Inside Dynamics of Integrodifference Equations with Mutations

Abstract

The method of inside dynamics provides a theory that can track the dynamics of neutral gene fractions in spreading populations. However, the role of mutations has so far been absent in the study of the gene flow of neutral fractions via inside dynamics. Using integrodifference equations, we develop a neutral genetic mutation model by extending a previously established scalar inside dynamics model. To classify the mutation dynamics, we define a mutation class as the set of neutral fractions that can mutate into one another. We show that the spread of neutral genetic fractions is dependent on the leading edge of population as well as the structure of the mutation matrix. Specifically, we show that the neutral fractions that contribute to the spread of the population must belong to the same mutation class as the neutral fraction found in the leading edge of the population. We prove that the asymptotic proportion of individuals at the leading edge of the population spread is given by the dominant right eigenvector of the associated mutation matrix, independent of growth and dispersal parameters. In addition, we provide numerical simulations to demonstrate our mathematical results, to extend their generality and to develop new conjectures about our model.

Appendices

Derivation of a General Mutation Matrix

Here, we show how one can generalize the assumption of a single locus with n different neutral alleles to m loci with two neutral alleles. Let there be m independent loci $a_i$, $1 \le i \le m$, where each loci has one of the two possible alleles: $a_i=0$ or $a_i=1$. Then we define the transition probabilities as follows:

$$\begin{aligned} \text {Pr}\{a_i=0 \rightarrow a_i=1\} = q_i \quad \text {and}\quad \text {Pr}\{a_i=1 \rightarrow a_i=0\} = r_i. \end{aligned}$$

(27)

We index this process by $t \in {\mathbb {N}}$ where t describes the number of possible transitions taken so far. There are $2^m$ possible states for this system. Let $n=2^m$ and let the probability of being in state j, $1 \le j \le n$, be given by $v^j$ where the state is

$$\begin{aligned} j = 1 + \sum _{i=1}^m a_i2^{i-1}. \end{aligned}$$

(28)

In the case when $m=2$, there are four total states. We denote our states in the following form,

$$\begin{aligned} \begin{pmatrix} a_1\\ a_2 \end{pmatrix}. \end{aligned}$$

(29)

Our indexing for j gives the following relationship between the state and the index as follows:

By letting $m_{jl} = \text {Pr}\{v^j \rightarrow v^l\}$, then the mutation matrix becomes

$$\begin{aligned} {\mathbf {M}} = \begin{bmatrix} (1-q_1)(1-q_2)&\quad r_1(1-q_2)&\quad (1-q_1)r_2&\quad r_1r_2\\ q_1(1-q_2)&\quad (1-r_1)(1-q_2)&\quad q_1r_2&\quad (1-r_1)r_2\\ (1-q_1)q_2&\quad r_1q_2&\quad (1-q_1)(1-r_2)&\quad r_1(1-r_2)\\ q_1q_2&\quad (1-r_1)q_2&\quad q_1(1-r_2)&\quad (1-r_1)(1-r_2) \end{bmatrix}. \end{aligned}$$

(30)

From this example, one can deduce how to generalize this process for more than two neutral alleles, making the structure of this mutation matrix quite general.

Proofs of the Theorems

Proof of Theorem 1

Without loss of generality, we can assume that neutral fraction i belongs to the mutation class q. Then, since ${\mathbf {M}}$ is block diagonal, we only need to consider the following equation

$$\begin{aligned} {\mathbf {v}}_{q,t+1}(x) = {\mathbf {M}}_q \int _{-\infty }^\infty k(x-y)g(u_t(y)){\mathbf {v}}_{q,t}(y)\,\text {d}y. \end{aligned}$$

(31)

Using the fact that $0<g(u) \le g(0)$ for all $u \in (0,1)$ we can use a comparison principle, see Lemma 2.1 of Li et al. (2005), to show that a new sequence ${\mathbf {w}}_{q,t}(x)$ defined by

$$\begin{aligned} {\mathbf {w}}_{q,t+1}(x) = g(0){\mathbf {M}}_q \int _{-\infty }^\infty k(x-y){\mathbf {w}}_{q,t}(y)\,\text {d}y \end{aligned}$$

