Sympy plot: plotting a function with multiple free variables (python 3.6)

Question

Firstly, can I plot something that I have declared as function? Because I'm getting loads of errors but it doesn't show where in my code the error is coming from e.g.:

import sympy as sy

rho = 1000 # Density of water [kg/m3]
A = 1.0 # Cross-sectional area of tank [m2]
c = 50 # Valve constant [kg/s per opening percent]
o = 100 # Percent valve opening 

h = sy.Function('h')
t = sy.symbols('t')

m_bal = sy.Eq(h(t).diff(t),c/(rho*A)*o)
gen_sol = sy.dsolve(m_bal,h(t))
C = gen_sol.subs([(t,0),(h(0),0)])
sol = gen_sol.subs([(C.rhs,C.lhs)])
sy.plot(t,h)

Then I get:

---------------------------------------------------------------------------
AssertionError                            Traceback (most recent call last)
<ipython-input-54-7aa8880d452a> in <module>()
     14 C = gen_sol.subs([(t,0),(h(0),0)])
     15 sol = gen_sol.subs([(C.rhs,C.lhs)])
---> 16 sy.plot(t,h)

~\Anaconda3\lib\site-packages\sympy\plotting\plot.py in plot(*args, **kwargs)
   1288     show = kwargs.pop('show', True)
   1289     series = []
-> 1290     plot_expr = check_arguments(args, 1, 1)
   1291     series = [LineOver1DRangeSeries(*arg, **kwargs) for arg in plot_expr]
   1292 

~\Anaconda3\lib\site-packages\sympy\plotting\plot.py in check_arguments(args, expr_len, nb_of_free_symbols)
   1791 
   1792         exprs = args[:i]
-> 1793         assert all(isinstance(e, Expr) for expr in exprs for e in expr)
   1794         free_symbols = list(set().union(*[e.free_symbols for expr in exprs
   1795                                         for e in expr]))

AssertionError:

Secondly, I want to plot that for t = 0s to t = 10s and I only want o = 100 for t = [2,7]. I have done this but I don't know where I'm going wrong:

import sympy as sy
sy.init_printing()

rho = 1000 # Density of water [kg/m3]
A = 1.0 # Cross-sectional area of tank [m2]
c = 50 # Valve constant [kg/s per opening percent]
o = 100

h = sy.Function('h')
t = sy.symbols('t')

m_bal = sy.Eq(h(t).diff(t),c/(rho*A)*o)
gen_sol = sy.dsolve(m_bal,h(t))
C = gen_sol.subs([(t,0),(h(0),0)])
sol = gen_sol.subs([(C.rhs,C.lhs)])
sy.plot(t,h)

Can I do this on sympy or do I have to use matplotlib and numpy and do:

o= np.zeros(100) # % Valve opening 
o[21:70] = 100 # % opening at times 2s to 7s

user6655984 · Accepted Answer · 2018-08-07 23:38:29Z

1

sy.plot(t, h) is problematic for multiple reasons. First, in SymPy's plot the function comes before the variable: for example, sy.plot(t**2, (t, 0, 3)).

Second, nothing you did with the ODE had any impact on t or h. You are just trying to plot something with sy.Function('h') with no indication of what h is. SymPy objects do not mutate; unless you assign something to h, it's not going to change.

The attempt to enforce the initial condition also looks strange. In the current version of SymPy (1.2) the initial conditions can be imposed by dsolve itself. This is how I would solve and plot the solution:

sol = sy.dsolve(m_bal, h(t), ics={h(0): 0})
sy.plot(sol.rhs, (t, 0, 3))

answered Aug 7, 2018 at 23:38

user6655984

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2 Comments

Shervin Rad Over a year ago

When I use ics={} and then print the solution, the integration constant is still not replaced by a value. But when I plot it, it takes the initial condition into account. How can I replace that constant? And thanks for your help btw

user6655984 Over a year ago

That's why I wrote "In the current version of SymPy (1.2)". In an older version such as yours, refer to stackoverflow.com/questions/41192748

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