Analytical Study of Giant Fluctuations and Temporal Intermittency

Abstract

We study analytically giant fluctuations and temporal intermittency in a stochastic one-dimensional model with diffusion and aggregation of masses in the bulk, along with influx of single particles and outflux of aggregates at the boundaries. We calculate various static and dynamical properties of the total mass in the system for both biased and unbiased movement of particles and different boundary conditions. These calculations show that (i) in the unbiased case, the total mass has a non-Gaussian distribution and shows giant fluctuations which scale as system size (ii) in all the cases, the system shows strong intermittency in time, which is manifested in the anomalous scaling of the dynamical structure functions of the total mass. The results are derived by taking a continuum limit in space and agree well with numerical simulations performed on the discrete lattice. The analytic results obtained here are typical of the full phase of a more general model with fragmentation, which was studied earlier using numerical simulations.

Appendices

Appendix A: Asymptotics

1.1 A.1 P(M) for Large M and Large L (Eq. (10), Sect. 3.1)

The generating function $Q(z)=\sum_{M=0}^{\infty} P(M)z^{M}$ of the probability distribution P(M) in Sect. 3.1 can be obtained by setting y=1 in Eq. (9). Q(z) can be inverted to obtain P(M) as follows:

$$\begin{aligned} Q(z) =& \operatorname {sech}\bigl[\sqrt{\lambda(1-z)} \bigr] \quad \text{where } \lambda=\frac{2\tilde{a}L}{D} \\ =&\sum _{n=0}^{\infty} \frac{E_{2n}}{(2n)!}(1-z)^{n} \lambda^{n} \quad\text{where $E_{2n}$ are Euler numbers} \\ =&\sum _{n=0}^{\infty} \frac{E_{2n}}{(2n)!} \lambda^{n}\sum _{M=0}^{n} {n\choose M}(-1)^{M}z^{M} \\ =& \sum _{M=0}^{\infty} \sum _{n=M}^{\infty} (-1)^{M}z^{M}\frac{E_{2n}}{(2n)!}\lambda^{n}{n \choose M} \end{aligned}$$

(68)

Thus, P(M) is given by:

$$ P(M)= (-1)^{M}\sum _{n=M}^{\infty} \frac{E_{2n}}{(2n)!}\lambda ^{n}{n\choose M} $$

(69)

This is exact so far, but can be further simplified for large M by using the large n form of E _2n :

$$ E_{2n} \sim8(-1)^{n}\sqrt{\frac{n}{\pi}} \biggl( \frac{4n}{\pi e} \biggr)^{2n} \sim(-1)^{n} \frac{4}{\pi} \biggl(\frac{4}{\pi^{2}} \biggr)^{n} (2n)! \quad \text{using Stirling's approximation} $$

(70)

Substituting from Eq. (70) into Eq. (69), we get:

$$\begin{aligned} P(M) \sim&\frac{4}{\pi}(-1)^{M}\sum _{n=M}^{\infty}{n\choose M} \biggl(\frac{-4\lambda}{\pi^{2}} \biggr)^{n} \\ =& \frac{4}{\pi} \biggl(\frac{4\lambda}{\pi^{2}} \biggr)^{M}\sum _{n=0}^{\infty}{n+M\choose n} \biggl(\frac{-4\lambda}{\pi^{2}} \biggr)^{n} \\ =&\frac{\frac{4}{\pi} (\frac{4\lambda}{\pi^{2}} )^{M}}{ (1+\frac{4\lambda}{\pi^{2}} )^{M+1}} \\ \sim&\frac{\pi}{\lambda} \biggl(1-\frac{\pi^{2}}{4\lambda} \biggr)^{M} \quad\text{for $M\gg1$ and $\frac{\pi^{2}}{4\lambda}\ll1$} \\ \sim&\frac{\pi}{\lambda}\exp \biggl(-\frac{\pi^{2}M}{4\lambda} \biggr) \end{aligned}$$

(71)

which is the same as Eq. (10).

