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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?

> restart;
> with(plots);
> Eq1 := diff(f(eta), eta, eta, eta)+f(eta)*(diff(f(eta), eta, eta))-2*(diff(f(eta), eta))^2-M^2*(diff(f(eta), eta)) = 0;
/ d / d / d \\\ / d / d \\
|----- |----- |----- f(eta)||| + f(eta) |----- |----- f(eta)||
\ deta \ deta \ deta /// \ deta \ deta //

2
/ d \ 2 / d \
- 2 |----- f(eta)| - M |----- f(eta)| = 0
\ deta / \ deta /
> Eq2 := 1+(4/3)*R*(diff(theta(eta), eta, eta))+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta)) = 0;
4 / d / d \\
1 + - R |----- |----- theta(eta)||
3 \ deta \ deta //

/ / d \ / d \ \
+ Pr |f(eta) |----- theta(eta)| - |----- f(eta)| theta(eta)| = 0
\ \ deta / \ deta / /
> bcs1 := f(0) = S, (D(f))(0) = 1+L*G, (D(D(f)))(0) = .1, f(6) = 0;
f(0) = S, D(f)(0) = 1 + L G, @@(D, 2)(f)(0) = 0.1, f(6) = 0
> fixedparameter := [S = .1, M = .1];
[S = 0.1, M = 0.1]
> Eq3 := eval(Eq1, fixedparameter);
/ d / d / d \\\ / d / d \\
|----- |----- |----- f(eta)||| + f(eta) |----- |----- f(eta)||
\ deta \ deta \ deta /// \ deta \ deta //

2
/ d \ / d \
- 2 |----- f(eta)| - 0.01 |----- f(eta)| = 0
\ deta / \ deta /
> fixedparameter := [R = .1, Pr = .7];
[R = 0.1, Pr = 0.7]
> Eq4 := eval(Eq2, fixedparameter);
/ d / d \\ / d \
1 + 0.1333333333 |----- |----- theta(eta)|| + 0.7 f(eta) |----- theta(eta)|
\ deta \ deta // \ deta /

/ d \
- 0.7 |----- f(eta)| theta(eta) = 0
\ deta /
> bcs2 := theta(0) = 1+T*B, (D(theta))(6) = B, theta(6) = 0;
theta(0) = 1 + T B, D(theta)(6) = B, theta(6) = 0

> T := .1; B := .1;
0.1
0.1
> L := [0., .1, .2, .3];
[0., 0.1, 0.2, 0.3]
> for k to 4 do R := dsolve(eval({Eq3, Eq4, bcs1, bcs2}, L = L[k]), [f(eta), theta(eta)], numeric, output = listprocedure); Y || k := rhs(R[2]); YL || k := rhs(R[3]) end do;
Error, (in dsolve/numeric/bvp/convertsys) too many boundary conditions: expected 6, got 7
> plot([YL || (1 .. 4)], 0 .. 6, 1 .. -.2, labels = [eta, diff(f(eta), eta)]);

 

i am solving 3 ODE with boundary condition.. with boundary condition

 

b.mw

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/b.mw .

Download b.mw

 

then i got this error

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

i dont know where i need to change.. could you help me..

 

i am solving 3 ODE question with boundary condition. when i running the programm i got this error.. any one could help me please.. :)

NULL

restart; with(plots); k := .1; E := 1.0; Pr := 7.0; Ec := 1.0; p := 2.0; blt := 11.5

Eq1 := diff(f(eta), eta, eta, eta)+f(eta)*(diff(f(eta), eta, eta))+Gr*theta(eta)-k*(diff(f(eta), eta))+2*E*g(eta) = 0;

diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))+Gr*theta(eta)-.1*(diff(f(eta), eta))+2.0*g(eta) = 0

(1)

Eq2 := diff(g(eta), eta, eta)+f(eta)*(diff(g(eta), eta))-k*g(eta)-2*E*(diff(f(eta), eta)) = 0;

diff(diff(g(eta), eta), eta)+f(eta)*(diff(g(eta), eta))-.1*g(eta)-2.0*(diff(f(eta), eta)) = 0

(2)

Eq3 := diff(theta(eta), eta, eta)+Pr*(diff(theta(eta), eta))*f(eta)+Pr*Ec*((diff(f(eta), eta, eta))^2+(diff(g(eta), eta))^2) = 0;

diff(diff(theta(eta), eta), eta)+7.0*(diff(theta(eta), eta))*f(eta)+7.00*(diff(diff(f(eta), eta), eta))^2+7.00*(diff(g(eta), eta))^2 = 0

(3)

bcs1 := f(0) = p, (D(f))(0) = 1, g(0) = 0, theta(0) = 1, theta(blt) = 0, (D(f))(blt) = 0, g(blt) = 0;

f(0) = 2.0, (D(f))(0) = 1, g(0) = 0, theta(0) = 1, theta(11.5) = 0, (D(f))(11.5) = 0, g(11.5) = 0

(4)

L := [10, 11, 12];

[10, 11, 12]

(5)

for k to 3 do R := dsolve(eval({Eq1, Eq2, Eq3, bcs1}, Gr = L[k]), [f(eta), g(eta), theta(eta)], numeric, output = listprocedure); Y || k := rhs(R[3]) end do

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

 

R

R

(6)

plot([Y || (1 .. 3)], 0 .. 10, labels = [eta, (D(f))(eta)]);

