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I have a $1D$ diffusion equation given by:

$$\frac{\partial}{\partial t}u(x,t) = D\frac{\partial^2}{\partial t^2}u(x,t) + Cu(x,t)$$

with boundary conditions $u \left(\frac{-L}{2},t \right) = u \left (\frac{L}{2},t \right)=0$.

By separation of variables, I have found what I believe is the correct general solution (verify?)

$$u(x,t)=\exp(-\lambda t) \left[A \cos \left( \sqrt{\frac{\lambda+C}{D}}x \right) + B \sin \left( \sqrt{\frac{\lambda+C}{D}} x \right) \right]$$

I have been asked to find the critical length $L$ which is defined as the point before the equation blows up exponentially but after the equation is ultimately quenched. Any hints or answers would be greatly appreciated.

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  • $\begingroup$ Note sorry that C, D and lambda are constants $\endgroup$
    – dahaka5
    Oct 17, 2017 at 14:05
  • $\begingroup$ Your general solution should be in the form of an infinite series. You should also be able to write out the general form on the eigenvalues through separation of variables. $\endgroup$
    – DaveNine
    Oct 17, 2017 at 17:52
  • $\begingroup$ I don't think I can because I don't have any initial conditions $\endgroup$
    – dahaka5
    Oct 17, 2017 at 18:02
  • $\begingroup$ That’s okay! Remember the role of the IC only pops up when you try to find the coefficients of said series. $\endgroup$
    – DaveNine
    Oct 17, 2017 at 18:03
  • $\begingroup$ Ah ok, but the fact that the boundary conditions are -L/2 and +L/2 make it so I don't know how to deal with the equation from that point, and I don't even know if that last equation is correct. $\endgroup$
    – dahaka5
    Oct 17, 2017 at 18:06

1 Answer 1

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Upon separation of variables assuming $u(x,t)=X(x)T(t)$, we arrive at

$$\frac{T'}{DT} = \frac{X'' + CX}{X} = -\lambda^2$$

So that we've got the system of DE's,

$\begin{cases} T' + D\lambda^2T=0 \\ X'' + (C+\lambda^2)X=0, X(\frac{-L}{2})=X(\frac{L}{2})=0 \end{cases} $

If we're careful, we arrive at the solutions:

$T(t) = a_1\exp{(-D\lambda^2 t)}$, and $X(x)=a_2 \sin{x\sqrt{(\lambda^2 + C)}}$

Because we require $X(\frac{L}{2})=0$, we see that $\lambda = \sqrt{\frac{4n^2\pi^2}{L^2} - C}$, $n=1,2,...$.

Thus we arrive to the eigenfunctions $u_{n}(x,t) = A_n \exp{\left(-D\left(\frac{4n^2\pi^2}{L^2}-C \right)t \right)}\sin{\left(\frac{2n\pi}{L}x\right)}$, and so a general solution, after superposition is

$$u(x,t) = \sum_{n=1}^{\infty} A_n \exp{\left(-D\left(\frac{4n^2\pi^2}{L^2}-C \right)t \right)}\sin{\left(\frac{2n\pi}{L}x\right)}$$

In order for solutions not to blow up exponentially, we require that

$$D\left(\frac{4n^2\pi^2}{L^2}-C \right)>0$$

So we need $D>0$ and $\frac{4n^2\pi^2}{L^2}-C>0$. Note that the second condition follows because it is exactly $\lambda^2$, which is always positive anyways, there's no need to consider when both could be negative as a result. Hence we see $$L < \frac{2n\pi}{\sqrt{C}} \text{ and } D>0$$ is the required condition on L (and consequently D)to ensure solutions don't blow up.

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  • $\begingroup$ How did you get X(x) to be that should there not be a cosine component too? $\endgroup$
    – dahaka5
    Oct 18, 2017 at 10:04
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    $\begingroup$ This comes from the condition that $X(\frac{L}{2}) = 0$ and $X(-\frac{L}{2}) = 0$. Alleviating any confusion, this isn't a periodic boundary condition. Initially you do have $X(x) = b_1 \cos{x \gamma} + b_2 \sin{x \gamma}$, where $\gamma = \sqrt{ \lambda^2 + C}$, but due to the boundary conditions the cosine term vanishes. $\endgroup$
    – DaveNine
    Oct 18, 2017 at 21:55

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