1.2 A.2 Small t Behaviour of S ₂(t) from Eq. (39) (Sect. 4.2)

Equation (39) expresses the structure function S ₂(t) as the following infinite sum (ignoring the sub-leading terms in L):

$$ S_{2}(t)=16\zeta^{2}\sum _{n=1,3,5,\ldots}^{\infty} \biggl\{ \frac {1}{(n \pi)^{4}} \biggl(n\pi\coth \biggl[\frac{n\pi}{2} \biggr]-1 \biggr) \biggr\} \biggl\{ 1- \exp \biggl[-\frac{D\pi^{2} n^{2}t}{L^{2}} \biggr] \biggr\} \quad\zeta=aL/D $$

(72)

To extract the small t (Dt/L ²≪1) behaviour of this expression, we approximate the sum over n by an integral, and then take the Dt/L ²→0 limit of this integral, so that:

$$\begin{aligned} S_{2}(t) \sim&16\zeta^{2}\int _{0}^{\infty} \frac{1-e^{-\pi ^{2}(2l+1)^{2}\tau}}{\pi^{3}(2l+1)^{3}} \biggl\{ \coth \biggl[\frac{\pi }{2}(2l+1) \biggr] - \frac{1}{\pi(2l+1)}\biggr\} dl \quad\text{where } \tau=\frac{Dt}{L^{2}} \\ =& \frac{8\zeta^{2}\tau}{\pi}\int _{\pi\sqrt{\tau}}^{\infty} \biggl[\frac{1-e^{-x^{2}}}{x^{3}} \biggr]\biggl\{ \coth \biggl[\frac{x}{2\sqrt {\tau}} \biggr]- \frac{\sqrt{\tau}}{x} \biggr\} dx \quad x=(2l+1)\pi\sqrt{\tau} \end{aligned}$$

(73)

In the limit τ→0, we have $\coth [x/2\sqrt{\tau } ]\sim1$, so that S ₂(t) becomes:

$$\begin{aligned} S_{2}(t) \sim&\frac{8\zeta^{2}\tau}{\pi}\int _{\pi\sqrt{\tau }}^{\infty} \biggl[\frac{1-e^{-x^{2}}}{x^{3}} \biggr] \biggl[1-\frac{\sqrt {\tau}}{x} \biggr]dx \\ =& \frac{8\zeta^{2}\tau}{\pi}\biggl\{ \biggl[\frac{1}{2\pi^{2}\tau}- \frac {e^{-\pi^{2}\tau}}{2\pi^{2}\tau}-\frac{1}{2}Ei\bigl[-\pi^{2}\tau\bigr] \biggr] \\ &{}+ \biggl[ \frac{e^{-\pi^{2}\tau}(1-2\pi^{2}\tau)-(1-2\pi^{7/2}\tau ^{3/2}\operatorname {Erfc}[\pi\sqrt{\tau}])}{3\pi^{3}\tau} \biggr]\biggr\} \quad\text{where } \\ Ei\bigl[-\pi^{2}\tau\bigr] =&-\int _{\pi^{2}\tau }^{\infty} \frac{e^{-u}}{u}du \quad\text{and}\quad \operatorname {Erfc}[\pi\sqrt{\tau}]=(2/\sqrt{\pi}) \int _{\pi\sqrt{\tau }}^{\infty} e^{-u^{2}} du \end{aligned}$$

(74)

In the limit τ→0, the complementary error function behaves as $\operatorname {Erfc}[\pi\sqrt{\tau}]\sim1-2\sqrt{\pi\tau}$ and the exponential integral has the asymptotic form Ei[−π ² τ]∼γ+log[π ² τ] where γ is the Euler-Mascheroni constant. Thus, as τ→0, the expression in Eq. (74) tends to:

$$ S_{2}(t) \sim\frac{4\zeta^{2}\tau}{\pi} \biggl[1-\gamma-\frac{2}{\pi }- \log\bigl[\pi^{2}\tau\bigr] \biggr] $$

(75)

This expression is not in very good agreement with numerics because of the significant corrections that appear while approximating the discrete sum with an integral. By taking these correction terms into account using the Euler-Maclaurin formula, better agreement with numerics is obtained. These correction terms basically modify the constants inside the square bracket in Eq. (75), so that the small t form of S ₂(t) is still given by:

$$ S_{2}(t) \sim\frac{4\zeta^{2}\tau}{\pi}\bigl(A_{0}-\log[\tau] \bigr)= -\frac {4}{\pi} \biggl(\frac{aL}{D} \biggr)^{2} \frac{Dt}{L^{2}}\log \biggl[A_{1}\frac{Dt}{L^{2}} \biggr] $$

(76)

where the simplest way of obtaining the constant A ₁ is by fitting to numerical data.

