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Hello!
Try to consider the following system of differential equations:

sys := diff(U(ksi), ksi) = (y(ksi)-(1.5*(1+0.01*ksi))*U(ksi)/ksi)/(1.5+0.015*ksi+0.002*ksi^2),
diff(y(ksi), ksi) = U(ksi)*(0.002+(1.5*(1+0.01*ksi))*(0.002*ksi^2)/(ksi^2*(1.5+0.015*ksi+0.002*ksi^2))-19.3^2)-y(ksi)*(0.002*ksi^2)/(ksi*(1.5+0.015*ksi+0.002*ksi^2))

with the boundary conditions: cond:= y(0.8) = 0, y(1) = 0

And Maple gives me zero solution for this system , i mean U(from 0.8  to 1) = 0 and y(from 0.8 to 1) = 0

How can i get some other solutions?


P.S. i need numerical solution => dsolve( [sys,cond],type= numeric)

Please, i need some help.

Dear all,

I am trying to solve the following partial differential equation (transport or advection equation) with given initial and boundary conditions:

restart: with(PDEtools):
sys := [v*diff(u(x,t), x) + diff(u(x,t), t) = 0, u(x,0) = exp(-x), u(0,t) = sin(t)];
pdsolve(sys);

But it does not work. The solution is (or should be): 

u(x, t) = exp(t*v-x)+Heaviside(t-x/v)*(sin(t-x/v)-exp(t*v-x))

I think the reason is that the interval for t (in [0, inf)) and x (in [0, 1]) is not specified. On the other hand, this works:

restart: with(PDEtools):
sys := [diff(u(x, t), t) = diff(u(x, t), x, x), u(0, t) = 0, u(1, t) = 0, u(x,0) = f(x)];
sol := pdsolve(sys);

How can I solve a PDE like the transport equation with given initial AND boundary conditions?

Thanks a lot

hello everyone. I have an undergradute project i'm currently working on and I'm stuck where I have to use the Differential Transforms Method to solve a problem with boundary conditions at infinity


restart;

Digits := 5;

F[0] := 0; F[1] := 0; F[2] := (1/2)*A; T[0] := 1; T[1] := B; M := 2; S := 1;

for k from 0 to 10 do F[k+3] := (2*(sum((r+1)*F[r+1]*(k+1-r)*F[k+1-r], r = 0 .. k))-T[k]-3*(sum((k+1-r)*(k+2-r)*F[r]*F[k+2-r], r = 0 .. k))-M*(k+1)*F[k+1])*factorial(k)/factorial(k+3);

T[k+2] := (-3*(sum((k+1-r)*F[r]*T[k+1-r], r = 0 .. k))-S*T[k])*factorial(k)/factorial(k+2)

end do; f := 0; t := 0;

for k from 0 to 10 do

f := f+F[k]*x^k;

t := t+T[k]*x^k end do;

print(f);
print(t);

but the problem is that i cant seem to evaluate

or higer diagonal pade-approximant. any help will be greatly appreciated. thank you.

(in dsolve/numeric/bvp) initial Newton iteration is not converging.

Hi,

     I'm trying to numerically solve a PDE in Maple for different boundary conditions, however I'm having trouble even getting Maple to numerically solve it for simple boundary conditions.

I have cylindrical coordinates, r, z, theta, and I treat r = r(z, theta) for convenience to plot my solution surface. The initial coundary condition is that at z = epsilon (z = 0 is singular) , r = constant and of course r is periodic in theta. This is just a circle, and the analytical solution is know to be a half-sphere  r = sqrt(R^2 - z^2). I entered my initial boundary conditions into Maple, but it doesn't like the periodic one

IBC := { r(epsilon, theta) = R - epsilon__r,
              r(z, 0) = r(z, 2*Pi) };

pdsolve(
  PDE,
  IBC,
  numeric,
  indepvars = [z, theta],
  time = z,
  range = 0..2*Pi);
Error, (in pdsolve/numeric/par_hyp) Incorrect number of boundary conditions, expected 2, got 1

I'm not sure how to make this work, and then generalize it to more arbitrary intial slices r(epsilon, theta) = f(theta).

Here's the attached worksheet, ForMaplePrimesSUbmission.mw

Any help is appreciated,

Thanks

How do i proceed to solve two differential equations?

Two equations two unknowns is easy to solve in polynomial algebraic equations. Example: x+y=5; x-y=3; The solution is x=4; y=1 by adding the equations we arrive at.

The two equations are second order differential equations with two variables say temperature T (x,y) and velocity c(x,y). Assume any simple equation (one dimensional as well i.e. T(x) and c(x) which you can demonstrate with ease, I have not formulated the exact equations and boundary conditions yet for SI Engine simulation.

Thanks for comments, suggestions and answers expected eagerly.

