solve a system of two equations for a derivative

solve
diff

asked 2016-03-23 08:03:07 +0100

sophia
77 ●4 ●5 ●12

I have the following defined:

sage: var('a, t');

sage: function('x, y, v, u')

sage: de1 = diff(x(t),t) - diff(y(t),t) == x(t)

sage: de2 = diff(y(t),t) == y(t)

I'd like to solve these two equations algebraically for diff(x(t),t) and diff(y(t),t).

sage: sol = solve(de1, diff(x(t),t)); works, and i get an explicit solution for diff(x(t),t), but when i try solving both equations using:

sage: sol = solve([de1, de2],diff(x(t),t),diff(y(t),t));

i get the following error:

Traceback (most recent call last): File "/projects/sage/sage-6.10/local/lib/python2.7/site-packages/smc_sagews/sage_server.py", line 905, in execute exec compile(block+'\n', '', 'single') in namespace, locals File "", line 1, in <module> File "/projects/sage/sage-6.10/local/lib/python2.7/site-packages/sage/symbolic/relation.py", line 822, in solve variables = tuple(args[0]) TypeError: 'sage.symbolic.expression.Expression' object is not iterable

=============================

I'd appreciate any help with this. thank you.

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Why not using desolve ? See documentation

ndomes ( 2016-03-23 12:13:07 +0100 )edit

I'm not trying to solve the differential equations just yet. i only want to solve the equations algebraically (explicitly) for the derivatives, so I can compute the Jacobian matrix of the right hand sides for further bifurcation analysis. the equations in my actual problem have parameters.

(this is pretty easy to do in Maple)

sophia ( 2016-03-23 21:45:55 +0100 )edit

kcrisman ( 2016-03-24 19:34:04 +0100 )edit

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answered 2016-03-23 22:47:39 +0100

calc314
4191 ●22 ●48 ●113

updated 2016-03-24 04:49:40 +0100

I see what you want to do, but apparently solve will not work with two equations and two derivatives in this way. I think this must have something to do with how Maxima works or how Sage is passing info to Maxima.

Another option would be to use x1 and y1 in place of your derivatives and then solve, since you are really not using the fact that you have functions of t in this solving procedure.

var('x y x1 y1')
de1 = x1 - y1 == x
de2 = y1 == y
sol = solve([de1,de2],[x1,y1])
sol

This gives:

[[x1 == x + y, y1 == y]]

To find the Jacobian, you need the appropriate vector field. You can get this from the solution with the following

F=map(lambda q: q.rhs(),sol[0])

Then, the Jacobian is:

jacobian(F,[x,y])

Does that resolve the issue?

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thanks for replying. i'd like to automate the solution process. i tried substituting x1 for diff(x(t),t), but got an error ( i don't want to manually type in the x1 and y1 part). i tried both:

sage: de1.substitute_function(diff(x(t),t),x1) and sage: de1.subs(diff(x(t),t)=x1)

any suggestions as to how i can sub x1 for diff(x(t),t) directly in de1?

thank you,

sophia ( 2016-03-24 02:27:30 +0100 )edit

Since you originally stated that you were hunting for the Jacobian, I will update my post to include what you need to get that matrix.

calc314 ( 2016-03-24 04:47:31 +0100 )edit

The following worked for me: de1.subs(diff(x(t),t)==x1,diff(y(t),t)==y1)

calc314 ( 2016-03-24 04:52:16 +0100 )edit

that helps me solve what i wanted to, thank you!

i had simplified the problem that i wanted to solve considerably. finding the Jacobian will require solving for the derivative of a functional w.r.t a function. Instead of crowding this post, i've asked a new question here:

http://ask.sagemath.org/question/32876/finding-the-derivative-of-a-functional-wrt-a-function/ (http://ask.sagemath.org/question/3287...)

thank you again for your help.

sophia ( 2016-03-24 07:51:23 +0100 )edit

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answered 2016-03-23 21:42:39 +0100

castor
324 ●1 ●5 ●16

updated 2016-03-23 21:43:10 +0100

In your case it would be something like this:

var('t');
x=function('x')(t)
y=function('y')(t)
de1 = diff(x,t) - diff(y,t) == x
de2 = diff(y,t) == y
desolve_system([de1, de2], [x,y])

and the result is as follows:

[x(t) == t*e^t*y(0) + e^t*x(0), y(t) == e^t*y(0)]

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thanks for replying. i don't want to use desolve yet. all i'd like to do is to solve the equations algebraically (explicitly) for the derivatives. thank you,

sophia ( 2016-03-24 02:28:57 +0100 )edit

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