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Infinite initial conditions in ODE and arbitrary constant

asked 2017-02-08 19:37:45 -0500

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Is it possible to include infinite initial conditions in an ODE? For example,

ode_soln = desolve(ode, q, ics[0, -Infity])

throws and error.

Alternatively, I can leave the initial condition blank:

ode_soln = desolve(ode, q)
soln = solve(ode_soln, q)
show(soln[0])

Then output Includes an arbitrary constant written C $$ q\left(t\right) = -\frac{p}{C {\left(p + 1\right)} + {\left(p + 1\right)} t} $$

I tried to set a value for $C$ via,

show(soln[0].substitute(C = 0))

but it made no difference. Here $p$ I declared as a variable, and I can do things like

show(soln[0].substitute(p = 1))

to produce the expected output

$$ q\left(t\right) = -\frac{1}{2 {\left(C + t\right)}} $$

So it seems that I need to refer to the arbitrary constant by some other name than $C$. How can I do this?

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answered 2017-02-09 05:43:50 -0500

eric_g gravatar image

The answer to your second question (substituting for $C$) is

soln[0].substitute({SR.var('_C'): 0})

This is because constants that appear in solutions of ODEs, like _C (which is displayed as $C$), are not put in the global namespace, hence they have to be recovered from their names in the Symbolic Ring (SR) by SR.var('_C').

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Excellent. That does the job, thank you.

Paul Bryan gravatar imagePaul Bryan ( 2017-09-07 06:42:11 -0500 )edit

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Asked: 2017-02-08 19:37:45 -0500

Seen: 40 times

Last updated: Feb 09