def unifying() -> None:
r"""
The Forward Euler, Backward Euler, and Crank-Nicolson schemes.
Notes
----------
The Forward Euler, Backward Euler, and Crank-Nicolson schemes can be
formulated as one scheme with varying parameter θ:
(uⁿ⁺¹ - uⁿ) / (t{n+1} - tn) = -a (θ uⁿ⁺¹ + (1 - θ) uⁿ)
Observe that
- θ = 0 gives the Forward Euler scheme.
- θ = 1 gives the Backward Euler scheme.
- θ = ½ gives the Crank-Nicolson scheme.
One may alternatively choose any other value of θ in [0, 1], but this is
not so common since the accuracy and stability of the scheme do not improve
compared to the values θ = 0, 1, ½. As before, uⁿ is considered known and
uⁿ⁺¹ unknown, so we solve for the latter:
uⁿ⁺¹ = (1 - (1 - θ) a (t{n+1} - tn)) / (1 + θ a (t{n+1} - tn))
This is known as the θ-rule, or alternatively written as the "theta-rule".
Compact notation: The θ-rule can be specified in operator notation by
[D̄ₜ u = -a ūᵗʼᶿ]ⁿ⁺ᶿ
We define a new time difference
[D̄ₜ u]ⁿ⁺ᶿ = (uⁿ⁺¹ - uⁿ) / (t{n+1} - tn)
To be applied at the time point t{n+θ} ≈ θ tn + (1 - θ) t{n+1}. This
weighted average gives rise to the weighted average operator
[ūᵗʼᶿ]ⁿ⁺ᶿ = (1 - θ) uⁿ + θ uⁿ⁺¹ ≈ u(t{n+θ})
where θ ∈ [0, 1] as usual. Note that for θ = ½, we recover the standard
centered difference and standard arithmetic mean. An alternative and
perhaps clearer notation is
[Dₜ u]ⁿ⁺¹⸍² = θ [-a u]ⁿ⁺¹ + (1 - θ) [-a u]ⁿ
"""
from utils.solver import solver_chap2 as solver
I = 1
a = 2
T = 8
dt = 0.8
# Write out a table of t and u values for each theta
for th in (0, 0.5, 1):
u, t = solver(I=I, a=a, T=T, dt=dt, theta=th)
print('theta = {:g}'.format(th))
for idx, t_i in enumerate(t):
print('t={0:6.3f} u={1:g}'.format(t_i, u[idx]))
print('-----')
def numerical_error() -> None:
r"""
Computing the numerical error as a mesh function.
Notes
----------
A natural way to compare the exact and discrete solutions is to calculate
their difference as a mesh function for the error:
eⁿ = uₑ(tn) - uⁿ, n = 0,1,...Nₜ
We may also compute the norm of the error mesh function, so that we can get
a single number expressing the size of the error. This is obtained by
taking the norm of the error function. Three common norms are
‖f‖_{L²} = √(∫₀ᵀ f(t)² dt)
‖f‖_{L¹} = ∫₀ᵀ |f(t)| dt
‖f‖_{L∞} = max_{t ∈ [0, T]} |f(t)|
The L² norm ("L-two norm") has nice mathematical properties and is the most
popular norm. Numerical computations involving mesh functions need
corresponding norms. Imagining that the mesh function is extended to vary
linearly between the mesh points, the Trapezoidal rule is in fact an exact
integration rule. A possible possible modification of the L² norm for a
mesh function fⁿ on a uniform mesh with spacing Δt is therefore the
well-known Trapezoidal formula. A common approximation of this, motivated
by convenience is
‖fⁿ‖_{ℓ²} = √(Δt ∑₀ᴺᵗ (fⁿ)²)
This is called the discrete L² norm and denoted by ℓ². If the square of
this norm is used instead of the Trapezoidal integration formula, the error
is Δt ((f⁰)² + (fᴺᵗ)²) / 2. This means that the weights at the end points
of the mesh function are perturbed, but as Δt -> 0, the error from this
perturbation goes to zero. The three discrete norms are then define by
‖fⁿ‖_{ℓ²} = √(Δt ∑₀ᴺᵗ (fⁿ)²)
‖fⁿ‖_{ℓ¹} = Δt ∑₀ᴺᵗ |fⁿ|
‖fⁿ‖_{ℓ∞} = max_{0 ≤ n ≤ Nₜ} |fⁿ|
Note that L², L¹, ℓ², and ℓ¹ norms depend on the length of the interval of
interest. In some applications it is convenient to think of the mesh
function as just a vector of function values without any relation to the
interval [0, T]. Then one can replace Δt by T / Nₜ and simply drop the T
(which is just a common scaling factor). Moreover, people prefer to divide
by the total length of the vector, Nₜ+1 instead of just Nₜ. This reasoning
gives rise to the vector norms for a vector f = (f₀,...,f_N):
‖f‖₂ = √((1 / (N + 1)) ∑₀ᴺᵗ (fₙ)²)
‖f‖₁ = (1 / (N + 1)) ∑₀ᴺ |fₙ|
‖f‖_{ℓ∞} = max_{0 ≤ n ≤ N} |fₙ|
We will mostly work with the mesh functions and use discrete ℓ² norm or the
max norm ℓ∞, but the vector norms are also much used in numerical
computations, so it is important to know the different norms and the
relations between them.
