Modified vib python file for consistency with notebook

Author: EdCaunt

src/vib/vib.py

@@ -1,4 +1,6 @@
 import numpy as np
+import sympy as sp
+from devito import Dimension, Constant, TimeFunction, Eq, solve, Operator
 #import matplotlib.pyplot as plt
 import scitools.std as plt
 
@@ -11,27 +13,33 @@ def solver(I, V, m, b, s, F, dt, T, damping='linear'):
     'quadratic', f(u')=b*u'*abs(u').
     F(t) and s(u) are Python functions.
     """
-    dt = float(dt); b = float(b); m = float(m) # avoid integer div.
+    dt = float(dt)
+    b = float(b)
+    m = float(m)
     Nt = int(round(T/dt))
-    u = np.zeros(Nt+1)
-    t = np.linspace(0, Nt*dt, Nt+1)
+    t = Dimension('t', spacing=Constant('h_t'))
+
+    u = TimeFunction(name='u', dimensions=(t,),
+                     shape=(Nt+1,), space_order=2)
+
+    u.data[0] = I
 
-    u[0] = I
     if damping == 'linear':
-        u[1] = u[0] + dt*V + dt**2/(2*m)*(-b*V - s(u[0]) + F(t[0]))
+        # dtc for central difference (default for time is forward, 1st order)
+        eqn = m*u.dt2 + b*u.dtc + s(u) - F(u)
+        stencil = Eq(u.forward, solve(eqn, u.forward))
     elif damping == 'quadratic':
-        u[1] = u[0] + dt*V + \
-               dt**2/(2*m)*(-b*V*abs(V) - s(u[0]) + F(t[0]))
+        # fd_order set as backward derivative used is 1st order
+        eqn = m*u.dt2 + b*u.dt*sp.Abs(u.dtl(fd_order=1)) + s(u) - F(u)
+        stencil = Eq(u.forward, solve(eqn, u.forward))
+    # First timestep needs to have the backward timestep substituted
+    stencil_init = stencil.subs(u.backward, u.forward-2*t.spacing*V)
+    op_init = Operator(stencil_init, name='first_timestep')
+    op = Operator(stencil, name='main_loop')
+    op_init.apply(h_t=dt, t_M=1)
+    op.apply(h_t=dt, t_m=1, t_M=Nt-1)
 
-    for n in range(1, Nt):
-        if damping == 'linear':
-            u[n+1] = (2*m*u[n] + (b*dt/2 - m)*u[n-1] +
-                      dt**2*(F(t[n]) - s(u[n])))/(m + b*dt/2)
-        elif damping == 'quadratic':
-            u[n+1] = (2*m*u[n] - m*u[n-1] + b*u[n]*abs(u[n] - u[n-1])
-                      + dt**2*(F(t[n]) - s(u[n])))/\
-                      (m + b*abs(u[n] - u[n-1]))
-    return u, t
+    return u.data, np.linspace(0, Nt*dt, Nt+1)
 
 def visualize(u, t, title='', filename='tmp'):
     plt.plot(t, u, 'b-')
@@ -46,8 +54,6 @@ def visualize(u, t, title='', filename='tmp'):
     plt.savefig(filename + '.pdf')
     plt.show()
 
-import sympy as sym
-
 def test_constant():
     """Verify a constant solution."""
     u_exact = lambda t: I
@@ -68,24 +74,24 @@ def test_constant():
 
 def lhs_eq(t, m, b, s, u, damping='linear'):
     """Return lhs of differential equation as sympy expression."""
-    v = sym.diff(u, t)
+    v = sp.diff(u, t)
     if damping == 'linear':
-        return m*sym.diff(u, t, t) + b*v + s(u)
+        return m*sp.diff(u, t, t) + b*v + s(u)
     else:
-        return m*sym.diff(u, t, t) + b*v*sym.Abs(v) + s(u)
+        return m*sp.diff(u, t, t) + b*v*sp.Abs(v) + s(u)
 
 def test_quadratic():
     """Verify a quadratic solution."""
     I = 1.2; V = 3; m = 2; b = 0.9
     s = lambda u: 4*u
-    t = sym.Symbol('t')
+    t = sp.Symbol('t')
     dt = 0.2
     T = 2
 
     q = 2  # arbitrary constant
     u_exact = I + V*t + q*t**2
-    F = sym.lambdify(t, lhs_eq(t, m, b, s, u_exact, 'linear'))
-    u_exact = sym.lambdify(t, u_exact, modules='numpy')
+    F = sp.lambdify(t, lhs_eq(t, m, b, s, u_exact, 'linear'))
+    u_exact = sp.lambdify(t, u_exact, modules='numpy')
     u1, t1 = solver(I, V, m, b, s, F, dt, T, 'linear')
     diff = np.abs(u_exact(t1) - u1).max()
     tol = 1E-13
@@ -94,8 +100,8 @@ def test_quadratic():
     # In the quadratic damping case, u_exact must be linear
     # in order exactly recover this solution
     u_exact = I + V*t
-    F = sym.lambdify(t, lhs_eq(t, m, b, s, u_exact, 'quadratic'))
-    u_exact = sym.lambdify(t, u_exact, modules='numpy')
+    F = sp.lambdify(t, lhs_eq(t, m, b, s, u_exact, 'quadratic'))
+    u_exact = sp.lambdify(t, u_exact, modules='numpy')
     u2, t2 = solver(I, V, m, b, s, F, dt, T, 'quadratic')
     diff = np.abs(u_exact(t2) - u2).max()
     assert diff < tol
@@ -127,11 +133,11 @@ def test_mms():
     """Use method of manufactured solutions."""
     m = 4.; b = 1
     w = 1.5
-    t = sym.Symbol('t')
-    u_exact = 3*sym.exp(-0.2*t)*sym.cos(1.2*t)
+    t = sp.Symbol('t')
+    u_exact = 3*sp.exp(-0.2*t)*sp.cos(1.2*t)
     I = u_exact.subs(t, 0).evalf()
-    V = sym.diff(u_exact, t).subs(t, 0).evalf()
-    u_exact_py = sym.lambdify(t, u_exact, modules='numpy')
+    V = sp.diff(u_exact, t).subs(t, 0).evalf()
+    u_exact_py = sp.lambdify(t, u_exact, modules='numpy')
     s = lambda u: u**3
     dt = 0.2
     T = 6
@@ -140,14 +146,14 @@ def test_mms():
     # Run grid refinements and compute exact error
     for i in range(5):
         F_formula = lhs_eq(t, m, b, s, u_exact, 'linear')
-        F = sym.lambdify(t, F_formula)
+        F = sp.lambdify(t, F_formula)
         u1, t1 = solver(I, V, m, b, s, F, dt, T, 'linear')
         error = np.sqrt(np.sum((u_exact_py(t1) - u1)**2)*dt)
         errors_linear.append((dt, error))
 
         F_formula = lhs_eq(t, m, b, s, u_exact, 'quadratic')
         #print sym.latex(F_formula, mode='plain')
-        F = sym.lambdify(t, F_formula)
+        F = sp.lambdify(t, F_formula)
         u2, t2 = solver(I, V, m, b, s, F, dt, T, 'quadratic')
         error = np.sqrt(np.sum((u_exact_py(t2) - u2)**2)*dt)
         errors_quadratic.append((dt, error))