Maple 2018 Questions and Posts

These are Posts and Questions associated with the product, Maple 2018

 

What do I need to do to the "2 + 3" in the attached Document in order to make it evaluate? I know about F5 to switch between Text and Math modes, but that's not enough to get me where I want to be. The "2 + 3" is already in Math mode, but that's not enough to get it to evaluate.

The Document: t.mw

 

Download Analisa_Dinamik_Limb_v1_(30).mwAnalisa_Dinamik_Limb_v1_(30).mw

So I have an equation that basically takes the component of vectors to be used as an equation. The variables that I after are FB1z, FB2x, and FB3y For example here is my equation: 

EOM1:=(AFB1[1]+AFB2[1]+AFB3[1])=TEOM[1]

EOM2:=(AFB1[2]+AFB2[2]+AFB3[2])=TEOM[2]

EOM3:=(AFB1[3]+AFB2[3]+AFB3[3])-TEOM[3]:

FBBp1:=FBPP1=(EulP1[1]+EulP2[1]+EulP3[1]):
FBBp2:=FBPP2=(EulP1[2]+EulP2[2]+EulP3[2]):
FBBp3:=FBPP3=(EulP1[3]+EulP2[3]+EulP3[3]):

However there are unknown variable in AFB2[1] named FB2x and AFB3[1] named FB3y. Then AFB1[2] has unknown equation named FB1z and AFB3[2] has FB3y and so on. While in my FBBp1,FBBp2,and FBBp3 holds all of the variable of FB1z, FB2x, and FB3x
I have tried to use 'solve' command to find the variable but my computer won't stop processing it:
sls:=solve({EOM1,EOM2,EOM3,FBBp1,FBBp2,FBBp3},{FB1z,FB2x,FB3y}):

I tried to use the Gauss-Elimination by forming a matrix but it doesn't work as well since I am really confused how to take out the variables out of the vector component.

zzz:=Matrix([0,AFB2[1],AFB3[1],jjj[1]],[AFB1[2],0,AFB3[2],jjj[2]],[AFB1[3],AFB2[3],0,jjj[3]],[FBP1[1],FBP2[1],FBP3[1],EulP[1]],[FBP1[2],FBP2[2],FBP3[2],EulP[2]],[FBP1[3],FBP2[3],FBP3[3],EulP[3]]):
GaussElimination:=(zzz)

I would be very grateful If someone could help me. Thankyou

 

Edit: here are the .txt files and .mpl files that required to run the program 

Inverse_Kinematics_ADRIAN2.mw

RotInertiax0_ADRIAN.txt

Download DisplacementXYZ.txt

inersia_platfrom.txt

There is an .mpl file that I couldn't upload so I will upload it in the comments

 

How to draw a circle transfomed by orthogonal affinity to oblique line and k ration ? Thank you.

