2023-09-01 FD Solutions#
Last time#
Discussion and scoping
Evaluating derivatives
Taylor series and truncation error
Today#
Discuss activity
Measuring errors
Stable discretizations
Solutions and matrix properties
using Plots
default(linewidth=3)
using LinearAlgebra
Finite Difference/collocation approach to solve \(u\):#
Represent the function \(u(x)\) by its values \(u_i = u(x_i)\) at a discrete set of points
\[ -1 = x_1 < x_2 < \dotsb < x_n = 1 . \]The FD framework does not uniquely specify the solution values at other points
Compute derivatives at \(x_i\) via differencing formulas involving a finite number of neighbor points (independent of the total number of points \(n\)).
FD methods ask for the differential equation to be satisfied pointwise at each \(x_i\) in the interior of the domain.
Evaluate the forcing term \(f\) pointwise at \(x_i\).
Approximate derivatives at discrete boundary points (\(x_n = 1\) above), typically using one-sided differencing formulas.
Activity on stability and accuracy computing derivatives#
A note on stable computation#
x = big(1e-15)
@show big(1+x)
@show log(1 + x)
big(1 + x) = 1.000000000000001000000000000000077705399876661079238307185601195015145492561714
log(1 + x) = 9.999999999999995777053998766613348662406422732737965069319591605458493216045864e-16
9.999999999999995777053998766613348662406422732737965069319591605458493216045864e-16
plot([h -> log(1+h), log1p], xlim=(-1e-15, 1e-15))
What happens as we zoom in?
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 = 40
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
default(legendfontsize=12)
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), 2)/sqrt(n))
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 hs .^ 2], label=["h" "\$h^2\$"])
What happens if we use a 1-norm or 2-norm?
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.
Consistency + Stability = Convergence#
[Lax Equivalence Theorem]#
Consistency#
When we apply the differential operator to the exact solution, we get a small residual.
The residual converges under grid refinement.
Hopefully fast as \(h \to 0\)
Stability#
There do not exist “bad” functions that also satisfy the equation.
The solution does not “blow up” for time-dependent problems.
Definition here is intentionally vague, and there are more subtle requirements for problems like incompressible flow.
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 = 20
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
)) # which norm?
end
error
end
refinement_error2 (generic function with 1 method)
fig = plot(xlabel="h", xscale=:log10, ylabel="Error", 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, label="\$h^2\$")
Both methods are second order accurate.
The
diff2b
method is more accurate thandiff2a
(by a factor of 4)The
diff2a
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
x = LinRange(-1, 1, 5)
diff1_mat(x)
5×5 Matrix{Float64}:
-2.0 2.0 0.0 0.0 0.0
-1.0 0.0 1.0 0.0 0.0
0.0 -1.0 0.0 1.0 0.0
0.0 0.0 -1.0 0.0 1.0
0.0 0.0 0.0 -2.0 2.0
n = 12
x = LinRange(-3, 3, n)
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?#
function my_spy(A)
cmax = norm(vec(A), Inf)
s = max(1, ceil(120 / size(A, 1)))
spy(A, marker=(:square, s), c=:diverging_rainbow_bgymr_45_85_c67_n256, clims=(-cmax, cmax))
end
D = diff1_mat(x)
my_spy(D)
svdvals(D)
12-element Vector{Float64}:
2.7718000318790152
2.7716788692501084
1.7708640148632901
1.732003875931066
1.5877132402714698
1.4519108222799693
1.2963624321753364
1.0397961559479483
0.9166666666666657
0.5413728320040285
0.4745015826879543
1.7546012240164312e-16
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)
my_spy(L)
cond(L)
56.42029373680045
Solutions#
L = laplacian_dirichlet(x)
f = one.(x)
f[1] = 0
f[end] = 0;
plot(x, f)
u = L \ f
plot(x, u)
Discrete “Green’s functions”#
plot(x, inv(L)[:, 2])
Ln = copy(L)
Ln[n, n-1] = -1
plot(x, inv(Ln)[:, 4])
Discrete eigenfunctions#
x = LinRange(-1, 1, 10)
L = laplacian_dirichlet(x)
Lambda, V = eigen(L)
plot(Lambda, marker=:circle)
plot(x, V[:, 1:4])
Outlook on our method#
Pros#
Consistent
Stable
Second order accurate (we hope)
Cons#
Only second order accurate (at best)
Worse than second order on non-uniform grids
Worse than second order at Neumann boundaries
Boundary conditions break symmetry