This is my original question, but I can't ask, so I put it here.
I have a system of pde, and it's hard to solve.
But I want to find the type of solutions which are polynomials of degree n of the several independent variables.
Is it possible using 'pdsolve'? Or there are some other commands in Maple to do that?
Thanks!
Download PDE_and_BC_Example_with_RootOf_eigenvalues.mw
Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft
Uxx + Uyy =0
y is less than Pi,x is greater than 0
B.c are u(0,y)=0 , u(Pi,y)=sinh*Pi*cosy
u(x,0)=sinx , u(x,Pi)=-sinhx
Any one can help me to solve the differential equations using maple to get the velocities u ,v and pressure p for the problem mentioned below
This is my program, I need to solve that differential equation but when i click on the pdsolve I don't get any result, do you know why?
thank you
This code used to work OK in an earlier version of Maple 2018. I do not know what changed and if it some change in Physics package that caused it.
restart; pde:=diff(u(r,theta),r$2)+1/r*diff(u(r,theta),r)+1/r^2*diff(u(r,theta),theta$2)=0; bc:=u(r,0)=0,u(r,Pi)=0,u(0,theta)=0,u(1,theta)=f(theta); sol:=pdsolve([pde,bc],u(r,theta),HINT = boundedseries(r = 0));
Now it gives the error
Error, (in assuming) when calling 'assume'. Received: 'invalid arguments'
Workaround is to remove the HINT above. So the following now works
restart; pde:=diff(u(r,theta),r$2)+1/r*diff(u(r,theta),r)+1/r^2*diff(u(r,theta),theta$2)=0; bc:=u(r,0)=0,u(r,Pi)=0,u(0,theta)=0,u(1,theta)=f(theta); sol:=pdsolve([pde,bc],u(r,theta));
Should this error message show up when using the HINT option?
Maple 2018.2 with Physics package cloud version 205.
Here is screen shot showing it was working in Physics 170
restart;
PDE := diff(y(x,t), x,x,x,x)+(diff(y(x, t), t,t))=0;
# Initial/boundary conditions
BCs:=y(0,t) = 0, y(1,t) = 0,D[1](y)(0,t)=0,D[1](y)(1,t)=0;
ICs:=y(x,0) =0, D[2](y)(x,0)=1 ;
num_solution := pdsolve(PDE, {BCs,ICs}, numeric);
num_solution :-plot3d(x=0..1, t=0..1);
1st Question:How to plot the 2D at point x=0.5?
2nd Question: How to use PDEplot?
restart;
#PDE
PDE := (diff(u(x,y), x,x))-1/x*(diff(u(x, y), x)) -x^2*(diff(u(x,y), y,y))= 2;
BCs:=u(0,t) = 0, u(1,t) = 0;
ICs:=u(x,0) =0, D[2](u)(x,0)=2 ;
num_sol := pdsolve(PDE, {BCs,ICs}, numeric);
BCs := u(0, t) = 0, u(1, t) = 0;
ICs := u(x, 0) = 0, D[2](u)(x, 0) = 2;
num_sol := pdsolve(PDE, {BCs,ICs}, numeric);
#ODE
ODE:=diff(y(x),x,x)+16*diff(y(x),x)=0;
BCs:=y(0) =0, y(2)=3 ;
ode_num_sol:=dsolve({ODE,BCs},numeric);
Suppose that we solved numerically an ODE or PDE like above.
How to calculate values of function y or u at special values x like analytical solutions?
For example; u(1,1)=? y(1)=? etc.
Trying Maple on a textbook problem to verify my hand solution.
But Maple pdsolve hangs with the mserver.exe tallomg almost 100% CPU and over 10 GB of RAM!
I waited for almost 20 minutes. Tried another time, same thing.
It is no problem if Maple can't solve this, but Maple seems to suffer from too many hangs when it is not able to solve a problem. I've had similar problems with dsolve also.