Etude d'un cas particulier a := 5: b := 7: k := 9: A := [a, 0]: B := [0, b]: #A et B fixes P := [t, 0]: Q := [0, k/t]:#P et Q 2 points mobiles cir := -a*x-b*y+x^2+y^2 = 0: sol := solve(subs(y = 5, cir), x): cen := [solve(diff(cir, x)), solve(diff(cir, y))]: x0 := sol[1]: y0 := 5: M := [x0, y0]: R := sqrt(cen[1]^2+cen[2]^2): beta := arctan(diff(solve(EQ(M, cen), y), x)): Recherche des valeurs de t pour que les 2 droites soient perpendiculaires eq := t^2*(y0-b)+t*(a*b-a*y0+b*x0-k)-x0*(a*b-k) = 0; sol := solve(eq, t); t := sol[1]; tp := sol[2]; P1 := [t, 0]; Q1 := [0, k/t]; PQ1 := simplify(x*(-a*b+b*t+k)+y*t*(t-a)-t*(-a*b+b*t+k)) = 0:#1ere tangente PQ2 := simplify(x*(-a*b+b*tp+k)+y*tp*(tp-a)-tp*(-a*b+b*tp+k)) = 0:#2ième tangente P2 := [tp, 0]; Q2 := [0, k/tp]; CIR := implicitplot(cir, x = -4 .. 8, y = -4 .. 12, color = red); Fig := proc (alpha) local Dr1, DR1, Dr2, DR2, N, u0, v0, Po, t, tp, sol; global a, b, k, cen, R; u0 := cen[1]+R*cos(alpha); v0 := cen[2]+R*sin(alpha); N := [u0, v0]; sol := solve(t^2*(v0-b)+t*(b*u0-a*v0+a*b-k)-u0*(a*b-k) = 0, t); t := sol[1]; tp := sol[2]; Dr1 := simplify(x*(-a*b+b*t+k)+y*t*(t-a)-t*(-a*b+b*t+k)) = 0; DR1 := implicitplot(Dr1, x = -4 .. 8, y = -4 .. 12, color = brown); Dr2 := simplify(x*(-a*b+b*tp+k)+y*tp*(tp-a)-tp*(-a*b+b*tp+k)) = 0; DR2 := implicitplot(Dr2, x = -4 .. 8, y = -4 .. 12, color = pink); Po := pointplot([N[]], symbol = solidcircle, color = [black], symbolsize = 8); display([Po, DR1, DR2]) end proc; DrPQ1 := implicitplot(PQ1, x = -4 .. 22, y = -4 .. 12, color = blue); DrPQ2 := implicitplot(PQ2, x = -4 .. 22, y = -4 .. 12, color = blue); Points := pointplot([A[], B[], M[], P1[], P2[], Q1[], Q2[], cen[]], symbol = solidcircle, color = [green], symbolsize = 10); T := plots:-textplot([[A[], "A"], [B[], "B"], [M[], "M"], [P1[], "P1"], [P2[], "P2"], [Q1[], "Q1"], [Q2[], "Q2"], [cen[], "cen"]], font = [times, 10], align = {below, left}); n := 19; display([seq(Fig(2*i*Pi/n), i = 0 .. n), Fig(beta), CIR, DrPQ1, DrPQ2, Points, T], scaling = constrained, size = [500, 500]); I would find out the focus of the ellipse. Thank you.

Hello

I use Maple 20018.2.

When I use "Data Set Search" and press Search I get following Error Message. I check that the network access to the internet is on enable. Does anybody has an Idea?

thank you

Murad

I'm trying to obtain the dynamical response of a simply-supported beam with a cantilever extension, coupled to a spring-mass system. In mathematical terms, this system is ruled by three PDEs (relative to each bare part of the main structure) and one ODE (relative to the spring-mass system). I think my mathemical model is finely formulated, but Maple keeps telling me this:

Error, (in pdsolve/numeric/process_IBCs) improper op or subscript selector

I believe it is because my PDEs depend on "x" and "t", while the ODE depends solely on "t". I have tried to transform my ODE into a "PDE", making it also dependent of "x", but without imposing any boundary conditions relative to "x". However, after this Maple points a new error message:

Error, (in pdsolve/numeric) initial/boundary conditions must be defined at one or two points for each independent variable

Could someone help me finding a solution? My algorythm in shown in the attached file below.

Worksheet.mw

Can anyone help me to frame the equations in Fractional Reduced Differential Transform Method 

system of nonlinear ordinary differential equations
ds/ dt = b−γ s(t)− (δ s(t)(i(t) + βa(t)) /N − ε s(t) m(t) 
de/ dt = δ (s(t)(i(t) + βa(t))/ N + ε s(t) m(t) − (1−ϑ) θ e(t) − ϑ α e(t) − γ e(t) 
di/ dt = (1−ϑ) θ e(t) − (ρ + γ) i(t)
da/ dt = ϑ α e(t) − (σ + γ) a(t)
dr /dt = ρ i(t) + σ a(t) − γ r(t)
dm /dt = τ i(t) + κ a(t) − ω m(t) 

Maple 2018  has recently has  become sluggish to start up  -and very slow to respond to input. Can anyone suggest remedies?  I have plenty of space and CPU. Other apps seem to start fine.  Can any suggest a diagnosis and/or solution?

Melvin


 

Analysis of the semiclassical (SC) momentum rate equations

Plotting the ICs and BCs and examining sensitivity to the Re and Im forces

MRB: 24/2/2020, 27/2/2020, 2/3/2020.