(32)

is always greater than the solution to any neutral fraction, ${\mathbf {v}}_{q,t}(x)$, with the same initial condition ${\mathbf {w}}_{q,0}(x)= {\mathbf {v}}_{q,0}(x)$. The solution of (32) is given by the t-fold convolution

$$\begin{aligned} {\mathbf {w}}_{q,t}(x) = \left[ g(0){\mathbf {M}}_q\right] ^t k^{*t}{\mathbf {w}}_{q,0}(x). \end{aligned}$$

(33)

Applying the reflected bilateral Laplace transform to (33) and using the convolution theorem, we obtain

$$\begin{aligned} {\mathcal {M}}[{\mathbf {w}}_{q,t}(x)](s)&= [g(0){\mathbf {M}}_q]^{t}\left[ {\mathcal {M}}\left[ k(x)\right] (s)\right] ^{t}{\mathcal {M}}\left[ {\mathbf {w}}_{q,0}(x)\right] (s) \end{aligned}$$

(34)

$$\begin{aligned}&= [g(0){\mathbf {M}}_q]^{t}\left[ \text {e}^{\frac{\sigma ^2s^2}{2}+\mu s}\right] ^{t}{\mathcal {M}}\left[ {\mathbf {w}}_{q,0}(x)\right] (s) \end{aligned}$$

(35)

$$\begin{aligned}&= [g(0){\mathbf {M}}_q]^{t}\text {e}^{\frac{\sigma ^2ts^2}{2}+\mu ts}{\mathcal {M}}\left[ {\mathbf {w}}_{q,0}(x)\right] (s) \end{aligned}$$

(36)

$$\begin{aligned}&= [g(0){\mathbf {M}}_q]^{t}{\mathcal {M}}\left[ \frac{1}{\sqrt{2\pi \sigma ^2t}}\text {e}^{-\frac{(x-\mu t)^2}{2\sigma ^2t}}\right] (s){\mathcal {M}}\left[ {\mathbf {w}}_{q,0}(x)\right] (s) \end{aligned}$$

(37)

$$\begin{aligned}&= [g(0){\mathbf {M}}_q]^{t}{\mathcal {M}}\left[ (k_t*{\mathbf {w}}_{q,0})(x)\right] (s), \end{aligned}$$

(38)

where $k_t$ is Gaussian with mean $\mu t$ and variance $\sigma ^2t$. Then applying the inverse transform yields

$$\begin{aligned} {\mathbf {w}}_{q,t}(x)&= [g(0){\mathbf {M}}_q]^t(k_t*{\mathbf {w}}_{q,0})(x) \end{aligned}$$

(39)

$$\begin{aligned}&= [g(0){\mathbf {M}}_q]^t\int _{-\infty }^\infty \frac{1}{\sqrt{2\pi \sigma ^2t}}\text {e}^{-\frac{(x-y-\mu t)^2}{2\sigma ^2t}}{\mathbf {w}}_{q,0}(y)\,\text {d}y. \end{aligned}$$

(40)

In the moving half-frame with fixed $A \in {\mathbb {R}}$, consider the element $x_0+ct$ with $c \ge c^* = \sqrt{2\sigma ^2\ln (g(0))} + \mu$. When we rewrite ${\mathbf {w}}_{q,t}(x)$ in this moving half-frame, we have

$$\begin{aligned} {\mathbf {w}}_{q,t}(x_0+ct)&= [g(0){\mathbf {M}}_q]^t\int _{-\infty }^\infty \frac{1}{\sqrt{2\pi \sigma ^2t}}\text {e}^{-\frac{(x_0+ct-y-\mu t)^2}{2\sigma ^2t}}{\mathbf {w}}_{q,0}(y)\,\text {d}y. \end{aligned}$$

(41)

