MaplePrimes Questions

I am trying Maple 2017 pdsolve for heat PDE in 1D. It seems Maple can solve now  heat PDE with homogeneous dirichlet boundary conditions (good). But when I set the boundary conditions to homogeneous neumann B.C. instead, I get an answer when this B.C. is prescribed to the left side. When this B.C. is on the right side, I get an error. Which is strange.

I am newbie in Maple, so may be I am doing something wrong in the syntax?  In addition, the answer I get when homogeneous neumann B.C. is on the left side, does not match my hand solution, which I know is correct. I'll show this below.

First, here is the case where it works. homogeneous dirichlet boundary conditions on both sides:

restart;
pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2);
bc:=u(0,t)=0,u(L,t)=0;
sol:=pdsolve([pde,bc]) assuming 0<L:

This answer is correct. Now when setting the right side to homogeneous neumann B.C. I get an error

restart;
pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2);
bc:=u(0,t)=0,D[1](u)(L,t)=0;
pdsolve([pde,bc]) assuming 0<L;

I think may be it does not like `L` there in the B.C. But how else to tell it this B.C.? The above is the only syntax I know. And finally, when using homogeneous neumann B.C. on the left side, I get this result

restart;
pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2);
bc:=D[1](u)(0,t)=0,u(L,t)=0;
pdsolve([pde,bc]) assuming 0<L;

The correct answer for this B.C. is

The answer should be series solution as well with eigenvalues. I think if I expand Maple solution in series may be I will get it to match my hand solution. I need to look at this more later.

my question is: Why do I get an error when homogeneous neumann B.C. is on the right side but not on the left side?

I suspect I am not entering the B.C. correctly? If so, How does one enter homogeneous neumann B.C. for this 1D heat PDE?

 

how can I sort a diff EQ by its degree.  Example 4th order with 4 being farthest left in normal reading format?

 

D4 + D3 + D2 +D + C = F(t)

 

my equation display is currently:

C + D + D2.......

Thanks,

Bill

Can I force Maple to solve the 2 variable recurrence from this page:

https://mathoverflow.net/questions/182188/limiting-probabilities-for-two-player-game-drawing-random-uniform-numbers?rq=1

How can I do that? Can I obtain a closed form from Maple?

I'm not quite sure how I should put this and the title might be a bit misleading. Basicly when I use variables which I have defined with a value I would like for the equation to be shown with the values of the variables before the results. Not sure if that made sense so I made this picture.

Basicly when I write as in line (1) it just gives me the result, which is fine but I'd like it converted too if that makes any sense as in line (2). I made line (2) myself (just changed the colour of standard mathtype). I'd like the calculations to be shown as in line (2) where first you see the formula: lambda=v/f, then it inserts the values of the variables and finally calculates this. Is there any way to do this?

I couldn't really find anything in the manual or using google.

Regards, Simon

 

  Assume that one has an equation P(a,b,c,d,e,f)=0. How do we solve this equation so that the parameters b and c are in terms of a, and the parameters d and e in terms of f?

hello . how can i get 7 given parameters(b1,a1,b1,a2,b2,.....) in this equation with maple. thanks

 

 

 

1.mw

Imagine we have an ODE system 

odeSys := {diff(x(t),t$2)+diff(x(t),t)+x(t)=f(t),diff(y(t),t$2)+2*diff(y(t),t)+3*y(t)=g(t)};

It is easy to transform this system into a first order form by hand. But for larger systems, the procedure by hand becomes very error prone. Is there an intelligent way to transform a system of n scalar ODEs (order m) into a first order system? I know that the first order form is not unique. It is only important to reduce the system to a system of first order equations.

 

Hi,

this forum has helped me a lot since I'm quite new at maple. There are lot of questions about the solution of _Z in roots but I still can't figure it out.

Here my problem:

f(x,y)= x*y - (V^(alpha)*W)/(V-a*x-b*y)^alpha

x>0, y>0, a>0, b>0, alpha>0, V>0, W>0, V-a*x-b*y>0

I need to find the maximum values for x and y. But all the solutions have got _Z.

How can I eliminate _Z? Thanks!

Best regards,

Won

 

I am learning pdsolve in Maple. When I try to solve a diffusion pde, I get this strange error message, and I do not understand what it means: This is using Maple 2017 on windows 7

restart;
bc:=f(t,0)=0,f(t,1)=1;
ic:=f(0,x)=piecewise(x=0,1,0);
pde:=diff(f(t,x),t)=diff(f(t,x),x$2);
pdsolve({pde,bc,ic},f(t,x));

The error is

Error, (in casesplit/K) this version of casesplit is not yet handling the function: piecewise

Am I writing the initial conditions (ic) wrong?  Maple help shows nothing on this. I think Maple does not like my initial conditions. But do not know how to correct it now.

What causes this error?

 

 

Hello.

I am asked to find all the candidate points to absolute maximum or minimum in a domain delimited by several lines or curves.

Let me show you an example:

I am having trouble trying to find the candidate points on the boundary of the domain with LagrangeMultipliers.

How should I express the line or curve equations so that I can use them as constraints of LagrangeMultipliers? Should I use another method?

Thanks a lot.

That the Bernoulli numbers has z/(exp(z)-1)) as an exponentional generating functions is of course very well known. I am wondering why the gfun package doesn't seem to be able to work this out? Naive approaches like

gfun[guessgf]([seq(bernoulli(n),n=0..100)],z);

give FAIL. I would expect gfun to be able to transform the given sequence corresponding to replacing A(z) by 1/A(z), easily recognize (exp(z)-1))/z, and transform back. Perhaps the package is in need of maintenance?

