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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

 

 

 

declare(W(x, y), Z(x, y));

 

sys := [-A*kappa3-`∂`(`∂`(W(x, y))/`∂`(x))*(2*G-A)/`∂`(x)-2*G*(`∂`(`∂`(W(x, y))/`∂`(y))/`∂`(y)+`∂`(`∂`(Z(x, y))/`∂`(x))/`∂`(y))+A*`∂`(`∂`(Z(x, y))/`∂`(x))/`∂`(y) = 0, `∂`(`∂`(Z(x, y))/`∂`(y))*(A-4*G)/`∂`(y)+`∂`(`∂`(W(x, y))/`∂`(x))*(A-2*G)/`∂`(y)-2*G*`∂`(`∂`(Z(x, y))/`∂`(x))/`∂`(x) = 0];

 

I have this system of coupled PDE and I wish to solve it using Maple.

It gives me error of this kind:

 

pdsolve(sys, [[W(x, y)], [Z(x, y)]]);


Error, (in pdsolve/sys) found functions depending on different variables in the given DE system: [`∂`(x), `∂`(y)]

 

Thanks a lot for help

hi

i have solved my equation as folllow :

 

pde:= diff(T(x, y), x)-1.555*10^(-7)*(diff(T(x, y), y, y))/ ...........

 

sol := pdsolve(pde, {T(0, y) = 0, (D[2](T))(x, 0) = 1325.754092, (D[2](T))(x, 0.25e-4) = 1970434.783}, numeric)

 

I wana know that maple has used which of numeric method to solve my equation ?

 

1.ForwardTime1Space[forward/backward]

2.CenteredTime1Space[forward/backward]

3.BackwardTime1Space[forward/backward]

4.ForwardTimeCenteredSpace or Euler

5.CenteredTimeCenteredSpace or CrankNicholson

6.BackwardTimeCenteredSpace or BackwardEuler

7.Box

8.LaxFriedrichs

or ... ?

 

Tahnks.

st6.mw,st7.mwI want to obtain the analytical solution for this PDE by maple   diff(T(x, y, z), x, x)+diff(T(x, y, z), y, y)+diff(T(x, y, z), z, z)+A*exp(-8*x^2/a^2)*cosh(alpha*((1/2)*b+y)) = 0    . But I can not do it. Please help me

((d^2)T/dx^2)+ ((d^2)T/dy^2)+ ((d^2)T/dz^2)=-A*Q(x,y,z)

Where   0 <x<a ,  0 <y<b  ,  0 <z<l

With the boundary conditions:

(dT(0,y,z)/dx)=n-T(0,y,z)    (dT(a,y,z)/dx)=n-T(a,y,z)   (dT(x,0,z)/dy)=n-T(x,0,z)

(dT(x,b,z)/dy)=n-T(x,b,z)     (dT(x,y,0)/dz)=n-T(x,y,0)    (dT(x,y,l)/dz)=n-T(x,y,l)

where n is constant and A is set of parameters.

Hi,

I have been trying to solve 2D Diffusion Equation with zero Neumann BC over the unit disk. If I use Gaussian type function with a sharp peak as initial condition, I get huge errors between initial values. Let's say u(r,phi,t) is the solution of the PDE and f(r,phi) is initial value function. The expectation is for the point (r*,phi*) ,  u(r*,phi*,0)=f(r*,phi*), but it is not.

Is Numerical integration in Maple not able to handle such sharp peak? I tried some of the built-in methods such as MonteCarlo,CubaVegas but no difference.

It might be a good idea to specify some nodes arround the peak. There is a command called "peaks", but I could not use it, error message says "invalid arguments".

Thanks in advance.

Soln_2D_Gaussian.mw

Hello,

How can I pde with maple?please explain completely,and other question :How can I solve pde with plot in maple because some questions dont have exact answer?

Hi,

 

I am having trouble solving the PDE. Can anyone shine some light?


``

sys:=-(1/2)*(diff(f(phi, s[1], s[2], w[1]), s[1]))*(s[1]-s[2])*s[1]/(s[2]*w[1])+(1/2)*(diff(f(phi, s[1], s[2], w[1]), s[2]))*(s[1]-s[2])*s[2]/((-1+w[1])*s[1])+diff(f(phi, s[1], s[2], w[1]), w[1]) = 0;

-(1/2)*(diff(f(phi, s[1], s[2], w[1]), s[1]))*(s[1]-s[2])*s[1]/(s[2]*w[1])+(1/2)*(diff(f(phi, s[1], s[2], w[1]), s[2]))*(s[1]-s[2])*s[2]/((-1+w[1])*s[1])+diff(f(phi, s[1], s[2], w[1]), w[1]) = 0

(1)

pdsolve(sys);

 

``


Download 1234.mw


Dear users,

In my attached file I have two PDES, (PDE1 and PDE2). PDE1 is a function of v(t) and w(x,t) and PDE2 is also a function of v(t) and w(x,t). I can solve PDE2 if I say v(t) is 1 for example and you can see the plot. But what if I put v(t) back in PDE2 and want to find v(t) and w(x,t) from PDE1 and PDE2 together? 

