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Fix broken Devito code examples in book introduction
The wave equation example in index.qmd used `Eq(u.dt2, c**2 * u.dx2)` which generates invalid C code (assigns to an expression). Fixed to use `solve()` to derive the explicit update stencil. The diffusion example in what_is_devito.qmd used `alpha` and `dt` without defining them. Fixed by adding parameter definitions and using the `solve()` pattern for consistency. Added test_book_code_verification.py to verify code examples are correct.
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chapters/devito_intro/what_is_devito.qmd

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Original file line numberDiff line numberDiff line change
@@ -49,29 +49,41 @@ This approach has several limitations:
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With Devito, the same problem becomes:
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```python
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from devito import Grid, TimeFunction, Eq, Operator
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from devito import Grid, TimeFunction, Eq, Operator, solve, Constant
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# Problem parameters
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Nx = 100
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L = 1.0
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alpha = 1.0 # diffusion coefficient
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F = 0.5 # Fourier number (for stability, F <= 0.5)
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# Compute dt from stability condition: F = alpha * dt / dx^2
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dx = L / Nx
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dt = F * dx**2 / alpha
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# Create computational grid
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grid = Grid(shape=(101,), extent=(1.0,))
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grid = Grid(shape=(Nx + 1,), extent=(L,))
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# Define the unknown field
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u = TimeFunction(name='u', grid=grid, time_order=1, space_order=2)
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# Set initial condition
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u.data[0, 50] = 1.0
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u.data[0, Nx // 2] = 1.0
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# Define the PDE update equation
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eq = Eq(u.forward, u + alpha * dt * u.dx2)
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# Define the PDE symbolically and solve for u.forward
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a = Constant(name='a')
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pde = u.dt - a * u.dx2
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update = Eq(u.forward, solve(pde, u.forward))
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# Create and run the operator
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op = Operator([eq])
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op(time=1000, dt=dt)
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op = Operator([update])
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op(time=1000, dt=dt, a=alpha)
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```
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This approach offers significant advantages:
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1. **Mathematical clarity**: The equation `u.forward = u + alpha * dt * u.dx2`
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directly mirrors the mathematical formulation
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1. **Mathematical clarity**: The PDE `u.dt - a * u.dx2 = 0` is written symbolically,
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and Devito derives the update formula automatically using `solve()`
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2. **Automatic optimization**: Devito generates C code with loop tiling,
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SIMD vectorization, and OpenMP parallelization
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3. **Dimension-agnostic**: The same code structure works for 1D, 2D, or 3D

index.qmd

Lines changed: 9 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -29,18 +29,21 @@ This work is licensed under [CC BY 4.0](https://creativecommons.org/licenses/by/
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Devito allows you to write PDEs symbolically and automatically generates optimized finite difference code:
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```python
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from devito import Grid, TimeFunction, Eq, Operator
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from devito import Grid, TimeFunction, Eq, Operator, solve, Constant
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grid = Grid(shape=(101,), extent=(1.0,))
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u = TimeFunction(name='u', grid=grid, time_order=2, space_order=2)
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c = 1.0 # wave speed
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c = Constant(name='c') # wave speed
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# Write the wave equation symbolically
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eq = Eq(u.dt2, c**2 * u.dx2)
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# Write the wave equation symbolically: u_tt = c^2 * u_xx
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pde = Eq(u.dt2, c**2 * u.dx2)
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# Solve for u at the next time step
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update = Eq(u.forward, solve(pde, u.forward))
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# Devito generates optimized C code
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op = Operator([eq])
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op.apply(time_M=100, dt=0.001)
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op = Operator([update])
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op.apply(time_M=100, dt=0.001, c=1.0)
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```
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## Book Structure {.unnumbered}
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@@ -0,0 +1,318 @@
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"""Test verification for code examples in the book.
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This module verifies that code examples shown in the book:
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1. Actually run without errors
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2. Produce correct results
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3. Follow Devito best practices
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Per CONTRIBUTING.md: All results must be reproducible with fixed random seeds,
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version-pinned dependencies, and automated tests validating examples.
