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$$\frac{\partial u}{\partial t}=\Delta u+\gamma f(u,v)\text{ for }x\in\Omega, t\geq 0$$ $$ \frac{\partial v}{\partial t}=d\Delta v +\gamma g(u,v)\text{ for }x\in\Omega,t\geq 0 $$ with Neumann boundary conditions $\frac{\partial u}{\partial n}=0=\frac{\partial v}{\partial n}$.

Suppose that $(u_0,v_0)\in\mathbb{R}^2$ is an equilibrium. Linearization in this equilibrium by using $w_1=u-u_0, w_2=v-v_0$, yields $$ \frac{\partial w_1}{\partial t}=\Delta w_1+\gamma(f_u(u_0,v_0)w_1+f_v(u_0,v_0)w_2),~~~~(1) $$ $$ \frac{\partial w_2}{\partial t}=d\Delta w_2+\gamma (g_u(u_0,v_0)w_1+g_v(u_0,v_0)w_2)~~~~~(2) $$ with linearized boundary conditions $\frac{\partial w_1}{\partial n}=0=\frac{\partial w_2}{\partial n}$.

Now, I would like to determine the general solution of this linearized boundary value problem and $\Omega=(0,a), a>0$.

I think I first have to consider (1). I think, the usual procedure now is to look at the associated homoneneous equation which should be $$ \frac{\partial w_1}{\partial t}=\Delta w_1+\gamma f_u(u_0,v_0)w_1. $$ Making the Ansatz $w_1=T(t)X(x)$, I get the eigenvalue problem $$ \Delta w_1+k^2w_1=0. $$ I think the characteristic equation is $$ \lambda^2+\frac{T''(t)}{T(t)}+k^2=0 $$ having solutions $$ \lambda_{1,2}=\pm\sqrt{-\frac{T''(t)}{T(t)}-k^2} $$

Now, I am not sure how to use the boundary conditions to determine the Eigenfunctions.

If I have the eigenfunctions I can solve the ODE with $T(t)$ and then write down the general solution for (1). Afterwards I would do the same for (2). And then I can put everything together?

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  • $\begingroup$ How did you go from $$\Delta w_{1} + k^{2} w_{1} = 0$$ to $$\lambda^{2} + T''/T + k^{2} = 0$$? $\endgroup$ May 30, 2016 at 14:21
  • $\begingroup$ I plugged in $w_1=X(x)T(t)$. Then $\Delta w_1=X''(x)T(t)+T''(t)X(x)$, or? $\endgroup$
    – H. Hawks
    May 30, 2016 at 14:22
  • $\begingroup$ No, $$\Delta w_{1} = \partial_{xx} w_{1} = X''T$$ There is no $T''X$ term. $\endgroup$ May 30, 2016 at 14:24
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    $\begingroup$ Yes, it is just $X''T$. Notice that $$\Delta w_{1} + k^{2} w_{1} = 0$$ is just an ODE in $x$ with solution $$w_{1} = c_{1} \cos(kx) + c_{2} \sin(kx)$$ $\endgroup$ May 30, 2016 at 14:37
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    $\begingroup$ I'll make a post in relation to your last question. Also, the eigenvalues are $k = n \pi / a$ not $k = nx$. $\endgroup$ May 30, 2016 at 14:56

1 Answer 1

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This is in response to your last question.

We have the solution to

$$\Delta w_{1} + k^{2} w_{1} = 0$$

is given by

$$w_{1} = c_{1} \cos(kx) + c_{2} \sin(kx)$$

Your boundary conditions are

$$w_{1} ' \lvert_{x = 0} = w_{1}' \lvert_{x = a} = 0$$

Now,

$$w_{1} ' = - k c_{1} \sin(kx) + k c_{2} \cos(kx)$$

so

\begin{align} w_{1}' \lvert_{x = 0} &= k c_{2} \\ &= 0 \\ \implies c_{2} &= 0 \quad \text{as $k \ne 0$} \\ \implies w_{1}' &= -k c_{1} \sin(kx) \end{align}

Also,

\begin{align} w_{1} ' \lvert_{x = a} &= - k c_{1} \sin(ka) \\ &= 0 \\ \implies c_{1} &= 0 \quad \text{or} \quad \sin(ka) = 0 \\ c_{1} &\ne 0 \quad \text{otherwise the solution $w_{1}$ is trivial} \\ \implies \sin(ka) &= 0 \\ \implies ka &= n \pi, \quad n \in \mathbb{N} \cup \{0\} \quad \text{as these are the roots of $\sin(x)$} \\ \implies k &= \frac{n \pi}{a}, \quad n \in \mathbb{N} \cup \{0\} \end{align}

Hence, the eigenvalues of your ODE are $\lambda = k^{2} = (n \pi/a)^{2}$.

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  • $\begingroup$ Why is $\frac{\partial w_1}{\partial n}=\nabla w_1(x) n(x)=w_1'(x)$? $\endgroup$
    – H. Hawks
    May 30, 2016 at 15:18
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    $\begingroup$ $\partial_{n} w_{1} \ne w_{1}'$. You'll have to do some research yourself. $\endgroup$ May 30, 2016 at 15:21
  • $\begingroup$ I think, we have by assumption that $\nabla w_1(0)n(0)=-w_1'(0)=0$ and $\nabla w_1(a)n(a)=w_1'(a)=0$, hence $w_1'(0)=w_1'(a)=0$. $\endgroup$
    – H. Hawks
    May 30, 2016 at 15:55

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