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

 

 

NULL

NULL


Download tyera(a).mw

With your help I have a solution to a system of three equations:

(parameters are calculated on the basis of the data (for different values) - one example below)
A1=0.00002072968491, A2=0, A3=0.001946449287, A4=0.01946449287

B1=, B2=0, B3=0.0004773383613, B4=0.00004773383613

C1=, C2=0, C3=, C4=0.00009087604510

 

eqa1: = A1 * (diff (Tg (x), x, x)) + A2 * (diff (Tg (x), x)) + (A3 + A4) * tan (x) + A3 * Tg (x) + A4 * Tw (x) = 0;

eqa2: = B1 * (diff (Tw (x), x, x)) + B2 * (diff (Tw (x), x)) + (B3 + B4) * Tw (x) + B3 * Tg (x) + B4 * tan (x) = 0;

eqa3: = C1 * (diff (Tz (x), x, x)) + (C3 + C4) * Tg (x) + C3 * tan (x) + C4 * Tw (x) = 0;

 

indets ({eqa1, eqa2, eqa3}) minus {x};

res: = Dsolve (eval ({eqa1, eqa2, eqa3}) union {boundary conditions ??}, numeric);

 

for k from 0 to 20 evalf (res (k), 4); from;

c1:= 0.524:

c2:=0.05:

m: = 0;

for m from 0 to 20 and

T (m): = c1 * rhs (op (6, res (m))) + c2 * rhs (op (2, res (m))) + (1-c1-c2) * rhs (op (4, res (m))); print (m, T (m)); end to:

 

How and what type boundary conditions (I was thinking about the simplest or third type) to be able to determine the values on the y-axis on the graph. For example, the values started at -10, and ended at 10 (at a point (x, -10), (x, 10) in the coordinate system for a predetermined x, for example, from 0 to 20 which start at the point (0, -10 ) and stop at the point (20,10)). My main purpose is to collect these three solutions  to one equation T (x) = az * Tz (x) + and * Tw (x) + ag * Tg (x), and the ends of the graph, they should be in the above-mentioned points (0, -10 ) - start and (20,10) - stop.

 

Now thank you very much for the advice.

Ewa.

I'm taking my first steps with maple and pdsolve, trying to run the example in the maplesoft support page:

http://www.maplesoft.com/support/help/Maple/view.aspx?path=examples/pdsolve_boundaryconditions

which reads

>
> restart; with(PDEtools);
> U := diff_table(u(x, t));
>

and I get a solution that is different from the web page, and when i run

Im using maple 13. Any tips about what's wrong?

 

regards

hi

 

please help me :

 

 

restart; eq := diff(T(x, y), x) = a*(diff(T(x, y), `$`(y, 2)))/u(y); u := proc (y) options operator, arrow; (-1)*1.218493440*10^11*y^2+4.244913600*10^6*y+0.33e-1 end proc; eq; ICs := (D[1](T))(x, 0) = 1000, (D[1](T))(x, 0.25e-4) = 2000, T(0, y) = 0; T_sol := pdsolve({ICs, eq}, T(x, y)); T_sol

diff(T(x, y), x) = a*(diff(diff(T(x, y), y), y))/u(y)

 

proc (y) options operator, arrow; (-1)*1.218493440*100000000000*(y^2)+4.244913600*1000000*y+0.33e-1 end proc

 

diff(T(x, y), x) = a*(diff(diff(T(x, y), y), y))/(-0.1218493440e12*y^2+4244913.600*y+0.33e-1)

 

(D[1](T))(x, 0) = 1000, (D[1](T))(x, 0.25e-4) = 2000, T(0, y) = 0

 

Error, (in PDEtools:-Library:-NormalizeBoundaryConditions) unable to isolate the functions {T(0, y), (D[1](T))(x, 0), (D[1](T))(x, 0.25e-4)} in the given boundary conditions {T(0, y) = 0, (D[1](T))(x, 0) = 1000, (D[1](T))(x, 0.25e-4) = 2000}

 

T_sol

(1)

NULL

``

BC1 = diff(T(x, 0), y)=1000

BC2 = diff(T(x, 0.000025), y)=2000

IC = T(0,y)=0

where :

u(y)=-1.218493440*10^11*y^2+4.244913600*10^6*y+0.033

Download PDE_Sol.mw

Hello everyone,

i'm trying to simulate a diffusion problem. It contains two connected regions in which a species is diffusing at different speeds. In one region (zeta) one boundary is set to be constant whereas in the other region (c) there is some oscillation at the boundary.The code i try to use is as follows:

sys1 := [diff(c(x, t), t) = gDiffusion*10^5*diff(c(x, t), x$2), diff(zeta(x, t), t) = KDiffusion*10^6*diff(zeta(x, t), x$2)]

pds := pdsolve(sys1, IBC, numeric, time = t, range = 0 .. 3000, spacestep = 3)

However the main problem are my boundary conditions:

IBC := {c(0, t) = 0, c(x > 0, 0) = 0, zeta(0, t) = .4, zeta(x > 0, 0) = .4, (D[1](c))(3000, t) = sin((1/100)*t), (D[1](zeta))(0, t) = 0}

Like this it principally works (however it is apparently ill-posed).

Now what i do like is that the two equations are coupled at x=2000 with the condition that c(2000,t)=zeta(2000,t). This however i dont seem to be able to implement.

I appreciate your comments

Goon

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

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