MaplePrimes Questions

I am trying to solve a diffusion equation with a potential term that has an integral in it. The equation has the following form: 

PDE := diff(g(x, t), t) = diff((beta(x, t)+diff(g(x, t), x)), x), 

with the function beta: 

beta := proc (x, t) options operator, arrow; int(exp(-abs(x-y))*g(y, t), y = -infinity .. +infinity) end proc

The boundary conditions for the function g(x,t) are simply assumed to be a zero-centered Gaussian in space (i.e. in x). So it is unity for x=0 and zero for the outer boundary that we can set as x=L. 

The problem is easily solved if the function beta is not an integral, but in the current form I get the following error: 
*******
Error, (in pdsolve/numeric/process_PDEs) inconsistent dependencies in PDEs: g(x, t) v.s. g(y, t)

*******

So it does not like the dummy variable in the function g.  

I can not write an additional PDE for beta because my Kernel is an exponential so the integral never goes away. Anyone with a way to solve this?

ADDENDUM: I have now copied the scriptPDE_DIFFUSION_INTEGRAL.mw
 

restart

L := 20; betaz := proc (x, t) options operator, arrow; int(exp(-abs(x-y))*g(y, t), y = 0 .. L) end proc

proc (x, t) options operator, arrow; int(exp(-abs(x-y))*g(y, t), y = 0 .. L) end proc

(1)

PDE := diff(g(x, t), t) = diff(-betaz(x, t)+diff(g(x, t), x), x)

diff(g(x, t), t) = -(int(-abs(1, x-y)*exp(-abs(x-y))*g(y, t), y = 0 .. 20))+diff(diff(g(x, t), x), x)

(2)

v__t := 1; v__d := 0; IBC := {g(0, t) = exp(-(0.-v__d)^2/v__t), g(L, t) = 0*exp(-(L-v__d)^2/v__t), g(x, 0) = exp(-(x-v__d)^2/v__t)}

{g(0, t) = 1., g(20, t) = 0, g(x, 0) = exp(-x^2)}

(3)

pds := pdsolve(PDE, IBC, numeric, time = 100, range = 0 .. L, spacestep = .1)

Error, (in pdsolve/numeric/process_PDEs) inconsistent dependencies in PDEs: g(x, t) v.s. g(y, t)

 

p0 := pds:-plot(t = 0, numpoints = 100, color = red); p1 := pds:-plot(t = 10, numpoints = 100, color = red); p2 := pds:-plot(t = 20, numpoints = 100, color = blue); p6 := pds:-plot(t = 60, numpoints = 100, color = blue); p5 := pds:-plot(t = 50, numpoints = 100, color = blue); p3 := pds:-plot(t = 30, numpoints = 100, color = blue); p4 := pds:-plot(t = 40, numpoints = 100, color = green); p7 := pds:-plot(t = 70, numpoints = 100, color = blue); p8 := pds:-plot(t = 80, numpoints = 100, color = black); p9 := pds:-plot(t = 90, numpoints = 100, color = blue); plots[display]({p0, p2, p4, p8})

NULL


 

Download PDE_DIFFUSION_INTEGRAL.mw

below. 

 

Hi everybody,

 

Written in Maple 2015
restart:
t := table([1=table([a=123]) ]):
save t, MyFile:
restart:
read MyFile:
t[1][a] := 321;

The answer is  t[1][a] =321 (here a double underscore)

Now I read MyFile from Maple 2018
restart:
read MyFile:
t[1][a] := 321;

The result is   (t[1])[a] = 321   (still a double undercsore but the first level is enclosed between parentheses).

For a more hierarchical table, all the entries but the deeper one are between parentheses.

Does this difference in representations mean something about the inner representation of a table ?

Thank you all

Hi everybody,

Maybe it's more a warning than a true question ?

I have written a rather heavy-duty code in Maple 2015. A few years ago I faced strong problems of increases in memory. After some investigations I found there came from a procedure P in which a loop realizes a few tenth of calls to  fsolve.
This same procedure  P was itself called a large number of times.