We examine solution of the SC version of the momentum rate equations, in which O`ℏ`^2 terms for u(x, t) are removed. A high level of sensitivity to ICs and BCs makes solution finding difficult.

restart;

with(PDETools): with(CodeTools):with(plots):

We set up the initial conditions:

ICu := {u(x, 0) = .1*sin(2*Pi*x)}; ICv := {v(x, 0) = .2*sin(Pi*x)};

{u(x, 0) = .1*sin(2*Pi*x)}

 

{v(x, 0) = .2*sin(Pi*x)}

(1)

plot([0.1*sin(2*Pi*x),0.2*sin(Pi*x)],x = 0..2, title="ICs:\n u(x,0) (red), v(x,0) (blue)",color=[red,blue],gridlines=true);  

 

The above initial conditions represent a positive velocity field v(x, 0) (blue) and a colliding momentum field u(x, t)(red).

 

Here are the BCs

BCu := {u(0,t) = 0.5*(1-cos(2*Pi*t))};

{u(0, t) = .5-.5*cos(2*Pi*t)}

(2)

BCv := {v(0,t) = 0.5*sin(2*Pi*t),v(2,t)=-0.5*sin(2*Pi*t)};  

{v(0, t) = .5*sin(2*Pi*t), v(2, t) = -.5*sin(2*Pi*t)}

(3)

plot([0.5*(1-cos(2*Pi*t)),0.5*sin(2*Pi*t),-0.5*sin(2*Pi*t)],t=0..1,color=[red,blue,blue],linestyle=[dash,dash,dot],title="BCs:\n u(0,t) (red-dash),\n v(0,t) (blue-dash), v(1,t) (blue-dot)",gridlines=true);

 

 

We can now set up the PDEs for the semiclassical case.

hBar:= 1:m:= 1:Fu:= 0.2:Fv:= 0.1:#1.0,0.2

pdeu := diff(u(x,t),t)+u(x,t)/m*(diff(u(x,t),x)) = Fu;

diff(u(x, t), t)+u(x, t)*(diff(u(x, t), x)) = .2

(4)

pdev := diff(v(x,t),t)+u(x,t)/m*(diff(v(x,t),x))-hBar*(diff(u(x,t),x$2))/(2*m)+v(x,t)*(diff(u(x,t),x))/m = Fv;  

diff(v(x, t), t)+u(x, t)*(diff(v(x, t), x))-(1/2)*(diff(diff(u(x, t), x), x))+v(x, t)*(diff(u(x, t), x)) = .1

(5)

ICu:={u(x,0) = 0.1*sin(2*Pi*x)};  

{u(x, 0) = .1*sin(2*Pi*x)}

(6)

ICv:={v(x,0) = 0.2*sin(Pi*x/2)};  

{v(x, 0) = .2*sin((1/2)*Pi*x)}

(7)

IC := ICu union ICv;  

{u(x, 0) = .1*sin(2*Pi*x), v(x, 0) = .2*sin((1/2)*Pi*x)}

(8)

BCu := {u(0,t) = 0.5*(1-cos(2*Pi*t)), D[1](u)(2,t) = 0.1*cos(2*Pi*t)};

{u(0, t) = .5-.5*cos(2*Pi*t), (D[1](u))(2, t) = .1*cos(2*Pi*t)}

(9)

BCv := {v(0,t) = 0.2*(1-cos(2*Pi*t))};  

{v(0, t) = .2-.2*cos(2*Pi*t)}

(10)

BC := BCu union BCv;  

{u(0, t) = .5-.5*cos(2*Pi*t), v(0, t) = .2-.2*cos(2*Pi*t), (D[1](u))(2, t) = .1*cos(2*Pi*t)}

(11)

We now set up the PDE solver:

pds := pdsolve({pdeu,pdev},{BC[],IC[]},time = t,range = 0..2,numeric);#'numeric' solution

_m2592591229440

(12)

Cp:=pds:-animate({[u, color = red, linestyle = dash],[v,color = blue,linestyle = dash]},t = 30,frames = 400,numpoints = 400,title="Semiclassical momentum equations solution for Re and Im momenta u(x,t) (red) and v(x,t) (blue) \n under respective constant positive forces [0.2, 0.1] \n with sinusoidal boundary conditions at x = 0, 1 and sinusoidal initial conditions: \n time = %f ", gridlines = true,linestyle=solid):Cp;

Error, (in pdsolve/numeric/animate) unable to compute solution for t>HFloat(0.0):
Newton iteration is not converging

 

Cp

(13)

Observations on the quantum case:

The classical equation for u(x, t) is independent of the equation for v(x, t).  u(x, t) (red) is a solution of the classical Burgers equation subject to a force 0.2, but u(x, t) is NOT influenced by v(x, t).  On the otherhand, v(x, t) (blue) is a solution of the quantum dynamics equation subject to force 0.1 and is influenced by u(x, t).   This one way causality (u " implies v")  is a feature of the semiclassical case, and it emphasises the controlling influence of the classical u(x, t), which modulates the quantum solution for v(x, t).  Causally, we have u" implies v".