Expanding in the exponential yields

$$\begin{aligned} \frac{(x_0+ct-y-\mu t)^2}{2\sigma ^2t}&=\frac{(x_0-y)^2}{2\sigma ^2t} + \frac{2(c-\mu )t(x_0-y)+(c-\mu )^2t^2}{2\sigma ^2t} \end{aligned}$$

(42)

$$\begin{aligned}&\ge \frac{(x_0-y)^2}{2\sigma ^2t} + \frac{c-\mu }{\sigma ^2}(x_0-y) + \ln (g(0))t. \end{aligned}$$

(43)

Thus,

$$\begin{aligned} {\mathbf {w}}_{q,t}(x_0+ct)&\le {\mathbf {M}}_q^t\frac{\text {e}^{\ln (g(0))t}}{\sqrt{2\pi \sigma ^2t}}\int _{-\infty }^\infty \text {e}^{-\frac{(x_0-y)^2}{2\sigma ^2t}}\text {e}^{-\frac{c-\mu }{\sigma ^2}(x_0-y)}\text {e}^{-\ln (g(0))t} {\mathbf {w}}_{q,0}(y)\,\text {d}y \end{aligned}$$

(44)

$$\begin{aligned}&={\mathbf {M}}_q^t\frac{1}{\sqrt{2\pi \sigma ^2t}}\int _{-\infty }^\infty \text {e}^{-\frac{(x_0-y)^2}{2\sigma ^2t}}\text {e}^{-\frac{c-\mu }{\sigma ^2}(x_0-y)}{\mathbf {w}}_{q,0}(y)\,\text {d}y \end{aligned}$$

(45)

$$\begin{aligned}&={\mathbf {M}}_q^t\frac{\text {e}^{-\frac{c-\mu }{\sigma ^2}x_0}}{\sqrt{2\pi \sigma ^2t}}\int _{-\infty }^\infty \text {e}^{-\frac{(x_0-y)^2}{2\sigma ^2t}}\text {e}^{\frac{c-\mu }{\sigma ^2}y}{\mathbf {w}}_{q,0}(y)\,\text {d}y. \end{aligned}$$

(46)

Since $x_0 \ge A$, we have

$$\begin{aligned} {\mathbf {w}}_{q,t}(x_0+ct)&\le {\mathbf {M}}_q^t\frac{\text {e}^{-\frac{A(c-\mu )}{\sigma ^2}}}{\sqrt{2\pi \sigma ^2t}}\int _{-\infty }^\infty \text {e}^{\frac{c-\mu }{\sigma ^2}y}{\mathbf {w}}_{q,0}(y)\,\text {d}y. \end{aligned}$$

(47)

Since ${\mathbf {M}}_q$ is a Markov matrix, we know that $\lim _{t\rightarrow \infty } {\mathbf {M}}_q^t = [{\mathbf {r}}_q, \ldots , {\mathbf {r}}_q]$ where ${\mathbf {r}}_q$ is the right eigenvector of ${\mathbf {M}}_q$ corresponding to eigenvalue 1 such that $\sum _{i=1} r_{q,i} = 1$. By Assumption A4, $\int _{-\infty }^\infty \text {e}^{\frac{c-\mu }{\sigma ^2}y}w^i_{q,0}(y)\,\text {d}y < \infty$ for every i in mutation class q we have $w^i_{q,t}(x_0+ct) \rightarrow 0$ uniformly as $t\rightarrow \infty$ in $[A,\infty )$. Recall that $w_q^i$ was constructed so that $0 \le v^i_{q,t}(x) \le w^i_{q,t}(x)$. This implies the uniform convergence of $v^i_t(x) \rightarrow 0$ as $t\rightarrow \infty$ in the moving half-frame $[A +ct, \infty )$ for each i in mutation class q. The proof of Theorem 1 is complete.$\square$

Proof of Theorem 2

Repeat the proof of Theorem 1 in the left moving half-frame with fixed $A \in {\mathbb {R}}$ and consider the element $x_0-ct$ with $c \ge c^*_- = \sqrt{2\sigma ^2\ln (g(0))}-\mu$. From this change, the result follows in the same manner as in Theorem 1.$\square$