1.3 A.3 Asymptotic Expressions for $\langle M_{i,j}^{n}\rangle$ (Eqs. (56a)–(56d), Sect. 5.1)

We start with the generating function G _u,j (z) in Eq. (54),

$$\begin{aligned} G_{u,j}(z) =&\sum _{i=1}^{\infty}F_{i,j}(z)u^{i}= \sum _{i=1}^{\infty} \Biggl[\sum _{M=0}^{\infty}P_{i,j}(M)z^{M} \Biggr]u^{i} \\ = &\biggl(\frac{u}{1-u} \biggr) \biggl(\frac{\eta^{2}(1-z)^{2}}{\eta ^{2}(1-z)^{2}-(1-u)} \biggr) \\ &{}\times \biggl[[1+\sqrt{1-u}]^{-j}- \biggl(\frac{1-u}{\eta^{2}(1-z)^{2}} \biggr)\bigl[1+ \eta(1-z)\bigr]^{-j} \biggr] \end{aligned}$$

(77)

The generating function $\sum_{i=1}^{\infty}\langle M_{i,j}^{2}\rangle u^{i}$ can be obtained by differentiating the above expression w.r.t. z and then setting z=1

$$ \sum _{i=1}^{\infty}\bigl\langle M_{i,j}^{2} \bigr\rangle u^{i}=\frac{\eta ^{2}u}{(1-u)^{2}} \bigl[2-2\{1+\sqrt{1-u} \}^{-j}+j(1+j) (1-u) \bigr]+\frac {\eta u j}{1-u} $$

(78)

Since we are interested in $\langle M_{i,j}^{2}\rangle$ in the limit i→∞, we consider the u→1 limit of the above equation. This can be obtained by Taylor expanding Eq. (78) in powers of the small parameter 1−u and retaining only the first few terms (terms that become asymptotically large) in 1−u. This gives:

$$ \sum _{i=1}^{\infty}\bigl\langle M_{i,j}^{2} \bigr\rangle u^{i}\sim u \biggl[\frac{2\eta^{2}j}{(1-u)^{3/2}} + \frac{\eta j}{1-u} + \frac{\eta ^{2}j(j^{2}+3j+2)}{3(1-u)^{1/2}}+\cdots \biggr] $$

(79)

Each of the above terms can be now Taylor expanded about u=0 to give:

$$ \bigl\langle M_{i,j}^{2}\bigr\rangle \sim2 \eta^{2}j(2i) \biggl(\frac{(2i)!}{2^{2i} (i!)^{2}} \biggr)+\eta j+ \frac{\eta^{2}j(j^{2}+3j+2)}{3} \biggl(\frac {2i}{2i-1} \biggr) \biggl(\frac{(2i)!}{2^{2i} (i!)^{2}} \biggr)+\cdots $$

(80)

By taking the limit i→∞ and using Stirling’s approximation for i!, Eq. (80) becomes:

$$ \bigl\langle M_{i,j}^{2}\bigr\rangle \sim\frac{4\eta^{2}j\sqrt{i}}{\sqrt{\pi}}+ \eta j +\frac{\eta^{2}j(j^{2}+3j+2)}{3\sqrt{\pi}\sqrt{i}}+\cdots $$

(81)