To try to  fix them I inserted a forget(fsolve, initialize=true) command after the fsolve command.
At the end this didn't prove to be very efficient and I decided to rewrite some part of the code in a more efficient way. Basically the  procedure  P is now called only a few times but its inner loop is executed about 200 hundred times.
This new version of the code no longer presents memory size problems, while being more efficient from a computational time point of view.

Today procedure  P still contains the forget(fsolve, initialize=true) command (I had forgotten to remove it)

Now I keep developping this same code under Maple 2018.
The Maple 2015 and Maple 2018 versions both return the same results, but the procedure  P runs in about 20 times the time it runs in Maple 2015 (40 seconds instead of 2)
This comes from the 200 forget(fsolve, initialize=true) calls which consumme about  38 seconds.

With Maple 2015 these same 200 calls to forget(fsolve, initialize=true) only consumme 0.5 second:


As a cure I remove the forget(fsolve, initialize=true) command, plain and simple.

But maybe this misadventure could reveal an unseen behaviour of Maple 2018 ?

 

Hi

I'm using solve,and i want to quantify the dimensions of the solution spaces of the output. For example

solve([a+b, c+b])

produces a singular 1 dimensional object

solve([a+b, -b^2+d^2])

produces 2 objects with dimension 2

EDIT:
my intuition is that the simplest way of doing this is to create a counter for equations of the form
variable=variable
and to run it on each of the lists that solve might produce- so far this kind of thing is beyond me

 

I am trying to solve an equation with respect to the variable w. However, although there seems to be a solution (see plot indicating a root), Maple produces the wron solution (0, when P = 1/2 and d = 1/10):


 

equations := S-P*S*((ln(-2*d*w+P+d)-ln(P-d))*(P-d)/(-1+w)+(P+d)*(ln(-P-d)-ln(d*(2*w-1)-P))/w)/(2*d^2) = 0

S-(1/2)*P*S*((ln(-2*d*w+P+d)-ln(P-d))*(P-d)/(-1+w)+(P+d)*(ln(-P-d)-ln(d*(2*w-1)-P))/w)/d^2 = 0

(1)

variables := w

w

(2)

solutions := solve(equations, variables)

(1/2)*(P+d-exp(RootOf(-P^2*exp(_Z)*ln(1/(P+d))-P*d*exp(_Z)*ln(1/(P+d))+P^3*ln(1/(P+d))-P*d^2*ln(1/(P+d))-ln(P-d)*P^2*exp(_Z)+ln(P-d)*P*d*exp(_Z)-2*P*d*exp(_Z)*_Z+ln(P-d)*P^3-ln(P-d)*P*d^2+d*(exp(_Z))^2-2*P*d*exp(_Z)+P^2*d-d^3)))/d

(3)

eval({solutions}, [P = 1/2, d = 1/10])

{3-5*exp(RootOf(-25*(exp(_Z))^2+75*exp(_Z)*ln(5/3)+50*ln(2/5)*exp(_Z)+25*exp(_Z)*_Z+25*exp(_Z)-30*ln(5/3)-30*ln(2/5)-6))}

(4)

evalf({%})

{{0.}}

(5)

plot(eval(S*(-P*(-2*d*w+P+d)*ln(-2*d*w+P+d)+P*w*(P-d)*ln(P-d)-(-1+w)*(P*(P+d)*ln(P+d)-2*d^2*w))/(2*d^2*(-1+w)*w), {P = 1/2, S = 1, d = 1/10}), w = 0 .. 1)

 

``


 

Download MaplePrimes_03072018.mw

This pde used to be solved in 2018 as far as I know. Now it gives a strange new error

restart;
interface(showassumed=0);
pde :=  diff(u(x,t),t)+k*diff(u(x,t),x$2)+sin(2*Pi*x/L);
ic  :=  u(x,0)=f(x);
bc  :=  D[1](u)(0,t)=0, D[1](u)(L,t)=0;
sol :=  pdsolve({pde,ic,bc},u(x,t)) assuming L>0,t>0,k>0;

Error, (in assuming) when calling 'dsolve'. Received: 'found differentiated functions with same name but depending on different arguments in the given DE system: {F0_0(L), F0_0(x)}'

I am using  Physics:-Version();     MapleCloud version: 72

Do others get this erorr? Why does it show up now when it worked OK before?

update

Iam running on Linux. Here is screen shot

Hello, can someone help me please?