Many Thanks,

Baharm31

 

Define PDE Euler-Bernoulli Beam

 

NULL

restart:

Parametrs of piezoelectric and cantilever beam

 

``

Ys := 70*10^9: # Young's Modulus structure

Yp := 11.1*10^10: # Young's Modulus pieazo

ha := -0.00125: # Position

hb := 0.001: # Position

hc := 0.0015: # Position

d31 := -180*10^(-12): # Piezoelectric constant

b := 0.01: #Width of the beam

tb := 0.002:

epsilon33 := 15.92*10^(-9):

hp :=0.00025: # Position

hpc := 0.00125: # Position

YI := b*(Ys*(hb^3- ha^3)+Yp*(hc^3-hb^3))/3: # Bending stiffness of the composit cross section

cs := 0.564: # The equivqlent coefficient of strain rate damping

ca := 0: # Viscous air damping coefficient

Ibeam := (b * tb^3 )/12: # The equivalent moment of inertia

m := 0.101: # Mass of the structure

upsilon := - Yp*d31*b*(hc^2-hb^2)/(2*hp): # Coupling term

lb := 0.57:# Length of the structure (Cantilever Beam)

lp := 0.05:# Length of the Piezoelectric

R:= 10000: # Shunted resistor

Electrical circuit equation

 

PDE1:=(epsilon33 * b*lp / hp) * diff(v(t), t) + (v(t)/R)+ int(d31*Yp*hpc*b* diff(w(x, t),$(x, 2))*diff(w(x, t), t),x = 0..lp)=0;

0.3184000000e-7*(diff(v(t), t))+(1/10000)*v(t)+int(-0.2497500000e-3*(diff(diff(w(x, t), x), x))*(diff(w(x, t), t)), x = 0 .. 0.5e-1) = 0

(1.1.1.1)

``

 

PDE Equation

 

fn := 3.8:# Direct Excitation frequency;

wb(x,t) := 0.01*sin(fn*2*Pi*t):#Direct Excitation;

plot(wb(x,t),t = 0 .. 0.25*Pi,labels = [t,wb], labeldirections = ["horizontal", "vertical"], labelfont = ["HELVETICA", 15], linestyle = [longdash], axesfont = ["HELVETICA", "ROMAN", 10], legendstyle = [font = ["HELVETICA", 10], location = right],color = black);

 

 

FunctionAdvisor(definition, Dirac(n,x));

[Dirac(n, x) = (1/2)*(Int((I*_k1)^n*exp(I*_k1*x), _k1 = -infinity .. infinity))/Pi, `with no restrictions on `(n, x)]

(1.2.1)

 

PDE2 := YI*diff(w(x, t),$(x, 4))+ cs*Ibeam*diff(w(x, t),$(x, 4))*diff(w(x, t), t)+ ca* diff(w(x, t), t) + m * diff(w(x, t),$(t, 2))+ upsilon*v(t)*(Dirac(1,x) -Dirac(1,x-lp) ) =-m*diff(wb(x, t),$(t, 2))-ca*diff(wb(x, t), t);#PDE

1.567812500*(diff(diff(diff(diff(w(x, t), x), x), x), x))+0.3760000000e-11*(diff(diff(diff(diff(w(x, t), x), x), x), x))*(diff(w(x, t), t))+.101*(diff(diff(w(x, t), t), t))+0.4995000000e-3*Dirac(1, x)-0.4995000000e-3*Dirac(1, x-0.5e-1) = 0.583376e-1*sin(7.6*Pi*t)*Pi^2

(1.2.2)

tmax := 0.3:

xmin := 0:

xmax := lb:

N := 20:#NUMBER OF NODE POINT

bc1 := dw(xmin, t) = 0:

bc2 := dw(xmax, t) = 0:

bc3 := w(xmin, t) = 0:

ic1 := wl(x, 0) = 0:

Maple's pdsolve command

 

 

 

bcs := { w(x,0)=0 , D[2](w)(x,0)=0 , w(0, t) = rhs(bc1), D[1](w)(0, t)= rhs(bc1), D[1,1](w)(lb,t) = rhs(bc2), D[1,1,1](w)(lb,t) = rhs(bc2)}; # Boundary conditions for PDE2.

{w(0, t) = 0, w(x, 0) = 0, (D[1](w))(0, t) = 0, (D[2](w))(x, 0) = 0, (D[1, 1](w))(.57, t) = 0, (D[1, 1, 1](w))(.57, t) = 0}

(2.1)

PDES := pdsolve(PDE2, bcs, numeric, time = t, range = 0 .. xmax, indepvars = [x, t], spacestep = (1/1000)*xmax, timestep = (1/1000)*tmax);

 

module () local INFO; export plot, plot3d, animate, value, settings; option `Copyright (c) 2001 by Waterloo Maple Inc. All rights reserved.`; end module

(2.2)

PDES:-plot3d(t = 0 .. tmax, x = 0 .. xmax, axes = boxed, orientation = [-120, 40], shading = zhue, transparency = 0.3);

 

 

NULL


Download Euler-Bernoulli_Beam-last_version.mw

Dear Maple users

 

I have a question about applying pdsolve MAPLE for solving two dimensional heat equations:

My codes have been provided but it shows to me this error:

Error, (in pdsolve/numeric/process_PDEs) can only numerically solve PDE with two independent variables, got {t, x, y}

If kindly is possible, please help me in this case.

