Rouben Rostamian

MaplePrimes Activity


These are replies submitted by Rouben Rostamian

@Paulo Baumbach  That's a good description of the problem and I have a better understanding of what you are attempting to do, but some details are still fuzzy.  For instance, I can't tell whether "phase 1" and "phase 2" refer to intervals in time or in space.  Also I don't see the differential equations.  I don't expect you to explain all that here because that would amount to copying  a good deal of the article you have referred to.

It is a hoiiday weekend in the United State and the university library is closed.  I will see if I can find the article next week and see what it is about.

 

I looked through your lengthy worksheet but was unable to understand the main idea, and therefore I am unable to offer a suggestion.

It will help if you could provide a statement of the problem in words and mathematical equations, separating them from the computational details.

 

@ahmadtalaei What is the purpose of introducing the variable eta into the operator Do?  I cannot follow that line of thought.  See if you can clarify what it is that you want to do.

As to D, D^2, D^3,  they are the equivalent of the notation w', w'', w''' in calculus.  Thus, the ODE that I have obtained looks like this:

A(η)*w''''(η) + B(η)*w'''(η) + C(η)*w'(η) = 0

After you have calculated the solution w(η), you may plug in r*sin(θ) for η. to obtain w(r*sin(θ)).  There is no differentiation involved in that operation, and therefore the question of "when taking the derivatives of w(r*sin(phi))" should not arise.

 

 

@Thomas Richard I didn't know about that MapleSim animation. Looks good!  Thanks for pointing it out.

 

@Preben Alsholm Oops, I was careless.  I have corrected the name now and thus have baptized him Dutch.  Sorry ;-)

 

@PhD_Wallyson I quickly looked over your worksheet.  Perhaps it's possible to fix it to make it work but altogether the approach there is somewhat disorganized and not very pleasant to work with.

I will do a writeup on how to apply the Krylov-Duncan functions in a systematic way to analyze multi-span beams, and will let you know when it is ready.  I may take a day or two.

 

You have a system of 15 first order PDEs in 15 unknowns, each being a function of z and t.  You discretize the z variable and obtain a system of 15n ODEs in 15n unknowns.  Even for moderate n that's a very inefficient way of solving the system, as well as being a very poor approximation.

 I suggest applying pdsolve() to solve the system of PDEs directly.  Have you tried that?

 

 

@Neel There are many different approaches to calculating the eigenfunctions.  I showed the simplest approach in my earlier reply.  Since you are asking for a solution with with matrices, here is the matrix version.  As to the forced vibration, I showed how to to that in my pdf writeup.

restart;

with(LinearAlgebra):

pde := diff(w(x,t),t,t) + c^2*diff(w(x,t),x$4) = 0;

diff(diff(w(x, t), t), t)+c^2*(diff(diff(diff(diff(w(x, t), x), x), x), x)) = 0

eval(pde, w(x,t) = u(x)*cos(omega*t));
de_tmp := expand(% / (c^2*cos(omega*t)));

-u(x)*omega^2*cos(omega*t)+c^2*(diff(diff(diff(diff(u(x), x), x), x), x))*cos(omega*t) = 0

-u(x)*omega^2/c^2+diff(diff(diff(diff(u(x), x), x), x), x) = 0

mu_def := omega = c*mu^2/L^2;

omega = c*mu^2/L^2

de := eval(de_tmp, mu_def);

-u(x)*mu^4/L^4+diff(diff(diff(diff(u(x), x), x), x), x) = 0

dsol_tmp := dsolve(de);

u(x) = _C1*exp(-mu*x/L)+_C2*exp(mu*x/L)+_C3*sin(mu*x/L)+_C4*cos(mu*x/L)

Convert to sinh and cosh, and make the result into a function:

subs(exp(mu*x/L) = cosh(mu*x/L)+sinh(mu*x/L),
     exp(-mu*x/L)= cosh(mu*x/L)-sinh(mu*x/L), dsol_tmp);
u_general := unapply(rhs(%), x);

u(x) = _C1*(cosh(mu*x/L)-sinh(mu*x/L))+_C2*(cosh(mu*x/L)+sinh(mu*x/L))+_C3*sin(mu*x/L)+_C4*cos(mu*x/L)

proc (x) options operator, arrow; _C1*(cosh(mu*x/L)-sinh(mu*x/L))+_C2*(cosh(mu*x/L)+sinh(mu*x/L))+_C3*sin(mu*x/L)+_C4*cos(mu*x/L) end proc

bc := u(0)=0, (D@@2)(u)(0)=0, u(L)=0, (D@@2)(u)(L)=0;

u(0) = 0, ((D@@2)(u))(0) = 0, u(L) = 0, ((D@@2)(u))(L) = 0

Apply the boundary conditions:

bc_applied := eval({bc}, u = u_general);