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"""
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import numpy as np
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import pytest
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try:
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from devito import Constant, Eq, Grid, Operator, TimeFunction, solve
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DEVITO_AVAILABLE = True
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except ImportError:
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DEVITO_AVAILABLE = False
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@pytest.mark.devito
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class TestIndexQmdWaveEquation:
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"""Test the wave equation code shown in index.qmd (the landing page)."""
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def test_symbolic_pde_form_needs_solve(self):
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"""Test that Eq(u.dt2, c**2 * u.dx2) needs solve() for correct behavior.
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The code in index.qmd line 39 uses:
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eq = Eq(u.dt2, c**2 * u.dx2)
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This is the symbolic PDE form. Devito needs an explicit update
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equation, so we should use solve() to derive the stencil:
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pde = Eq(u.dt2, c**2 * u.dx2)
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update = Eq(u.forward, solve(pde, u.forward))
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"""
37+
if not DEVITO_AVAILABLE:
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pytest.skip("Devito not available")
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grid = Grid(shape=(101,), extent=(1.0,))
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u = TimeFunction(name='u', grid=grid, time_order=2, space_order=2)
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c = 1.0
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# Set initial condition (Gaussian pulse)
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x_coords = np.linspace(0, 1.0, 101)
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u.data[0, :] = np.exp(-((x_coords - 0.5) ** 2) / (2 * 0.1**2))
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u.data[1, :] = u.data[0, :]
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# Method 1: Original code from index.qmd (symbolic PDE form)
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# This may not work correctly as it's not an explicit update
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eq_symbolic = Eq(u.dt2, c**2 * u.dx2)
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# Method 2: Correct approach using solve()
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dt_const = Constant(name='dt')
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pde = Eq(u.dt2, c**2 * u.dx2)
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update = Eq(u.forward, solve(pde, u.forward))
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# Both should create operators, but behavior differs
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try:
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op_symbolic = Operator([eq_symbolic])
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# This test documents that the symbolic form may not work as expected
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# The operator might be created but not produce a useful update
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except Exception as e:
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pytest.fail(f"Symbolic PDE form failed to create operator: {e}")
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op_correct = Operator([update])
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# Run the correct version and verify it works
69+
u.data[0, :] = np.exp(-((x_coords - 0.5) ** 2) / (2 * 0.1**2))
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u.data[1, :] = u.data[0, :]
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op_correct.apply(time_M=10, dt=0.001)
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# Solution should still be bounded (not exploding)
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assert np.max(np.abs(u.data[0, :])) < 10.0, "Solution exploded"
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def test_correct_wave_equation_pattern(self):
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"""Test the recommended pattern for wave equation in Devito.
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This is the pattern that should be used in index.qmd:
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pde = Eq(u.dt2, c**2 * u.dx2)
81+
update = Eq(u.forward, solve(pde, u.forward))
82+
op = Operator([update])
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"""
84+
if not DEVITO_AVAILABLE:
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pytest.skip("Devito not available")
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# Problem setup
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L = 1.0
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Nx = 100
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c = 1.0
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C = 0.5 # Courant number
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dx = L / Nx
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dt = C * dx / c
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Nt = 100
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97+
grid = Grid(shape=(Nx + 1,), extent=(L,))
98+
u = TimeFunction(name='u', grid=grid, time_order=2, space_order=2)
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# Initial condition: Gaussian pulse
101+
x_coords = np.linspace(0, L, Nx + 1)
102+
u.data[0, :] = np.exp(-((x_coords - 0.5 * L) ** 2) / (2 * 0.1**2))
103+
u.data[1, :] = u.data[0, :]
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105+
# Correct pattern: Define PDE symbolically, then solve for u.forward
106+
c_const = Constant(name='c')
107+
pde = Eq(u.dt2, c_const**2 * u.dx2)
108+
update = Eq(u.forward, solve(pde, u.forward))
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110+
op = Operator([update])
111+
op.apply(time_M=Nt, dt=dt, c=c)
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113+
# Verify solution is reasonable
114+
max_val = np.max(np.abs(u.data[0, :]))
115+
assert max_val < 2.0, f"Solution amplitude {max_val} too large"
116+
assert max_val > 0.01, f"Solution amplitude {max_val} too small"
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118+
119+
@pytest.mark.devito
120+
class TestWhatIsDevitoQmdDiffusion:
121+
"""Test the diffusion code shown in what_is_devito.qmd."""