The same procedure can be followed to compute $\langle M_{i,j}^{3}\rangle$ and $\langle M_{i,j}^{4}\rangle$ in the i→∞ limit: first, $\sum_{i=1}^{\infty}\langle M_{i,j}^{3}\rangle u^{i}$ and $\sum_{i=1}^{\infty}\langle M_{i,j}^{4}\rangle u^{i}$ are expanded in powers of 1−u and next, each of the (1−u)^α terms in this expansion is further expanded about u=0. Finally, by using Stirling approximation for i! etc. in the i→∞ limit, we get:

$$\begin{aligned} &\bigl\langle M_{i,j}^{3}\bigr\rangle \sim6 \eta^{3}ji+\frac{12\eta^{2}}{\sqrt{\pi }}j\sqrt{i}+\eta j+\eta^{3}j \bigl(j^{2}+3j+2\bigr)+ \frac{\eta ^{2}j(j^{2}+3j+2)}{\sqrt{\pi}\sqrt{i}}+\cdots \end{aligned}$$

(82a)

$$\begin{aligned} &\bigl\langle M_{i,j}^{4}\bigr\rangle \sim\frac{32\eta^{4}}{\sqrt{\pi }} \bigl(ji^{3/2}\bigr)+36\eta^{3}(ji) + \frac{28\eta^{2} j+8\eta^{4}j(j^{2}+3j+2)}{\sqrt{\pi}} \sqrt{i} + 6\eta ^{3}j\bigl(j^{2}+3j+2\bigr) \\ &\phantom{\bigl\langle M_{i,j}^{4}\bigr\rangle \sim\,\,}{}+\eta j+\cdots \end{aligned}$$

(82b)

The expressions for $\langle M_{i,j}^{n}\rangle$ simplify further in the limit j→∞, i→∞ with $j/\sqrt {i}$ finite. In this limit, terms that are $\mathcal{O} (j/i)$ etc. go to zero, so that $\langle M_{i,j}^{n}\rangle$ are given by:

$$\begin{aligned} &\bigl\langle M_{i,j}^{2}\bigr\rangle \sim i \mathcal{R} _{2} \biggl(\frac{j}{\sqrt{i}} \biggr) \sim \frac{4\eta^{2}i}{\sqrt{\pi}} \biggl[ \biggl(\frac{j}{\sqrt{i}} \biggr)+\frac{1}{12} \biggl(\frac{j}{\sqrt{i}} \biggr)^{3}+\cdots \biggr] \end{aligned}$$

(83a)

$$\begin{aligned} &\bigl\langle M_{i,j}^{3}\bigr\rangle \sim i^{3/2} \mathcal {R} _{3} \biggl(\frac{j}{\sqrt{i}} \biggr) \sim6 \eta^{3}i^{3/2} \biggl[ \biggl(\frac{j}{\sqrt{i}} \biggr)+ \frac{1}{6} \biggl(\frac{j}{\sqrt{i}} \biggr)^{3}+\cdots \biggr] \end{aligned}$$

(83b)

$$\begin{aligned} &\bigl\langle M_{i,j}^{4}\bigr\rangle \sim i^{2} \mathcal{R} _{4} \biggl(\frac{j}{\sqrt{i}} \biggr) \sim \frac{32\eta^{4}{i^{2}}}{\sqrt{\pi }} \biggl[ \biggl(\frac{j}{\sqrt{i}} \biggr)+\frac{1}{4} \biggl(\frac{j}{\sqrt {i}} \biggr)^{3}+\cdots \biggr] \end{aligned}$$

(83c)

Appendix B: Solution of Various Partial Differential Equations

2.1 B.1 Laplace Equation in a Right Isosceles Triangle (Eqs. (11a)–(11b) and (35a)–(35b))

Consider a function p(x,y) which:

• (i) satisfies the Laplace equation in a right isosceles triangle with vertices (0,0), (0,1) and (1,0).

• (ii) is equal to zero uniformly on the hypotenuse.³

• (iii) satisfies some specified boundary conditions (mixed B.C. for Eqs. (11a)–(11b) and Neumann B.C. for Eqs. (35a)–(35b)) on the other two sides of the triangle.

The key step in solving this equation is the folding transformation introduced in [44] for the Laplace equation in the equilateral triangle, and adapted to the right isosceles triangle in [43]. This transformation, as applied to Eqs. (11a)–(11b) and (35a)–(35b), is described below and also shown schematically in Fig. 8.