{_C1+_C2+_C4 = 0, _C1*mu^2/L^2+_C2*mu^2/L^2-_C4*mu^2/L^2 = 0, _C1*(mu^2*cosh(mu)/L^2-mu^2*sinh(mu)/L^2)+_C2*(mu^2*cosh(mu)/L^2+mu^2*sinh(mu)/L^2)-_C3*mu^2*sin(mu)/L^2-_C4*mu^2*cos(mu)/L^2 = 0, _C1*(cosh(mu)-sinh(mu))+_C2*(cosh(mu)+sinh(mu))+_C3*sin(mu)+_C4*cos(mu) = 0}

bc_matrix_form := GenerateMatrix(bc_applied, [_C1, _C2, _C3, _C4])[1];

Matrix(4, 4, {(1, 1) = 1, (1, 2) = 1, (1, 3) = 0, (1, 4) = 1, (2, 1) = mu^2/L^2, (2, 2) = mu^2/L^2, (2, 3) = 0, (2, 4) = -mu^2/L^2, (3, 1) = mu^2*cosh(mu)/L^2-mu^2*sinh(mu)/L^2, (3, 2) = mu^2*cosh(mu)/L^2+mu^2*sinh(mu)/L^2, (3, 3) = -mu^2*sin(mu)/L^2, (3, 4) = -mu^2*cos(mu)/L^2, (4, 1) = cosh(mu)-sinh(mu), (4, 2) = cosh(mu)+sinh(mu), (4, 3) = sin(mu), (4, 4) = cos(mu)})

eq := Determinant(bc_matrix_form);

-8*mu^4*sin(mu)*sinh(mu)/L^4

solve(eq, allsolutions);

{L = L, mu = 0}, {L = L, mu = Pi*_Z1}, {L = L, mu = I*Pi*_Z2}

about(_Z1);

Originally _Z1, renamed _Z1~:
  is assumed to be: integer
 

We don't want the mu = 0 solution (first group), nor do we want the complex solution (third group),
therefore we pick the middle groups which says
mu = n*Pi for integer n.  Plug this into the matrix.
The system's nonzero solutions correspond to the matrix's null space:

eval(bc_matrix_form, mu=n*Pi) assuming n::integer;
N := NullSpace(%);

Matrix(4, 4, {(1, 1) = 1, (1, 2) = 1, (1, 3) = 0, (1, 4) = 1, (2, 1) = n^2*Pi^2/L^2, (2, 2) = n^2*Pi^2/L^2, (2, 3) = 0, (2, 4) = -n^2*Pi^2/L^2, (3, 1) = n^2*Pi^2*cosh(n*Pi)/L^2-n^2*Pi^2*sinh(n*Pi)/L^2, (3, 2) = n^2*Pi^2*cosh(n*Pi)/L^2+n^2*Pi^2*sinh(n*Pi)/L^2, (3, 3) = 0, (3, 4) = -n^2*Pi^2*(-1)^n/L^2, (4, 1) = cosh(n*Pi)-sinh(n*Pi), (4, 2) = cosh(n*Pi)+sinh(n*Pi), (4, 3) = 0, (4, 4) = (-1)^n})

{Vector[column](%id = 18446883856494723910)}

Evaluate the coefficients, then the general solution to obtain the eigenfunction.

convert(<_C1, _C2, _C3, _C4> =~ op(N), set);
subs(%, u_general(x));

{_C1 = 0, _C2 = 0, _C3 = 1, _C4 = 0}

sin(mu*x/L)

where mu = n*Pi/L.
 

Download eigenfunction-calculation2.mw

 

@Neel See if this helps.

restart;

pde := diff(w(x,t),t,t) + c^2*diff(w(x,t),x$4) = 0;

diff(diff(w(x, t), t), t)+c^2*(diff(diff(diff(diff(w(x, t), x), x), x), x)) = 0

(1)

eval(pde, w(x,t) = u(x)*cos(omega*t));
de_tmp := expand(% / (c^2*cos(omega*t)));

-u(x)*omega^2*cos(omega*t)+c^2*(diff(diff(diff(diff(u(x), x), x), x), x))*cos(omega*t) = 0

 

-u(x)*omega^2/c^2+diff(diff(diff(diff(u(x), x), x), x), x) = 0

(2)

mu_def := omega = c*mu^2/L^2;

omega = c*mu^2/L^2

(3)

de := eval(de_tmp, mu_def);

-u(x)*mu^4/L^4+diff(diff(diff(diff(u(x), x), x), x), x) = 0

(4)

bc := u(0)=0, (D@@2)(u)(0)=0, u(L)=0, (D@@2)(u)(L)=a;

u(0) = 0, ((D@@2)(u))(0) = 0, u(L) = 0, ((D@@2)(u))(L) = a

(5)

dsol := dsolve({de,bc}, u(x));

u(x) = (1/2)*a*L^2*exp(mu*x/L)/((exp(mu)-exp(-mu))*mu^2)-(1/2)*a*L^2*exp(-mu*x/L)/((exp(mu)-exp(-mu))*mu^2)-(1/2)*a*L^2*sin(mu*x/L)/(sin(mu)*mu^2)