A single number that expresses the size of the numerical error will be
taken as ‖eⁿ‖_{ℓ²} and called E:
E = √(Δt ∑₀ᴺᵗ (eⁿ)²)
"""
from utils.solver import solver_chap2 as solver
I = 1
a = 2
T = 8
th_dict = {0: ('Forward Euler', 'fe'), 1: ('Backward Euler', 'be'),
0.5: ('Crank-Nicolson', 'cn')}
for th in th_dict:
print('theta = {:g}'.format(th))
print('dt error')
for dt in (0.4, 0.04):
# Solve
u, t = solver(I=I, a=a, T=T, dt=dt, theta=th)
# Calculate exact solution
u_exact = lambda t, I, a: I * np.exp(-a * t)
u_e = u_exact(t, I, a)
# Calculate the error and its discrete norm
e = u_e - u
E = np.sqrt(dt * np.sum(e**2))
print('{0:6.2f} {1:12.3E}'.format(dt, E))
# Plot with red dashes w/ circles
plt.figure()
plt.errorbar(t, u, e, 0, 'r--o', label='numerical')
# Plot with blue line
plt.plot(t, u_e, 'b-', label='exact')
# Save figure
plt.xlabel('t')
plt.ylabel('u')
plt.title('{}, dt={:g}'.format(th_dict[th][0], dt))
plt.legend()
plt.savefig(IMGDIR + '{0}_{1:.2f}_err.png'.format(
th_dict[th][1], dt), bbox_inches='tight')
print('-----')
def forward_euler() -> None:
r"""
Solve ODE (1) by the Forward Euler (FE) finite difference scheme.
Notes
----------
Solving an ODE like (1) by a finite difference method consists of the
following four steps:
1) discretising the domain,
2) requiring fulfillment of the equation at discrete time points,
3) replacing derivatives by finite differences,
4) formulating a recursive algorithm.
Step 1: Discretising the domain.
We represent the time domain [0, T] by a finite number of points Nₜ+1,
called a "mesh" or "grid".
0 = t0 < t1 < t2 < ... < t{Nₜ-1} < t{Nₜ} = T
The goal is to find the solution u at the mesh points: u(tn) := uⁿ,
n=0,1,...,Nₜ. More precisely, we let uⁿ be the numerical approximation
to the exact solution u(tn) at t=tn.
We say that the numerical approximation constitutes a mesh function,
which is defined at discrete points in time. To compute the solution
for some solution t ∈ [tn, tn+1], we can use an interpolation method,
e.g. the linear interpolation formula
u(t) ≈ uⁿ + (uⁿ⁺¹ - uⁿ) / (t{n+1} - tn) * (t - tn)
Step 2: Fulfilling the equation at discrete time points.
We relax the requirement that the ODE hold for all t ∈ (0, T] and
require only that the ODE be fulfilled at discrete points in time. The
mesh points are a natural (but not the only) choice of points. The
original ODE is reduced to the following
u'(tn) = -a u(tn) a > 0 n = 0,...,Nₜ u(0) = I
Step 3: Replace derivative with finite differences.
The next most essential step is to replace the derivative u' by the
finite difference approximation. Here the forward difference
approximation is
u'(tn) ≈ (uⁿ⁺¹ - uⁿ) / (t{n+1} - tn)
The name forward relates to the face that we use a value forward in
time, uⁿ⁺¹, together with the value uⁿ at the point tn. Therefore,
(uⁿ⁺¹ - uⁿ) / (t{n+1} - tn) = -a uⁿ, n=0,1,...,Nₜ-1
This is often referred to as a finite difference scheme or more
generally as the discrete equations of the problem. The fundamental
feature of these equations is that they are algebraic and can hence be
straightforwardly solved to produce the mesh function uⁿ.
Step 4: Formulating a recursive algorithm
The final step is to identify the computational algorithm to be
implemented in a program. In general, starting from u⁰ = u(0) = I, uⁿ
can be assumed known, and then we can easily solve for the unknown uⁿ⁺¹
uⁿ⁺¹ = uⁿ - a (t{n+1} - tn) uⁿ
From a mathematical point of view, these type of equations are known as
difference equations since they express how differences in the
dependent variable, here u, evolve with n.
Interpretation: We have computed some point values on the solution curve,
and the question is how we reason about the next point. Since we know u and
t at the most recently computed point, the differential equation gives us
the slope of the solution u' = -a u. We can continue the solution curve
along that slop. As soon as we have chosen the next point on this line, we
have a new t and u value and can compute a new slope and continue this
process.
Derivation: The key tool for this is the Taylor series.
f(x) ≈ f(a) + f'(a)(x - a) + ½f''(a)(x - a)² + ⅙f'''(a)(x - a)³ + ...
+ (1 / m!) dᵐf/dxᵐ(a)(x - a)ᵐ
For a function of time, f(t), related to mesh spacing Δt,
f(tn + Δt) ≈ f(tn) + f'(tn)Δt + ½f''(tn)Δt² + ⅙f'''(tn)Δt³ + ...
+ (1 / m!) dᵐf/dtᵐ(tn)Δtᵐ
Now, by rearranging for f'(tn)
f'(tn) ≈ (f(tn + Δt) - f(tn)) / Δt - ½f''(tn)Δt - ⅙f'''(tn)Δt² - ...
- (1 / m!) dᵐf/dtᵐ(tn)Δtᵐ⁻¹
Now, in the limit Δt -> 0,
f'(tn) ≈ (f(tn + Δt) - f(tn) / Δt
An interesting point is that we have a measure of the error as seen by the
O(Δtᵐ) terms. A leading order, the error can be given as ½f''(tn)Δt.
