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I would like to obtain the solution of the diffusion equation $$ \frac{\partial p(x,t)}{\partial t} = D \frac{\partial^2 p(x,t)}{\partial x^2}$$ for reflecting boundary conditions at $x=-a$ and absorbing ones at $x=b$. These boundary conditions are $$ \frac{\partial p(x,t)}{\partial x}\Bigg|_{x=-a}=0 $$ and $$ p(b,t) =0.$$

Using separation of variables I found $$ p(x,t) = A \sum_{l=1,3,5,\dots} e^{-\lambda_l^2 D t}\Big[\cos(\lambda_l x) + \tan(\lambda_l a)\sin(\lambda_l x)\Big],$$ where $\lambda_l = \pi l/[2(a+b)]$ and $A$ is a coefficient I have not yet been able to determine. Because one boundary is absorbing, the probability distribution is not normalized at all times. It is only normalized when $t=0$. I would like to use this condition to calculate $A$.

Obviously I could set $t=0$ and integrate over $x \in [-a,b]$, but this gives an infinite series I'm unable to sum. I wonder if there is a trick. I know the initial condition is $\delta(x) = p(x,0)$. If I set $$\delta(x) = \frac{1}{2\pi}\sum_{n=-\infty}^\infty e^{i n x} = A \sum_{l=1,3,5,\dots} \Big[\cos(\lambda_l x) + \tan(\lambda_l a)\sin(\lambda_l x)\Big], $$ is there a way to manipulate the RHS to match factors of $e^{inx}$ and thereby solve for $A$?

kevinkayaks
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  • Your series ought to have an $A_l$ associated with each term. As it stands your system is underdetermined. It's common to see an initial condition $p(x, t) = p_0(x)$ specified. – A rural reader Aug 10 '21 at 18:15
  • An $A_l$ for each term is not consistent with my mathematics or with any similar problems in chapter 8 of this book: https://www.springer.com/gp/book/9783030622329. If you can demonstrate this I'd like to see it as an answer. As it stands there is one unknown and one equation (the normalization at $t=0$). This is not underdetermined. – kevinkayaks Aug 10 '21 at 19:13
  • In fact $A = \Big[\sum_{n=0}^\infty \frac{(-1)^n}{\lambda_{2n+1} \cos(\lambda_{2n+1}a)}\Big]^{-1}$ if I remember correctly from last night. – kevinkayaks Aug 10 '21 at 19:20

3 Answers3

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I have a piece of notes which happens to solve this problem. As we know, the solution to the heat equation is the transition density of a Brownian motion.

When the BM on $[a,b)$ is reflected at $a$ and killed at $b$ its transition density $p(t;x,y)$ is obtained from the known transition density $q(t;x,y)$ of the BM on $(2a-b,b)$ which is killed at $2a-b$ or $b$ as follows:

From [1] p. 105 it follows that

\begin{eqnarray} q(t;x,y)&=&\frac{1}{\sqrt{2\pi t}}\sum_{n=-\infty}^\infty \Bigg(\exp\Bigg(-\frac{(x-y-4n(b-a))^2}{2t}\Bigg)\nonumber\\ &&~~~~~~~~~~~~~~~~-\exp\Bigg(-\frac{(x+y+2b-4a+4n(b-a))^2}{2t}\Bigg)\Bigg)\,. \end{eqnarray} In this equation we don't have a factor $1/2$ because in contrast to [1] we assume that the transition density is w.r.t. the Lebesgue measure, not the speed measure $2dx\,.$ The Brownian motion reflected at $a$ can be written as $a+|W_t-a|$ with a Brownian motion $W$ starting at $x\ge a\,.$ Also, if $W_t$ is killed at $2a-b$ or $b$ the reflected Browninan motion is killed at $b\,.$ The transition density we are looking for is therefore that of $a+|W_t-a|$ which is