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123+
def test_diffusion_missing_variables(self):
124+
"""Test that the diffusion code in what_is_devito.qmd has issues.
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126+
The code at lines 51-69 uses `alpha` and `dt` without defining them:
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eq = Eq(u.forward, u + alpha * dt * u.dx2)
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129+
This test verifies the corrected version works.
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"""
131+
if not DEVITO_AVAILABLE:
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pytest.skip("Devito not available")
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# Create computational grid
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grid = Grid(shape=(101,), extent=(1.0,))
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137+
# Define the unknown field
138+
u = TimeFunction(name='u', grid=grid, time_order=1, space_order=2)
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# Set initial condition
141+
u.data[0, 50] = 1.0
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# CORRECTED: Define alpha and dt properly
144+
alpha = 1.0 # diffusion coefficient
145+
Nx = 100
146+
L = 1.0
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dx = L / Nx
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F = 0.5 # Fourier number for stability
149+
dt = F * dx**2 / alpha # Time step from stability
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# Define the PDE update equation (CORRECTED)
152+
eq = Eq(u.forward, u + alpha * dt * u.dx2)
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154+
# Create and run the operator
155+
op = Operator([eq])
156+
op(time=100, dt=dt)
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# Verify solution is bounded and diffusing
159+
max_val = np.max(u.data[0, :])
160+
assert max_val < 1.0, "Diffusion should reduce peak"
161+
assert max_val > 0.0, "Solution should not go to zero immediately"
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163+
def test_diffusion_with_solve_pattern(self):
164+
"""Test the recommended pattern using solve() for diffusion.
165+
166+
Better approach:
167+
a_const = Constant(name='a')
168+
pde = u.dt - a_const * u.dx2
169+
stencil = Eq(u.forward, solve(pde, u.forward))
170+
"""
171+
if not DEVITO_AVAILABLE:
172+
pytest.skip("Devito not available")
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174+
# Parameters
175+
L = 1.0
176+
Nx = 100
177+
alpha = 1.0
178+
F = 0.5
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180+
dx = L / Nx
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dt = F * dx**2 / alpha
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183+
grid = Grid(shape=(Nx + 1,), extent=(L,))
184+
u = TimeFunction(name='u', grid=grid, time_order=1, space_order=2)
185+
186+
# Initial condition: sinusoidal
187+
x_coords = np.linspace(0, L, Nx + 1)
188+
u.data[0, :] = np.sin(np.pi * x_coords / L)
189+
190+
# Use solve() pattern
191+
a_const = Constant(name='a')
192+
pde = u.dt - a_const * u.dx2
193+
stencil = Eq(u.forward, solve(pde, u.forward))
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195+
# Boundary conditions
196+
bc_left = Eq(u[grid.stepping_dim + 1, 0], 0)
197+
bc_right = Eq(u[grid.stepping_dim + 1, Nx], 0)
198+
199+
op = Operator([stencil, bc_left, bc_right])
200+
201+
# Run for a short time
202+
T = 0.1
203+
Nt = int(T / dt)
204+
for _ in range(Nt):
205+
op.apply(time_m=0, time_M=0, dt=dt, a=alpha)
206+
u.data[0, :] = u.data[1, :]
207+
208+
# Verify against exact solution
209+
exact = np.exp(-alpha * (np.pi / L) ** 2 * T) * np.sin(np.pi * x_coords / L)
210+
error = np.max(np.abs(u.data[0, :] - exact))
211+
212+
assert error < 0.01, f"Error {error:.4f} too large"
213+
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@pytest.mark.devito
216+
class TestFirstPdeQmdWaveEquation:
217+
"""Test the wave equation code shown in first_pde.qmd."""