I have to find the minimum (x,y) over the domain [0;2Pi] of the following function

f(x,y):=1+8 cos(1/2 x-1/2 y) cos(1/2 x) cos(1/2 y) which i plotted to have an idea where its minimum is located.

plot3d(f(x,y),x=0..2Pi,y=0..2Pi)

 

I tried to use the command  ' minimize(f(x,y)) '

 

Thank you in advance,

best regards.

 

 Ps: the function normaly attains its minimum at (2Pi/3,-2Pi/3) and (-2Pi/3,2Pi/3).

Hello dear users!

Hope you would be fine. I want to fine the roots of the following cubic equation

u^3+u*d[1]+d[0];

when the discriment is zero, positive and negative. I am waiting your positive response. Thanks

 

@acer @Carl Love @Kitonum @Preben Alsholm

I want to calculate and reproduce this question  in Maple.

with(Optimization):

f := (x, y) -> op(1, NLPSolve(sin(a*x*y), a = 1 .. 5));

int(`if`(f(x, y) > 0, 1, 0), [x = 0 .. 1, y = 0 .. 1], numeric);

# 0 

,but I FAIL.

It should give me: 0.922105

 

Thanks.

I have three columns of data (real numbers) in Excel that have about one thousand rows and I wish to plot their relationships in Maple. The source data simply comprises three columns of numbers and so, are not ordered in triples, [x,y,z]. Does anyone know of a simple routine available that allows me to import the raw data and output a 3-D wire / solid plot in Maple?

Thank you! 

a1:=odeplot(p,[t,x(t)],0..5,color=black,thickness=2):

a2:=plot((2/125)*t^3-(3/25)*t^2+t+1,t=0..5,color=red,style=point,thickness=4):
display(a1,a2,labels=[x1,x2]);
Error, (in plots:-display) expecting plot structure but received: a1
 

I have an explicit mathematical expresssion in the form

m = f[1](x,y)*A + f[2](x,y)*B + f[3](x,y)*C

I wonder if it is possible to use the collect function on the vector z = [A,B,C]^T and write the resulting expression as the dot product

m = f^T.z 

Basically, I am interested in obtaining the vector f = [f[1](x,y),f[2](x,y),f[3](x,y)]^T as a Maple variable in closed-form.

 

Thank you in advance for your help.

 

Marco

I'm referancing the maple help page: https://www.maplesoft.com/support/help/maple/view.aspx?path=Groebner in my thesis, but I am not sure what fields to include, and what to put in them- particularly for the year.

I'm a newbie.I am trying to solve a differential equation symbolically.But I'm having  a\some problem.

First 3 lines works fine....but 4th line gives no output...

Secondly How I'm supposed to simplify the output of 3rd line

 

1.>ode := P*(diff(y(x), x, x))+diff( K(x)*(diff(y(x), x, x)), x, x ) = 0

2.>ics := y(0) = 0, y(L) = 0, ((D@@2)(y))(0) = 0, ((D@@2)(y))(L) = 0

3.>dsolve(ode, y(x))

3rd lines output--y(x) = _C2*x+_C1+Int(DESol({diff(_Y(x), x, x)+2*(diff(K(x), x))*(diff(_Y(x), x))/K(x)+(P+diff(K(x), x, x))*_Y(x)/K(x)}, {_Y(x)}), [x, x]) 

 

4.>dsolve({ics,ode},y(x))

5.>

4th lines gives no output.After hitting "Enter",I get a new line in my Worksheet(line 5).Why..can anyone explain please?

 

I want to match   a*x^b only if "a" and "b" are both not the symbol "y"

Therefore 5*x^y should fail to match, but this below matches.

restart;
expr:=5*x^y;
if patmatch(expr,conditional(a::anything*x^(b::anything),(a<>y and b<>y)),'la') then
   print("it matched, why??");
   map(assign,la):
   print("a = ",a, " b=",b);
fi;

it gives

                      "it matched, why??"
                      "a = ", 5, " b=", y

What did I do wrong? is my conditional wrong above?

 

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