Compact notation: For a function u(t), a forward difference approximation
is denoted by the Dₜ⁺ operator and written as
[Dₜ⁺ u]ⁿ = (uⁿ⁺¹ - uⁿ) / Δt (≈ du(tn)/dt)
This notation consists of an operator that approximates differentiation wrt
an independent variable, here t. The operator is built on the symbol D,
with the independent variable as subscript and a superscript denoting the
type of difference. The superscript ⁺ indicates a forward difference. We
place square brackets around the operator and the function it operates on
and specify the mesh point, where the operator is acting, by a superscript
after the closing bracket. In our compact notation, the Forward Euler
scheme can be written as
[Dₜ⁺ u]ⁿ = -a uⁿ
In difference equations, we often place the square brackets around the
whole equation, to indicate which mesh point the equation applies, since
each term by be approximated at the same point:
[Dₜ⁺ u = -a u]ⁿ
"""
# Define solver function
def solver(I: float, a: float, T: float, dt: float):
"""Solve u'=-a*u, u(0)=I, for t in (0,T] with steps of dt using FE.
Parameters
----------
I : Initial condition.
a : Constant coefficient.
T : Maximum time to compute to.
dt : Step size.
Returns
----------
u : Mesh function.
t : Mesh points.
"""
# Initialise data structures
Nt = int(T / dt) # Number of time intervals
u = np.zeros(Nt + 1) # Mesh function
t = np.linspace(0, Nt * dt, Nt + 1) # Mesh points
# Calculate mesh function using difference equation
# uⁿ⁺¹ = uⁿ - a (t{n+1} - tn) uⁿ
u[0] = I
for n_idx in range(Nt):
u[n_idx + 1] = (1 - a * dt) * u[n_idx]
return u, t
I = 1
a = 2
T = 8
dt = 0.8
u, t = solver(I=I, a=a, T=T, dt=dt)
# Write out a table of t and u values
for idx, t_i in enumerate(t):
print('t={0:6.3f} u={1:g}'.format(t_i, u[idx]))
# Plot with red dashes w/ circles
plt.figure()
plt.plot(t, u, 'r--o', label='numerical')
# Calculate exact solution
u_exact = lambda t, I, a: I * np.exp(-a * t)
t_e = np.linspace(0, T, 1001)
u_e = u_exact(t_e, I, a)
# Plot with blue line
plt.plot(t_e, u_e, 'b-', label='exact')
# Save figure
plt.xlabel('t')
plt.ylabel('u')
plt.title('Forward Euler, dt={:g}'.format(dt))
plt.legend()
plt.savefig(IMGDIR + 'fe.png', bbox_inches='tight')
def crank_nicolson() -> None:
r"""
Solve ODE (1) by the Crank-Nicolson (CN) finite difference scheme.
Notes
----------
The finite difference schemes derived in `forward_euler` and
`backward_euler` are both one-sided differences. Such one-sided differences
are known to be less accurate than central (or midpoint) differences, where
we use information both forward and backward in time. A natural next step
is therefore to construct a central difference approximation.
The central difference approximation to the derivative is sought at the
point t{n+½} = ½(tn + t{n+1}). The approximation reads
u'(t{n + ½}) ≈ (uⁿ⁺¹ - uⁿ) / (t{n+1} - tn)
With this formula, it is natural to demand the ODE be fulfilled at the time
points between the mesh points:
u'(t{n + ½}) = -a u(t{n + ½}), n=0,...,Nₜ-1
Combining these results results in the approximate discrete equation
(uⁿ⁺¹ - uⁿ) / (t{n+1} - tn) = -a uⁿ⁺¹⸍², n=0,1,...,Nₜ-1
However, there is a fundamental problem with the right-hand side of the
equation. We aim to compute uⁿ for integer n, which means that uⁿ⁺¹⸍² is
not a quantity computed by our method. One possibility is to approximate
uⁿ⁺¹⸍² as an arithmetic mean of the u values at the neighbouring mesh
points
uⁿ⁺¹⸍² ≈ ½(uⁿ + uⁿ⁺¹)
We then obtain the approximate discrete equation
(uⁿ⁺¹ - uⁿ) / (t{n+1} - tn) = -a ½(uⁿ + uⁿ⁺¹), n=0,1,...,Nₜ-1
There are three approximation steps leading to this formula. First, the ODE
is only valid at discrete points. Second, the derivative is approximated by
finite differences, and third, the value of u between mesh points is
approximated by an arithmetic mean value. Despite one more approximation
than for the Backward and Forward Euler schemes, the use of a centered
difference leads to a more accurate method.
To formulate a recursive method, we assume that uⁿ is already computed so
that uⁿ⁺¹ is the unknown, which we can solve for:
uⁿ⁺¹ = (1 - ½ a (t{n+1} - tn) / (1 + ½ a (t{n+1} - tn) * uⁿ
Derivation: The centered difference approximates the derivative at
tn + ½Δt. The Taylor expansions of f(tn) and f(t{n+1}) around tn + ½Δt are
f(tn) ≈ f(tn + ½Δt) - f'(tn + ½Δt)½Δt + ½f''(tn + ½Δt)(½Δt)² -
⅙f'''(tn + ½Δt)(½Δt)³ + ... + (1 / m!) dᵐf/dtᵐ(tn + ½Δt)(½Δt)ᵐ
f(t{n+1}) ≈ f(tn + ½Δt) + f'(tn + ½Δt)½Δt + ½f''(tn + ½Δt)(½Δt)² +
⅙f'''(tn + ½Δt)(½Δt)³ + ... + (1 / m!) dᵐf/dtᵐ(tn + ½Δt)(½Δt)ᵐ
Subtracting the first from the second gives
f(t{n+1}) - f(tn) ≈ f'(tn + ½Δt)Δt + ⅓f'''(tn + ½Δt)(½Δt)³ + ...