(6)

We recognize the e^mu-e^(-mu) combination as 2*sinh(mu).  Same with the e^(-mu*x/L)thing.

algsubs( exp(mu) - exp(-mu) = 2*sinh(mu), rhs(dsol));
algsubs( exp(mu*x/L) - exp(-mu*x/L) = 2*sinh(mu*x/L), %);
my_u_tmp := %;

(1/4)*a*L^2*exp(mu*x/L)/(mu^2*sinh(mu))-(1/4)*a*L^2*exp(-mu*x/L)/(mu^2*sinh(mu))-(1/2)*a*L^2*sin(mu*x/L)/(sin(mu)*mu^2)

 

(1/2)*a*L^2*(sinh(mu*x/L)*sin(mu)-sin(mu*x/L)*sinh(mu))/(mu^2*sinh(mu)*sin(mu))

 

(1/2)*a*L^2*(sinh(mu*x/L)*sin(mu)-sin(mu*x/L)*sinh(mu))/(mu^2*sinh(mu)*sin(mu))

(7)

We know that any constant multiple of the solution u(x)of our differential equation is also a solution.
Therefore we simplify the solution obtained above by getting rid of some irrelevant multipliers:

my_u := expand(my_u_tmp/(-a*L^2)*(2*mu^2*sin(mu)));

-sinh(mu*x/L)*sin(mu)/sinh(mu)+sin(mu*x/L)

(8)

Let's apply the boundary condition u__xx(L) = 0.

diff(my_u, x, x);
eq := eval(%, x=L);

-mu^2*sinh(mu*x/L)*sin(mu)/(sinh(mu)*L^2)-mu^2*sin(mu*x/L)/L^2

 

-2*mu^2*sin(mu)/L^2

(9)

We need to solve eq=0.  We don't want the mu = 0 solution since that leads to the trivial
solution
u(x) = 0. Therefore we are left with sin(mu) = 0, the solution to which is mu = n*Pi for
any integer
n.

 

Aside:  It is overkill to ask Maple to solve that equation, but if you insist, we do:

solve(sin(mu)=0, allsolutions);

Pi*_Z1

(10)

What is `~`[_Z1]?  Let's ask Maple:

about(_Z1);

Originally _Z1, renamed _Z1~:
  is assumed to be: integer
 

 

There we have it.  The solution is n*Pi for any integer n, as asserted earlier.  Let's plug that
value of
mu into our solution:

eval(my_u, mu=n*Pi);
Eigenfunction := simplify(%) assuming n::integer;

-sinh(n*Pi*x/L)*sin(n*Pi)/sinh(n*Pi)+sin(n*Pi*x/L)

 

sin(n*Pi*x/L)

(11)

That's the eigenfunction that you had asked for.

 

But please note:  This worksheet's calculations are best done by hand.  Maple can be useful in many
situations, but the problem worked out here is not one of those.

 

 

Download eigenfunction-calculation.mw

 

@Neel See if the writeup in the attached PDF answers your questions.

By the way, in your line of work you will need to go beyond writing mathematics by hand.  Look into learning LaTeX if you haven't done so already.  My writeup is done in LaTeX.

steady-state-oscillations.pdf

Note added later:
I spotted a few (harmless) typos in the previous PDF.  Here is one with corrections.

steady-state-oscillations2.pdf

You are suffering from culture shock.  You must already know that when in a foreign country, speaking louder in your own language does not make you any better understood.  Learn the local language and customs instead of complaining that they do not agree with yours.

 

@Neel In equation (9.B.20) you have the force expressed as a delta function.  The picture below that equation also indicates that you do mean a delta function.  But now you are saying q(x,t)= Fo*(1-(x/L))* cos(omega*t) without a delta function.  You need to decide which one you want.

 

 

@Neel I looked quickly through your PDF.  The calculations up to (9.B.19) appear to be correct, but after that point I felt lost.

In (9.B.22) we have the equation of motion:

ρ A wtt + E I wxxxx = q(x,t)

which is correct, but we have no clear statement of what q(x,t) is.  You need to say
q(x,t) = something.

What is that something?

The boundary conditions
w(0,t) = wxx(0,t) = w(L,t) = wxx(L,t) = 0.
for pinned ends are good.

I also don't see what the initial condition are.  You need to specify the initial displacement and initial velocity as

w(x,0) = something
wt(x,0) = something

Can you tell me what those "something"s are?

 

@Earl   The Harvard Weekly Problems is a good collection math and physics puzzles.  The one you are referring to is Problem #15.  Beware though that the shape discussed in that problem is the 3D solid obtained by rotating your lamina about the horizontal axis.

 

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