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Rouben Rostamian

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19 years, 34 days
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Try a simpler problem first...

Rouben Rostamian 8501
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 8501
October 15 2021
0 0

@vv That's very clever.  Vote up!

What is optimal?...

Rouben Rostamian 8501
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 8501
September 30 2021

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

 

Good example...

Rouben Rostamian 8501
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 8501
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 8501
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 8501
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 8501
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 8501
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

 

It's better done in your head...

Rouben Rostamian 8501
September 18 2021
0 0

You wrote:

Is there a way to get the function f(x) to adhere to the rules imposed on it by the text in the book? If that can be done, the answering of the questions will be a lot easier!

Sure, it will be a lot easier with Maple, but that's not the goal—there is nobody out there holding their breath waiting for your answer.  The point of the exercise is to process it in your head so that it becomes a second nature to you.

That reminds me of the athlete who considered riding a motorcycle along the length of the 10 km running track as sufficient training.  He was able to reach the end of the course.  Wasn't that the goal?

Nice analysis!...

Rouben Rostamian 8501
August 17 2021
0 0

@Preben Alsholm Hello again, Preben.  Your calculations and their explanations are insightful and thorough.  You have extracted more juice from this problem than I had thought possible.

I have been late in responding to your posts due to an overload of tasks that need immediate attention.  I have made a bookmark of this page so that I may return to it to read your analysis in a more leisurely manner.

 

@Preben Alsholm Hello Preben, thank...

Rouben Rostamian 8501
August 13 2021
0 0

@Preben Alsholm Hello Preben, thanks for your remarks. I will be away from my computer this evening.  Will go through the details of your post tomorrow.

Linear vs nonlinear...

Rouben Rostamian 8501
August 12 2021
0 0

@dtn "Linear" and "nonlinear" are very specific technical terms.  Merely calling something "nonlinear" does not make it nonlinear. I am afraid that you have a malformed idea of what a nonlinear differential equation means.

Just about every elementary book on differential equations has a section/chapter on linear first order equations.  These are equations of the form x'(t) = p(t) + q(t)*x(t).  Your equation falls in that category, that is, it is a linear first order equation.

The article that you have cited provides conditions under which a nonlinear equation may be approximated by a linear equation. You don't have a nonlinear equation, therefore the methods of that article do no apply in your case.

 

What do you mean by linear and nonlinear...

Rouben Rostamian 8501
August 11 2021
0 0

@dtn One linearizes a nonlinear equation. Your equation is already linear. I can make no sense of your request of linearizing a linear equation.

I suggest that you attempt to solve your problem the best you can—after all, you must have read and understood the article that you have cited—and then post the Maple worksheet of your attempt.  That will tell in less vague terms what it is that you are trying to achieve, and people will then provide advice if you need help with Maple.

 

 

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