Any help would be much appreciated.

Best,

Soren

 

 

I want to solve a system of algebraic equations in maple using as follows:

 

equations:={b[1](2)*a[1](2, 1) = 1/2, b[1](2)*a[1](2, 1)^2 = 1/3, b[1](2)*a[2](2, 1) = 1/2, b[1](2)*a[2](2, 1)^2 = 1/3, b[2](2)*a[1](2, 1) = 1/2, b[2](2)*a[2](2, 1) = 1/2, b[1](2)*(a[3](2, 1)+a[3](2, 2))*a[1](2, 1) = 1/3, b[1](2)*(a[3](2, 1)+a[3](2, 2))*a[2](2, 1) = 1/3, b[1](2)*a[2](2, 1)*a[1](2, 1) = 1/3, b[1](1)*(a[3](1, 1)+a[3](1, 2))+b[1](2)*(a[3](2, 1)+a[3](2, 2)) = 1/2, b[1](1)*(a[3](1, 1)+a[3](1, 2))^2+b[1](2)*(a[3](2, 1)+a[3](2, 2))^2 = 1/3, b[2](1)*(a[3](1, 1)+a[3](1, 2))+b[2](2)*(a[3](2, 1)+a[3](2, 2)) = 1/2, b[1](1)+b[1](2) = 1, b[2](1)+b[2](2) = 1}

variables:={a[1](2, 1), a[2](2, 1), a[3](1, 1), a[3](1, 2), a[3](2, 1), a[3](2, 2), b[1](1), b[1](2), b[2](1), b[2](2), b[3](1), b[3](2)}

 

solve(equations,variables);

 

But I get an error mesage that the equations is not valid equation or expression.

Any help?

 

Let

z := Diff(x(t),t)*y(t) + x(t)*Diff(y(t),t);

Is there a way to tell Maple to collapse that into Diff(x(t)*y(t), t) ?

I tried factor, combine, simplify, but none of them worked.

 

 

Hi all,

 

I am looking for a boolean logic check to see if a rational expression is simplified.

For example: x/x^2  =  1/x  I want something along the lines of issimplified(x/x^2)=FALSE

Similarily, (x^2-x-12)/(x-4) = x+3  so I would like some logic test to say (x^2-x-12)/(x-4) is NOT simplified.

 

Thanks in advance,

Mark

[[1000, 20], [2000, 25], [3000, 24], [4000, 23], [5000, 24]];
  [[1000, 20], [2000, 25], [3000, 24], [4000, 23], [5000, 24]]
data1 := [[1000, 20], [2000, 21], [3000, 32], [4000, 23], [5000, 23]]; 'data1';
                             data1
a*x^3+b*x^2+c*x+d;
                        3      2          
                     a x  + b x  + c x + d
x;
                               x
Equn1 := CurveFitting[LeastSquares]([[1000, 20], [2000, 25], [3000, 24], [4000, 23], [5000, 24]], x, curve = a*x^3+b*x^2+c*x+d);
plot(Equn1,x= 1000..5000)


 

Least Squares Approximation

 

 

Calculate a least squares approximation using specified data points.

 

 

Theoretical Curves for the Two-Stroke Engines and Four-Stroke Engines Brake Power Vs Brake Efficiency

List of Data Points:

[[1000, 20], [2000, 25], [3000, 24], [4000, 23], [5000, 24]]

[[1000, 20], [2000, 25], [3000, 24], [4000, 23], [5000, 24]]

(1)

data1 := [[1000, 20], [2000, 21], [3000, 32], [4000, 23], [5000, 23]]; 'data1'

data1

(2)

Fitting Curve:

a*x^3+b*x^2+c*x+d

a*x^3+b*x^2+c*x+d

(3)

Independent Variable:

x

x

(4)

Least Squares Curve:

Equn1 := CurveFitting[LeastSquares]([[1000, 20], [2000, 25], [3000, 24], [4000, 23], [5000, 24]], x, curve = a*x^3+b*x^2+c*x+d)

plot(Equn1,x= 1000..5000)

 

 

 

 

NULL

Equn1

31/5+(83/4200)*x-(23/3500000)*x^2+(1/1500000000)*x^3

(5)

 

 

Least Squares Fit of Data by a Specified Curve

List of Data Points:

[[3, -1], [5, 3], [6, -7], [7, 5], [9, -2]]

[[3, -1], [5, 3], [6, -7], [7, 5], [9, -2]]

(6)

Fitting Curve:

a*x^2+b*x+c

a*x^2+b*x+c

(7)

Independent Variable:

x

x

(8)

Least Squares Curve:

CurveFitting[LeastSquares]([[3, -1], [5, 3], [6, -7], [7, 5], [9, -2]], x, curve = a*x^2+b*x+c)

-901/210+(213/140)*x-(11/84)*x^2

(9)
 

 

a*x^2+b*x+c

a*x^2+b*x+c

(10)

plot(sin(x), x = 0 .. 4*Pi)

 

``


 

Download LeastSquareApproximation_2nd_and_3rd_Order.mw

The above command plots one curve alright. I want four such curves to go in the same figure using command like

plot(Equn1,Equn2,Equn3,Equn4,view(x=1000..5000)

I am not getting by the above command what I want. Can any one help. A shortcut method is required for me to repeat many times.

Thanks for help.

Ramakrishnav V
 

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