Ramakrishnan

I'm trying to execute the program, which can be found here http://www.maplesoft.com/support/help/Maple/view.aspx?path=examples/pdsolve_boundaryconditions , but it does not work. I copied exactly what is written there:

restart; with(PDEtools):
U := diff_table(u(x,t)):
pde[1] := U[t]+c*U[x]=-lambda*U[];
bc[1] := eval(U[], t=0) = phi(x);
sys[1] := [pde[1], bc[1]];
pdsolve(sys[1]);

But after last command it just sais that

Error, (in pdsolve/sys/info) found functions with same name but depending on different arguments in the given DE system: [u(x,t), u(x,0)]

What's wrong?

hai everyone. i am currently trying to solve an integration of the following ∫g(η)dη . integrate from 0 to 10.

from the following odes.

f ''' +1-(f ')2 +ff ''=0,

g''-gf'+fg'=0,

with boundary conditions f(0)=0, f'(0)=λ, f'(∞)=1, g(0)=1,g(∞)=0

First, i solve the odes using the shooting method. then i used the trapezoidal rule to solve for the integration of g(eta) using the following codes

> with(student);
> trapezoid(g(eta), eta = 0 .. 10, 10);
> evalf(%);

it seems that it can not read the data from the shooting method. can anyone suggest why it is happening?

thank you verymuch for your concern :)

I try to solve a set of differential equations with boundary conditions by dsolve command but I got this error:

--------------------------------------------------------------------------------------------------------------------------

odes := D2*(diff(P(x), x))/((3*D1*a+4*D2)*P(x))-(diff(S(x), x))/(q*S(x)-1) = 0, diff(S(x), `$`(x, 2))+(diff(S(x), x))*cotx+4*pi*(3*D1*a+4*D1)*P(x)/((q*S(x)-1)*D2) = 0

ics := P((1/2)*pi) = 1, S((1/2)*pi) = -1, (D(S))((1/2)*pi) = 0, (D(P))((1/2)*pi) = 0

sol := dsolve({ics, odes}, numeric);
Error, (in dsolve/numeric/process_input) input system must be an ODE system, got independent variables {x, (1/2)*pi}

--------------------------------------------------------------------------------------------------------------------------

Does any body knows what the problem is??

 

 

Hi everybody!

It's nice to join in this forum.

I'm trying to get the analytic solution of the Bernouilli-Euler beam equation, with the next boundary conditions:


w(x,t) = displacements.

w(0,t) = 0   -> It's a cantilever beam. At the x=0 it's clamped.

diff(w(x,t),x) = 0.   -> the gyro in the clamp is zero.

E*I*diff(w(L,t),x,x) = 0  -> the moment at the end of the beam (x=L) is zero.

E*I*diff(w(L,t),x,x,x) = 0  -> the shear at the end of the beam (x=L) is zero too.


I'm not able to introduce the second and the third derivatives as boundary conditions in the pdsolve equation. I post the hole code:

restart;
ode := I*E*(diff(w(x, t), x, x, x, x))+m*(diff(w(x, t), t, t)) = 0;

s := pdsolve(ode, w(x, t));

ode1 := op([2, 1, 1], s);

ode2 := op([2, 1, 2], s);

f1 := op(4, rhs(ode1));

f2 := op(2, rhs(ode2));

sol1 := dsolve(ode1, f1);

sol2 := dsolve(ode2, f2);

sol := rhs(sol1)*rhs(sol2);

conds := [w(0, t) = 0, (D[1](w))(0, t) = 0, eval(I*E*(D[1, 1](w))(x, t), x = L) = 0, eval(I*E*(D[1, 1, 1](w))(x, t), x = L) = 0];

pde := [ode, conds];

pdsolve(pde, w(u, t));


And I get this error:

"Error, (in PDEtools/pdsolve) invalid input: `pdsolve/sys` expects its 1st argument, SYS, to be of type {list({`<>`, `=`, algebraic}), set({`<>`, `=`, algebraic})}, but received [I*E*(diff(diff(diff(diff(w(x, t), x), x), x), x))+m*(diff(diff(w(x, t), t), t)) = 0, [w(0, t) = 0, (D[1](w))(0, t) = 0, I*E*(D[1, 1](w))(L, t) = 0, I*E*(D[1, 1, 1](w))(L, t) = 0]]"


It's seems I'm introducing the Boundary conditions of the second and third derivatives in a wrong way, but I can't discover how to do it.

Thanks very much in advance to everybody!!

Ger89



P.D. - I have use this "tutorial" to write the code ( http://homepages.math.uic.edu/~jan/mcs494f02/Lec34/pde.html ). Thanks very much again. 

 

I've got the following double integral over a region A:

e^(1/x*y)/(y^2)*(x+1)^2 where A={(x,y):1/2<=x*y<=2,1<=x<=3}

to evaluate this:

I've tried :

int6:=int(int(e^(1/x*y)/((y^2)*(x+1)^2),x=1..3),y=(1/2)..(2/3));

since the largest lower bound and smallest upper bound for y based on 1/2<=xy<=2 are 1/2 and 2/3 respectively.

