Diffusion with decay

Comments

• AbS

  March 2021 edited March 2021

  1. $u_t = u_{xx}$ is the classic diffusion equation, representing flow from areas of higher density to areas of lower density. The $-ku$ term represents decay, with the rate of decay of $u$ directly proportional to $u$ by a factor of $k.$
  2. See the attached image below. (Unfortunately, the website does not seem to allow placing images within ordered lists, and attempting to resume the list after the image automatically converts 3. to 1.) Given that the graph of $u$ decays over time, and $k$ is the proportionality constant of the rate of decay of $u$ per $u$, the graph would decay more with each time interval as $k$ increases.
  3. Using separation of variables, our differential equation becomes $$(XT)_t = (XT)_{xx} - kXT,$$ or $$XT' = X''T - kXT.$$ Rearranging, $$\frac{T'}{T} + k = \frac{X''}{X} = -\lambda.$$ Thus, $T' = -(\lambda + k)T,$ so $$T = e^{-(\lambda + k)t},$$ while $\frac{X''}{X} = -\lambda,$ so $$X = a\cos\left(\sqrt{\lambda}x\right) + b\sin\left(\sqrt{\lambda}x\right).$$ From our initial conditions, we know $$X(0) = a + 0 = 0,$$ so $a = 0,$ and $$X(1) = b\sin\left(\sqrt{\lambda}\right) = 0,$$ so $$\sqrt{\lambda} = n\pi.$$ Thus, our full solution is the Fourier sine series $$u(x,t) = \sum_{n=1}^{\infty}b_ne^{-(n^2\pi^2 + k)t}\sin(n\pi x),$$ where $$b_n = 2\int_{0}^{L} 4(x - x^2)\sin(n\pi x)dx.$$ While this integral can be evaluated by parts, we will use Mathematica to do it for us: $$b_n = \frac{16\left(1-(-1)^n\right)}{n^3\pi^3} = \frac{32}{n^3\pi^3},$$ where n is odd. Hence, $$u(x,t) = 32\sum_{n=1}^{\infty}\frac{1}{(2n-1)^3\pi^3}e^{-\left((2n - 1)^2\pi^2 + k\right)t}\sin((2n - 1)\pi x).$$ As can be seen in the equation, the exponential rate of decay increases with $k.$

  

  
  
• AbS

  March 2021

  $$b_n = 2\int_{0}^{L} 4(x - x^2)\sin(n\pi x)dx$$

  should be

  $$b_n = 2\int_{0}^{1} 4(x - x^2)\sin(n\pi x)dx.$$