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substituting new functions

Posted on 2008-03-21 11:18 By Sandor Szabo (Maple Rating 2 112)
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I have an  ode system, and use  2 D Math  (I don't know how to copy here, so I type by hand)


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y'(t) = x(t) + y(t) (1 - x^2(t) - y^2(t)).

I would like to introduce  x(t) = r(t) cos(Theta(t)), y(t) = r(t) sin(Theta(t)) and manipulate the equations further.

The problem is the left hand side. In one case I obtain Dx(t), in the other case (r(t) cos(Theta(t))' without calculation on it.

I ask some help. Thanks.

                                                  Sandor

 

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Format for dsolve or DEtools

Comment on 2008-03-21 13:43 from jaytreiman (Maple Rating 1 41)

When I work with differential equations I use the following format for the DE,

diff(y(t),t) = x(t) + y(t) (1 - x(t)^2 - y(t)^2);

 

One can then substitute for x(t) and y(t) very easily with

eval(%,{x(t) = r(t)*cos(Theta(t)), y(t) = r(t)*sin(Theta(t)) });

 

Since you did not give the whole problem, I cannot go much further.

 

Format for dsolve

Comment on 2008-03-21 14:37 from Sandor Szabo (Maple Rating 2 112)

Sorry, i was not able to type in the other equation.

I tried earlier your suggestion, but the result is terrible

> eval((D(x))(t) = -y(t)+(x(t))(1-x(t)^2-y(t)^2), [x(t) = r(t)*cos(Theta(t)), y(t) = r(t)*sin(Theta(t))]);

                                          /        2              2
      D(x)(t) = -r(t) sin(Theta(t)) + r(t)\1 - r(t)  cos(Theta(t))

               2              2\              /        2              2
         - r(t)  sin(Theta(t)) / cos(Theta(t))\1 - r(t)  cos(Theta(t))

               2              2\
         - r(t)  sin(Theta(t)) /
 

I didn' t type in eval by hand, I used the Expression palette,  the last row, if you move the cursor  on it, the eval tooltip appears.

Sandor

 

diff and D

Comment on 2008-03-21 15:15 from Sandor Szabo (Maple Rating 2 112)

Sorry again. After reading Robert's response I realized that you used diff and in Math 2D there is D.  Sandor

work with diffeq

Comment on 2008-03-21 13:53 from Alex Smith (Maple Rating 3 294)

I would approach the problem this way:

 

eqn:=proc(x,y)
  diff(y(t),t)=x(t)+y(t)*(1-x(t)^2-y(t)^2);
end proc;

eqn(x,y);

X:=t->r(t)*cos(theta(t));
Y:=t->r(t)*sin(theta(t));

 eqn(X,Y);
 

Set up the diffeq as a proc that takes general functions x and y. Then pass specific functions X and Y.

Robert Israel's picture

Using D

Comment on 2008-03-21 14:16 from Robert Israel (Maple Rating 4 1460)

y' in 2D math is really D(y).  So instead of substituting for y(t) you have to substitute for the function y.  Thus:

(D(y))(t) = x(t)+y(t)*(1-x(t)^2-y(t)^2)

> eval(%,{x=(t -> r(t)*cos(Theta(t))),y=(t->r(t)*sin(Theta(t)))});

(D(r))(t)*sin(Theta(t))+r(t)*cos(Theta(t))*(D(Theta))(t) = r(t)*cos(Theta(t))+r(t)*sin(Theta(t))*(1-r(t)^2*cos(Theta(t))^2-r(t)^2*sin(Theta(t))^2)

D and eval

Comment on 2008-03-21 15:44 from Sandor Szabo (Maple Rating 2 112)

Thanks. A little bit can be improved your answer.

eval(%, {x=r*cos@Theta, y=r*sin@Theta})

also works. (You said: we have to substitute the function, so I substituted with functions. You did the same,

but it can be omitted the t variable.)                 Sandor

D and eval

Comment on 2008-03-21 16:19 from Sandor Szabo (Maple Rating 2 112)

Errhh,  I'm a bit tired. I forgot brackets.    x = r*(cos@Theta), y = r*(sin@Theta).             Sandor

Doug Meade's picture

and, in 1D, with simplify also

Comment on 2008-03-21 14:31 from Doug Meade (Maple Rating 4 722)

Robert's response is good, in particular the part about how 2D math represents the derivative. The way this has to be handled is not very attractive (at least to me). I much prefer the 1D approach, shown below:

ODE := diff(y(t),t)=x(t)+y(t)*(1-x(t)^2-y(t)^2);
                  d                     /        2       2\
                 --- y(t) = x(t) + y(t) \1 - x(t)  - y(t) /
                  dt                                       
SUBS := [ x(t)=r(t)*cos(Theta(t)), y(t)=r(t)*sin(Theta(t)) ];
           [x(t) = r(t) cos(Theta(t)), y(t) = r(t) sin(Theta(t))]

q := eval( ODE, SUBS );
/ d      \                                    / d          \                   
|--- r(t)| sin(Theta(t)) + r(t) cos(Theta(t)) |--- Theta(t)| = r(t) cos(Theta(t
\ dt     /                                    \ dt         /                   

                          /        2              2       2              2\
  )) + r(t) sin(Theta(t)) \1 - r(t)  cos(Theta(t))  - r(t)  sin(Theta(t)) /

Moreover, hitting this result with simplify yields some additional benefits:

simplify( q );
   / d      \                                    / d          \         /
   |--- r(t)| sin(Theta(t)) + r(t) cos(Theta(t)) |--- Theta(t)| = -r(t) \
   \ dt     /                                    \ dt         /          
                                        2              \
   -cos(Theta(t)) - sin(Theta(t)) + r(t)  sin(Theta(t))/

Either way, these are the basic tools to do what you want.

Doug

---------------------------------------------------------------------
Douglas B. Meade  <><
Math, USC, Columbia, SC 29208  E-mail: mailto:meade@math.sc.edu       
Phone:  (803) 777-6183         URL:    http://www.math.sc.ed

 

1D and 2D

Comment on 2008-03-21 15:56 from Sandor Szabo (Maple Rating 2 112)

Thanks, your answer is also good. I make an exam and I want to show the Maple's solution

to my colleagues and they not necesserally  understand Maple codes, so this is why I prefer the Robert's solution.

Sandor

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