This is on windows 10, 64 bit with Maple 2018.2 With Physics updates version 218
restart;
PackageTools:-IsPackageInstalled("Physics Updates");218
interface(showassumed=0);
infolevel[pdsolve]:=2;
pde := diff(u(x,t),t)=k*diff(u(x,t),x$2)+Q(x,t);
ic := u(x,0)=f(x);
bc := eval(diff(u(x,t),x),x=0)=A(t),eval(diff(u(x,t),x),x=1)=B(t);
sol:=pdsolve({pde,ic,bc},u(x,t)) assuming t>0,k>0;then
* trying method "_Fn" for 2nd order PDEs -> trying "linear_in_xt" -> trying "BC_equal_0" * trying method "_Cn_cn" for 2nd order PDEs Trying travelling wave solutions as power series in tanh ... Trying travelling wave solutions as power series in ln ... Trying travelling wave solutions as power series in tanh ... Trying travelling wave solutions as power series in ln ... * trying method "Wave" for 2nd order PDEs -> trying "Cauchy" -> trying "SemiInfiniteDomain" -> trying "WithSourceTerm" * trying method "Heat" for 2nd order PDEs -> trying "SemiInfiniteDomain" -> trying "WithSourceTerm" * trying method "Series" for 2nd order PDEs -> trying "ThreeBCsincos" -> trying "FourBC" -> trying "ThreeBC" -> trying "ThreeBCPeriodic" -> trying "WithSourceTerm" * trying method "_Fn" for 2nd order PDEs -> trying "linear_in_xt" -> trying "BC_equal_0" * trying method "_Cn_cn" for 2nd order PDEs
And here is hangs. Notice that because both ends are Neumann, there is no unique solution to this problem. So the solution will contain arbitrary constant. May be this is what made pdsolve hang?
No it is not. Trying with only one end nonhomogeneous Neumann, and the other end Dirichlet, it still hangs. The problem seems to be with one end is nonhomogeneous Neumann, which is a function of time. So this hangs also (same place)
restart;
interface(showassumed=0);
infolevel[pdsolve]:=2;
pde := diff(u(x,t),t)=k*diff(u(x,t),x$2)+Q(x,t);
ic := u(x,0)=f(x);
bc := eval(diff(u(x,t),x),x=0)=A(t),eval(diff(u(x,t),x),x=1)=0;
sol:=pdsolve({pde,ic,bc},u(x,t)) assuming t>0,k>0;And this also
restart;
interface(showassumed=0);
infolevel[pdsolve]:=2;
pde := diff(u(x,t),t)=k*diff(u(x,t),x$2)+Q(x,t);
ic := u(x,0)=f(x);
bc := eval(diff(u(x,t),x),x=0)=sin(t),eval(diff(u(x,t),x),x=1)=0;
sol:=pdsolve({pde,ic,bc},u(x,t)) assuming t>0,k>0;But this does not hang
interface(showassumed=0);
infolevel[pdsolve]:=2;
pde := diff(u(x,t),t)=k*diff(u(x,t),x$2)+Q(x,t);
ic := u(x,0)=f(x);
bc := eval(diff(u(x,t),x),x=0)=1,eval(u(x,t),x=1)=0;
sol:=pdsolve({pde,ic,bc},u(x,t)) assuming t>0,k>0;The issue seems to be when one end is nonhomogeneous Neumann which is function of time.
Is there a workaround so it does not hang? The complaint is that Maple hangs, and not that Maple unable to solve the PDE.
Using Latest Physics updates (I am not sure when this started), pdsolve gives Error, (in PDEtools/eval/2) numeric exception: division by zero on the following problem from a HW from text book.
restart;
PackageTools:-IsPackageInstalled("Physics Updates");"220"
pde:=diff(w(x,t),t)+3*t*diff(w(x,t),x)=w(x,t);
ic:=w(x,0)=f(x);
sol:=pdsolve([pde,ic],w(x,t));
Error, (in PDEtools/eval/2) numeric exception: division by zero
Mathematica answer btw is
pde = D[w[x, t], t] + 3 t D[w[x, t], x] == w[x, t];
ic = w[x, 0] == f[x];
sol = Simplify[DSolve[{pde, ic}, w[x, t], {x, t}]]
This is on Maple 2018.2 on windows 10 64 bit.