Proof of Theorem 3

We begin by decomposing ${\mathbb {R}}^n$ according to the eigenspace of the matrix ${\mathbf {M}}_q$. By assumption, since all the blocks in ${\mathbf {M}}$ are primitive, we know that the principle eigenvalue is simple and equal to 1 with nonnegative eigenvector ${\mathbf {r}}_q$ because ${\mathbf {M}}_q$ is Markov. With a small abuse of notation, we call ${\mathbf {r}}_q$ the eigenvector of ${\mathbf {M}}$ associated with the eigenvalue 1 coming from the matrix ${\mathbf {M}}_q$. In a similar manner, we define $\varvec{\ell }_q$ be the eigenvector of ${\mathbf {M}}^T$ associated with the eigenvalue 1 from the matrix ${\mathbf {M}}_q^T$. Moreover, since the mutation matrix ${\mathbf {M}}$ is block diagonal, we can decompose the space ${\mathbb {R}}^n$ as follows

$$\begin{aligned} {\mathbb {R}}^n = \bigoplus _{q=1}^{n_q}{\mathbf {r}}_q{\mathbb {R}}\bigoplus _{q=1}^{n_q}\left( {\mathbf {r}}_q{\mathbb {R}}\right) ^{\perp }, \end{aligned}$$

(48)

where $\left( {\mathbf {r}}_q{\mathbb {R}}\right) ^{\perp } = \left\{ {\mathbf {v}} \in {\mathbb {R}}^n, {\mathbf {v}}^T\varvec{\ell }_q={\mathbf {0}}\right\}$. Let ${\mathbf {v}}_0(x)$ satisfy $0 \le v^i_0(x) \le {u}_0(x)$ in ${\mathbb {R}}$ for $i=1,\ldots , n$. Then, we can decompose ${\mathbf {v}}_t(x)$ as follows

$$\begin{aligned} {\mathbf {v}}_t(x) = \sum _{q=1}^{n_q}a_t^q(x){\mathbf {r}}_q + \sum _{q=1}^{n_q}b_t^q(x){\mathbf {h}}_t^q, \end{aligned}$$

(49)

where $a_t^q(x)$ and $b_t^q(x)$ are functions from ${\mathbb {R}}$ to ${\mathbb {R}}$ and ${\mathbf {h}}_t^q$ are in $\left( {\mathbf {r}}_q{\mathbb {R}}\right) ^{\perp }$ with $\Vert {\mathbf {h}}_t^q\Vert =1$. Then by applying our decomposition to (3), we can see that

$$\begin{aligned} {\mathbf {v}}_{t+1}(x)&= \int _{-\infty }^\infty k(x-y)g(u_t(y)){\mathbf {M}}{\mathbf {v}}_t(y)\,\text {d}y \end{aligned}$$

(50)

$$\begin{aligned}&= \int _{-\infty }^\infty k(x-y)g(u_t(y)){\mathbf {M}}\left( \sum _{q=1}^{n_q}a_t^q(y){\mathbf {r}}_q + \sum _{q=1}^{n_q}b_t^q(y){\mathbf {h}}_t^q\right) \,\text {d}y \end{aligned}$$

(51)

$$\begin{aligned}&= \sum _{q=1}^{n_q}\int _{-\infty }^\infty k(x-y)g(u_t(y))a_t^q(y)\,\text {d}y {\mathbf {M}}{\mathbf {r}}_q \nonumber \\&\quad + \sum _{q=1}^{n_q}\int _{-\infty }^\infty k(x-y)g(u_t(y))b_t^q(y)\,\text {d}y {\mathbf {M}}{\mathbf {h}}_t^q. \end{aligned}$$

(52)

Since ${\mathbf {M}}{\mathbf {r}}_q ={\mathbf {r}}_q$ and ${\mathbf {M}}$ stabilizes the space $\left( {\mathbf {r}}_q{\mathbb {R}}\right) ^{\perp }$, we can see from (49) and (52) that for all $q \in \{1, \ldots , n_q\}$ and $t>0$

$$\begin{aligned} a_{t+1}^q(x)&= \int _{-\infty }^\infty k(x-y)g(u_t(y))a_t^q(y)\,\text {d}y \text { and } \end{aligned}$$