Solving with respect to f'(tn + ½Δt) results in
f'(tn + ½Δt) ≈ (f(t{n+1}) - f(tn)) / Δt - (1/24)f'''(tn + ½Δt)Δt² + ...
The error measure goes like O(Δt²), which means the error here goes faster
to zero compared to the forward and backward differences for small Δt.
Compact notation: The centered difference operator notation reads
[Dₜ u]ⁿ = (uⁿ⁺¹⸍² - uⁿ⁻¹⸍²) / Δt (≈ du(tn)/dt)
Note here that no superscript implies a central differences. An averaging
operator is also convenient to have:
[ūᵗ]ⁿ = ½(uⁿ⁻¹⸍² + uⁿ⁺¹⸍²) ≈ u(tn)
The superscript ᵗ indicates that the average is taken along the time
coordinate. The common average (uⁿ + uⁿ⁺¹⸍²) / 2 can now be expressed as
[ūᵗ]ⁿ⁺¹⸍².
Now the Crank-Nicolson scheme to our ODE can be written as
[Dₜ u = -a ūᵗ]ⁿ⁺¹⸍²
"""
# Define solver function
def solver(I: float, a: float, T: float, dt: float):
"""Solve u'=-a*u, u(0)=I, for t in (0,T] with steps of dt using BE.
Parameters
----------
I : Initial condition.
a : Constant coefficient.
T : Maximum time to compute to.
dt : Step size.
Returns
----------
u : Mesh function.
t : Mesh points.
"""
# Initialise data structures
Nt = int(T / dt) # Number of time intervals
u = np.zeros(Nt + 1) # Mesh function
t = np.linspace(0, Nt * dt, Nt + 1) # Mesh points
# Calculate mesh function using difference equation
# uⁿ⁺¹ = 1 / (1 + a (t{n+1} - tn)) * uⁿ
u[0] = I
for n_idx in range(Nt):
u[n_idx + 1] = (1 - 0.5 * a * dt) / (1 + 0.5 * a * dt) * u[n_idx]
return u, t
I = 1
a = 2
T = 8
dt = 0.8
u, t = solver(I=I, a=a, T=T, dt=dt)
# Write out a table of t and u values
for idx, t_i in enumerate(t):
print('t={0:6.3f} u={1:g}'.format(t_i, u[idx]))
# Plot with red dashes w/ circles
plt.figure()
plt.plot(t, u, 'r--o', label='numerical')
# Calculate exact solution
u_exact = lambda t, I, a: I * np.exp(-a * t)
t_e = np.linspace(0, T, 1001)
u_e = u_exact(t_e, I, a)
# Plot with blue line
plt.plot(t_e, u_e, 'b-', label='exact')
# Save figure
plt.xlabel('t')
plt.ylabel('u')
plt.title('Crank-Nicolson, dt={:g}'.format(dt))
plt.legend()
plt.savefig(IMGDIR + 'cn.png', bbox_inches='tight')
def backward_euler() -> None:
r"""
Solve ODE (1) by the Backward Euler (BE) finite difference scheme.
Notes
----------
There are several choices of difference approximations in step 3 of the
finite difference scheme presenting in `forward_euler`. Another alternative
is
u'(tn) ≈ (uⁿ - uⁿ⁻¹) / (tn - t{n-1})
Since this difference is going backward in time (t{n-1}) for information,
it is known as a backward difference, also called Backward Euler
difference. Inserting our equation yields
(uⁿ - uⁿ⁻¹) / (tn - t{n-1}) = -a uⁿ, n=1,...,Nₜ
For direct similarity to the Forward Euler scheme, we replace n by n+1 and
solve for the unknown value uⁿ⁺¹
uⁿ⁺¹ = 1 / (1 + a (t{n+1} - tn)) * uⁿ, n=0,...,Nₜ-1
Derivation: Here we use the Taylor series around f(fn - Δt)
f(tn - Δt) ≈ f(tn) - f'(tn)Δt + ½f''(tn)Δt² - ⅙f'''(tn)Δt³ + ...
+ (1 / m!) dᵐf/dtᵐ(tn)Δtᵐ
Solving with respect to f'(tn) gives
f'(tn) ≈ (f(tn) - f(tn - Δt)) / Δt + ½f''(tn)Δt - ⅙f'''(tn)Δt² + ...
- (1 / m!) dᵐf/dtᵐ(tn)Δtᵐ⁻¹
Then term ½f''(tn)Δt can be taken as a simple measure of the approximation
error.
Compact notation: The backward difference reads
[Dₜ⁻ u]ⁿ = (uⁿ - uⁿ⁻¹) / Δt (≈ du(tn)/dt)
Note the subscript ⁻ denotes the backward difference. The Backward Euler
scheme to our ODE can be written as
[Dₜ⁻ u = -a u]ⁿ
"""
# Define solver function
def solver(I: float, a: float, T: float, dt: float):
"""Solve u'=-a*u, u(0)=I, for t in (0,T] with steps of dt using BE.
Parameters
----------
I : Initial condition.
a : Constant coefficient.
T : Maximum time to compute to.
dt : Step size.
Returns
----------
u : Mesh function.
t : Mesh points.