 

The initial conditions are of low momentum amplitude:"+/-"0.1 for the classical u(x, 0) (red) field and`&+-`(0).2 for v(x, 0) (blue)  but their influence is soon washed out by the boundary conditions "u(0,t) ~1, v(0,t)~0.5" and "v(1,t)~0.5" that drive the momentum dynamics.

 

The temporal frequency of the boundary condition on the v-field is twice that of the classical u-field. This is evident in the above blue transient plot. Moreover, the">=0" boundary condition on the classical u-momentum (red), drives that field in the positive direction, initially overtaking the quantum v(x, t) field, as consistent with the applied forces [0.2, 0.1]. NULLAlthough initially of greater amplitude than the classical u(x, t)field, the v(x, t) momentum field is asymptotically of the same amplitude as the u(x, t) field, but has greater spatial and temporal frequency, owing to the boundary conditions.

 

Referring to the semiclassical momentum rate equations, we note that the classical field u(x, t) (red) modulates the quantum momentum rate equation for v(x, t).

``

 

 

 


 

Download SC-plots.mw

I am having difficulty getting solutions to a pair of PDEs.  Would anyone like to cast an eye over the attached file, incase I am missing something.

Thanks

Melvin

Hello, 

For a few days Maple crashs everytime i try to use the command "plot3d()". 

I had'nt this problem befor and I have no idea what the reason could be. It ist irrelevant what Funktion I try to visualize,  the window just get closed evertime.

I hope someone can help me.

Thank you!

Tom

How do I simplify KdV equation in Maple by using =fxt))xx)?)


 

``

   I am by using =2*difffxtxx)
    My aim is to get the form
   diff((f*(diff(f, x, t))-(diff(f, x))*(diff(f, t))+f*(diff(f, x, x, x))-4*(diff(f, x, x, x))*(diff(f, x))+3*(diff(f, x, x))^2)/f^2, x) = 0

NULL

``

restart; with(PDEtools); with(DEtools)

``

alias(u = u(x, t)); declare(u(x, t)); alias(f = f(x, t)); declare(f(x, t))

u

 

` u`(x, t)*`will now be displayed as`*u

 

u, f

 

` f`(x, t)*`will now be displayed as`*f

(1)

KdV := diff(u, t)+6*u.(diff(u, x))+diff(u, x, x, x) = 0

diff(u, t)+6*(u.(diff(u, x)))+diff(diff(diff(u, x), x), x) = 0

(2)

KdV_f := eval(KdV, u = 2*(diff(ln(f), x, x)))

2*(diff(diff(diff(f, t), x), x))/f-2*(diff(diff(f, x), x))*(diff(f, t))/f^2-4*(diff(f, x))*(diff(diff(f, t), x))/f^2+4*(diff(f, x))^2*(diff(f, t))/f^3+6*((2*(diff(diff(f, x), x))/f-2*(diff(f, x))^2/f^2).(2*(diff(diff(diff(f, x), x), x))/f-6*(diff(diff(f, x), x))*(diff(f, x))/f^2+4*(diff(f, x))^3/f^3))+2*(diff(diff(diff(diff(diff(f, x), x), x), x), x))/f-10*(diff(diff(diff(diff(f, x), x), x), x))*(diff(f, x))/f^2+40*(diff(diff(diff(f, x), x), x))*(diff(f, x))^2/f^3-20*(diff(diff(diff(f, x), x), x))*(diff(diff(f, x), x))/f^2-120*(diff(diff(f, x), x))*(diff(f, x))^3/f^4+60*(diff(diff(f, x), x))^2*(diff(f, x))/f^3+48*(diff(f, x))^5/f^5 = 0

(3)

df := collect(KdV_f, f)

