2-d problem with Dirichlet boundary conditions

Let us consider the solution of the diffusion equation in two dimensions. Suppose that

$$\frac{\partial T(x,y,t)}{\partial t} = D\,\frac{\partial^2 T... ...)}{\partial x^2}+D\,\frac{\partial^2 T(x,y,t)} {\partial y^2},$$ (214)

for $x_l\leq x\leq x_h$, and $0\leq y \leq L$. Suppose that $T(x,y,t)$ satisfies mixed boundary conditions in the $x$-direction:

$$\alpha_l(t)\,T(x,y,t) + \beta_l(t)\,\frac{\partial T(x,y,t)}{\partial x} = \gamma_l(y,t),$$ (215)

at $x=x_l$, and

$$\alpha_h(t) \,T(x,y,t)+ \beta_h(t)\,\frac{\partial T(x,y,t)}{\partial x} = \gamma_h(y,t),$$ (216)

at $x=x_h$. Here, $\alpha_l$, $\beta_l$, etc., are known functions of $t$, whereas $\gamma_l$, $\gamma_h$ are known functions of $y$ and $t$. Furthermore, suppose that $T(x,y,t)$ satisfies the following simple Dirichlet boundary conditions in the $y$-direction:

$$T(x,0,t) = T(x,L,t) = 0.$$ (217)

As before, we discretize in time on the uniform grid $t_n=t_0+n\,\delta t$, for $n=0,1,2,\cdots$. Furthermore, in the $x$-direction, we discretize on the uniform grid $x_i = x_l + i\,\delta x$, for $i=0,N+1$, where $\delta x = (x_h-x_l)/(N+1)$. Finally, in the $y$-direction, we discretize on the uniform grid $y_j = j\,\delta y$, for $j=0,J$, where $\delta y = L/J$. Adopting the Crank-Nicholson temporal differencing scheme discussed in Sect. 6.6, and the second-order spatial differencing scheme outlined in Sect. 5.2, Eq. (214) yields

$$\frac{T_{i,j}^{n+1}-T_{i,j}^n}{\delta t} - \frac{D}{2}\, \frac{T_... ... x)^2} - \frac{D}{2}\left(\frac{\partial^2 T}{\partial y^2}\right)_{i,j}^{n+1}=$$

$$\frac{D}{2}\, \frac{T_{i-1,j}^{n}-2\,T_{i,j}^{n}+T_{i+1,j}^{n}}{(... ...ta x)^2} + \frac{D}{2}\left(\frac{\partial^2 T}{\partial y^2}\right)_{i,j}^{n},$$ (218)

where $T_{i,j}^n\equiv T(x_i,y_j,t_n)$. The discretized boundary conditions take the form

$$T_{0,j}^n = \frac{\gamma_{l\,j}^n\,\delta x-\beta_l^n\,T_{1,j}^n}{\alpha_l^n\,\delta x -\beta_l^n},$$ (219)

$$T_{N+1,j}^n = \frac{\gamma_{h\,j}^n\,\delta x + \beta_h^n\,T_{N,j}^n}{\alpha_h^n\,\delta x +\beta_h^n},$$ (220)

plus

$$T_{i,0}^n = T_{i,J}^n = 0.$$ (221)

Here, $\alpha_l^n\equiv \alpha_l(t_n)$, etc., and $\gamma_{l\,j}^n\equiv \gamma_l(y_j,t_n)$, etc.

Adopting the Fourier method introduced in Sect. 5.7, we write the $T_{i,j}^n$ in terms of their Fourier-sine harmonics:

$$T_{i,j}^n = \sum_{k=0}^{J} \hat{T}_{i,k}^n\,\sin(j\,k\,\pi/J),$$ (222)

which automatically satisfies the boundary conditions (221). The above expression can be inverted to give (see Sect. 5.9)

$$\hat{T}_{i,j}^n = \frac{2}{J}\sum_{k=0}^{J} T_{i,k}^n\,\sin(j\,k\,\pi/J).$$ (223)

When Eq. (218) is written in terms of the $\hat{T}_{i,j}^n$, it reduces to

$$-\frac{C}{2}\,\hat{T}_{i-1,j}^{n+1} + \left\{1+C\,(1+j^2\,\kappa^2/2)\right\}\, \hat{T}_{i,j}^{n+1}-\frac{C}{2}\,\hat{T}_{i+1,j}^{n+1} =$$

$$\frac{C}{2}\,\hat{T}_{i-1,j}^{n} + \left\{1-C\,(1+j^2\,\kappa^2/2)\right\}\, \hat{T}_{i,j}^{n}+\frac{C}{2}\,\hat{T}_{i+1,j}^{n},$$ (224)

for $i=1,N$, and $j=0,J$. Here, $C= D\,\delta t/(\delta x)^2$, and $\kappa = \pi\,\delta x/L$. Moreover, the boundary conditions (219) and (220) yield

$$\hat{T}_{0,j}^n = \frac{\Gamma_{l\,j}^n\,\delta x-\beta_l^n\,\hat{T}_{1,j}^n}{\alpha_l^n\,\delta x -\beta_l^n},$$ (225)

$$\hat{T}_{N+1,j}^n = \frac{\Gamma_{h\,j}^n\,\delta x + \beta_h^n\,\hat{T}_{N,j}^n}{\alpha_h^n\,\delta x +\beta_h^n},$$ (226)

where

$$\hat{\Gamma}_{l\,j}^n = \frac{2}{J}\sum_{k=0}^{J} \gamma_{l,k}^n\,\sin(j\,k\,\pi/J),$$ (227)

etc. Equations (224)--(226) constitute a set of $J+1$ uncoupled tridiagonal matrix equations for the $\hat{T}_{i,j}^{n+1}$, with one equation for each separate value of $j$.

In order to advance our solution by one time-step, we first Fourier transform the $T_{i,j}^n$ and the boundary conditions, according to Eqs. (223) and (227). Next, we invert the $J+1$ tridiagonal equations (224)--(226) to obtain the $\hat{T}_{i,j}^{n+1}$. Finally, we reconstruct the $T_{i,j}^n$ via Eq. (222).