2021-08-27 Finite Difference 2¶
Last time¶
Evaluating derivatives
Taylor series and truncation error
Activity on stability of computing derivatives
Today¶
Stability
Derivatives as matrices
Boundary conditions
Discrete eigenvalues/eigenvectors
using Plots
Discuss activity on stability and accuracy computing derivatives¶
A note on stable computation¶
x = 1e-15
@show x
@show log(1 + x)
x = 1.0e-15
log(1 + x) = 1.110223024625156e-15
1.110223024625156e-15
plot([x -> log(1 + x), log1p], xlims=(-1e-15, 1e-15))
plot!(x -> x)
A few methods on grids¶
diff1l(x, u) = x[2:end], (u[2:end] - u[1:end-1]) ./ (x[2:end] - x[1:end-1])
diff1r(x, u) = x[1:end-1], (u[2:end] - u[1:end-1]) ./ (x[2:end] - x[1:end-1])
diff1c(x, u) = x[2:end-1], (u[3:end] - u[1:end-2]) ./ (x[3:end] - x[1:end-2])
difflist = [diff1l, diff1r, diff1c]
n = 20
h = 2 / (n - 1)
x = LinRange(-3, 3, n)
u = sin.(x)
fig = plot(cos, xlims=(-3, 3))
for d in difflist
xx, yy = d(x, u)
plot!(fig, xx, yy, marker=:circle, label=d)
end
fig
Measuring error on grids¶
using LinearAlgebra
grids = 2 .^ (2:10)
hs = 1 ./ grids
function refinement_error(f, fprime, d)
error = []
for n in grids
x = LinRange(-3, 3, n)
xx, yy = d(x, f.(x))
push!(error, norm(yy - fprime.(xx), Inf))
end
error
end
refinement_error (generic function with 1 method)
fig = plot(xscale=:log10, yscale=:log10)
for d in difflist
error = refinement_error(sin, cos, d)
plot!(fig, hs, error, marker=:circle, label=d)
end
plot!(fig, hs, hs .^ 2)
Stability¶
Are there “rough” functions for which these formulas estimate \(u'(x_i) = 0\)?
x = LinRange(-1, 1, 9)
f_rough(x) = cos(.1 + 4π*x)
fp_rough(x) = -4π*sin(.1 + 4π*x)
plot(x, f_rough, marker=:circle)
plot!(f_rough)
fig = plot(fp_rough, xlims=(-1, 1))
for d in difflist
xx, yy = d(x, f_rough.(x))
plot!(fig, xx, yy, label=d, marker=:circle)
end
fig
If we have a solution \(u(x)\), then \(u(x) + f_{\text{rough}}(x)\) is indistinguishable to our FD method.
Second derivatives¶
We can compute a second derivative by applying first derivatives twice.
function diff2a(x, u)
xx, yy = diff1c(x, u)
diff1c(xx, yy)
end
function diff2b(x, u)
xx, yy = diff1l(x, u)
diff1r(xx, yy)
end
diff2list = [diff2a, diff2b]
n = 10
x = LinRange(-3, 3, n)
u = - cos.(x);
fig = plot(cos, xlims=(-3, 3))
for d2 in diff2list
xx, yy = d2(x, u)
plot!(fig, xx, yy, marker=:circle, label=d2)
end
fig
How fast do these approximations converge?¶
grids = 2 .^ (3:10)
hs = 1 ./ grids
function refinement_error2(f, f_xx, d2)
error = []
for n in grids
x = LinRange(-3, 3, n)
xx, yy = d2(x, f.(x))
push!(error, norm(yy - f_xx.(xx), Inf))
end
error
end
refinement_error2 (generic function with 1 method)
fig = plot(xscale=:log10, yscale=:log10)
for d2 in diff2list
error = refinement_error2(x -> -cos(x), cos, d2)
plot!(fig, hs, error, marker=:circle, label=d2)
end
plot!(fig, hs, hs .^ 2)
Both methods are second order accurate.
The
diff2b
method is more accurate thandiff2a
(by a factor of 4)The
diff2b
method can’t compute derivatives at points adjacent the boundary.We don’t know yet whether either is stable
Differentiation matrices¶
All our diff*
functions thus far have been linear in u
, therefore they can be represented as matrices.
function diff1_mat(x)
n = length(x)
D = zeros(n, n)
h = x[2] - x[1]
D[1, 1:2] = [-1/h 1/h]
for i in 2:n-1
D[i, i-1:i+1] = [-1/2h 0 1/2h]
end
D[n, n-1:n] = [-1/h 1/h]
D
end
diff1_mat (generic function with 1 method)
x = LinRange(-3, 3, 10)
plot(x, diff1_mat(x) * sin.(x), marker=:circle)
plot!(cos)
How accurate is this derivative matrix?¶
fig = plot(xscale=:log10, yscale=:log10, legend=:topleft)
error = refinement_error(sin, cos, (x, u) -> (x, diff1_mat(x) * u))
plot!(fig, hs, error, marker=:circle)
plot!(fig, hs, hs, label="\$h\$")
plot!(fig, hs, hs .^ 2, label="\$h^2\$")
Can we study it as a matrix?¶
D = diff1_mat(x)
spy(D, marker=(:square, 10), c=:bwr)
svdvals(D)
10-element Vector{Float64}:
2.268133218393964
2.2674392839412794
1.4265847744427302
1.368373796830966
1.2135254915624205
1.0228485194005286
0.8816778784387096
0.5437139466339259
0.46352549156242107
3.7873060138463766e-17
Second derivative with Dirichlet boundary conditions¶
The left endpoint in our example boundary value problem has a Dirichlet boundary condition,
This matrix is not symmetric even if \(A\) is.
function laplacian_dirichlet(x)
n = length(x)
D = zeros(n, n)
h = x[2] - x[1]
D[1, 1] = 1
for i in 2:n-1
D[i, i-1:i+1] = (1/h^2) * [-1, 2, -1]
end
D[n, n] = 1
D
end
laplacian_dirichlet (generic function with 1 method)
Laplacian as a matrix¶
L = laplacian_dirichlet(x)
spy(L, marker=(:square, 10), c=:bwr)
cond(L)
36.5545720479542
Discrete “Green’s functions”¶
plot(x, inv(L)[:, 2])
Ln = copy(L)
Ln[n, n-1] = -1
plot(x, inv(Ln)[:, 4])