# Hopf Bifurcation of Reaction-Diffusion System

I'm considering the following reaction-diffusion system:

$\frac{\partial u}{\partial t} = f(u,v)+ D_1 \frac{d^2 u}{dx^2}$
$\frac{\partial v}{\partial t} = g(u,v)+ D_2 \frac{d^2 v}{dx^2}$

where f and g describe the reaction kinetics, $D_1$ and $D_2$ are positive constants.

I have been looking at the conditions for diffusion-driven instability which are as follows:

$f_u + g_v < 0$
$f_u g_v - f_v g_u > 0$
$D_2 f_u - D_1 g_v > 0$
$(D_2 f_u - D_1 g_v)^2 > 4D_1 D_2 (f_u g_v - f_v g_u)$

My notes then say that under these conditions bifurcation to solutions oscillating in time as well as space (called Hopf bifurcation) is not possible.

I don't really understand this and really appreciate someone explaining why this is the case / proving it.

Thanks

## 1 Answer

The idea that Turin Propose is that it was possible to chose a reaction-diffusion system in which the kinetics yielded a steady state which was stable in the absence of diffusion, but introducing diffusion, which we might think as a stabilizing process, causes instability.

Without loss of generality, assume that $f_u = a, f_v = b, g_u = c$ and $g_v =d$, then the sufficient and necessary conditions for diffusion-driven instability are given by: $$a + d < 0$$ $$ad - bc > 0$$ $$(D_2 a + D_1 d)^2 > 4D_1 D_2 (ad-bc)$$ It is important to consider the equilibrium point of the system without diffusion, that is $$\frac{\partial u}{\partial t} = f(u,v),$$ $$\frac{\partial v}{\partial t} = g(u,v).$$ Assume that $(u_0,v_0)^T$ is a steady-state solution, that is $$f(u_0,v_0)=0$$ $$g(u_0,v_0)=0$$ Now, consider the linear system about this equilibrium of the system with reaction-diffusion,this yields $$u_t(x,t) = D_1 u_{xx}(x,t) + au + bv,$$ $$v_t(x,t) = D_1 v_{xx}(x,t) + cu + dv,$$ Our next step is to assume that the solution is given by $$u(x,t) = U(x)T(t) \text{ and } v(x,t) = V(x)T(x),$$ then, substituting in the linear system, we obtain: $$\frac{T'}{T} = D_1\frac{U''}{U} +a + b\frac{V}{U}=D_2\frac{V''}{V} +d + c\frac{U}{V} = \lambda$$ Therefore, we obtain the following equations: $$T'=\lambda T,$$ $$D_1 U'' + (a-\lambda)U + bV=0,$$ $$D_2 V'' + (d-\lambda)U + cU=0,$$ For the last two equations, we seek solutions of the form: $$U = c_1\cos(px) + c_2\sin(px) \quad \quad V = c_3\cos(px) + c_4\sin(px),$$ If we substitute back into the equations above and noticing the fact that $\sin(x)$ and $\cos(x)$ are linearly independent, we obtain the following set of equations, $$[(a-\lambda-p^2D_1)c_1 + bc_3]\cos(px) = 0,$$ $$[(a-\lambda-p^2D_1)c_2 + bc_4]\sin(px) = 0,$$ $$[cc_1 + (d-\lambda-p^2D_2)c_3 ]\cos(px) = 0,$$ $$[cc_2 + (d-\lambda-p^2D_2)c_4 ]\sin(px) = 0.$$ These four equations are satisfied with $c_1,c_2,c_3,c_4$ not all zero if the two equations, $$(a-\lambda-p^2D_1)c_1 + bc_3=0,$$ $$cc_1 + (d-\lambda-p^2D_2)c_3=0,$$ have a solution with $c_1$ and $c_3$ not both zero.The conditions is that the determinant of the matrix, $$\begin{bmatrix} a-p^2D_1-\lambda & b\\ c & d-p^2D_2 -\lambda \end{bmatrix}$$ be zero. In other words, $\lambda$ is an eigenvalue of the matrix, $$D = \begin{bmatrix} a-p^2D_1 & b\\ c & d-p^2D_2 \end{bmatrix}.$$ Now, the system without the diffusion terms, $D_1=D_2=0$ is, $$A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}$$ and asymptotically stable if and only if: $$Tr(A) = a+d < 0,$$ $$\det(A) = ad-bc > 0.$$ For the system with diffusion, we require that: $$Tr(D) = a+d -p^2(D_1+D_2)<0,$$ $$\det(D) = (a-p^2D_1)(d-p^2D_2)-bc>0.$$ In particular, if $Tr(A) < 0 \Rightarrow Tr(D) < 0$, thus we only need to check conditions on the determinant.First, we have that $\det(A) = ad-bc>0$ and $$\det(D) = p^4D_1D_2 - p^2(aD_2 + dD_1) + ad-bc,$$ we require that $aD_2 + dD_1>0$, this implies that $D_1 \neq D_2$. Now, let $z = p^2$ and define: $$H(z) = z^2D_1D_2 - z(aD_2 + dD_1) + ad-bc.$$ We want to obtain the minimum of this function, that is $$H'(z) = 2zD_1D_2 -(aD_2 + dD_1) = 0 \Rightarrow z = \frac{aD_2 + dD_1}{2D_1D_2}.$$ Substituting back in $H(z)$, we obtain that the minimum is given by $$ad-bc - \frac{aD_2 + dD_1}{2D_1D_2}.$$ Returning back to the variable $p$, one can wee that to make such quantity negative, we require $$ad-bc < \frac{(aD_2 + dD_1)^2}{4D_1D_2} \Rightarrow aD_2 + dD_1 > 2\sqrt{D_1D_2(ad-bc)}.$$ Now, if $ad>0$, then the condition $a+d < 0$, implies that $a<0$ and $d<0$, but then $D_1d + D_2a < 0$, which gives a contradiction.Thus diffusive instability requires $ad<0$, and then $ad-bc > 0$ which implies that $bc<0$. we may assume without loss of generality that $a < 0$, but then $d <0$. On the other hand, since $bc<0$, the matrix $A$ have one of the two forms, $$(i)\begin{bmatrix} + & -\\ + & - \end{bmatrix}, \quad \quad (ii)\begin{bmatrix} + & +\\ - & - \end{bmatrix}$$ As a remark, notice that the condition that you propose is: $$D_2 f_u - D_1 g_v > 0,$$ $$(D_2 f_u - D_1 g_v)^2 > 4D_1 D_2 (f_u g_v - f_v g_u),$$ and I proved that we require: $$aD_2 + dD_1>0,$$ $$\frac{(aD_2 + dD_1)^2}{4D_1D_2} > ad-bc,$$ this can be justified by using the fact the $d < 0,$ so $d = -\bar{d}$, where $\bar{d} > 0$.