"""
# Initialise data structures
Nt = int(T / dt) # Number of time intervals
u = np.zeros(Nt + 1) # Mesh function
t = np.linspace(0, Nt * dt, Nt + 1) # Mesh points
# Calculate mesh function using difference equation
# uⁿ⁺¹ = 1 / (1 + a (t{n+1} - tn)) * uⁿ
u[0] = I
for n_idx in range(Nt):
u[n_idx + 1] = 1 / (1 + a * dt) * u[n_idx]
return u, t
I = 1
a = 2
T = 8
dt = 0.8
u, t = solver(I=I, a=a, T=T, dt=dt)
# Write out a table of t and u values
for idx, t_i in enumerate(t):
print('t={0:6.3f} u={1:g}'.format(t_i, u[idx]))
# Testing Backward difference
for idx in reversed(range(len(u) - 1)):
print('calc u(n-1) = {:.6f}'.format(u[idx + 1] + a * dt * u[idx + 1]))
print('actu u(n-1) = {:.6f}'.format(u[idx]))
# Plot with red dashes w/ circles
plt.figure()
plt.plot(t, u, 'r--o', label='numerical')
# Calculate exact solution
u_exact = lambda t, I, a: I * np.exp(-a * t)
t_e = np.linspace(0, T, 1001)
u_e = u_exact(t_e, I, a)
# Plot with blue line
plt.plot(t_e, u_e, 'b-', label='exact')
# Save figure
plt.xlabel('t')
plt.ylabel('u')
plt.title('Backward Euler, dt={:g}'.format(dt))
plt.legend()
plt.savefig(IMGDIR + 'be.png', bbox_inches='tight')
def scaling() -> None:
"""
Scaling and dimensionless variables.
Notes
----------
Real applications of a model u' = -a u + b will often involve a lot of
parameters in the expressions for a and b. It can be quite challenging to
find relevant values for all parameters. In simple problems, however, it
turns out that it is not always necessary to estimate all parameters
because we can lump them into one or a few dimensionless numbers by using a
very attractive technique called scaling. It simply means to stretch the u
and t axis in the present problem - and suddenly all parameters in the
problem are lumped into one parameter if b ≠ 0 and no parameter when b = 0!
Scaling means we introduce a new function ū(ť), with
ū = (u - uᵐ) / uᶜ ť = t / tᶜ
where uᵐ is a characteristic value of u, uᶜ is a characteristic size of the
range of u values, and tᶜ is a characteristic size of the range of t where
u shows significant variation. Choosing uᵐ, uᶜ, and tᶜ is not always easy
and is often an art in complicated problems. We just state one choice
first:
uᶜ = I, uᵐ = b / a, tᶜ = 1 / a
Inserting u = uᵐ + uᶜ ū and t = tᶜ ť in the problem u' = a u + b, assuming
a and b are constants, results (after some algebra) in the scaled problem
dū/dť = -ū, ū(0) = 1 - β
where β = b / (I a)
The parameter β is a dimensionless number. An important observation is that
ū depends on ť and β. That is, only the special combination of b / (I a)
matters, not what the individual values of b, a, and I are. The original
unscaled function depends on t, b, a, and I. A second observation is
striking: if b = 0, the scaled problem is independent of a and I! In
practice, this means that we can perform a single numerical simulation of
the scaled problem and recover the solution of any problems for any given a
and I by stretching the axis in the plot: u = I ū and t = ť / a. For any
b ≠ 0, we simulate the scaled problem for a few β values and recover the
physical solution u by translating and stretching the t axis.
In general, scaling combines the parameters in a problem to a set of
dimensionless parameters. The number of dimensionless parameters is usually
much smaller than the number of original parameters.
This scaling breaks down if I = 0. In that case, we may choose uᵐ = 0,
uᶜ = b / a, and tᶜ = 1 / b, resulting in the slightly different scaled
problem
dū/dť = 1 - ū, ū(0) = 0
As with b = 0, the case I = 0 has a scaled problem with no physical
parameters! It is common to drop the bars after scaling and write the
scaled problem as u' = u, u(0) = 1 - β, or u' = 1 - u, u(0) = 0. Any
implementation of the problem u' = -a u + b, u(0) = I can then be reused
for the scaled problem by setting a = 1, b = 0, and I = 1 - β if I ≠ 0, or
one sets a = 1, b = 1, and I = 0 when the physical I is zero.
