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Rouben Rostamian

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19 years, 2 days
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Memory issue likely...

Rouben Rostamian 7711
October 28 2021
0 0

@Pets71 For the record, I am running Maple 2021 on Ubuntu 20.04 LTS installed on a 10 year old laptop with 4GB of memory. I use it just about every day in my classes for teaching, without a single problem.

I agree with Thomas Richard that what you have described smells very much like a bad RAM in your laptop. That's a hardware problem and has nothing to do with Maple.

See https://linuxhint.com/run_memtest_ubuntu/ on how to test your RAM in Ubuntu.

 

Sij versus S[i,j]...

Rouben Rostamian 7711
October 26 2021
0 0

Isn't it more natural to name your variables S[i,j] rather than Sij?  If you do the former, then you may do

local S;
seq(seq(assign(S[i,j]=Vector(datatype=float)),i=1..9),j=1..9);

 

Homework...

Rouben Rostamian 7711
October 26 2021
0 0

It will be good if you showed some attempt toward solving this homework problem.  You don't want someone else to do your homework for you, do you?

Numeric calculation...

Rouben Rostamian 7711
October 23 2021
0 0

@rlopez My original example required a symbolic solution (x(p) and y(p)). Should a symbolic solution be unattainable, we may resort to a numerical solution, which turns out to be even simpler than the symbolic, as illustrated in the following worksheet.

> 

restart;
> 

with(plots):

Here is a family of function of x defined in terms of a parameter p.

> 

f := (x,p) -> (x-p)^2 + cos(p);

proc (x, p) options operator, arrow; (x-p)^2+cos(p) end proc

The family's envelope is obtained by solving the system
f(x, p) = y

diff(f(x, p), p) = 0.``

Here are two options:

 

Option 1. We solve the system for x and y in terms of p, we obtain a

parametric representation x(p), y(p)of the envelope.

 

Option 2. For a given x, we solve the second equation for p and obtain
p(x).  Then evaluate y = f(x, p(x)), thus obtaining a mapping from
x to y of the points on the envelope.

NULL

I followed Option 1 in my original post.  That option works best if we

are able to solve the system symbolically.

NULL

Option 2 works best if the system is not solvable symbolically and we

need to resort to numerical calculations.

NULL

Here I will solve the original problem numerically to show that we

obtain the same result.NULL

 

Option 2 is implemented in the following proc.  It receives x and
returns y so that (x,y) is on the envelope.

> 

elim := proc(x) option remember;
        if not type (x, realcons) then return 'procname'(args) end if;
        y = f(x,p), D[2](f)(x,p) = 0;
        fsolve({%});
        eval(y, %);
end proc:

Here is the plot of the envelope. It agrees with the result of the previous

symbolic calculation.

> 

plot(elim(x), x=-8..8, color="Blue");

As before, we calculate the area between the central arch of

that graph and the x axis.  The arch's rightmost endpoint is at x = L, where

> 

L := fsolve(elim(x), x=0..2);

1.323245517

Therefore, the area under the arch is

> 

A := int(elim(x), x=-L..L, numeric);

1.586776257

which agrees with the previous result.

 

Download area-under-envelope-numeric.mw

 

Regarding the Method of Lines...

Rouben Rostamian 7711
October 20 2021
0 0

@mmcdara As rlopez has noted, the method of lines (MoL) is unrelated to the method of characteristics. When a PDE has both space and time variables, in MoL we discretize space through finite differences, while treating the time variable as continuous.  Thus, the PDE is approximated by a system of ODEs.

About a year ago, I illustrated the MoL in the post https://www.mapleprimes.com/posts/212624-Solving-Multispan-Euler-Beams-With by applying it to the equation of the deflection of a multispan Euler beam.

Try a simpler problem first...

Rouben Rostamian 7711
October 18 2021
0 0

A wise man once said: "If you don't know how to solve a problem, there is a simpler problem that you don't know how to solve. Try that one first".

In your case, the simpler problem would be the one-dimensional case of the heat equation. Do you know how to do that one?

@vv That's very clever.  V...

Rouben Rostamian 7711
October 15 2021
0 0

@vv That's very clever.  Vote up!

What is optimal?...

Rouben Rostamian 7711
October 10 2021
0 0

Your homework asks you to define a function and then find the optimal solution.  But optimal solution to what? Perhaps you have abbreviated the question too much to be comprehensible.

You attempt to solve f(x,a,b,c)=0 for x.  The x that you would obtain there is called the root of f. In what sense is that related to the optimum of anything?

