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which numeric method ?...

Yesterday at 12:21 PM wo0olf 20

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.

hello, i went solve these equation ,with a & b take any value

b*x*ln(x)-x*ln(a)+a=0

thank you

hi..i am a problem with solving following ....please help me ....thanks alot

dsys3 := {10*f2(x)+12*(diff(f1(x), x))+14*f3(x) = 0, 2*(diff(f1(x), x, x))+4*(diff(f2(x), x))+6*(diff(f3(x), x)) = 0, 16*(diff(f3(x), x, x, x, x))+19*(diff(f3(x), x, x))+22*(diff(f1(x), x))+25*f2(x)+27*f3(x)+29*f3(x)+31+32 = 0, f1(0) = 0, f1(1) = 0, f2(0) = 0, f2(1) = 0, f3(0) = 0, f3(1) = 0, ((D@@1)(f1))(0) = 0, ((D@@1)(f1))(1) = 0, ((D@@1)(f2))(0) = 0, ((D@@1)(f2))(1) = 0, ((D@@1)(f3))(0) = 0, ((D@@1)(f3))(1) = 0}; dsol5 := dsolve(dsys3, 'maxmesh' = 500, numeric, range = 0 .. 1, abserr = .1, output = listprocedure); fy3 := eval(f3(x), dsol5); fy2 := eval(f2(x), dsol5); fy1 := eval(f1(x), dsol5)

ERROR.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.

Having solution of an inequations system, is there a way/function/algorithm to find a particular numeric solution (as simplex[minimize] can do) ?

ex:

Q := {1 < x - y, x + y < 1};

R := solve(Q);

      { x < 1 - y, y < 0, y + 1 < x }

manually it's easy to find some numeric solutions:


      y = -1, x = 1
      y = -2, x = 0

but I need an automatic way.

Thank you for your help
s.py

 

Hi all;

I have following program for plotting numerous function using hybrid functions.

if g1(t) is arbitrary function and g2(t) is its approximate by hybrid functions, I want to have a table of g1(t)-g2(t) for different value of t. but the result is without numeric values. what part is wrong????

best wishes

OHB.mws

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

I got a problem in using dsolve.
In my real question, some functions are quite complicated.
so here is a simple example.
F1:=x(t)^2;F2:=piecewise(t>=0,y(t)^3,t>=0.1,exp(y(t)));
eq1:=diff(x(t),t$2)=F1;eq2:=diff(y(t),t$2)=subs(x=y,F1)-F2;
ic1:=x(0)=1.2,D(x)(0)=0;ic2:=y(0)=MM(tf),D(y)(0)=NN(tf);
#tf is the point where x(tf)=30.
dsolve({eq1,eq2,ic1,ic2,x(tf)=30},numeric);

above command can't give an answer.

how to use dsolve solve this problem?

any ideas is appreciated.

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}


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