(53)

$$\begin{aligned} b_{t+1}^q(x) {\mathbf {h}}_{t+1}^q&= \int _{-\infty }^\infty k(x-y)g(u_t(y))b_t^q(y)\,\text {d}y {\mathbf {M}}{\mathbf {h}}_t^q. \end{aligned}$$

(54)

We first focus our attention on (54). By the properties of ${\mathbf {M}}$ through the matrices ${\mathbf {M}}_q$, there exists $\delta \in (0,1)$ such that for any $q \in \{1, \ldots , n_q\}$ and ${\mathbf {h}} \in \left( {\mathbf {r}}_q{\mathbb {R}}\right) ^{\perp }$ then $\Vert {\mathbf {M}}{\mathbf {h}}\Vert \le \delta \Vert {\mathbf {h}}\Vert$. Thus, from (54) we can see that

$$\begin{aligned} |b_{t+1}^q(x)|&=\Vert b_{t+1}^q(x) {\mathbf {h}}_{t+1}^q\Vert \end{aligned}$$

(55)

$$\begin{aligned}&\le \int _{-\infty }^\infty k(x-y)g(u_t(y))|b_t^q(y)|\,\text {d}y \Vert {\mathbf {M}}{\mathbf {h}}_t^q\Vert \end{aligned}$$

(56)

$$\begin{aligned}&\le \delta \int _{-\infty }^\infty k(x-y)g(u_t(y))u_t(y)\frac{|b_t^q(y)|}{u_t(y)}\,\text {d}y. \end{aligned}$$

(57)

Since $0 \le v^i_0(x) \le {u}_0(x)$ for all $x \in {\mathbb {R}}$ for $i=1,\ldots , n$, it is clear that $|b_0^q(x)| \le \delta 'u_0(x)$ for all $x\in {\mathbb {R}}$ where $0 < \delta ' \le 1$. From iteration of (57), we have that

$$\begin{aligned} |b_{t+1}^q(x)| \le \delta ^t\delta ' \max _{x\in {\mathbb {R}}}\Vert u_0(x)\Vert . \end{aligned}$$

(58)

Since $\delta \in (0,1)$,

$$\begin{aligned} \lim _{t\rightarrow \infty }|b_{t+1}^q(x)|&\le \lim _{t\rightarrow \infty }\delta ^t\delta ' \max _{x\in {\mathbb {R}}}\Vert u_0(x)\Vert \end{aligned}$$

(59)

$$\begin{aligned}&=0. \end{aligned}$$

(60)

Thus, $b_{t+1}^q(x)$ converges uniformly to 0 on ${\mathbb {R}}$. Next, we turn our attention to the remaining piece of our decomposition for $a_t^q(x)$. First, it is important to note that $a_0^q(x)$ is a projection of ${\mathbf {v}}_0(x)$ on ${\mathbf {r}}_q$. Thus, it satisfies

$$\begin{aligned} a_0^q(x) = \frac{{\mathbf {v}}_0(x)^T\varvec{\ell }_q}{{\mathbf {r}}_q^T\varvec{\ell }_q}. \end{aligned}$$

(61)

From our assumption that the mutation class q is present at the leading edge of the front, we have

$$\begin{aligned} \frac{a_0^q(x)}{u_0(x)} \rightarrow p_0^q>0 \text { as } x\rightarrow \infty \text { and } \int _{-\infty }^\infty \text {e}^{\frac{c-\mu }{\sigma ^2}y}\left| a_0^q(y) - p_0^qu_0(y)\right| \,\text {d}y <\infty . \end{aligned}$$

(62)

Next, we consider the sequence $|z_t(x)| = |a_t^q(x)-p_0^qu_t(x)|$ that satisfies

$$\begin{aligned} |z_{t+1}(x)| \le \int _{-\infty }^\infty k(x-y)g(u_t(y))|z_t(y)|\,\text {d}y \end{aligned}$$

