nm

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These are questions asked by nm

What is the correct way to specify region where solution of a PDE is needed? For example, I am trying to verify my hand solution to this HW problem: Solve Poisson PDE in 2D 

The above is the only information given in the textbook. So it is only in the upper half plane. If I type this

interface(showassumed=0);
pde := VectorCalculus:-Laplacian(u(x,y),[x,y])=-1/(1+y);
bc:=u(x,0)=0;
pdsolve([pde,bc],u(x,y)) assuming y>0

Maple gives 

But I get by hand using method of images is

So not exactly the same. I think I made mistake in my solution. But I am also not sure that just saying "assuming y>0" is doing what I think it is supposed to do. For example, suppose we want to solve the same PDE say in the first quadrant. Typing

pdsolve([pde,bc],u(x,y)) assuming y>0,x>0

Gives same solution. But the solution should be different. And typing

pdsolve([pde,bc],u(x,y)) 

Gives same answer as well.

So I think I need another way to tell Maple the region of the solution. i.e. I need to tell Maple to use the Laplacian for the upper half plane only and not the Laplacian in the whole 2D space.

Any suggestions what to do and how to handle such problems?

Thank you

Hello;

What trick if any is needed to obtain zero for this sum in Maple?

sum(sin(Pi*n/2)*sin(n*Pi*(x + 1)/2)*cos(n*Pi*t/2),n=1..infinity)

The above sum, according to Mathematica is zero. I am trying to see if same result can be obtained by Maple in order to verify this result. It is possible ofcourse that Mathematica result is not correct. I am also trying to verify the sum is zero by hand, but no success so far.

mySum:=sum(sin(Pi*n/2)*sin(n*Pi*(x + 1)/2)*cos(n*Pi*t/2),n=1..infinity)

sum(sin((1/2)*Pi*n)*sin((1/2)*n*Pi*(x+1))*cos((1/2)*n*Pi*t), n = 1 .. infinity)

value(mySum)

sum(sin((1/2)*Pi*n)*sin((1/2)*n*Pi*(x+1))*cos((1/2)*n*Pi*t), n = 1 .. infinity)

simplify(mySum)

sum(sin((1/2)*Pi*n)*sin((1/2)*n*Pi*(x+1))*cos((1/2)*n*Pi*t), n = 1 .. infinity)

 


Here is Mathematica result

Using Maple 2019.2 on windows 10. 

Thanks

Download q2.mw

This is another problem I just found in Maple 2019.2 on windows 10. professional.

I wanted to close Maple, so did  File->Exit 

 

But Maple did nothing. It did not close.  Also Alt-F4 did not close Maple. I had to click on the little X on top right corner of the open window to close Maple.  

In earlier version this used to work to close Maple.

Do others see this as well?  To reproduce, simply start Maple, and do File->Exit.

Here is a movie also

 

Maple 2019.2.

These two expressions are mathematically equivalent:

But simplify(expr1-expr2) does not give zero where simplify(convert(expr1,trig)-expr2) does.

Is this normal behavior or can be expected sometimes? As a user I would have expected Maple internally to figure all of this itself. Compare to Mathematica:

Is there a different command in Maple that will show mathematical equivalence of two expressions to try other than simplify?

Thanks
 

restart;

expr1:=(-exp(n*Pi*(2*b - y)/a) + exp(n*Pi*y/a))/((exp(2*n*Pi*b/a) - 1)):
expr2:= sinh(n*Pi/a*y)/tanh(n*Pi/a*b)-cosh(n*Pi/a*y):
simplify(expr1-expr2);

(-sinh(n*Pi*b/a)*exp(n*Pi*(2*b-y)/a)+(-sinh(n*Pi*y/a)*cosh(n*Pi*b/a)+cosh(n*Pi*y/a)*sinh(n*Pi*b/a))*exp(2*n*Pi*b/a)+(exp(n*Pi*y/a)-cosh(n*Pi*y/a))*sinh(n*Pi*b/a)+sinh(n*Pi*y/a)*cosh(n*Pi*b/a))/((exp(2*n*Pi*b/a)-1)*sinh(n*Pi*b/a))

simplify(convert(expr1,trig)-expr2);

0

 


 

Download q.mw

 

from help, it says about option of 'spacestep'=numeric to pdsolve numerical solver the following

Specifies the spacing of the spatial points on the discrete mesh on which the solution 
is computed, and defaults to 1/20th of the spatial range of the problem, 
r-L, where L is the left boundary, and r is the right. Note: The spacing must 
be small enough that a sufficient number of points are in the spatial domain 
for the given method, boundary conditions, and spatial interpolation (see below). 
If the given value of spacestep does not fit into the input domain an integral 
number of times, the closest smaller step size that does is used. For problems 
using error estimates, or an adaptive approach, the total number of spatial 
points must be odd, so again, if this is not the case for the specified 
spacestep, then it is reduced to satisfy this requirement.

In this problem, the domain is [-1,1]. So the default is 1/20 of this which is 1/10.  Why is it when using spacestep smaller than 1/11, the animator generated from the solution does not work right?  i.e. playing the animate does not produce correct result as the case when using spacestep=1/11 or spacestep=1/10?   

When using spacestep=1/16 or 1/15 or 1/14 or 1/13 or 1/12, it all produce bad animation.  I do not see why that is. 

Any ideas?  Am I doing something wrong?
 

pde := diff(u(x,t),t$2)=diff(u(x,t),x$2):
bc  := u(-1,t)=u(1,t),D[1](u)(-1,t)=D[1](u)(1,t):
f:=x->piecewise(-1/2<x and x<0,x+1/2,0<x and x<1/2,1/2-x,true,0):
plot(f(x),x=-1..1):
ic  := u(x,0)=f(x),D[2](u)(x,0)=0:
sol:=pdsolve(pde,{bc,ic},u(x,t),numeric,time=t,range=-1..1,  timestep=1/16,spacestep=1/16)

_m1793827573184

sol:-animate(t=1,frames=50,title="time = %f");

sol:=pdsolve(pde,{bc,ic},u(x,t),numeric,time=t,range=-1..1,  timestep=1/16,spacestep=1/12):
sol:-animate(t=1,frames=50,title="time = %f");

sol:=pdsolve(pde,{bc,ic},u(x,t),numeric,time=t,range=-1..1,  timestep=1/16,spacestep=1/11):
sol:-animate(t=1,frames=50,title="time = %f");

 

 

Download solving_wave_pde_2.mw

 

 

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