This statement however, only evaluates the inner integral; is my approach correct?

Dear All,

I am solving 6 ODE equations with boundary conditions using Runge kutta Felbergh 45 (Maple 12). then, i got this problem.. any suggestion??

Thank you :)

ISPC3.mw

``

restart; with(plots); M := 3; k = .2; blt := 6; r := 2; l := .1; Pr := 6.8; Ec := 2; N := .5; rho := .5; Tv := .5; Tt := .5; c := 1; cm := .1; cp := .1

Eq1 := diff(f(eta), eta, eta, eta)+f(eta)*(diff(f(eta), eta, eta))-(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta))) = 0;

diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-3*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta))) = 0

(1)

Eq2 := G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0;

G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0

(2)

Eq3 := G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0;

G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0

(3)

Eq4 := G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0;

G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0

(4)

Eq5 := diff(theta(eta), eta, eta)+Pr*(f(eta)*(diff(theta(eta), eta))-2*(diff(f(eta), eta))*theta(eta))+N*Pr*(theta1(eta)-theta(eta))/(rho*c*Tt)+N*Pr*Ec*(F(eta)-(diff(f(eta), eta)))^2/(rho*Tv) = 0;

diff(diff(theta(eta), eta), eta)+6.8*f(eta)*(diff(theta(eta), eta))-13.6*(diff(f(eta), eta))*theta(eta)+13.60000000*theta1(eta)-13.60000000*theta(eta)+27.20000000*(F(eta)-(diff(f(eta), eta)))^2 = 0

(5)

Eq6 := 2*F(eta)*theta1(eta)+G(eta)*(diff(theta1(eta), eta))+cp*(theta1(eta)-theta(eta))/(c*cm*Tt) = 0;

2*F(eta)*theta1(eta)+G(eta)*(diff(theta1(eta), eta))+2.000000000*theta1(eta)-2.000000000*theta(eta) = 0

(6)

bcs1 := f(0) = r, (D(f))(0) = -1, (D(f))(blt) = 0, F(blt) = 0, G(blt) = -f(blt), H(blt) = k, theta(0) = 1, theta(blt) = 0, theta1(blt) = 0;

f(0) = 2, (D(f))(0) = -1, (D(f))(6) = 0, F(6) = 0, G(6) = -f(6), H(6) = k, theta(0) = 1, theta(6) = 0, theta1(6) = 0

(7)

L := [0.1e-2];

[0.1e-2]

(8)

for k to 1 do R := dsolve(eval({Eq1, Eq2, Eq3, Eq4, Eq5, Eq6, bcs1}, B = L[k]), [f(eta), F(eta), G(eta), H(eta), theta(eta), theta1(eta)], numeric, output = listprocedure); Y || k := rhs(R[2]); YP || k := rhs(R[3]); YR || k := rhs(R[4]); YQ || k := rhs(R[5]) end do

Error, (in dsolve/numeric/bvp) initial Newton iteration is not converging

 

R

R

(9)

print([(YP || (1 .. 1))(0)]);

[YP1(0)]

(10)

``

P1 := plot([YP || (1 .. 1)], 0 .. 14, labels = [eta, (D(f))(eta)]):

Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct

 

plots:-display([P1]);

 

``

``


Download ISPC3.mw

Hello guys

I have a linear differentional equation which is in the 4th order. It is shown in the below:

P:=phi(x):
eq:=a11*diff(p,x,x,x,x)+a22*diff(p,x,x)+a33*p:
eq:=0:
where a11 and a22 and a33 are constant coefficients. The boundary value for this equation is:

phi(a)=sigma1 , phi(-a)=sigma1 , diff(p,x)(a)=0 , diff(p,x)=0

Now consider :

a11:=2.731e-10:
a22:=-1.651e-9:
a33:=3.09027e-10:
a:=35.714:
sigma1:=200e6:

when I use dsolve for deriving a good answer in this equation. there are four real roots .How can I solve it with these boundary condition?

I need to extract phi(x) from this equation.

Thanks

Hello,

Since I was working in Matlab with Galerkin method which implies periodic boundary conditions I was wondering how to implement this in maple.

I tried this:

restart;

pde2 := diff(u(x, t), t)+3*(diff(u(x, t)^2, x))+diff(u(x, t),x$3) = 0

IBC := {u(0, t) = u(2, t), u(x, 0) = sech(50*(x-1/2))^2+2*sech(30*(x-1))^2, (D[1](u))(0, t) = (D[1](u))(2, t), (D[2](u))(0, t) = (D[2](u))(2, t)}

pds := pdsolve(pde2, IBC, numeric, time = t, range = 0 .. 2)

But it's telling me: 

Error, (in pdsolve/numeric/process_IBCs) initial/boundary conditions can only contain derivatives which are normal to the boundary, got (D[2](u))(0, t)

So what's wrong?

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