218+
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def test_first_pde_manual_stencil_correct(self):
220+
"""Verify the manual stencil form in first_pde.qmd works.
221+
222+
The code uses the manually derived stencil:
223+
eq = Eq(u.forward, 2*u - u.backward + (c*dt)**2 * u.dx2)
224+
225+
This is correct because it explicitly defines u.forward.
226+
"""
227+
if not DEVITO_AVAILABLE:
228+
pytest.skip("Devito not available")
229+
230+
# Problem parameters
231+
L = 1.0
232+
c = 1.0
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T = 1.0
234+
Nx = 100
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C = 0.5 # Courant number
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237+
# Derived parameters
238+
dx = L / Nx
239+
dt = C * dx / c
240+
Nt = int(T / dt)
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# Create the computational grid
243+
grid = Grid(shape=(Nx + 1,), extent=(L,))
244+
245+
# Create a time-varying field
246+
u = TimeFunction(name='u', grid=grid, time_order=2, space_order=2)
247+
248+
# Set initial condition: a Gaussian pulse
249+
x_coord = 0.5 * L
250+
sigma = 0.1
251+
x_coords = np.linspace(0, L, Nx + 1)
252+
u.data[0, :] = np.exp(-((x_coords - x_coord) ** 2) / (2 * sigma**2))
253+
u.data[1, :] = u.data[0, :] # Zero initial velocity
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# Define the update equation (manual stencil form)
256+
eq = Eq(u.forward, 2 * u - u.backward + (c * dt) ** 2 * u.dx2)
257+
258+
# Create the operator
259+
op = Operator([eq])
260+
261+
# Run the simulation
262+
op(time=Nt, dt=dt)
263+
264+
# Verify solution is reasonable
265+
max_val = np.max(np.abs(u.data[0, :]))
266+
assert max_val < 2.0, f"Solution amplitude {max_val} too large"
267+
268+
269+
@pytest.mark.devito
270+
class TestDiffu1DDevitoQmd:
271+
"""Test the diffusion code shown in diffu1D_devito.qmd."""
272+
273+
def test_diffu1d_devito_dt_defined(self):
274+
"""Verify dt is properly defined before use in the diffusion example.
275+
276+
The code should define dt from the Fourier number:
277+
dt = F * dx**2 / a
278+
"""
279+
if not DEVITO_AVAILABLE:
280+
pytest.skip("Devito not available")
281+
282+
# Domain and discretization
283+
L = 1.0
284+
Nx = 100
285+
a = 1.0 # Diffusion coefficient
286+
F = 0.5 # Fourier number
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288+
dx = L / Nx
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dt = F * dx**2 / a # Time step from stability condition
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# Verify dt is defined and reasonable
292+
assert dt > 0, "dt must be positive"
293+
assert dt < 1.0, "dt seems too large"
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295+
# Create Devito grid
296+
grid = Grid(shape=(Nx + 1,), extent=(L,))
297+
u = TimeFunction(name='u', grid=grid, time_order=1, space_order=2)
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299+
# Set initial condition
300+
x_coords = np.linspace(0, L, Nx + 1)
301+
u.data[0, :] = np.sin(np.pi * x_coords)
302+
303+
# Using solve() pattern
304+
a_const = Constant(name='a_const')
305+
pde = u.dt - a_const * u.dx2
306+
stencil = Eq(u.forward, solve(pde, u.forward))
307+
308+
op = Operator([stencil])
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310+
# Run one time step to verify operator works
311+
op.apply(time_m=0, time_M=0, dt=dt, a_const=a)
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313+
# Solution should be bounded
314+
assert np.all(np.isfinite(u.data[1, :])), "Solution contains NaN/Inf"
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if __name__ == "__main__":
318+
pytest.main([__file__, "-v"])

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