Fig. 8

Folding transformation used to solve Laplace equation in the right isosceles triangle

We first consider the case, where the boundary conditions at x=0 and y=1 are of the mixed kind, as in Eqs. (11a)–(11b):

$$\begin{aligned} &\frac{\partial^{2}{p(x,y)}}{\partial{x^{2}}}+\frac{\partial ^{2}{p(x,y)}}{\partial{y^{2}}}=0, \quad 0\leq x\leq y\leq1 \end{aligned}$$

(84a)

$$\begin{aligned} &p(x,y=x)=0, \qquad \frac{\partial{p}}{\partial{y}}\bigg\vert _{y=1}=0, \quad\quad p(0,y)= a(y) \end{aligned}$$

(84b)

Consider another function q(x,y) on the square [0≤x≤1, 0≤y≤1] which satisfies:

$$\begin{aligned} &\frac{\partial^{2}{q(x,y)}}{\partial{x^{2}}}+\frac{\partial ^{2}{q(x,y)}}{\partial{y^{2}}}=0 \end{aligned}$$

(85a)

$$\begin{aligned} &q(0,y)= a(y) \qquad q(x,0)= -a(x) \qquad\frac{\partial{q}}{\partial {y}}\bigg\vert _{y=1}= \frac{\partial{q}}{\partial{x}}\bigg\vert _{x=1}=0 \end{aligned}$$

(85b)

Then, it can be seen that the function $w(x,y)=\frac {1}{2}[q(x,y)-q(y,x)]$ is the required solution of Eqs. (84a)–(84b) in the triangular region as:

$$\begin{aligned} &\nabla^{2}w(x,y)=\frac{1}{2}\bigl[\nabla^{2}q(x,y)- \nabla^{2}q(y,x)\bigr]=0 \end{aligned}$$

(86a)

$$\begin{aligned} &\begin{aligned} &w(x,y=x)=\frac{1}{2}\bigl[q(x,x)-q(x,x) \bigr]=0 \\ &\frac{\partial{w}}{\partial{y}}\bigg\vert _{y=1}=\frac{1}{2} \biggl( \frac {\partial{q(x,y)}}{\partial{y}}-\frac{\partial{q(y,x)}}{\partial {y}} \biggr)\bigg\vert _{y=1} = \frac{1}{2} \biggl(\frac{\partial{q(x,y)}}{\partial{y}}\bigg\vert _{y=1}- \frac{\partial{q(x,y)}}{\partial{x}}\bigg\vert _{x=1} \biggr)\\ &\phantom{\frac{\partial{w}}{\partial{y}}\bigg\vert _{y=1}}=0 \\ &w(0,y)= \frac{1}{2}\bigl[q(0,y)-q(y,0)\bigr]=\frac{1}{2} \bigl[a(y)-\bigl(-a(y)\bigr)\bigr]=a(y) \end{aligned} \end{aligned}$$

(86b)

Equations (85a)–(85b) can be solved by a standard application of the superposition method [45] i.e. by decomposing q(x,y) as q(x,y)=u(x,y)+v(x,y), such that:

$$\begin{aligned} &\nabla^{2} u(x,y)=0 \qquad\frac{\partial{u}}{\partial{y}}\bigg\vert _{y=1}= \frac{\partial{u}}{\partial{x}}\bigg\vert _{x=1}=0 \qquad u(x,0)=0 \qquad u(0,y)=a(y) \end{aligned}$$

(87a)

$$\begin{aligned} &\nabla^{2} v(x,y)=0 \qquad\frac{\partial{v}}{\partial{y}}\bigg\vert _{y=1}= \frac{\partial{v}}{\partial{x}}\bigg\vert _{x=1}=0 \qquad v(x,0)=-a(x) \qquad v(0,y)=0 \end{aligned}$$

(87b)

Each of the functions u(x,y) and v(x,y) can be solved for by separation of variables. Their sum gives the function q(x,y), which in turn gives p(x,y).