 

With kind regards,

Emran Tohidi.

 

> restart;
> with(plots);
print(??); # input placeholder
> with(PDEtools);
print(??); # input placeholder
> declare(u(x, y, t));
print(`output redirected...`); # input placeholder
                    u(x, y, t) will now be displayed as u
> S := 1/100; tR := 0 .. 1; xR := 0 .. 1; yR := 0 .. 1; NF := 30; NP := 100;
print(??); # input placeholder
> N := 3; L1 := [red, blue, green]; L2 := [0, 1/2, 1]; Ops := spacestep = S, timestep = S;
print(??); # input placeholder
> Op1 := frames = NF, numpoints = NP;
print(??); # input placeholder
> PDE1 := diff(u(x, y, t), t)-(diff(u(x, y, t), `$`(x, 2)))-(diff(u(x, y, t), `$`(y, 2))) = 0;
print(??); # input placeholder
> IC := {u(x, y, 0) = exp(x+y)}; BC := {u(0, y, t) = exp(2*t+y), u(1, y, t) = exp(2*t+y+1), u(x, 0, t) = exp(2*t+x), u(x, 1, t) = exp(2*t+x+1)};
print(??); # input placeholder
> Sol := pdsolve(PDE1, `union`(IC, BC), numeric, u(x, t), Ops);
Error, (in pdsolve/numeric/process_PDEs) can only numerically solve PDE with two independent variables, got {t, x, y}


Dear all,

I tried to use pdsolve to solve the parabolic pde but get the unexpected answer:

Is it the PDESolStruc or the other structure? Where can I find the description about this kind of structure.

Thanks.

Dear all,

It's very convenient to define a DE or PDE through Differential Operator D, for example,

((D[1, 1]+D[1, 2]+D[2, 2])(z))(x, y) = exp(x)*sin(y)

Is it possible to realize Inverse Operator Method of Operator D? How to solve the following equation if we rewrite the pde through inverse operator method?

(z)(x, y)=((D[1, 1]+D[1, 2]+D[2, 2])^(-1))exp(x)*sin(y)

 

Thanks a lot.

How can I solve this problem on Maple?
Can anyone help me please ... I wrote another post before but I can not solve the problem.

lambda is an experimental parameter. I have this initial condition n(x,0)=0.4, c(x,0)=0.

Thanks to everyone

Dear Maple Users,

I'm beginner in Maple.

I have this system of Pde:

with lambda experimental parameter and n,c,v dependent variables. I write this on Maple but I read on internet that the solution "float(undefined)" is an error.

I will insert this initial condition: c(x,0)=0,n(x,0)=0.4

Thanks everybody

Is it possible to solve piecewise differential equations directly instead of separating the pieces and solving them separately.

like for example if i have a two dimensional function f(t,x) whose dynamics is as follows:

dynamics:= piecewise((t,x) in D1, pde1, pde2); where D1 is some region in (t,x)-plane

now is it possible to solve this system with one pde call numerically?

pde(dynamics, boundary conditions, numeric); doesnot work

hi,

i obtained values for h(t) and u(t) and i wana obtaine s(t) , i don't konw how can i do it

 

for h(t) and u(t) ,

 

sol1 := dsolve([4.00000000000000*u(t)^2*h(t)*(diff(diff(h(t), t), t))-8.00000000000000*h(t)^2*(427.2460938*u(t)+385620.1174/u(t)-25671.38673)^2+16.0000000000000*u(t)*h(t)*(427.2460938*u(t)+385620.1174/u(t)-25671.38673)*(diff(h(t), t))+(16406.2500000000*(-29.8200000000000+u(t)))*h(t)^2*(427.2460938*u(t)+385620.1174/u(t)-25671.38673)+7171.87500000000*u(t)*h(t)*(diff(h(t), t))-16406.2500000000*h(t)^2*(30-u(t))*(427.2460938*u(t)+385620.1174/u(t)-25671.38673)+u(t)^2*h(t)-0.225000000000000e-1+0.225000000000000e-1*(.997494986604054+.106105802501554*sin(t-1.5))^2+4.00000000000000*(h(t)*(427.2460938*u(t)+385620.1174/u(t)-25671.38673)+u(t)*(diff(h(t), t)))^2 = 0, diff(u(t), t) = 427.2460938*u(t)+385620.1174/u(t)-25671.38673, h(1.44) = 0.145e-4, (D(h))(1.44) = .12093, u(1.44) = .62], numeric)

 

and s(t) is , s(t)=1/r*d/dt(u(t)*h(t))

 

thanks.

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