Rouben Rostamian

MaplePrimes Activity


These are questions asked by Rouben Rostamian

I can derive a symbolic solution by hand for the following ODE, but cannot get Maple to do it for me.  Any tricks?

restart;

Velocity field, -infinity < x and x < infinity,  t > 0.

v := (x,t) -> piecewise(x <= -t, 0, x < t, 1 - 1/2*(1 - x/t));

v := proc (x, t) options operator, arrow; piecewise(x <= -t, 0, x < t, 1/2+(1/2)*x/t) end proc

Position x(t):

de := diff(x(t),t) = v(x(t),t);

de := diff(x(t), t) = piecewise(x(t) <= -t, 0, x(t) < t, 1/2+x(t)/(2*t))

Initial condition, assuming a > 0

ic := x(0) = -a;

x(0) = -a

Symbolic solution, calculated by hand:

x__exact := t -> piecewise(t < a, -a, t - 2*sqrt(t)*sqrt(a));

x__exact := proc (t) options operator, arrow; piecewise(t < a, -a, t-2*sqrt(t)*sqrt(a)) end proc

Verify exact solution by comparing it against the numeric solution for some a > 0:

a := 3;  # any a>0 should do
dsol := dsolve({de,ic}, numeric):
plots:-odeplot(dsol, t=0..5);   # dsolve solution
plot(x__exact(t), t=0..5);      # symbolic solution (calculated by hand)
a := 'a';

3

a

Can Maple's dsolve find the exact solution?  This one returns empty in Maple 2022:

dsolve({de, ic}) assuming a > 0, t > 0;

Download ode-piecewise.mw

 

I expect that there must exist a Maple proc that does the equivalent of the following but I couldn't find it.  Can it be in the combinat package?

And if there isn't one, can the following be improved?  It seems to be horribly inefficient to me, although efficiency is not a major concern for me right now since I need it only for small values of n.

restart;

Proc produces all lists of length n consisting of the

two distinct symbols a and b.

doit := proc(a, b, n::posint)
        local p := 1, L := [ [a], [b] ];
        for p from 1 to n-1 do
                 L := [ map( x -> [a,op(x)], L)[], map( x -> [b,op(x)], L)[] ];
        end do:
        return L;
end proc:

doit(a,b,1);

[[a], [b]]

doit(p,q,3);

[[p, p, p], [p, p, q], [p, q, p], [p, q, q], [q, p, p], [q, p, q], [q, q, p], [q, q, q]]

doit(5,7,3);

[[5, 5, 5], [5, 5, 7], [5, 7, 5], [5, 7, 7], [7, 5, 5], [7, 5, 7], [7, 7, 5], [7, 7, 7]]
 

Download mw.mw

I have been unable to add a directed light source to my plot3d drawings.  Perhaps I am misreading the documentation.  In the following experiment, I attempt to shine a red light from the direction (phi,theta) but changing the values of phi and theta don't seem to change the scene's lighting.  Can you change the lighting in your Maple?

restart;
with(plottools):
with(plots):
Explore(display(sphere(), style=surface, color=yellow, light=[phi,theta,1,0,0]),
	phi=0..Pi, theta=0..Pi);

 

restart;

q := (u,v) -> u^2 + v^2;

proc (u, v) options operator, arrow; v^2+u^2 end proc

D[3](q);

Error, (in D/procedure) index out of range: function takes only 2 arguments

Question: How does D know that q takes two arguments?

In general, if I pass q to another proc, how can I find out, within
that proc, that q takes only two arguments?

download number-of-arguments.mw

Given a vector-valued function z(u,v), I want to calculate the derivative of z with respect to its first argument by applying the D operator to it. I don't see how.  Any suggestions?

restart;

z := (u,v) -> < a(u,v), b(u,v) >;

proc (u, v) options operator, arrow; `<,>`(a(u, v), b(u, v)) end proc

Calculate the derivative of z with respect to its first argument:

P := diff(z(u,v), u);

Vector(2, {(1) = diff(a(u, v), u), (2) = diff(b(u, v), u)})

Express P through the D operator:

Q := convert(P, D);

Vector(2, {(1) = (D[1](a))(u, v), (2) = (D[1](b))(u, v)})

Question:  How do we obtain Q directly by applying the D operator to z

without the help of diff?  This one doesn't work:

D[1](z)(u,v);
type(%, Vector);

(D[1](`<,>`))(a(u, v), b(u, v))*(D[1](a))(u, v)+(D[2](`<,>`))(a(u, v), b(u, v))*(D[1](b))(u, v)

false

Download diff.mw

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