6*((2*(diff(diff(f, x), x))/f-2*(diff(f, x))^2/f^2).(2*(diff(diff(diff(f, x), x), x))/f-6*(diff(diff(f, x), x))*(diff(f, x))/f^2+4*(diff(f, x))^3/f^3))+(2*(diff(diff(diff(f, t), x), x))+2*(diff(diff(diff(diff(diff(f, x), x), x), x), x)))/f+(-2*(diff(diff(f, x), x))*(diff(f, t))-20*(diff(diff(diff(f, x), x), x))*(diff(diff(f, x), x))-4*(diff(f, x))*(diff(diff(f, t), x))-10*(diff(diff(diff(diff(f, x), x), x), x))*(diff(f, x)))/f^2+(60*(diff(diff(f, x), x))^2*(diff(f, x))+4*(diff(f, x))^2*(diff(f, t))+40*(diff(diff(diff(f, x), x), x))*(diff(f, x))^2)/f^3-120*(diff(diff(f, x), x))*(diff(f, x))^3/f^4+48*(diff(f, x))^5/f^5 = 0

(4)

factor(simplify(df, size))

2*(12*(((diff(diff(f, x), x))*f-(diff(f, x))^2)/f^2.(((diff(diff(diff(f, x), x), x))*f^2-3*(diff(diff(f, x), x))*(diff(f, x))*f+2*(diff(f, x))^3)/f^3))*f^5+f^4*(diff(diff(diff(f, t), x), x))+f^4*(diff(diff(diff(diff(diff(f, x), x), x), x), x))-f^3*(diff(diff(f, x), x))*(diff(f, t))-10*f^3*(diff(diff(f, x), x))*(diff(diff(diff(f, x), x), x))-2*f^3*(diff(f, x))*(diff(diff(f, t), x))-5*f^3*(diff(f, x))*(diff(diff(diff(diff(f, x), x), x), x))+30*f^2*(diff(diff(f, x), x))^2*(diff(f, x))+2*f^2*(diff(f, t))*(diff(f, x))^2+20*f^2*(diff(diff(diff(f, x), x), x))*(diff(f, x))^2-60*(diff(diff(f, x), x))*(diff(f, x))^3*f+24*(diff(f, x))^5)/f^5 = 0

(5)

``

``

``

``


 

Download KdV_simplify

Hello
    In this example, we have the KdV equation    
         t] - 6 uux] + xxx] = 0                
    I would like to find the Lax pair for the KdV equation, which are    
               Lψ=λψ                
               ψ[t] = Mψ                
        
              Lt+ML-LM = 0  called a compatibility condition               
    So, I will start from this purpose    
    Then we will assume M in the form   
    will assume M in the form   
              M := a3*Dx^3+a^2+a1*Dx+a0              
    thenb using M and L in the for L[tL-LM = 0can find   
      Dx^5+( ) Dx^4+( ) Dx^3+( ) Dx^2+( ) Dx+( )=0              
    then wean find a_i =0,1,2,3   
  In the following maple code to do that 
  my question is    
   .How I canoue the soluo get a_i2,3 usinmaple code  
    any maple packge to find Lax pair for PDE -  


 

restart; with(DEtools); with(PDEtools)

     in this exampile we have KdV equation

      u[t]-6*uu[x]+u[xxx] = 0

    I would likeind the Lax pair for the KdV equation, which are

       Lψ=λψ

    psi[t] = M*psi

   where``

    L[t]+ML-LM = 0    called  apatibility  condition

    So, I  will start this purpose

     L:=-Dx^2+u;

    then we will assume M the m

    Ma3*Dx^3+a2*Dx^2+Dx+a0

    then busing in the form L[t]+ML-LM = 0 can find

  ( ) Dx^5+( ) Dx^4+( ) Dx^3+( ) Dx^2+( ) Dx+( )=0

 then we can find a_i ;i=,2,3

  

the fllowile code to

 my queion is ;

  1) How I can continue the solution  to get a_i ;i=0,1,2,3 using maple code  ?

  2) isir any maple packge to find  Lax pair for PDE ?

