Items tagged with pde pde Tagged Items Feed

Greetings,

I need some help/advice because I think I am going crazy..!

I am trying to solve numerically a system of two PDE's. My initial condition is a gaussian and I apply cyclic boundaries.

With Maple 18 and pdsolve/numeric I am waiting a considerable amount of time and the result is a bell shaped travelling wave that increases its height over time. The clush between two such waves leaves them almost unaffected (the big pass through the small one).

 

With Mathematica 10 and NDSolve the waiting time is almost zero and the result (after few timesteps) is a shock wave travelling with steady height. Moreover, the clush between two of them is fully plastic (the big "eat" the small one)

 

I double-triple check the equations and the boundary conditions and I assure you that are the same..Any ideas what is hapening? Is there any way to check that the Maple results are correct?

The model of fixed-bed adsorption column

Fluid phase:

PDE:= diff(U(x, tau),tau)+ psi*Theta*diff(U(x, tau),x)-(1/Pe)*psi*Theta*diff(U(x, tau),$(x, 2))=-3*psi*xi*(U(x, tau)-Q/K);

 

IBC:={U(x, 0) = 0,U(0, tau) = 1+(1/Pe)*(D[1](U))(0, tau),(D[1](U))(1, tau)=0};

Particle:

PDE:= diff(Q(r, tau), tau) = diff(Q(r, tau), $(r, 2))+(2/r)*diff(Q(r, tau),r);

IBC:={Q(r, 0) = 0,(D[1](Q))(0, tau) = 0,(1/K)*(D[1](Q))(1, tau)=xi*(U-Q(1, tau)/K)};

Pe:=0.01:

psi:=6780:

Theta:=3.0:

xi:=10000:

I will really appreciate your help. Thanks in anticipation.

now the equation is

d2u/dt2-(2*d2u/x2)+d2u/dxdt=0    

initial condition: u(x,0)=1-(xsign(x)), abslute x<1,0 otherwise. Assume sign(x)=-1 for x<0, 1for x>0 

 Ut(x,0)=cos(pix), bslute x<1, 0 otherwise , he didnt give any B.Cs

so I would like to know the analytical and numerical sols, and plots for the wave at t=2,4

for Numerical:   delta x=0.1, delta t=0.025, range 0..4

I am currently learning about nonlinear waves and and am having problems with my maple coding where I am plotting the characteristics of the initial value problem

$u_[t] + uu_[x] = 0$ where $u(x,t) = a$ if $x < -1$

                                               = $b$ if $ -1 < x < 1$

                                               = $c$ if $ x > 1$

where $a$, $b$ and $c$ are unequal constants. Also, there are 6 cases to this, $a > b > c$, $a > c > b$, $b > a >c$, $b > c > a$, $c > a > b$ and $c > b > a$. In each case I need to plot the characteristics (and where possible plot the rarefaction waves, but this bit is not necessary right now). My problem is that I am highly certain that my code is incorrect as I am unsure where/how to implement the fact that $a$, $b$ and $c$ vary since there are different cases. My current code is down below

 

 

restart: with(plots): with(plottools): with(PDEtools):
ploti:= implicitplot({seq(t*((k/2))+k/2-x,k=-10..-1)},x=-10..10,t=0..10,view=[-5..5,0..5],color=blue,thickness=2):
ploti2:= implicitplot({seq(t*(k/2)+k/2-x,k=-1..1)},x=-10..10,t=0..10,view=[-5..5,0..5],color=red,thickness=2):
ploti3:= implicitplot({seq(t*((k/2))+k/2-x,k=1..10)},x=-10..10,t=0..10,view=[-5..5,0..5],color=yellow,thickness=2):
display(ploti,ploti2,ploti3);

Any help would be much appreciated

 

Good day everyone,

Please I do get numerical output/values in this solution U1.mw

Best regards

Hi, 

     I have a question regarding pdsolve, or Solve from the PDEtools package. I have a set of equations relating partial derivatives, and I'd like to isolate certain terms without explicitly known the functions. I can do this for a single equation, but not multiple ones. I'm curious if Maple can currently handle a system of eqns like these easily, since I will be increasing the number of eqns in the future. Here's the code 

 

 

restart;

PDEtools:-declare(H=H(x,y,t)):

H(x, y, t)*`will now be displayed as`*H

(1)

eq1:= H[tt](x,y,t) = H[xx](x,y,t) + H[yy](x,y,t);

H[tt](x, y, t) = H[xx](x, y, t)+H[yy](x, y, t)

(2)

eq2 := diff(H[tt](x,y,t), t) = diff(H[tx](x,y,t), x) + diff(H[ty](x,y,t), y);

diff(H[tt](x, y, t), t) = diff(H[tx](x, y, t), x)+diff(H[ty](x, y, t), y)

