# Diffusion Equation finding the critical length?

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.

• Note sorry that C, D and lambda are constants Oct 17, 2017 at 14:05
• 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. Oct 17, 2017 at 17:52
• I don't think I can because I don't have any initial conditions Oct 17, 2017 at 18:02
• That’s okay! Remember the role of the IC only pops up when you try to find the coefficients of said series. Oct 17, 2017 at 18:03
• 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. Oct 17, 2017 at 18:06

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.

• How did you get X(x) to be that should there not be a cosine component too? Oct 18, 2017 at 10:04
• 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. Oct 18, 2017 at 21:55