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Dear all,

restart:with(plots):
eq1:=diff(f(y), y$4)-(diff(f(y), y$2));

bcs:=f(h1) = (1/2), f(h2) = -(1/2), (D(f))(h1) = -1, (D(f))(h2) = -1:

h1:= 1+cos(x):h2:=-1-cos(x+g):

db:=eq1,bcs:
d1 := subs(g=1,[db]):
P1:= eval(diff(diff(f(y),y$2)-f(y),y));

for x from 0 to 1 by 0.1 do
F2[x]:=dsolve(d1, numeric,maxmesh=25500,output=listprocedure): 
P2[x]:=subs(F2[x],P1); # subing values into P1 
end do:
Vls:=Vector([seq(P2[x],x=0..1,0.1)]):
XX := `<|>`(`<,>`(seq(x, x = 0..1, 0.1))):
plot(<<XX>|<Vls>>, color=red);

I'm trying to plot P1 vs x but getting empty plot. Please help me out. 

Thanks

 

Good day, can any one help in writing maple programme for the finite difference (FD) formulae define to solve this coupled non-linear  ODEs. See it here FDM_programme.mw Thank you

NOTE: please disregard the earlier link.

Hi

I'm dealing with 2nd order ODE on Maple. By using " infolevel 5" Maple tell me that it use Kovacic's algorithm to find the solution. Could anybody tell me how or at least some idea so that I can go on this my self. Following here my ODE

Thank you so much

Chaimongkol

Hi:
i solved nonlinear ode in terms of t (y(t)) with dsolve command,how i will evaluate value function(y(t)) in points t=0..1 with delta t=0.01 and results(t and y(t)) inside a excel file?

eq := diff(y(t), t, t)-y(t)^2 = 1
res := dsolve({eq, y(0) = 0, (D(y))(0) = 0}, {y(t)}, numeric)

 

Hello! How can I find extremes of numeric solution of ODE system obtained using "dsolve"? Can I use something like "extrema" function?

I failed to solve the ODE system shown as follows, where y1(x) and y2(x) are functions of x, ranging from -L/2 to L/2. All the other parameters are constants (A,B,C,F,G). The analytic or numeric solution of y1(x) and y2(x) are wanted.Really appreciate for you experts' help and time!!!

dsys:={diff(y1(x),x$2)-A*x^2*y1=B*diff(y2(x),x$3),diff(y2(x),x$2)-C*diff(y1(x),x)=F*x^2+G}

boundary conditions:y1(0)=0, diff(y2(L/2),x$2)=0, D(y2)(0)=0, y2(L/2)=0

 

 


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}


restart:with(plots):with(PDEtools):

 pde:=1/r*diff(r*diff(U(r,z),r),r)+1/b^2*diff(U(r,z),z,z)=0;

ans := pdsolve(pde);

ics:=(D[1](U))(0, z) = 0,(D[2](U))(r, 0)-B*U(r,0) = 0;

bcs:=(D[2](U))(r, 1)+B*U(r,1) = B,(D[1](U))(1, z)+B*U(1,z) = 0;

B:=1:b:=1:

S:= pdsolve(pde, {bcs, ics}, numeric);

Error, (in pdsolve/numeric) unable to handle elliptic PDEs

anyway around this?

 

RIZPDE.mw

Hi:

when use the dsolve,numeric,I see error,why?

f := (x, t) -> piecewise(t < 10, 0.480e9*(1-(1/10)*t)*sin(Pi*x), 10 < t, 0)

eq1 := diff(y(t), t, t)-y(t)^2-f(x,t) = 0:
eq2 :=simplify( int(lhs(eq1)*sin(Pi*x), x = 0 .. 1) = 0):

dsolve({eq2, y(0) = 0, (D(y))(0) = 0}, numeric)

initial conditions are zero.

 

 

Good day,

What scheme does midrich method is using in solving BVP?

Thanks.

I am a problem with solve differential equation, please help me: THANKS 

g := (y^2-1)^2; I4 := int(g^4, y = -1 .. 1); I5 := 2*(int(g^3*(diff(g, y, y)), y = -1 .. 1)); I6 := int(g^3*(diff(g, y, y, y, y)), y = -1 .. 1); with(Student[Calculus1]); I10 := ApproximateInt(6/(1-f(x)*g)^2, y = -1 .. 1, method = simpson);

dsys3 := {I4*f(x)^2*(diff(f(x), x, x, x, x))+I5*f(x)^2*(diff(f(x), x, x))+I6*f(x)^3 = I10, f(-1) = 0, f(1) = 0, ((D@@1)(f))(-1) = 0, ((D@@1)(f))(1) = 0};

dsol5 := dsolve(dsys3, numeric, output = array([0.]));

              Error, (in dsolve/numeric/bvp) system is singular at left endpoint, use midpoint method instead

****************FORMAT TWO ********************************************************

g := (y^2-1)^2; I4 := int(g^4, y = -1 .. 1); I5 := 2*(int(g^3*(diff(g, y, y)), y = -1 .. 1)); I6 := int(g^3*(diff(g, y, y, y, y)), y = -1 .. 1); with(Student[Calculus1]); I10 := ApproximateInt(6/(1-f(x)*g)^2, y = -1 .. 1, method = simpson);
dsys3 := {I4*f(x)^2*(diff(f(x), x, x, x, x))+I5*f(x)^2*(diff(f(x), x, x))+I6*f(x)^3 = I10, f(-1) = 0, f(1) = 0, ((D@@1)(f))(-1) = 0, ((D@@1)(f))(1) = 0};

dsol5 := dsolve(dsys3, method = bvp[midrich], output = array([0.]));
%;
                                   Error, (in dsolve) too many levels of recursion

I DONT KNOW ABOUT THIS ERROR

PLEASE HELP ME

THANKS A LOT

 

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

hello

we have an exam next week and I want to know

how I can write (fordo) in maple for numerical integration

in different methods such as trapezoid , newton cotes and so on.

thanks

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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