"""
from utils.solver import solver_chap4 as solver
# Definitions
# Solving y'(x) = λ y + α x ∈ (0, X] y(0) = I ≠ 0
I = 1
X = 4
# lamda = -0.8
# alpha = 0.5
# ylims = (0.5, 1)
lamda = 0.8
alpha = 0.5
ylims = (0, 40)
y_exact = lambda x: np.exp(x * lamda) * I + \
((-1 + np.exp(x * lamda)) * alpha) / lamda
u_exact = lambda I, t: I * np.exp(t)
# Scaling
# ỹ = (y + yᵐ) / yᶜ, 𝐱 = x / xᶜ
# yᶜ = I, yᵐ = α / λ, xᶜ = 1 / λ
# therefore,
# dỹ/d𝐱 = ỹ, 𝐱 ∈ (0, λ T], ỹ(0) = 1 + β = 1 + α / (I λ)
#
# y(x) = (ỹ(λ x) - β) * I
beta = alpha / (I * lamda)
ulims = list(map(lambda x: (x / I) + beta, ylims))
th_dict = {
0: ('Forward Euler', 'fe', 'r-s'),
1: ('Backward Euler', 'be', 'g-v'),
0.5: ('Crank-Nicolson', 'cn', 'b-^')
}
fig1 = plt.figure(figsize=(14, 10))
fig3 = plt.figure(figsize=(14, 10))
fig1.suptitle(
r'$\frac{dy(x)}{dx}=\lambda y(x)+\alpha\:{\rm where}\:\lambda=' +
f'{lamda:g}' + r',\alpha=' + f'{alpha:g}' + r'$',
y=0.95)
fig3.suptitle(r'$\frac{du(t)}{dt}=u(t)\:{\rm where}\:\beta=' +
f'{beta:g}' + r'$',
y=0.95)
axs1 = []
axs3 = []
gs1 = gridspec.GridSpec(2, 2)
gs3 = gridspec.GridSpec(2, 2)
gs1.update(hspace=0.2, wspace=0.2)
gs3.update(hspace=0.2, wspace=0.2)
for th_idx, th in enumerate(th_dict):
fig2, axs2 = plt.subplots(2,
2,
figsize=(10, 8),
gridspec_kw={'hspace': 0.3})
for dx_idx, dx in enumerate((1.5, 1.25, 0.75, 0.1)):
# Calculate mesh points
Nx = int(X / dx)
x = np.linspace(0, Nx * dx, Nx + 1)
# Calculate scaled mesh points
T = lamda * X
t = lamda * x
dt = lamda * dx
if th_idx == 0:
gssub1 = gs1[dx_idx].subgridspec(2,
1,
height_ratios=(2, 1),
hspace=0)
ax1 = fig1.add_subplot(gssub1[0])
ax2 = fig1.add_subplot(gssub1[1])
gssub2 = gs3[dx_idx].subgridspec(2,
1,
height_ratios=(2, 1),
hspace=0)
axs1.append([ax1, ax2])
ax3 = fig3.add_subplot(gssub2[0])
ax4 = fig3.add_subplot(gssub2[1])
axs3.append([ax3, ax4])
ax1.set_title('$\Delta x={:g}$'.format(dx))
ax1.set_ylim(*ylims)
ax1.set_xlim(0, X)
ax1.set_ylabel('y(x)')
ax1.set_xticks([])
ax1.set_yticks(ax1.get_yticks()[1:])
ax2.grid(c='k', ls='--', alpha=0.3)
ax2.set_yscale('log')
ax2.set_xlim(0, X)
ax2.set_xlabel('x')
ax2.set_ylabel('log err')
ax3.set_title('$\Delta t={:g}$'.format(dt))
ax3.set_ylim(*ulims)
ax3.set_xlim(0, T)
ax3.set_ylabel('u(t)')
ax3.set_xticks([])
ax3.set_yticks(ax3.get_yticks()[1:])
ax4.grid(c='k', ls='--', alpha=0.3)
ax4.set_yscale('log')
ax4.set_xlim(0, T)
ax4.set_xlabel('t')
ax4.set_ylabel('log err')
# Calculate exact solution
x_e = np.linspace(0, X, 1001)
y_e = y_exact(x_e)
t_e = np.linspace(0, T, 1001)
u_e = u_exact(1 + beta, t_e)
# Plot with black line
ax1.plot(x_e, y_e, 'k-', label='exact')
ax3.plot(t_e, u_e, 'k-', label='exact')
ax1, ax2 = axs1[dx_idx]
ax3, ax4 = axs3[dx_idx]
ax = axs2.flat[dx_idx]
ax.set_title('{}, dx={:g}'.format(th_dict[th][0], dx))
# ax.set_ylim(0, 15)
ax.set_xlim(0, X)
ax.set_xlabel(r'x')
ax.set_ylabel(r'$y(x)$')
# Solve dỹ/d𝐱 = ỹ using Crank-Nicolson scheme
u = solver(I=1 + beta, a=-1, t=t, theta=th)
# Convert back to y
y = (u - beta) * I
# Calculate exact solution
y_e = y_exact(x)
u_e = u_exact(1 + beta, t)
# Plot
ax.plot(x, y, 'r--o', label='numerical')
ax1.plot(x, y, th_dict[th][2], label=th_dict[th][0])
ax1.legend()
ax3.plot(t, u, th_dict[th][2], label=th_dict[th][0])
ax3.legend()
# Plot exact solution
ax.plot(x, y_e, 'b-', label='exact')
ax.legend()
err = np.abs(y_e - y)
ax2.plot(x, err, th_dict[th][2])
err = np.abs(u_e - u)
ax4.plot(t, err, th_dict[th][2])
# Save figure
fig2.savefig(IMGDIR + f'{th_dict[th][1]}_growth.png',
bbox_inches='tight')
# Save figure
fig1.savefig(IMGDIR + f'comparison.png', bbox_inches='tight')
fig3.savefig(IMGDIR + f'comparison_scaled.png', bbox_inches='tight')
def generalisation() -> None:
"""
Generalisation: including a variable coefficient and a source term.
Notes
----------
We now start to look at the generalisations u' = -a(t) u and
u' = -a(t) u + b(t). Verification can no longer make use of an exact
solution of the numerical problem, so we make use of manufactured
solutions, for deriving an exact solution of the ODE problem, and then we
can compute empirical convergence rates for the method and see if these
coincide with the expected rates from theory.
We start by considered the case where a depends on time
u'(t) = -a(t) u(t) t ∈ (0, T] u(0) = I (2)
A Forward Euler scheme consists of evaluating (2) at t = tn, and
approximating with a forward difference [Dₜ⁺ u]ⁿ.