I can guess what the question is about but that would be only a guess and I don't want to put words in your mouth.  It will be good if you formulate your question in a clearer way.

Thanks!...

Rouben Rostamian 7711
September 30 2021

@C_R Thanks for the data.  I'll see what I can do.

 

Good example...

Rouben Rostamian 7711
September 28 2021
0 0

@vv That's a very clever construction. I see that the deviation of the answer from a rational in this case is of the order of exp(-10^28)  which is too small to be detected by fsolve().

 

Thanks for the demo...

Rouben Rostamian 7711
September 28 2021
0 0

@Mariusz Iwaniuk Thanks for this demo which clearly shows the trend.  If find Ronan's explanation is more insightful.

Ah, of course......

Rouben Rostamian 7711
September 28 2021
0 0

@Ronan Yes, that's it.  Now that you have pointed this out, it looks so obvious to me.  I have converted your "Reply" to "Answer" in order to give it a Vote Up.

 

Very nice!...

Rouben Rostamian 7711
September 28 2021

A very nice demo.  Vote up!

I would be interested in reproduding this in Maple as I don't have MapleSim.  Can you post the complete set of this demo's parameters?

Need to be specific...

Rouben Rostamian 7711
September 26 2021
0 0

@Muhammad Usman XJTU As tomleslie as noted, your ode1 with the intitial condition phi(0)=0 admits a unique solution.  Isn't that all you need?  If you have something else in mind, then you should state it clearly, because I find it impossible to understand what you are aiming to do by looking over your worksheet.  It will help if you present your question in words, not in Maple.

A more compact solution...

Rouben Rostamian 7711
September 19 2021
0 0

@mmcdara That's a good alternative.  But in going over both your solution and my previous solution, I see that it's possible to streamline the process quite a bit. Here it is.

> 

restart;

We have five differential equations:

> 

de1 := diff(u__1(y),y,y) + 1/2*diff(N(y),y) + 1/2*theta__1(y) = 0;

diff(diff(u__1(y), y), y)+(1/2)*(diff(N(y), y))+(1/2)*theta__1(y) = 0

> 

de2 := diff(N(y),y,y) - 2/3 * (2*N(y) + diff(u__1(y),y)) = 0;

diff(diff(N(y), y), y)-(4/3)*N(y)-(2/3)*(diff(u__1(y), y)) = 0

> 

de3 := diff(theta__1(y),y,y) = 0;

diff(diff(theta__1(y), y), y) = 0

> 

de4 := diff(u__2(y),y,y) + theta__2(y) = 0;

diff(diff(u__2(y), y), y)+theta__2(y) = 0

> 

de5 := diff(theta__2(y),y,y) = 0;

diff(diff(theta__2(y), y), y) = 0

Find the general solution:

> 

dsol__general := dsolve({de1,de2,de3,de4,de5});

{N(y) = exp(y)*_C2-exp(-y)*_C3+(1/3)*_C10*y+(1/6)*_C9*y^2+_C4, u__1(y) = -2*_C4*y-(1/3)*_C10*y^2-(1/9)*_C9*y^3-(1/2)*exp(y)*_C2-(1/2)*exp(-y)*_C3+(1/2)*_C9*y+_C1, u__2(y) = -(1/6)*_C7*y^3-(1/2)*_C8*y^2+_C5*y+_C6, theta__1(y) = _C9*y+_C10, theta__2(y) = _C7*y+_C8}

How many integrations constants?

> 

indets(dsol__general, name);

{_C1, _C10, _C2, _C3, _C4, _C5, _C6, _C7, _C8, _C9, y}

So there are 10 integration constants, which is good since we have 10 boundary and interface conditions.

 

Let's evaluate the solutions and their first derivatives at y = -1, y = 0, y = 1:

> 

dsol__general union diff~(dsol__general, y):
convert(%, D):
conditions := eval(%, y=-1) union eval(%, y=0) union eval(%, y=1):

Now let us look at the boundary and interface conditionsj.

