SymPy dsolve with a parameter gives a wrong answer when the parameter is zero
My Code
I have a program that calculates the solutions to a 2nd-order differential equation, like in the code snippet below:
import sympy as sp
print('sympy version:', sp.__version__)
t = sp.symbols('t', real=True, nonnegative=True)
n = sp.symbols('n', integer=True, nonnegative=True)
f = sp.symbols('f', cls=sp.Function)
diff_eq = sp.Eq(f(t).diff(t, 2) + n**2*f(t), 0)
print('general solution:', sp.dsolve(diff_eq, f(t)))
print('solution at n=0 (pre-subs):', sp.dsolve(diff_eq.subs(n, 0), f(t)))
print('solution at n=0 (post-subs):', sp.dsolve(diff_eq, f(t)).subs(n, 0))
Results:
sympy version: 1.3
general solution: Eq(f(t), C1*sin(n*t) + C2*cos(n*t))
solution at n=0 (pre-subs): Eq(f(t), C1 + C2*t)
solution at n=0 (post-subs): Eq(f(t), C2)
My Problem
The solution form for a general n does not seem to accurately describe the specific solution form for n=0. Specifically, using dsolve first and subs(n, 0) second produces different results than using subs(n, 0) first and dsolve second, even though the two should be logically equivalent.
Can somebody explain the reason for the discrepancy in my results? Am I doing something wrong, or is this a bug?
python sympy
asked Nov 14 '18 at 7:28
CrepeGoatCrepeGoat
893914
add a comment |
My Code
I have a program that calculates the solutions to a 2nd-order differential equation, like in the code snippet below:
import sympy as sp
print('sympy version:', sp.__version__)
t = sp.symbols('t', real=True, nonnegative=True)
n = sp.symbols('n', integer=True, nonnegative=True)
f = sp.symbols('f', cls=sp.Function)
diff_eq = sp.Eq(f(t).diff(t, 2) + n**2*f(t), 0)
print('general solution:', sp.dsolve(diff_eq, f(t)))
print('solution at n=0 (pre-subs):', sp.dsolve(diff_eq.subs(n, 0), f(t)))
print('solution at n=0 (post-subs):', sp.dsolve(diff_eq, f(t)).subs(n, 0))
Results:
sympy version: 1.3
general solution: Eq(f(t), C1*sin(n*t) + C2*cos(n*t))
solution at n=0 (pre-subs): Eq(f(t), C1 + C2*t)
solution at n=0 (post-subs): Eq(f(t), C2)
My Problem
The solution form for a general n does not seem to accurately describe the specific solution form for n=0. Specifically, using dsolve first and subs(n, 0) second produces different results than using subs(n, 0) first and dsolve second, even though the two should be logically equivalent.
Can somebody explain the reason for the discrepancy in my results? Am I doing something wrong, or is this a bug?
python sympy
asked Nov 14 '18 at 7:28
CrepeGoatCrepeGoat
893914
add a comment |
My Code
I have a program that calculates the solutions to a 2nd-order differential equation, like in the code snippet below:
import sympy as sp
print('sympy version:', sp.__version__)
t = sp.symbols('t', real=True, nonnegative=True)
n = sp.symbols('n', integer=True, nonnegative=True)
f = sp.symbols('f', cls=sp.Function)
diff_eq = sp.Eq(f(t).diff(t, 2) + n**2*f(t), 0)
print('general solution:', sp.dsolve(diff_eq, f(t)))
print('solution at n=0 (pre-subs):', sp.dsolve(diff_eq.subs(n, 0), f(t)))
print('solution at n=0 (post-subs):', sp.dsolve(diff_eq, f(t)).subs(n, 0))
Results:
sympy version: 1.3
general solution: Eq(f(t), C1*sin(n*t) + C2*cos(n*t))
solution at n=0 (pre-subs): Eq(f(t), C1 + C2*t)
solution at n=0 (post-subs): Eq(f(t), C2)
My Problem
The solution form for a general n does not seem to accurately describe the specific solution form for n=0. Specifically, using dsolve first and subs(n, 0) second produces different results than using subs(n, 0) first and dsolve second, even though the two should be logically equivalent.
Can somebody explain the reason for the discrepancy in my results? Am I doing something wrong, or is this a bug?
python sympy
asked Nov 14 '18 at 7:28
CrepeGoatCrepeGoat
893914
My Code
I have a program that calculates the solutions to a 2nd-order differential equation, like in the code snippet below:
import sympy as sp
print('sympy version:', sp.__version__)
t = sp.symbols('t', real=True, nonnegative=True)
n = sp.symbols('n', integer=True, nonnegative=True)
f = sp.symbols('f', cls=sp.Function)
diff_eq = sp.Eq(f(t).diff(t, 2) + n**2*f(t), 0)
print('general solution:', sp.dsolve(diff_eq, f(t)))
print('solution at n=0 (pre-subs):', sp.dsolve(diff_eq.subs(n, 0), f(t)))
print('solution at n=0 (post-subs):', sp.dsolve(diff_eq, f(t)).subs(n, 0))
Results:
sympy version: 1.3
general solution: Eq(f(t), C1*sin(n*t) + C2*cos(n*t))
solution at n=0 (pre-subs): Eq(f(t), C1 + C2*t)
solution at n=0 (post-subs): Eq(f(t), C2)
My Problem
The solution form for a general n does not seem to accurately describe the specific solution form for n=0. Specifically, using dsolve first and subs(n, 0) second produces different results than using subs(n, 0) first and dsolve second, even though the two should be logically equivalent.