(uⁿ⁺¹ - uⁿ) / Δt = -a(tn) uⁿ
The Backward scheme becomes
(uⁿ - uⁿ⁻¹) / Δt = -a(tn) uⁿ
The Crank-Nicolson scheme builds on sampling the ODE at t{n+½}. We can
evaluate at t{n+½} and use an average u at times tn and t{n+1}
(uⁿ⁺¹ - uⁿ) / Δt = -a(t{n+½}) ½(uⁿ + uⁿ⁺¹)
Alternatively, we can use an average for the product a u
(uⁿ⁺¹ - uⁿ) / Δt = -½(a(tn) uⁿ + a(t{n+1}) uⁿ⁺¹)
The θ-tule unifies the three mentioned schemes. One version is to have a
evaluated at the weighted time point (1 - θ) tn + θ t{n+1}
(uⁿ⁺¹ - uⁿ) / Δt = -a((1 - θ) tn + θ t{n+1})((1 - θ) uⁿ + θ uⁿ⁺¹)
Another possibility is to apply a weighted average for the product a u
(uⁿ⁺¹ - uⁿ) / Δt = -(1 - θ) a(tn) uⁿ + θ a(t{n+1})uⁿ⁺¹
With the finite difference operator notation, the Forward Euler and
Backward Euler can be summarised as
[Dₜ⁺ u = -a u]ⁿ
[Dₜ⁻ u = -a u]ⁿ
The Crank-Nicolson and θ schemes depend on whether we evaluate a at the
sample point for the ODE or if we use an average.
[Dₜ u = -a ūᵗ]ⁿ⁺¹⸍²
[Dₜ u = -ā ūᵗ]ⁿ⁺¹⸍²
[D̄ₜ u = -a ūᵗʼᶿ]ⁿ⁺ᶿ
[D̄ₜ u = -ā ūᵗʼᶿ]ⁿ⁺ᶿ
A further extension of the model ODE is to include a source term b(t):
u'(t) = a(t) u(t) + b(t) t ∈ (0, T] u(0) = I
The time point where we sample the ODE determines where b(t) is evaluated.
For the Crank-Nicolson scheme and the θ-rule, we have a choice of whether
to evaluate a(t) and b(t) at the correct point, or use an average. The
chosen strategy becomes particularly clear if we write up the schemes in
operation notation:
[Dₜ⁺ u = -a u + b]ⁿ
[Dₜ⁻ u = -a u + b]ⁿ
[Dₜ u = -a ūᵗ + b]ⁿ⁺¹⸍²
[Dₜ u = -ā ū + b̄ᵗ]ⁿ⁺¹⸍²
[D̄ₜ u = -a ūᵗʼᶿ + b]ⁿ⁺ᶿ
[D̄ₜ u = -ā ū + b̄ᵗʼᶿ]ⁿ⁺ᶿ
Deriving the θ-rule formula when averaging over a and b, we get
(uⁿ⁺¹ - uⁿ) / Δt = θ(-aⁿ⁺¹ uⁿ⁺¹ + bⁿ⁺¹) + (1 - θ)(-aⁿ uⁿ + bⁿ)
Solving for uⁿ⁺¹,
uⁿ⁺¹ = ((1 - Δt(1 - θ)aⁿ) uⁿ + Δt(θ bⁿ⁺¹ + (1 - θ)bⁿ)) / (1 + Δt θ aⁿ⁺¹)
Here, we start by verifying a constant solution, where u = C. We choose any
a(t) and set b(t) = a(t) C and I = C.
"""
# Definitions
u_const = 2.15
I = u_const
Nt = 4
dt = 4
theta = 0.4
a = lambda t: 2.5 * (1 + t**3)
b = lambda t: a(t) * u_const
# Solve
u, t = solver(I=I, a=a, b=b, T=Nt * dt, dt=dt, theta=theta)
# Calculate exact solution
u_exact = lambda t: u_const
u_e = u_exact(t)
# Assert that max deviation is below tolerance
tol = 1E-14
diff = np.max(np.abs(u_e - u))
if diff > tol:
raise AssertionError(f'Tolerance not reached, diff = {diff}')
def convergence() -> None:
"""
Computing convergence rates.
Notes
----------
We expect that the error E in the numerical solution is reduced if the mesh
size Δt is decreased. More specifically, many numerical methods obey a
power-law relation between E and Δt
E = C Δtʳ
where C and r are (usually unknown) constant independent of Δt. The
parameter r is known as the convergence rate. For example, if the
convergence rate is 2, halving Δt reduces the error by a factor of 4.
Diminishing Δt then has a greater impact on the error compared with methods
that have r = 1. For a given value of r, we refer to the method as of r-th
order.
There are two alternative ways of estimating C and r based on a set of m
simulations with corresponding pairs (Δtᵢ, Eᵢ), i = 0,...,m-1, and
Δtᵢ < Δtᵢ₋₁.
1) Take the logarithm of E = C Δtʳ and fit a straight line to the data
points (Δtᵢ, Eᵢ).
2) Consider two consecutive experiments (Δtᵢ, Eᵢ) and (Δtᵢ₋₁, Eᵢ₋₁).
Dividing the equation Eᵢ₋₁ = C Δtʳᵢ₋₁ by Eᵢ = C Δtʳᵢ and solving for r
rᵢ₋₁ = ln(Eᵢ₋₁ / Eᵢ) / ln(Δtᵢ₋₁ / Δtᵢ)
for i = 1,...,m-1. Note that we have introduced a subindex i - 1 on r
because r estimated from a pair of experiments must be expected to change
with i.
