MDL: direct discretization scheme, theory in doc and dirty implementation in scripts
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Marco De Lucia
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@ -316,6 +316,8 @@ be rewritten:
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\frac{\partial C }{\partial t} & = \frac{\partial}{\partial x} \left(\alpha(x) \frac{\partial C }{\partial x} \right)
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\end{align}
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** Discretization of the equation using chain rule
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\noindent From the product rule for derivatives we obtain:
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#+NAME: eqn:product
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@ -381,4 +383,59 @@ terms (at time level $t+1/2$) on the left hand side we obtain:
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\noindent If the diffusion coefficients are constant,
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$A_{i,j}=B_{i,j}=\alpha$ and the scheme reverts to the homogeneous
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case.
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case. Problem with this discretization is that the terms in $A_{ij}$
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and $B_{ij}$ can be negative depending on the derivative of the
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diffusion coefficient, resulting in unphysical values for the
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concentrations.
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** Direct discretization
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As noted in literature (LeVeque and Numerical Recipes) a better way is
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to discretize directly the physical problem ([[eqn:hetdiff]]) at points
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halfway between grid points:
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\begin{align*}
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\begin{cases}
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\displaystyle \alpha(x_{i+1/2}) \frac{\partial C }{\partial x}(x_{i+1/2}) & \displaystyle = \alpha_{i+1/2} \left( \frac{C_{i+1} -C_{i}}{\Delta x} \right) \\
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\displaystyle \alpha(x_{i-1/2}) \frac{\partial C }{\partial x}(x_{i-1/2}) & \displaystyle = \alpha_{i-1/2} \left( \frac{C_{i} -C_{i-1}}{\Delta x} \right)
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\end{cases}
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\end{align*}
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\noindent A further differentiation gives us the centered
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approximation of $\frac{\partial}{\partial x} \left(\alpha(x)
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\frac{\partial C }{\partial x}\right)$:
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\begin{align*}
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\frac{\partial}{\partial x} \left(\alpha(x)
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\frac{\partial C }{\partial x}\right)(x_i) & \simeq \frac{1}{\Delta x}\left[\alpha_{i+1/2} \left( \frac{C_{i+1} -C_{i}}{\Delta x} \right) - \alpha_{i-1/2} \left( \frac{C_{i} -C_{i-1}}{\Delta x} \right) \right]\\
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&\displaystyle =\frac{1}{\Delta x^2} \left[ \alpha_{i+1/2}C_{i+1} - (\alpha_{i+1/2}+\alpha_{i-1/2}) C_{i} + \alpha_{i-1/2}C_{i-1}\right]
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\end{align*}
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\noindent The ADI scheme with this approach becomes:
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#+NAME: eqn:genADI_hetdir
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\begin{equation}
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\left\{
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\begin{aligned}
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\frac{C^{t+1/2}_{i,j}-C^{t }_{i,j}}{\Delta t/2} = & \frac{1}{\Delta x^2} \left[ \alpha_{i+1/2,j} C^{t+1/2}_{i+1,j} - (\alpha_{i+1/2,j}+\alpha_{i-1/2,j}) C^{t+1/2}_{i,j} + \alpha_{i-1/2,j} C^{t+1/2}_{i-1,j}\right] + \\
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& \frac{1}{\Delta y^2} \left[ \alpha_{i,j+1/2}C^{t}_{i,j+1} - (\alpha_{i,j+1/2}+\alpha_{i,j-1/2}) C^t_{i} + \alpha_{i,j-1/2}C^{t}_{i,j-1}\right]\\
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\frac{C^{t+1 }_{i,j}-C^{t+1/2}_{i,j}}{\Delta t/2} = & \frac{1}{\Delta x^2} \left[ \alpha_{i+1/2,j}C^{t+1/2}_{i+1,j} - (\alpha_{i+1/2,j}+\alpha_{i-1/2,j}) C_{i} + \alpha_{i-1/2,j}C^{t+1/2}_{i-1,j}\right] + \\
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& \frac{1}{\Delta y^2} \left[ \alpha_{i,j+1/2}C^{t+1}_{i,j+1} - (\alpha_{i,j+1/2}+\alpha_{i,j-1/2}) C^{t+1}_{i,j} + \alpha_{i,j-1/2}C^{t+1}_{i,j-1}\right]
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\end{aligned}
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\right.
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\end{equation}
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\noindent Doing the usual algebra and separating implicit from
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explicit terms, the two sweeps become:
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#+NAME: eqn:sweepX_hetdir
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\begin{equation}
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\left\{
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\begin{aligned}
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-S_x \alpha^x_{i+1/2,j} C^{t+1/2}_{i+1,j} + (1 + S_x(\alpha^x_{i+1/2,j}+ \alpha^x_{i-1/2,j}))C^{t+1/2}_{i,j} - S_x \alpha^x_{i-1/2,j} C^{t+1/2}_{i-1,j} = & \\
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S_y \alpha^y_{i,j+1/2} C^{t }_{i,j+1} + (1 - S_y(\alpha^y_{i,j+1/2}+ \alpha^y_{i,j-1/2}))C^{t }_{i,j} + S_y \alpha^y_{i,j-1/2} C^{t }_{i,j-1} & \\[1em]
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-S_y \alpha^y_{i,j+1/2} C^{t+1 }_{i,j+1} + (1 + S_y(\alpha^y_{i,j+1/2}+ \alpha^y_{i,j-1/2}))C^{t+1 }_{i,j} - S_y \alpha^y_{i,j-1/2} C^{t+1 }_{i,j-1} = & \\
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S_x \alpha^x_{i+1/2,j} C^{t+1/2}_{i+1,j} + (1 - S_x(\alpha^x_{i+1/2,j}+ \alpha^x_{i-1/2,j}))C^{t+1/2}_{i,j} + S_x \alpha^x_{i-1/2,j} C^{t+1/2}_{i-1,j}
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\end{aligned}
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\right.