The case where both the x=0 and y=1 boundary conditions are of the Neumann kind (as in Eqs. (35a)–(35b)) can be dealt with similarly using the folding transformation. Suppose, the boundary condition at x=0 in Eq. (84b) is given by $\frac{\partial{p}}{\partial{x}} \vert _{x=0}=b(y)$, then the corresponding Laplace equation in the square that needs to be solved is:

$$ \nabla^{2} q(x,y)=0, \qquad\frac{\partial{q}}{\partial{y}}\bigg\vert _{y=1}= \frac{\partial{q}}{\partial{x}}\bigg\vert _{x=1}=0, \qquad\frac{\partial{q}}{\partial{x}}\bigg\vert _{x=0}=b(y), \qquad \frac{\partial{q}}{\partial{y}}\bigg\vert _{y=0}=-b(x) $$

(88)

Note that all the boundary conditions for this equation are of the Neumann kind. The Laplace equation with Neumann boundary conditions has a solution only if $\int_{0}^{1} dy\frac{\partial{q}}{\partial {x}}\vert _{x=1} - \int_{0}^{1} dy\frac{\partial{q}}{\partial {x}}\vert _{x=0} +\int_{0}^{1} dx\frac{\partial{q}}{\partial{y}} \vert _{y=1} -\int _{0}^{1} dx\frac{\partial{q}}{\partial{y}} \vert _{y=0} =0$ i.e. there is no net flux through the boundaries [45, 48]. This is simply because the solution of the Laplace equation is the steady state solution of a heat equation and for a steady state solution to exist, the net heat flux through the boundaries must be zero. Thus, while decomposing q(x,y) as q(x,y)=u(x,y)+v(x,y), both u(x,y) and v(x,y) must individually satisfy the condition of no net flux through the boundaries. This can be done [48] by defining a new function:

$$ s(x,y)=q(x,y)+(K/2) \bigl[(1-x)^{2}-(1-y)^{2} \bigr] $$

(89)

The function s(x,y) now satisfies:

$$ \begin{aligned} &\nabla^{2}s(x,y)=0, \qquad \frac{\partial{s}}{\partial{y}}\bigg\vert _{y=1}= \frac{\partial {s}}{\partial{x}}\bigg\vert _{x=1}=0, \\ &\frac{\partial{s}}{\partial {x}}\bigg\vert _{x=0}=b(y)-K, \qquad\frac{\partial{s}}{\partial{y}}\bigg\vert _{y=0}=-b(x)+K \end{aligned} $$

(90)

Choosing $K=\int_{0}^{1} b(y)dy$ ensures that the net flux of s(x,y) through each side of the square is zero. Then s(x,y) can be decomposed as s(x,y)=u(x,y)+v(x,y) such that:

$$\begin{aligned} &\nabla^{2} u(x,y)=0, \qquad\frac{\partial{u}}{\partial{y}}\bigg\vert _{y=1}= \frac{\partial{u}}{\partial{x}}\bigg\vert _{x=1}=\frac{\partial {u}}{\partial{y}}\bigg\vert _{y=0}=0, \qquad\frac{\partial{u}}{\partial{x}}\bigg\vert _{x=0}=b(y)-K \end{aligned}$$

(91a)

$$\begin{aligned} &\nabla^{2} v(x,y)=0, \qquad\frac{\partial{v}}{\partial{y}}\bigg\vert _{y=1}= \frac{\partial{v}}{\partial{x}}\bigg\vert _{x=1}=\frac{\partial {v}}{\partial{x}}\bigg\vert _{x=0}=0, \qquad\frac{\partial{v}}{\partial{y}}\bigg\vert _{y=0}=-b(x)+K \end{aligned}$$

(91b)

The functions u(x,y) and v(x,y) both satisfy the condition of zero net flux at the boundaries. Thus, solutions to Eqs. (91a) and (91b) exist and can be found by separation of variables.