Can somebody explain the reason for the discrepancy in my results? Am I doing something wrong, or is this a bug?
python sympy
python sympy
asked Nov 14 '18 at 7:28
CrepeGoatCrepeGoat
893914
asked Nov 14 '18 at 7:28
CrepeGoatCrepeGoat
893914
edited Nov 15 '18 at 1:18
user6655984
asked Nov 14 '18 at 7:28
CrepeGoatCrepeGoat
893914
asked Nov 14 '18 at 7:28
CrepeGoatCrepeGoat
893914
asked Nov 14 '18 at 7:28
CrepeGoatCrepeGoat
893914
893914
add a comment |
add a comment |
1 Answer
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It can be considered a bug in dsolve logic: it finds two eigenvalues n and -n and treats them as different without considering the special case n=0 when they are equal. Ideally it would output a Piecewise like the following code does.
sol_nonzero = sp.dsolve(diff_eq, f(t)).rhs
sol_zero = sp.dsolve(diff_eq.subs(n, 0), f(t)).rhs
sol_complete = sp.Piecewise((sol_nonzero, sp.Ne(n, 0)), (sol_zero, True))
print('general solution:', sol_complete)
print('solution at n=0:', sol_complete.subs(n, 0))
This prints
general solution: Piecewise((C1*sin(n*t) + C2*cos(n*t), Ne(n, 0)), (C1 + C2*t, True))
solution at n=0: C1 + C2*t
A more familiar mathematical form is provided by sp.pprint(sol_complete).
⎧C₁⋅sin(n⋅t) + C₂⋅cos(n⋅t) for n ≠ 0
⎨
⎩ C₁ + C₂⋅t otherwise
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It can be considered a bug in dsolve logic: it finds two eigenvalues n and -n and treats them as different without considering the special case n=0 when they are equal. Ideally it would output a Piecewise like the following code does.
sol_nonzero = sp.dsolve(diff_eq, f(t)).rhs
sol_zero = sp.dsolve(diff_eq.subs(n, 0), f(t)).rhs
sol_complete = sp.Piecewise((sol_nonzero, sp.Ne(n, 0)), (sol_zero, True))
print('general solution:', sol_complete)
print('solution at n=0:', sol_complete.subs(n, 0))
This prints
general solution: Piecewise((C1*sin(n*t) + C2*cos(n*t), Ne(n, 0)), (C1 + C2*t, True))
solution at n=0: C1 + C2*t
A more familiar mathematical form is provided by sp.pprint(sol_complete).
⎧C₁⋅sin(n⋅t) + C₂⋅cos(n⋅t) for n ≠ 0
⎨
⎩ C₁ + C₂⋅t otherwise
add a comment |
It can be considered a bug in dsolve logic: it finds two eigenvalues n and -n and treats them as different without considering the special case n=0 when they are equal. Ideally it would output a Piecewise like the following code does.
sol_nonzero = sp.dsolve(diff_eq, f(t)).rhs
sol_zero = sp.dsolve(diff_eq.subs(n, 0), f(t)).rhs
sol_complete = sp.Piecewise((sol_nonzero, sp.Ne(n, 0)), (sol_zero, True))
print('general solution:', sol_complete)
print('solution at n=0:', sol_complete.subs(n, 0))
This prints
general solution: Piecewise((C1*sin(n*t) + C2*cos(n*t), Ne(n, 0)), (C1 + C2*t, True))
solution at n=0: C1 + C2*t
A more familiar mathematical form is provided by sp.pprint(sol_complete).
⎧C₁⋅sin(n⋅t) + C₂⋅cos(n⋅t) for n ≠ 0
⎨
⎩ C₁ + C₂⋅t otherwise
add a comment |
It can be considered a bug in dsolve logic: it finds two eigenvalues n and -n and treats them as different without considering the special case n=0 when they are equal. Ideally it would output a Piecewise like the following code does.
sol_nonzero = sp.dsolve(diff_eq, f(t)).rhs
sol_zero = sp.dsolve(diff_eq.subs(n, 0), f(t)).rhs
sol_complete = sp.Piecewise((sol_nonzero, sp.Ne(n, 0)), (sol_zero, True))
print('general solution:', sol_complete)
print('solution at n=0:', sol_complete.subs(n, 0))
This prints
general solution: Piecewise((C1*sin(n*t) + C2*cos(n*t), Ne(n, 0)), (C1 + C2*t, True))
solution at n=0: C1 + C2*t
A more familiar mathematical form is provided by sp.pprint(sol_complete).
⎧C₁⋅sin(n⋅t) + C₂⋅cos(n⋅t) for n ≠ 0
⎨
⎩ C₁ + C₂⋅t otherwise
It can be considered a bug in dsolve logic: it finds two eigenvalues n and -n and treats them as different without considering the special case n=0 when they are equal. Ideally it would output a Piecewise like the following code does.
sol_nonzero = sp.dsolve(diff_eq, f(t)).rhs
sol_zero = sp.dsolve(diff_eq.subs(n, 0), f(t)).rhs
sol_complete = sp.Piecewise((sol_nonzero, sp.Ne(n, 0)), (sol_zero, True))
print('general solution:', sol_complete)
print('solution at n=0:', sol_complete.subs(n, 0))
This prints
general solution: Piecewise((C1*sin(n*t) + C2*cos(n*t), Ne(n, 0)), (C1 + C2*t, True))
solution at n=0: C1 + C2*t
A more familiar mathematical form is provided by sp.pprint(sol_complete).
⎧C₁⋅sin(n⋅t) + C₂⋅cos(n⋅t) for n ≠ 0
⎨
⎩ C₁ + C₂⋅t otherwise
answered Nov 15 '18 at 1:26
user6655984
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