(63)

with $|z_0(x)| = |a_0^q(x)-p_0^qu_0(x)|$. By the assumption that $0 < g(u) \le g(0)$ for all $u\in (0,1)$, we obtain a super-solution

$$\begin{aligned} |z_{t+1}(x)| \le g(0)\int _{-\infty }^\infty k(x-y)|z_t(y)|\,\text {d}y \end{aligned}$$

(64)

with the same initial condition. The solution of (64) is bounded by the t-fold convolution

$$\begin{aligned} |z_{t}(x)| \le \left[ g(0)\right] ^t k^{*t}|z_0(x)|. \end{aligned}$$

(65)

Applying the reflected bilateral Laplace transform to (65) and using the convolution theorem, we obtain

$$\begin{aligned} {\mathcal {M}}[|z_{t}(x)|](s)&\le [g(0)]^{t}\left[ {\mathcal {M}}\left[ k(x)\right] (s)\right] ^{t}{\mathcal {M}}\left| z_0(x)|\right] (s) \end{aligned}$$

(66)

$$\begin{aligned}&= [g(0)]^{t}\left[ \text {e}^{\frac{\sigma ^2s^2}{2}+\mu s}\right] ^{t}{\mathcal {M}}\left[ |z_0(x)|\right] (s) \end{aligned}$$

(67)

$$\begin{aligned}&= [g(0)]^{t}\text {e}^{\frac{\sigma ^2ts^2}{2}+\mu ts}{\mathcal {M}}\left[ |z_0(x)|\right] (s) \end{aligned}$$

(68)

$$\begin{aligned}&= [g(0)]^{t}{\mathcal {M}}\left[ \frac{1}{\sqrt{2\pi \sigma ^2t}}\text {e}^{-\frac{(x-\mu t)^2}{2\sigma ^2t}}\right] (s){\mathcal {M}}\left[ |z_0(x)|\right] (s) \end{aligned}$$

(69)

$$\begin{aligned}&= [g(0)]^{t}{\mathcal {M}}\left[ (k_t*|z_0|)(x)\right] (s), \end{aligned}$$

(70)

where $k_t$ is Gaussian with mean $\mu t$ and variance $\sigma ^2t$. Then applying the inverse transform yields

$$\begin{aligned} |z_{t}(x)|&\le [g(0)]^t(k_t*|z_0|)(x) \end{aligned}$$

(71)

$$\begin{aligned}&= [g(0)]^t\int _{-\infty }^\infty \frac{1}{\sqrt{2\pi \sigma ^2t}}\text {e}^{-\frac{(x-y-\mu t)^2}{2\sigma ^2t}}|z_0(y)|\,\text {d}y. \end{aligned}$$

(72)

In the moving half-frame with fixed $A \in {\mathbb {R}}$, consider the element $x_0+ct$ with $c \ge c^* = \sqrt{2\sigma ^2\ln (g(0))} + \mu$. When we rewrite $|z_{t}(x)|$ in this moving half-frame, we have

$$\begin{aligned} |z_{t}(x_0+ct)|&\le [g(0)]^t\int _{-\infty }^\infty \frac{1}{\sqrt{2\pi \sigma ^2t}}\text {e}^{-\frac{(x_0+ct-y-\mu t)^2}{2\sigma ^2t}}|z_{0}(y)|\,\text {d}y. \end{aligned}$$

(73)

Expanding in the exponential yields

$$\begin{aligned} \frac{(x_0+ct-y-\mu t)^2}{2\sigma ^2t}&=\frac{(x_0-y)^2}{2\sigma ^2t} + \frac{2(c-\mu )t(x_0-y)+(c-\mu )^2t^2}{2\sigma ^2t} \end{aligned}$$

(74)

$$\begin{aligned}&\ge \frac{(x_0-y)^2}{2\sigma ^2t} + \frac{c-\mu }{\sigma ^2}(x_0-y) + \ln (g(0))t. \end{aligned}$$