The folding transformation described above, can also be used to solve the 3D Laplace equation in the region 0≤x≤y≤z≤1, which comes up in the calculation of multi-sector moments such as $\langle M^{2}_{xy}M_{yz}\rangle$ (see Sect. 4.1). For example, suppose g(x,y,z) satisfies the Laplace equation in the region 0≤x≤y≤z≤1 with the boundary conditions g(x,x,z)=g(x,y,y)=0 and some specified Dirichlet or Neumann boundary conditions on the other two surfaces. Then, we can find g(x,y,z) by solving for a function h(x,y,z) which satisfies the Laplace equation inside a cube [0≤x≤1, 0≤y≤1, 0≤z≤1] and appropriately chosen boundary conditions on the six faces. If h(x,y,z) is known, then g(x,y,z) can be obtained as the antisymmetric combination:

$$g(x,y,z)=(1/6)\bigl[h(x,y,z)-h(y,x,z)+h(y,z,x)-h(z,y,x)+h(z,x,y)-h(x,z,y)\bigr] $$

This procedure can be generalised to solve Laplace equations on higher dimensional regions as well.

2.2 B.2 Inhomogeneous Heat Equation in 1 Spatial Dimension (Eqs. (25a)–(25c) and (26a)–(26c))

Equations (25a)–(25c) and (26a)–(26c) constitute a set of equations of the type:

$$ \begin{aligned} &\frac{\partial{p(x,t)}}{\partial{t}}=\gamma\frac{\partial ^{2}{p(x,t)}}{\partial{x^{2}}}+a(x,t) \\ &p(0,t)=0, \qquad\frac{\partial{p}}{\partial{x}}\bigg\vert _{x=1}=0, \qquad p(x,0)=b(x) \end{aligned} $$

(92)

The inhomogeneous heat equation can be solved [45] by expressing p(x,t) as the sum of the complementary and particular solutions. The complementary solution, which is the solution of the homogeneous equation corresponding to Eq. (92), has a variable separable form given by:

$$ p_{c}(x,t)=\sum _{n=0}^{\infty}u_{n} \exp\bigl(-\gamma\alpha _{n}^{2}t\bigr)\sin[ \alpha_{n}x] \quad\text{where } \alpha_{n}= \biggl(n+ \frac {1}{2} \biggr)\pi $$

(93)

By assuming the particular solution to be of the form $\sum _{n=0}^{\infty}B_{n}(t)\sin[\alpha_{n}x]$, the general solution can be expressed as $p(x,t)=\sum_{n=0}^{\infty }[B_{n}(t)+u_{n}\exp(-\gamma\alpha_{n}^{2}t)]\sin[\alpha_{n}x] =\sum_{n=0}^{\infty}C_{n}(t)\sin[\alpha_{n}x]$. The inhomogeneous source term a(x,t) in Eq. (92) can also be written in the same eigenbasis as $a(x,t)=\sum_{n=0}^{\infty}A_{n}(t)\sin[\alpha_{n}x]$. Then, it follows that C _n (t) satisfies:

$$ \dot{C}_{n}(t)=-\gamma\alpha_{n}^{2}C_{n}(t)+A_{n}(t) $$

(94)

This can be solved to give $C_{n}(t)= C_{n}(0)\exp[-\gamma\alpha _{n}^{2}t]+\int_{0}^{t} A_{n}(t')\exp[-\gamma\alpha _{n}^{2}(t-t')]dt'$ where C _n (0) can be obtained from the initial condition $p(x,0)=\sum_{n=0}^{\infty}C_{n}(0)\sin[\alpha _{n}x]=b(x)$.

2.3 B.3 Heat Equation in 2 Spatial Dimensions on a Right Isosceles Triangle with Neumann Boundary Conditions (Eqs. (38a)–(38c) and (41a)–(41c))

Consider a function p(x,y,t) which satisfies the heat equation in the triangular region 0≤x≤y≤1:

$$\begin{aligned} &\frac{\partial{p(x,y,t)}}{\partial{t}}=\gamma \biggl[\frac{\partial ^{2}{p(x,y,t)}}{\partial{x^{2}}}+\frac{\partial^{2}{p(x,y,t)}}{\partial {y^{2}}} \biggr] \end{aligned}$$