The disadvantage of method 1 is that it may not be valid for the coarsest
meshes (largest Δt values). Fitting a line to all the data points is then
misleading. Method 2 computes the convergence rates for pairs of
experiments and allows us to see if the sequence rᵢ converges to some
values as i → m - 2. The final rₘ₋₂ can then be taken as the convergence
rate.
The strong practical application of computing convergence rates is for
verification: wrong convergence rates point to errors in the code, and
correct convergence rates provide strong support for a correct
implementation. Bugs in the code can easily destroy the expected
convergence rate.
The example here used the manufactured solution uₑ(t) = sin(t) exp{-2t} and
a(t) = t². This implies we must fit b as b(t) = u'(t) + a(t) u(t). We first
compute with SymPy expressions and then convert the exact solution, a, and
b to Python functions that we can use in subsequent numerical computing.
"""
# Created manufactured solution with SymPy
t = sym.Symbol('t')
u_e_sym = sym.sin(t) * sym.exp(-2 * t)
a_sym = t**2
b_sym = sym.diff(u_e_sym, t) + a_sym * u_e_sym
# Turn SymPy expressions into Python functions
u_exact = sym.lambdify([t], u_e_sym, modules='numpy')
a = sym.lambdify([t], a_sym, modules='numpy')
b = sym.lambdify([t], b_sym, modules='numpy')
T = 6
I = u_exact(0)
dt_values = [1 * 2**(-i) for i in range(6)]
print(STR_FMT.format('dt_values', f'{dt_values}'))
th_dict = {
0: ('Forward Euler', 'fe', 'r-s'),
1: ('Backward Euler', 'be', 'g-v'),
0.5: ('Crank-Nicolson', 'cn', 'b-^')
}
for theta in th_dict:
print(f'{th_dict[theta][0]}')
E_values = []
for dt in dt_values:
# Solve for mesh function
u, t = solver(I=I, a=a, b=b, T=T, dt=dt, theta=theta)
# Compute error
u_e = u_exact(t)
e = u_e - u
E = np.sqrt(dt * np.sum(e**2))
E_values.append(E)
# Compute convergence rates
r = compute_rates(dt_values, E_values)
print(STR_FMT.format('r', f'{r}'))
# Test final entry with expected convergence rate
expected_rate = 2 if theta == 0.5 else 1
tol = 0.1
diff = np.abs(expected_rate - r[-1])
if diff > tol:
raise AssertionError(f'Tolerance not reached, diff = {diff}')
def verification() -> None:
"""
Verification via manufactured solutions.
Notes
----------
Following the idea above, we choose any formula as the exact solution,
insert the formula in the ODE problem and fit the data a(t), b(t), and I to
make the chosen formula fulfill the equation. This powerful technique for
generating exact solutions is very useful for verification purposes and
known as the method of manufactured solutions, often abbreviated MMS.
One common choice of solution is a linear function in the independent
variable(s). The rationale behind such a simple variation is that almost
any relevant numerical solution method for differential equation problems
is able to reproduce a linear function exactly to machine precision. The
linear solution also makes some stronger demands to the numerical solution
and the implementation than the constant solution one.
We choose a linear solution u(t) = c t + d. From the initial condition, it
follows that d = I. Inserting this u in the left-hand side of (1), we get
c = -a(t) u + b(t)
Any function u = c t + I is then a correct solution if we choose
b(t) = c + a(t)(c t + I)
Therefore, we must check that uⁿ = c a(tn)(c tn + I) fulfills the discrete
equations. For these equations, it is convenient to compute the action of
the difference operator on a linear function t:
[Dₜ⁺ t]ⁿ = (tⁿ⁺¹ - tⁿ) / Δt = 1
[Dₜ⁻ t]ⁿ = (tⁿ - tⁿ⁻¹) / Δt = 1
[Dₜ t]ⁿ = (tⁿ⁺¹⸍² - tⁿ⁻¹⸍²) / Δt
= ((n + ½)Δt - (n - ½)Δt) / Δt = 1
Clearly, all three difference approximations to the derivative are exact
for u(t) = t or its mesh function counterpart uⁿ = tn. The difference
equation in the Forward Euler scheme
[Dₜ⁺ u = -a u + b]ⁿ
with aⁿ = a(tn), bⁿ = c + a(tn)(c tn + I), and uⁿ = ctn + I then results in
c = -a(tn)(c tn + I) + c + a(tn)(c tn + I) = c
which is always fulfilled. Similar calculations can be done for the Forward
Euler and Crank-Nicolson schemes. Therefore, we expect that uⁿ - uₑ(tn) = 0
mathematically and |uⁿ - uₑ(tn)| less than a small number about the machine
precision.
"""
# Definitions
I = 0.1
T = 4
dt = 0.1
Nt = int(T / dt)
theta = 0.4
c = -0.5
u_exact = lambda t: c * t + I
a = lambda t: t**0.5
b = lambda t: c + a(t) * u_exact(t)
# Solve
u, t = solver(I=I, a=a, b=b, T=Nt * dt, dt=dt, theta=theta)
# Calculate exact solution
u_e = u_exact(t)
# Assert that max deviation is below tolerance
tol = 1E-14
diff = np.max(np.abs(u_e - u))
if diff > tol:
raise AssertionError(f'Tolerance not reached, diff = {diff}')
