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I have been struggling to find the general solution of the following BVP of reaction-diffusion equation: $$\frac{\partial N}{\partial t}=\frac{\partial^2 N}{\partial x^2}+N(1-N)-\sigma N$$ $$N(0,x)=N_0(x)$$ $$N(t,0)=\alpha \hspace{2mm}, N(t,L)=\beta; t>0,0<x<L$$

I am trying to have some mathematical analysis to find stability of this problem from this BVP.And analyse the bifurcation digram.I linearized the PDE and found the fixed points $(N_{1}^*,N_{2}^*)=(0,0)\hspace{2mm} and\hspace{2mm} (N_{1}^*,N_{2}^*)=(1-\sigma)$.

For $(N_{1}^*,N_{2}^*)=(0,0)$ the eigenvalues are $\lambda_{1,2}=\frac{-c\pm\sqrt{c^2-4(1-\sigma)}}{2}$ which is asymptotically stable.For $(N_{1}^*,N_{2}^*)=(1-\sigma,0)$ the eigenvalues are $\lambda_{1,2}=\frac{-c\pm \sqrt{c^2+4(1-\sigma)}}{2}$

But what I really need is that how can I get a general solution from this BVP which satisfies the boundary condition? I tried both the separation of variable method and fourier sine function but I failed completely. I seems hard for me.And can I easily have general solution from eigenvalues?I really need help!

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    $\begingroup$ Do you mean $\partial_{xx} N$ rather than $\partial_{tt} N$? $\endgroup$ Commented May 13, 2023 at 8:24
  • $\begingroup$ yes. I just edited my question.thank you.@MatthewCassell $\endgroup$ Commented May 13, 2023 at 9:22
  • $\begingroup$ It's nonlinear, so you shouldn't really expect to be able to find the solution explicitly. But it's an equation that has been studied before: en.wikipedia.org/wiki/Fisher%27s_equation. $\endgroup$ Commented May 13, 2023 at 9:52

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The equation is quadratic in N. There is no time indepent solution, so the search for eigenfunctions is probably useless.

But one can set up a solution scheme by expanding in products of $e^{-\lambda t} f_\lambda(x) $

$$\ \ \text{pd}(f) = \partial_t f- \partial_{x,x} f -N(1-N)- \sigma N $$

 Coefficient[ 
   pd[E^(-\[Lambda]  t) f[x] +  E^(-2 \[Lambda] t) g[x]] // 
       Expand,  Exp[-\[Lambda] t Range[1, 6]]] // TableForm

$$ -f''(x)-\lambda f(x)-f(x),\ \ f(x)^2-g''(x)-2 \lambda g(x)-g(x), \ \ 2 f(x) g(x),g(x)^2,\ \ 0, \ \ 0$$

demonstrating, that the series can be solved serially.

  DSolveValue[-f[x] - \[Lambda] f[x] - (f^\[Prime]\[Prime])[x] == 0,f[x], x]

   E^(x Sqrt[-1 - \[Lambda]]) C[1] + E^(-x Sqrt[-1 - \[Lambda]]) C[2]


    Coefficient[pd[E^(-\[Lambda]  t) (E^(x Sqrt[-1 - \[Lambda]]) C[1] + 
    E^(-x Sqrt[-1 - \[Lambda]]) C[2]) +  
 E^(-2 \[Lambda] t) g[x]] // Expand,  Exp[-\[Lambda] t Range[1, 6]]] // TableForm

$$\left( \begin{array}{c} 0 \\ -g''(x)-2 \lambda g(x)-g(x)+c_1{}^2 e^{2 \sqrt{-\lambda -1} x}+c_2{}^2 e^{-2 \sqrt{-\lambda -1} x}+2 c_2 c_1 \\ 2 c_1 g(x) e^{\sqrt{-\lambda -1} x}+2 c_2 g(x) e^{\sqrt{-\lambda -1} (-x)} \\ g(x)^2 \\ 0 \\ 0 \\ \end{array} \right)$$

You get a series in $\exp^{-i n \ \sqrt (1 + \lambda) }$ by their coefficient isolation.

The exponentials in $t$ are independent. Their coefficients have to vanish each .

This gives you a Fourier series in the $x$ - exponentials . Numerically at least, not a problem . The roots of $\lambda$ are imaginary, so the constant coefficients can be determined to match the start distribution a $t = 0$. Replace $\lambda$ by $1 + k^2$

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  • $\begingroup$ I dont want time independent solution.this problem expects to have a time dependent solution something as N(x,t). @RolandF $\endgroup$ Commented May 13, 2023 at 10:47

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