Similar to
Fisher's equation, a traveling wave solution can be found for this problem. Suppose the wave to be traveling from right to left with a constant velocity U, then in the coordinate attached to the wave, i.e., z=x+Ut, the problem becomes steady. The ZFK equation reduces to U\frac{d\theta}{dz} = \frac{d^2\theta}{dz^2} + \frac{\beta^2}{2} \theta(1-\theta) e^{-\beta(1-\theta)} satisfying the boundary conditions \theta(-\infty) = 0 and \theta(+\infty) = 1. The boundary conditions are satisfied sufficiently smoothly so that the derivative d\theta/dz also vanishes as z \to \pm\infty. Since the equation is translationally invariant in the z direction, an additional condition, say for example \theta(0)=1/2, can be used to fix the location of the wave. The speed of the wave U is obtained as part of the solution, thus constituting a nonlinear eigenvalue problem. Numerical solution of the above equation, \theta, the eigenvalue U and the corresponding reaction term \omega are shown in the figure, calculated for \beta=15.
Asymptotic solution Source: The ZFK regime as \beta \to \infty is formally analyzed using
activation energy asymptotics. Since \beta is large, the term e^{-\beta(1-\theta)} will make the reaction term practically zero, however that term will be non-negligible if 1-\theta \sim 1/\beta. The reaction term will also vanish when \theta = 0 and \theta = 1. Therefore, it is clear that \omega is negligible everywhere except in a thin layer close to the right boundary \theta = 1. Thus the problem is split into three regions, an inner diffusive-reactive region flanked on either side by two outer convective-diffusive regions.
Outer region The problem for outer region is given by U\frac{d\theta}{dz} = \frac{d^2\theta}{dz^2}. The solution satisfying the condition \theta(-\infty) = 0 is \theta=e^{Uz}. This solution is also made to satisfy \theta(0) = 1 (an arbitrary choice) to fix the wave location somewhere in the domain because the problem is translationally invariant in the z direction. As z \to 0^-, the outer solution behaves like \theta = 1 + Uz + \cdots which in turn implies d\theta/dz = U + \cdots. The solution satisfying the condition \theta(+\infty) = 1 is \theta = 1. As z \to 0^+, the outer solution behaves like \theta = 1 and thus d\theta/dz = 0. We can see that although \theta is continuous at z = 0, d\theta/dz has a jump at z = 0. The transition between the derivatives is described by the inner region.
Inner region In the inner region where 1-\theta \sim 1/\beta, reaction term is no longer negligible. To investigate the inner layer structure, one introduces a stretched coordinate encompassing the point z=0 because that is where \theta is approaching unity according to the outer solution and a stretched dependent variable according to \eta = \beta z, \, \Theta = \beta(1-\theta). Substituting these variables into the governing equation and collecting only the leading order terms, we obtain 2\frac{d^2\Theta}{d\eta^2} = \Theta e^{-\Theta}. The boundary condition as \eta \to -\infty comes from the local behaviour of the outer solution obtained earlier, which when we write in terms of the inner zone coordinate becomes \Theta \to -U\eta = +\infty and d\Theta/d\eta=-U. Similarly, as \eta \to+\infty. we find \Theta = d\Theta/d\eta = 0. The first integral of the above equation after imposing these boundary conditions becomes \begin{align} \left.\left(\frac{d\Theta}{d\eta}\right)^2\right |_{\Theta=\infty} - \left.\left(\frac{d\Theta}{d\eta}\right)^2\right |_{\Theta=0} &= \int_0^\infty \Theta e^{-\Theta}d\Theta\\ U^2 &= 1 \end{align} which implies U=1. It is clear from the first integral, the wave speed square U^2 is proportional to integrated (with respect to \theta) value of \omega (of course, in the large \beta limit, only the inner zone contributes to this integral). The first integral after substituting U=1 is given by \frac{d\Theta}{d\eta} = - \sqrt{1-(\Theta+1)\exp(-\Theta)}. ==KPP–ZFK transition==