(95a)

$$\begin{aligned} &p(x,y=x,t)=0, \qquad\frac{\partial{p}}{\partial{y}}\bigg\vert _{y=1}=0, \qquad \frac{\partial{p}}{\partial{x}}\bigg\vert _{x=0}=a(y,t) \quad p(x,y,0)=b(x,y) \end{aligned}$$

(95b)

As in the case of the Laplace equation, this can be solved by first solving the corresponding heat equation in a square:

$$\begin{aligned} &\frac{\partial{q(x,y,t)}}{\partial{t}}=\gamma \biggl[\frac{\partial ^{2}{q(x,y,t)}}{\partial{x^{2}}}+\frac{\partial^{2}{q(x,y,t)}}{\partial {y^{2}}} \biggr] \end{aligned}$$

(96a)

$$\begin{aligned} &\frac{\partial{q}}{\partial{y}}\bigg\vert _{y=1}=\frac{\partial{q}}{\partial {x}}\bigg\vert _{x=1}=0, \qquad \frac{\partial{q}}{\partial{x}}\bigg\vert _{x=0}=a(y,t) \qquad\frac{\partial {q}}{\partial{y}}\bigg\vert _{y=0}=-a(x,t) \end{aligned}$$

(96b)

where the initial condition at t=0 is unspecified. If we can solve Eqs. (95a)–(95b) (upto undetermined constant coefficients corresponding to the unknown initial condition), then the function p(x,y,t) can be obtained from this solution using $p(x,y,t)=\frac{1}{2}[q(x,y,t)-q(y,x,t)]$. The constant coefficients can now be determined from the initial condition p(x,y,0)=b(x,y) by using appropriate Fourier transforms.

If a(x,t)=0 (as in Eqs. (38a)–(38c)), then the boundary conditions are homogeneous and Eqs. (96a)–(96b) can be solved simply by separation of variables. If however, a(x,t) is non-zero (as in Eqs. (41a)–(41c)), then an additional transformation is required. This involves defining a new function:

$$ s(x,y,t)=q(x,y,t)+\frac{(1-x)^{2}}{2}a(y,t)-\frac{(1-y)^{2}}{2}a(x,t) $$

(97)

It can be checked that the function s(x,y,t) now satisfies an inhomogeneous heat equation (with a source term) and homogeneous boundary conditions:

$$\begin{aligned} \frac{\partial{s(x,y,t)}}{\partial{t}} =&\gamma \biggl[ \frac{\partial ^{2}{s(x,y,t)}}{\partial{x^{2}}}+\frac{\partial^{2}{s(x,y,t)}}{\partial {y^{2}}} \biggr] \\ &{}+\frac{(1-x)^{2}}{2} \biggl[\frac{\partial{a(y,t)}}{\partial{t}}-\gamma \frac{\partial^{2}{a(y,t)}}{\partial{y^{2}}} \biggr] -\frac{(1-y)^{2}}{2} \biggl[\frac{\partial{a(x,t)}}{\partial{t}}-\gamma \frac{\partial^{2}{a(x,t)}}{\partial{x^{2}}} \biggr] \\ &{}+\gamma\bigl[a(x,t)-a(y,t)\bigr] \end{aligned}$$

(98a)

$$\begin{aligned} \frac{\partial{s}}{\partial{y}}\bigg\vert _{y=1} =&\frac{\partial{s}}{\partial {x}}\bigg\vert _{x=1}= \frac{\partial{s}}{\partial{y}}\bigg\vert _{y=0}=\frac{\partial{s}}{\partial {x}} \bigg\vert _{x=0}=0 \end{aligned}$$

(98b)

The transformation in Eq. (97) is not a general prescription for solving Eqs. (96a)–(96b) for an arbitrary function a(x,t). It works only when $\frac{\partial{a(x,t)}}{\partial{x}} \vert _{x=1}=\frac{\partial {a(x,t)}}{\partial{x}} \vert _{x=0}=0$, which is the case for Eqs. (41a)–(41c).

Equations (98a)–(98b) is the heat equation with an inhomogeneous term. It can be solved, as in the 1D case, by first obtaining the complementary solution of the corresponding homogeneous equation and then obtaining the particular solution by appropriate Fourier transforms of the inhomogeneous source terms.