\begin{eqnarray}\label{eReflectedKilled} p(t;x,y)&=& \frac{d}{dy}P\Big(a+|W_t-a|\le y\Big) = \frac{d}{dy}P\Big(-y+2a\le W_t\le y\Big)\nonumber\\ &=&\frac{d}{dy}\int_{-y+2a}^{y} q(t;x,z)\,dz=q(t;x,y)+q(t;x,-y+2a)\nonumber\\ &=& \frac{1}{\sqrt{2\pi t}}\sum_{n=-\infty}^\infty \Bigg(\exp\Bigg(-\frac{(x-y-4n(b-a))^2}{2t}\Bigg)\nonumber\\ &&~~~~~~~~~~~~~~~~-\exp\Bigg(-\frac{(x+y+2b-4a+4n(b-a))^2}{2t}\Bigg)\nonumber\\ &&~~~~~~~~~~~~~~~~+\exp\Bigg(-\frac{(x+y-2a-4n(b-a))^2}{2t}\Bigg)\nonumber\\ &&~~~~~~~~~~~~~~~~-\exp\Bigg(-\frac{(x-y+2b-2a+4n(b-a))^2}{2t}\Bigg)\Bigg)\,. \end{eqnarray}

[1] A.N. Borodin, P. Salminen. Handbook of Brownian Motion - Facts and Formulae. Birkhauser Verlag, Basel, Boston, Berlin 1996.

Kurt G.
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  • Thank you for sharing @Kurt ! I discovered to solve it my way and reproduce your solution I just needed to use the Poisson summation formula. With this, the sums are actually not so bad and my answer becomes yours. – kevinkayaks Aug 10 '21 at 17:46
  • Right. I remember vaguely that the different summation formulas each have their pros and cons probably depending on $|b-a|$. – Kurt G. Aug 10 '21 at 17:47
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This is for the interval $[0, 1]$. The system is $p_t = D p_{xx}$ subject to $p_x(0, t) = 0$, $p(1, t) = 0$. Standard separation of variables give solutions of the form $e^{-D\lambda^2 t}(A \cos(\lambda x) + B\sin(\lambda x))$.

The boundary condition at $x = 0$ requires $B = 0$, and from the one at $x = 1$, $\lambda = (l + \tfrac{1}{2})\pi$ for $l = 0, \pm 1, \dots$. Set $\lambda_l := (l + \tfrac{1}{2})\pi/2$.

This suggests a general solution of the form \begin{equation*} p(x, t) = \sum_{l = 0}^\infty\, A_l\, e^{-D\lambda_l^2 t} \cos(\lambda_l x). \end{equation*} If you have an initial distribution $p(x, 0)$ you can determine the coefficients $A_0, A_1, \dots$ by its Fourier expansion.

A rural reader
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  • Bit confusing to me - you determined that $\lambda$ depended on $l$, but not $A$. Where does the math demonstrate that $A$ depends on $l$? In particular for my problem $p(x,0) = \delta(x)$. What are the $A_l$ in this case? Arent they all $A_l = \Big[ \sum_{n=0}^\infty \frac{(-1)^n}{\lambda_{2n+1}}\Big]^{-1}$ , independent of $l$? – kevinkayaks Aug 10 '21 at 19:45
  • Could I be skipping a step assuming $A$ is independent of $l$ but finding this works regardless since the initial distribution is a delta function? Or do I have boundary conditions and normalization met but not the initial distribution? Hard for me to tell – kevinkayaks Aug 10 '21 at 19:51
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    If you know $\delta(x) = \sum_l, A_l \cos(\lambda_l x)$, multiply both sides by $\cos(\lambda_m x)$ and integrate over $[0, 1]$ taking advantage of the orthogonality of the cosines. This'll give you an expression for each $A_l$. – A rural reader Aug 10 '21 at 20:05
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The article by W. Schwarz (1992) ``The Wiener process between a reflecting and an absorbing barrier'', Journal of Applied Probability, 29, 597-604 explains in detail why some of the eigenvalues considered here can be complex; eq. (15) in this article gives a complete answer to the problem posed.

lupus
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