arrival from below. Hence $w$ decreases strictly and has a limit $\ell \in [0,k)$. To identify the limit, its omega-limit set lies on the line $w=\ell$ and is invariant. Any trajectory remaining on this line has $p=w^{\prime }=0$; its second equation then forces $\ell (k-\ell )=0$. Since $\ell <k$, it follows that $\ell =0$, and the only possible omega-limit set is $(0,0)$. Backward in $z$, the unstable manifold tends to $(k,0)$ by its construction. We have therefore obtained a positive decreasing heteroclinic orbit and hence a nonnegative travelling wave for every $c\ge 2\sqrt{k}$.

For completeness, the threshold is necessary for such a front. When $0<c<2\sqrt{k}$, put $\nu =\sqrt{k-c^2/4}$, $X=w$, and $Y=(p+cw/2)/\nu$. Near the zero equilibrium point the system becomesFor a nonzero trajectory its polar angle satisfiesA trajectory approaching zero therefore makes infinitely many rotations and must cross $X=w=0$. It cannot be a nonnegative front. At $c=0$, the first integralis conserved. It has values $k^3/6$ and $0$ at the two desired end states, so no heteroclinic orbit connects them. ConsequentlyThis is the phase-plane construction of a Fisher travelling front, including the critical speed. The statement for all $c\ge 0$ must distinguish these nonconstant fronts from the two constant travelling waves.

For the intended nontrivial interface, take the limiting amplitude at the patterned end to be $R_{-}>0$; the identically zero solution is otherwise a trivial exception. Splitting the stationary real-coefficient cubic-quintic amplitude equation into real and imaginary parts givesThe imaginary equation implies $R^2{\varphi }^{\prime }=J$, a constant. The radial first integral isA regular solution has finite $E$. As $R\to 0$, the nonnegative $J^2/(2R^2)$ term would diverge unless $J=0$. Hence ${\varphi }^{\prime }=0$ wherever the nontrivial front has positive amplitude. This conclusion does not assume in advance that the patterned end has no phase gradient.

For a heteroclinic orbit connecting two constant states, $R^{\prime }$ tends to zero at both ends. Thus $E=V(0)=0$, while the nonzero end must obey both $\mu +\alpha q-q^2=0$ and $V(\sqrt{q})=0$. Eliminating $\mu$ givesThis is the stationary front of a cubic-quintic amplitude equation: a unique equal-potential parameter in the bistable interval.

Existence, not just a necessary condition, follows from the first integral. At ${\mu }_M$, $V(R)=-R^2(R^2-q{)}^2/6$. The descending solution obeys $R^{\prime }=-R(q-R^2)/\sqrt{3}$, which integrates toIt connects the stable upper pattern at $x\to -\infty$ to conduction at $x\to +\infty$, with arbitrary translation $x_0$ and constant phase ${\varphi }_0$.

For $u_t=u_{xx}+u(k-u)$ with $k>0$, a decreasing travelling wave connecting $k$ to $0$ satisfies $w^{\prime }=p$, $p^{\prime }=-cp-w(k-w)$. If $c\ge 2\sqrt{k}$, choose $a>0$ with $a(c-a)\ge k$. The triangle $0\le w\le k$, $-aw\le p\le 0$ is forward invariant: on its sloping side, $(p+aw{)}^{\prime }=w[a(c-a)-k+w]\ge 0$. The descending unstable manifold of the saddle equilibrium $(k,0)$ stays in this triangle and tends to $(0,0)$, producing a positive heteroclinic orbit. Below the threshold the origin is a stable spiral for positive $c$, forcing sign changes; at $c=0$ conserved energy also excludes the connection.

Using $\mathbf{u}=(-{\psi }_y,{\psi }_x)$, the displayed stream function gives ${\mathbf{u}}_0=(\sin x\cos y,-\cos x\sin y)$. Its streamlines are level sets of ${\psi }_0$. At $(n\pi ,m\pi )$, the Jacobian matrix is $\mathrm{diag}((-1{)}^{n+m},-(-1{)}^{n+m})$, so the points are saddle equilibria. At $(\pi /2+n\pi ,\pi /2+m\pi )$ the eigenvalues are $\pm i$, giving center equilibria surrounded by closed streamlines. The grid lines $x=n\pi$ and $y=m\pi$ form heteroclinic orbits between adjacent saddles; circulation alternates between cells.