Any idea what is causing this and any workaround? Do others get the same exception?
This is a Wave PDE inside disk. with fixed edge of disk, and no theta dependency. initial position and velocity given.
When using "c^2" for the wave propagation speed, Maple only gives solution when also using assumptions to tell that c is positive.
restart; Physics:-Version()[2]; `2018, November 28, 1:35 hours, version in the MapleCloud: 224, version installed in this computer: 224`
And the PDE is
pde := diff(u(r, t), t$2) = c^2*( diff(u(r,t), r$2) + (1/r)* diff(u(r,t),r) ) ; ic:=u(r,0)=1, eval( diff(u(r,t),t),t=0)=r/3; bc:=u(1,t)=0; sol:=pdsolve([pde, ic,bc], u(r, t)) assuming t>0;
sol:=()
Now adding assuming c>0, or "c::positive", or "c>=0", it solves it
restart; pde := diff(u(r, t), t$2) = c^2*( diff(u(r,t), r$2) + (1/r)* diff(u(r,t),r) ) ; ic:=u(r,0)=1, eval( diff(u(r,t),t),t=0)=r/3; bc:=u(1,t)=0; sol:=pdsolve([pde, ic,bc], u(r, t)) assuming t>0,c>=0;
Any idea why the assumption "c>0" is needed when the speed is given as "c^2" ? I also tried assumption "c::real", but it still did not solve it. It seems related to Maple using Laplace transform to solve it.
It seems to me the assumption c>0 should not needed here. But if it is, I'd like to learn why.
Maple 2018.2, windows 10
Update
Thanks to comment below
" so u(r, 0)=1 (for all r), and u(1, t)=0 (for all t). So what is u(1,0)???"
I should have added "r>0,r<1". The reason I did not, is that I copied the code to solve this from Mathematica. In Mathematica, it did not need this assumption to solve it. It seems to depend on the method of solution used by Mathematica vs. Maple to cause this difference.
But now I see it helps to have it there. With the above assumption, now, there is no need to do any assumption on "c" at all and it gives solution. adding c>0 or t>0 or c>=0, makes no change now.
restart; pde := diff(u(r, t), t$2) = c^2*( diff(u(r,t), r$2) + (1/r)* diff(u(r,t),r) ) ; ic:=u(r,0)=1, eval( diff(u(r,t),t),t=0)=r/3; bc:=u(1,t)=0; sol:=pdsolve([pde, ic,bc], u(r, t)) assuming t>0,r>0,r<1;
Hello,
I have this problem with pdsolve , I need to know the value of dTf(x,y)/dy, but I knew just the value of Tf
Pleas find attached the code
Thank you
Hello, i am trying to solve the PDE
(E*I) d4/dx4 [w(x,t)] + (rho*A) d2/dt2 [w(x,t)]=0
where w(x,t) = W(x) * cos (ω *t)
W(x) is the amplitude of the beam, the PDE is solved for W(x) by seperation of variables.
my first question is how to simplify the equations by replacing the e^ terms with alpha as done in the lecture, so that things look cleaner. And the next question is how do i proceed to input the Boundary conditions to solve the problem? I have looked in quite a lot of places but there is no example of a second or third order boundary condition(if that makes sense)
Hello,
given is the lower, partial differential equation. I have solved this (hopefully) correctly. I was given the tip to do in Maple a simple control in which the solution is inserted into the original equation. Since I'm quite inexperienced in maple, I would like to ask you for help on how I do that best!
Thanks a lot!