(75)

Thus,

$$\begin{aligned} |z_{t}(x_0+ct)|&\le \frac{\text {e}^{\ln (g(0))t}}{\sqrt{2\pi \sigma ^2t}}\int _{-\infty }^\infty \text {e}^{-\frac{(x_0-y)^2}{2\sigma ^2t}}\text {e}^{-\frac{c-\mu }{\sigma ^2}(x_0-y)}\text {e}^{-\ln (g(0))t} |z_0(y)|\,\text {d}y \end{aligned}$$

(76)

$$\begin{aligned}&=\frac{1}{\sqrt{2\pi \sigma ^2t}}\int _{-\infty }^\infty \text {e}^{-\frac{(x_0-y)^2}{2\sigma ^2t}}\text {e}^{-\frac{c-\mu }{\sigma ^2}(x_0-y)}|z_0(y)|\,\text {d}y \end{aligned}$$

(77)

$$\begin{aligned}&=\frac{\text {e}^{-\frac{c-\mu }{\sigma ^2}x_0}}{\sqrt{2\pi \sigma ^2t}}\int _{-\infty }^\infty \text {e}^{-\frac{(x_0-y)^2}{2\sigma ^2t}}\text {e}^{\frac{c-\mu }{\sigma ^2}y}|z_0(y)|\,\text {d}y. \end{aligned}$$

(78)

Since $x_0 \ge A$, we have

$$\begin{aligned} |z_{t}(x_0+ct)|&\le \frac{\text {e}^{-\frac{A(c-\mu )}{\sigma ^2}}}{\sqrt{2\pi \sigma ^2t}}\int _{-\infty }^\infty \text {e}^{\frac{c-\mu }{\sigma ^2}y}|z_0(y)|\,\text {d}y. \end{aligned}$$

(79)

From our assumption that the mutation class q is initially present at the leading edge of the front, we know that $\int _{-\infty }^\infty \text {e}^{\frac{c-\mu }{\sigma ^2}y}|z_0(y)|\,\text {d}y<\infty$ Thus, we have that $|z_t(x)| \rightarrow 0$ uniformly as $t \rightarrow \infty$ in the moving half-frame $[A+ct,\infty )$ with speed $c\ge c^*$. Returning to the definition of $|z_t(x)|$, we can see that

$$\begin{aligned} |a_t^q(x)-p_0^qu_t(x)| \rightarrow 0 \;\text {uniformly as } t \rightarrow \infty \end{aligned}$$

(80)

in the moving half-frame $[A+ct,\infty )$. Then by putting together all the pieces, we can see from (49) that

$$\begin{aligned} \left\| {\mathbf {v}}_t(x)- \sum _{q=1}^{n_q}a_t^q(x){\mathbf {r}}_q\right\|&=\left\| \sum _{q=1}^{n_q}b_t^q(x){\mathbf {h}}_t^q\right\| \end{aligned}$$

(81)

$$\begin{aligned}&\le \sum _{q=1}^{n_q}|b_t^q(x)|\Vert {\mathbf {h}}_t^q\Vert \end{aligned}$$

(82)

$$\begin{aligned}&=\sum _{q=1}^{n_q}|b_t^q(x)|. \end{aligned}$$

(83)

Therefore, from (60) and (80) we can conclude that

$$\begin{aligned} \max _{[A+ct,\infty )}\left\| {\mathbf {v}}_t(x)- \sum _{q=1}^{n_q}p_0^qu_t(x){\mathbf {r}}_q\right\| \rightarrow 0 \text { as } t\rightarrow \infty . \end{aligned}$$

(84)

The proof of Theorem 3 is complete. $\square$

Proof of Theorem 4

Repeat the proof of Theorem 3 in the left moving half-frame with fixed $A \in {\mathbb {R}}$ and consider the element $x_0-ct$ with $c\ge c^*_- = \sqrt{2\sigma ^2\ln (g(0))}-\mu$. From this change, the result follows in the same manner as in Theorem 3. $\square$