(3)

eq3 := diff(H[tx](x,y,t), t) = diff(H[xx](x,y,t), x) + diff(H[xy](x,y,t), y);

diff(H[tx](x, y, t), t) = diff(H[xx](x, y, t), x)+diff(H[xy](x, y, t), y)

(4)

eq4 :=diff(H[ty](x,y,t), t) = diff(H[xy](x,y,t), x) + diff(H[yy](x,y,t), y);

diff(H[ty](x, y, t), t) = diff(H[xy](x, y, t), x)+diff(H[yy](x, y, t), y)

(5)

PDEtools:-Solve(eq3, H[xy]);

H[xy](x, y, t) = Int(diff(H[tx](x, y, t), t)-(diff(H[xx](x, y, t), x)), y)+_F1(x, t)

(6)

PDEtools:-Solve({eq1, eq2, eq3, eq4}, H[xy]);

Error, (in pdsolve/sys) the input system cannot contain equations in the arbitrary parameters alone; found equation depending only on {H[tt](x,y,t), H[xx](x,y,t), H[yy](x,y,t)}: H[tt](x,y,t)-H[xx](x,y,t)-H[yy](x,y,t)

 

 

 

 

Download PDESolveHelp.mw

I have a system of pdes and solved numerically using pdsolve (numeric) command.

The system consists of four first order partial differentia equations.

for example u(x,t), R(x,t)....

what command should I give to the Maple and get the graph of u(x,t) at a specific point x_0?

For example, I need a plot for u(30,t).

Is it possible with the maple plot?

I really appreciate your help.

Thank you for reading this post. :)

 

Trying to solve the 1-dimensional heat equation with maple with constant boundary temperatures:

restart;

with(PDETools):
U := diff_table(u(x,t)):
pde := U[t]=U[x,x];
bc := u(0, t) =0, u(1, t) = 1, u(x,0)=x;
pdsolve([pde,bc]);

The solution of this equation is u(x,t)=x , but pdsolve(...) does not return anything at all! What is going wrong? Is it too hard PDE for maple? And if it is too hard, where can be found the types of equations, which are too hard and not too hard? Thank you.

In another experiment with pdsolve, I am solving a PDE with two sets of boundary conditions. Unfortunately, after invoking pdsolve, I get no result at all. What can be wrong here?

Can someone help me figure out what's going on? Here's the PDE I'm trying to solve, and I'm clearly getting the wrong answer.

 

I'm trying to solve a system of four pdes and I know that the Newton method won't converge.

Are there other numerical methods that I can use?

Any help would be greatly appreciated!

Thanks,

Eve

Hello,

I would like to solve this kind of system with any numeric method.


With any kind of IBCs.

 

My code :

Maple says : Error, (in pdsolve/numeric/plot) unable to compute solution for t<HFloat(0.0):
matrix is singular
Error, (in pdsolve/numeric/plot) unable to compute solution for t>HFloat(0.0):
matrix is singular

Any idea why ? Any help ?

Thank you

test.mw

In this file, I tried my best to solve the pde. But the answer is still rather non-informative. I need some help to simplify it.

I did notice that my Maple might need reinstallation, due to a "bug" in the 18.02 update.

 

My ultimate aim is try to use some similar techniques to solve this, test2.mw, which has a similar type pde.

 

The standard pdsolve(pde) would just not work.

 

UPDATE:

I used the same file in Maple 17 on a differnt Machine, which can be solved by pdsolve. So I guess it's just that  the 18.02 update package is broken itself. I have tried to uninstall and reinstall twice.

 

 

Thanks,

 

casper

Hi

My equations are:

diff(Th(z, t), t) = 7.1428*(diff(Th(z, t), z))-1397941.885*(279-Tw(z, t))-0.2160487e-1*(diff(Th(z, t), z, z))

diff(Tc(z, t), t) = -7.1428*(diff(Tc(z, t), z))+1298990.852*(Tw(z, t)-291)+0.189366e-1*(diff(Tc(z, t), z, z))

diff(Tw(z, t), t) = 3.3024901*(Th(z, t)-2*Tw(z, t)+Tc(z, t))+8.0029*10^(-4)*(diff(Tw(z, t), z, z))

and boundry conditions are:

Th(z, 0) = 296, (D[1](Th))(0, t) = 0, Th(1, t) = 296

Tc(z, 0) = 275, (D[1](Tc))(1, t) = 0, Tc(0, t) = 275

Tw(z, 0) = 0, (D[1](Tw))(1, t) = 0,Tw(0, t) = 0

I shoud solve this equations numerically and plot Th vs.t(and Th vs.z),

please help me.

thanks

Hi

I have this PDE and was wondering how I can get Maple to solve it

utt+2ut-uxx=18sin(3πx/l)

with conditions u(0,t)=u(l,t)=0 and u(x,0)=ut(x,0)=0

Thanks

James

 

 

 

1 2 3 4 5 6 7 Page 1 of 8