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\end{equation}
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@ -1,4 +1,4 @@
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## Time-stamp: "Last modified 2022-12-20 21:17:32 delucia"
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## Time-stamp: "Last modified 2022-12-21 13:47:00 delucia"
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## Brutal implementation of 2D ADI scheme
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## Square NxN grid with dx=dy=1
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@ -124,7 +124,7 @@ plot(adi2[[length(adi2)]], ref2, log="xy", xlab="ADI", ylab="ode.2D (reference)"
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abline(0,1)
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## Test heterogeneous scheme
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## Test heterogeneous scheme, chain rule
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ADIHet <- function(field, dt, iter, alpha) {
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if (!all.equal(dim(field), dim(alpha)))
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@ -246,3 +246,104 @@ par(mfrow=c(1,2))
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image(alphas)
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image(adih1[[length(adih1)]])
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points(0.5,0.5, col="red",pch=4)
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## Test heterogeneous scheme, direct discretization
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ADIHetDir <- function(field, dt, iter, alpha) {
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if (!all.equal(dim(field), dim(alpha)))
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stop("field and alpha are not matrix")
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## now both field and alpha must be nx*ny matrices
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nx <- ncol(field)
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ny <- nrow(field)
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dx <- dy <- 1
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## find out the center of the grid to apply conc=1
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cenx <- ceiling(nx/2)
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ceny <- ceiling(ny/2)
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field[cenx, ceny] <- 1
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## prepare containers for computations and outputs
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tmpX <- tmpY <- res <- field
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out <- vector(mode="list", length=iter)
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for (it in seq(1, iter)) {
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for (i in seq(2, ny-1)) {
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Aij <- cbind((alpha[i,]+alpha[i-1,])/2, (alpha[i,]+alpha[i+1,])/2)
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Bij <- cbind((alpha[,i]+alpha[,i-1])/2, (alpha[,i]+alpha[,i+1])/2)
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tmpX[i,] <- SweepByRowHetDir(i, res, dt=dt, Aij, Bij)
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}
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resY <- t(tmpX)
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for (i in seq(2, nx-1))
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tmpY[i,] <- SweepByRowHetDir(i, resY, dt=dt, Bij, Aij)
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res <- t(tmpY)
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out[[it]] <- res
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}
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return(out)
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}
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## Direct discretization, Workhorse function to fill A, B and solve
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## for a given *row* of the grid matrix
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SweepByRowHetDir <- function(i, field, dt, Aij, Bij) {
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dx <- 1 ## fixed in our test
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Sx <- Sy <- dt/2/dx/dx
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## diagonal of A at once
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A <- matrix(0, nrow(field), ncol(field))
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diag(A) <- 1 + Sx*(Aij[,1]+Aij[,2])
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## adjacent diagonals "Sx"
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for (ii in seq(1, nrow(field)-1)) {
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A[ii+1, ii] <- -Sx*Aij[ii,1]
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A[ii, ii+1] <- -Sx*Aij[ii,2]
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}
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B <- numeric(ncol(field))
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for (ii in seq_along(B))
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B[ii] <- Sy*Bij[ii,2]*field[i+1,ii] + (1 - Sy*(Bij[ii,1]+Bij[ii,2]))*field[i, ii] + Sy*Bij[ii,1]*field[i-1,ii]
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x <- solve(A, B)
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x
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}
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## adi2 <- ADI(n=51, dt=10, iter=200, alpha=1E-3)
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## ref2 <- DoRef(n=51, alpha=1E-3, dt=10, iter=200)
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n <- 51
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field <- matrix(0, n, n)
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alphas <- matrix(1E-3*runif(n*n, 1,1.2), n, n)
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## for (i in seq(1,nrow(alphas)))
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## alphas[i,] <- seq(1E-7,1E-3, length=n)
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#diag(alphas) <- rep(1E-2, n)
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adih <- ADIHetDir(field=field, dt=10, iter=100, alpha=alphas)
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adi2 <- ADI(n=n, dt=10, iter=100, alpha=1E-3)
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par(mfrow=c(1,3))
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image(adi2[[length(adi2)]])
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image(adih[[length(adih)]])
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points(0.5,0.5, col="red",pch=4)
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plot(adih1[[length(adih1)]], adi2[[length(adi2)]], pch=4, log="xy")
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abline(0,1)
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sapply(adih, sum)
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sapply(adi2, sum)
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adi2
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par(mfrow=c(1,2))
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image(alphas)
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image(adih1[[length(adih1)]])
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points(0.5,0.5, col="red",pch=4)
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