Basis equation:
Solution:
I've been using the following syntax to set boundary condition which is a derivative, when passing it to pdsolve. Say we want to set u(r,theta,t) to have insulated boundary conditions at r=1. So the BC will be
For example, to set derivative of u w.r.t. "r" to zero when r=1
eval( diff(u(r,theta,t),r), r=1) = 0; #(1)
or using this syntax
D[1](u)(1, theta, t) = 0; #(2)
But now I find, on one example below, that the above no longer works. I have to use this syntax (which I did not know about) for it to work
D[1]*u(1, theta, t) = 0; #(3)
Has something changed? why when using (3) pdsolve now gives result, but when using (2) or (1) it returns unevaluated? are the three semantically equivalent? when to use which syntax?
I am using Physics updates 265, Latest Maple 2018.2
Here is an example showing the (1,2) syntax no longer works, but the (3) syntax works
#articolo example 6.9.2 restart; #using (1) syntax pde := diff(u(r, theta, t), t) = (diff(u(r, theta, t), r)+r*(diff(u(r, theta, t), r, r))+(diff(u(r, theta, t), theta, theta))/r)/(25*r); bc_on_r := eval(diff(u(r,theta,t),r), r=1) = 0; bc_on_theta:= u(r,0,t)=0, u(r,Pi,t)=0; ic := u(r,theta,0)=(r-1/3*r^3)*sin(theta); pdsolve([pde, bc_on_r,bc_on_theta,ic], u(r, theta, t), HINT = boundedseries(r = [0]))
does not solve it.
restart; #using (2) syntax pde := diff(u(r, theta, t), t) = (diff(u(r, theta, t), r)+r*(diff(u(r, theta, t), r, r))+(diff(u(r, theta, t), theta, theta))/r)/(25*r); bc_on_r := D[1](u)(1, theta, t) = 0; bc_on_theta:= u(r,0,t)=0, u(r,Pi,t)=0; ic := u(r,theta,0)=(r-1/3*r^3)*sin(theta); pdsolve([pde, bc_on_r,bc_on_theta,ic], u(r, theta, t), HINT = boundedseries(r = [0]))
does not solve it.
restart; #using(3) syntax pde := diff(u(r, theta, t), t) = (diff(u(r, theta, t), r)+r*(diff(u(r, theta, t), r, r))+(diff(u(r, theta, t), theta, theta))/r)/(25*r); bc_on_r := D[1]*u(1,theta,t)=0; bc_on_theta:= u(r,0,t)=0, u(r,Pi,t)=0; ic := u(r,theta,0)=(r-1/3*r^3)*sin(theta); pdsolve([pde, bc_on_r,bc_on_theta,ic], u(r, theta, t), HINT = boundedseries(r = [0]))
I've used syntax (1) before many times and it works. Here is an example where all three syntax work
pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2); bc:=eval(diff(u(x,t),x),x=0)=0,u(L,t)=0; ic:=u(x,0)=f(x); sol:=pdsolve([pde,bc,ic],u(x,t));
pdsolve gives
Using syntax (2)
pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2); bc:=D[1](u)(0,t)=0,u(L,t)=0; ic:=u(x,0)=f(x); sol:=pdsolve([pde,bc,ic],u(x,t));
gives same answer as (1) and using syntax (3)
pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2); bc:=D[1]*u(0,t)=0,u(L,t); ic:=u(x,0)=f(x); sol:=pdsolve([pde,bc,ic],u(x,t));
The answer also looks like different and simpler, but I assume they are equivalent for now without looking too much into it.
Which syntax should one use as now I am really confused.
It looks like (3) is the one that should be used? Why the others did not work on first example? i.e. pdsolve did not give an answer at all? And if (1,2,3) syntax are supposed to be equivalent, whysecond example gives slightly different looking answer when using one syntax vs. the other?