> 

bc := {
             # conditions at y=-1
             u__1(-1) = 0,
                N(-1) = 0,
         theta__1(-1) = 1,

             # conditions at y=0
             u__1(0) = u__2(0),
         theta__1(0) = theta__2(0),
      D(theta__1)(0) = D(theta__2)(0),
             D(N)(0) = 0,
          D(u__1)(0) + 1/2*N(0) = 1/2*D(u__2)(0),

             # conditions at y=+1
             u__2(1) = 0,
         theta__2(1) = 0
};

{(D(u__1))(0)+(1/2)*N(0) = (1/2)*(D(u__2))(0), N(-1) = 0, u__1(-1) = 0, u__1(0) = u__2(0), u__2(1) = 0, theta__1(-1) = 1, theta__1(0) = theta__2(0), theta__2(1) = 0, (D(N))(0) = 0, (D(theta__1))(0) = (D(theta__2))(0)}

Apply the previously calculated expressions for u__1(-1), (D(u__1))(-1), etc., to the boundary conditions:

> 

eqns := eval(bc, conditions);

{_C10 = _C8, _C9 = _C7, _C10-_C9 = 1, -(3/2)*_C4+(1/2)*_C9 = (1/2)*_C5, _C7+_C8 = 0, _C2+_C3+(1/3)*_C10 = 0, -(1/2)*_C2-(1/2)*_C3+_C1 = _C6, -(1/6)*_C7-(1/2)*_C8+_C5+_C6 = 0, exp(-1)*_C2-exp(1)*_C3-(1/3)*_C10+(1/6)*_C9+_C4 = 0, 2*_C4-(1/3)*_C10-(7/18)*_C9-(1/2)*exp(-1)*_C2-(1/2)*exp(1)*_C3+_C1 = 0}

Solve this system of 10 equations in the 10 unknown coefficients:

> 

the_coeffs := solve(eqns);

{_C1 = -(1/6)*(6*exp(-1)*exp(1)-21*exp(-1)-5*exp(1))/(11*exp(-1)+9*exp(1)), _C10 = 1/2, _C2 = -(1/18)*(27*exp(1)-67)/(11*exp(-1)+9*exp(1)), _C3 = -(1/18)*(33*exp(-1)+67)/(11*exp(-1)+9*exp(1)), _C4 = -(1/36)*(12*exp(-1)*exp(1)+35*exp(-1)+53*exp(1))/(11*exp(-1)+9*exp(1)), _C5 = (1/12)*(12*exp(-1)*exp(1)-31*exp(-1)-exp(1))/(11*exp(-1)+9*exp(1)), _C6 = -(1/12)*(12*exp(-1)*exp(1)-53*exp(-1)-19*exp(1))/(11*exp(-1)+9*exp(1)), _C7 = -1/2, _C8 = 1/2, _C9 = -1/2}

Apply the calculated coefficients to obtain the final form of the solutions to the ODEs:

> 

dsol := eval(dsol__general, the_coeffs);

{N(y) = -(1/18)*exp(y)*(27*exp(1)-67)/(11*exp(-1)+9*exp(1))+(1/18)*exp(-y)*(33*exp(-1)+67)/(11*exp(-1)+9*exp(1))+(1/6)*y-(1/12)*y^2-(1/36)*(12*exp(-1)*exp(1)+35*exp(-1)+53*exp(1))/(11*exp(-1)+9*exp(1)), u__1(y) = (1/18)*(12*exp(-1)*exp(1)+35*exp(-1)+53*exp(1))*y/(11*exp(-1)+9*exp(1))-(1/6)*y^2+(1/18)*y^3+(1/36)*exp(y)*(27*exp(1)-67)/(11*exp(-1)+9*exp(1))+(1/36)*exp(-y)*(33*exp(-1)+67)/(11*exp(-1)+9*exp(1))-(1/4)*y-(1/6)*(6*exp(-1)*exp(1)-21*exp(-1)-5*exp(1))/(11*exp(-1)+9*exp(1)), u__2(y) = (1/12)*y^3-(1/4)*y^2+(1/12)*(12*exp(-1)*exp(1)-31*exp(-1)-exp(1))*y/(11*exp(-1)+9*exp(1))-(1/12)*(12*exp(-1)*exp(1)-53*exp(-1)-19*exp(1))/(11*exp(-1)+9*exp(1)), theta__1(y) = -(1/2)*y+1/2, theta__2(y) = -(1/2)*y+1/2}

> 

plots:-display(
        plot(eval(u__1(y),dsol), y=-1..0, color=red),
        plot(eval(u__2(y),dsol), y=0..1, color=blue),
labels=[y, u(y)]);

> 

plots:-display(
        plot(eval(theta__1(y),dsol), y=-1..0, color=red),
        plot(eval(theta__2(y),dsol), y=0..1, color=blue),
labels=[y, theta(y)]);

> 

plot(eval(N(y),dsol), y=-1..1, color="Green", labels=[y, N(y)]);

 

 

Download mw2.mw

 

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