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
 A: 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 $\bar{d} > 0$.