And finally to make things more confusing, here is an example where syntax (3) does not work, but syntax 1 and 2 work:
#articolo example 8.4.3 restart; pde := diff(u(x, t), t) = (1/20)*(diff(u(x, t), x, x))+t; bc := u(0, t) = 5, (u(1, t)+ D[1](u)(1, t)) = 10; ic:= u(x, 0) = -40*x^2*(1/3)+45*x*(1/2)+5; pdsolve([pde, bc,ic], u(x, t))
gives answer.
restart; pde := diff(u(x, t), t) = (1/20)*(diff(u(x, t), x, x))+t; bc := u(0, t) = 5, (u(1, t)+eval(diff(u(x,t),x),x=1)) = 10; ic:= u(x, 0) = -40*x^2*(1/3)+45*x*(1/2)+5; pdsolve([pde, bc,ic], u(x, t))
gives same answer. But
restart; pde := diff(u(x, t), t) = (1/20)*(diff(u(x, t), x, x))+t; bc := u(0, t) = 5, (u(1, t)+ D[1]*u(1, t)) = 10; ic:= u(x, 0) = -40*x^2*(1/3)+45*x*(1/2)+5; pdsolve([pde, bc,ic], u(x, t))
does not work.
Clearly there is something I do not understand between these 3 syntaxes and when to use which.
Using 265 version
Physics:-Version() "C:\Maple_updates\Physics+Updates.maple", 2018, December 22, 10:41 hours, version in the MapleCloud: 265, version installed in this computer: not installed
downloaded today.
trying some examples from 209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018
I noticed some answer change when adding assumptions which should not really cause the answer to change. Numerically, both answers seem to be the same. But one is series solution and the other with assumptions added uses integrals.
In wave pde on string where left end at x=0 and adding assumptions x>0,t>0 to pdsolve should produce the same form and it does on some problems but no on others.
Compare
#example 19, wave pde, on string left end fixed, right end moving, with source, no initial velocity restart; pde := diff(u(x, t), t$2) = 4*diff(u(x, t), x$2)+(1+t)*x; bc := u(0,t)=0,u(Pi,t)=sin(t); ic := u(x,0)=0,eval(diff(u(x,t),t),t=0)=0; sol:=pdsolve([pde, bc,ic],u(x,t)); #1 sol:=pdsolve([pde, bc,ic],u(x,t)) assuming t>0; #2 sol:=pdsolve([pde, bc,ic],u(x,t)) assuming x>0; #3 sol:=pdsolve([pde, bc,ic],u(x,t)) assuming t>0,x>0; #4
(1,2,3) above give same answer which is
but (4) gives
Tried random points and numerically they are the same (set the sum to n=100 to evaluate).
my question is why would adding no assuming, or adding t>0 on its own or adding x>0 on its own give same answer, but adding t>0,x>0 at same time give such a different looking soution.
I typically adding such assumptions to be safe, but here it seems to had some side effect I have not noticed before.
Physics:-Version(); "C:\Maple_updates\Physics+Updates.maple", 2018, December 22, 10:41 hours, version in the MapleCloud: 265, version installed in this computer: not installed kernelopts(version) Maple 2018.2, X86 64 WINDOWS, Nov 16 2018, Build ID 1362973
Maple 2018.2.1, using Physicsupdates 266.
I undertsand method=Fourier needs boundary conditions to work, but I do not think this error message is right. Compare
restart; pde := diff(u(r, theta), r, r)+diff(u(r, theta), theta, theta) = 0; iv := u(2, theta) = 3*sin(2*theta)+1; pdsolve([pde,iv], u(r,theta), method = Fourier)
With
restart; pde := diff(u(r, theta), r, r)+diff(u(r, theta), theta, theta) = 0; pdsolve(pde, u(r,theta), method = Fourier)
Error, (in pdsolve/info) wrong extra arguments: {method = Fourier}
pdsolve should return no solution instead. The way it is above, I thought at first I had wrong syntax with the "method = Fourier" settings and I think this error message can be misleading to a user.