 

alias(u = u(x, t)); declare(u(x, t)); alias(a3 = a3(x, t)); declare(a3(x, t)); alias(a2 = a2(x, t)); declare(a2(x, t)); alias(a1 = a1(x, t)); declare(a1(x, t)); alias(a0 = a0(x, t)); declare(a0(x, t))

u

 

` u`(x, t)*`will now be displayed as`*u

 

u, a3

 

` a3`(x, t)*`will now be displayed as`*a3

 

u, a3, a2

 

` a2`(x, t)*`will now be displayed as`*a2

 

u, a3, a2, a1

 

` a1`(x, t)*`will now be displayed as`*a1

 

u, a3, a2, a1, a0

 

` a0`(x, t)*`will now be displayed as`*a0

(1)

_Envdiffopdomain := [Dx, x]

[Dx, x]

(2)

L := -Dx^2+u

-Dx^2+u

(3)

M := Dx^3*a3+Dx^2*a2+Dx*a1+a0

a3*Dx^3+a2*Dx^2+a1*Dx+a0

(4)

 

 

 

LM := expand(mult(L, M))

-a3*Dx^5-2*Dx^4*(diff(a3, x))-a2*Dx^4+Dx^3*u*a3-Dx^3*(diff(diff(a3, x), x))-2*Dx^3*(diff(a2, x))-Dx^3*a1+Dx^2*u*a2-Dx^2*(diff(diff(a2, x), x))-2*Dx^2*(diff(a1, x))-Dx^2*a0+Dx*u*a1-Dx*(diff(diff(a1, x), x))-2*Dx*(diff(a0, x))+u*a0-(diff(diff(a0, x), x))

(5)

ML := expand(mult(M, L))

-a3*Dx^5-a2*Dx^4+Dx^3*u*a3-Dx^3*a1+3*Dx^2*a3*(diff(u, x))+Dx^2*u*a2-Dx^2*a0+3*Dx*a3*(diff(diff(u, x), x))+2*Dx*a2*(diff(u, x))+Dx*u*a1+a3*(diff(diff(diff(u, x), x), x))+a2*(diff(diff(u, x), x))+a1*(diff(u, x))+u*a0

(6)

Commutator := simplify(ML-LM)

a3*(diff(diff(diff(u, x), x), x))+(3*Dx*a3+a2)*(diff(diff(u, x), x))+diff(diff(a0, x), x)+Dx*(diff(diff(a1, x), x))+Dx^2*(diff(diff(a2, x), x))+Dx^3*(diff(diff(a3, x), x))+(3*Dx^2*a3+2*Dx*a2+a1)*(diff(u, x))+2*Dx^4*(diff(a3, x))+2*Dx^3*(diff(a2, x))+2*Dx^2*(diff(a1, x))+2*Dx*(diff(a0, x))

(7)

sol := diff(L, t)-Commutator = 0

diff(u, t)-a3*(diff(diff(diff(u, x), x), x))-(3*Dx*a3+a2)*(diff(diff(u, x), x))-(diff(diff(a0, x), x))-Dx*(diff(diff(a1, x), x))-Dx^2*(diff(diff(a2, x), x))-Dx^3*(diff(diff(a3, x), x))-(3*Dx^2*a3+2*Dx*a2+a1)*(diff(u, x))-2*Dx^4*(diff(a3, x))-2*Dx^3*(diff(a2, x))-2*Dx^2*(diff(a1, x))-2*Dx*(diff(a0, x)) = 0

(8)

collect(sol, [Dx, x])

-2*Dx^4*(diff(a3, x))+(-(diff(diff(a3, x), x))-2*(diff(a2, x)))*Dx^3+(-3*a3*(diff(u, x))-(diff(diff(a2, x), x))-2*(diff(a1, x)))*Dx^2+(-2*a2*(diff(u, x))-3*a3*(diff(diff(u, x), x))-(diff(diff(a1, x), x))-2*(diff(a0, x)))*Dx-a1*(diff(u, x))-a2*(diff(diff(u, x), x))-a3*(diff(diff(diff(u, x), x), x))-(diff(diff(a0, x), x))+diff(u, t) = 0

(9)

 

 

 

 

``

NULL


 

Download find_lax_pair.mw

I want to define the co-ordianates (phi, PI, ....)  as functions of some variable eg:- x,y.

 

I am studying the motion of a beam coupled to piezoelectric strips. This continuous system is modelled by two DE:

YI*diff(w(x,t), x$4)-N[0]*cos(2*omega*t)*diff(w(x,t), x$2)+c*diff(w(x,t), t)+`ρA`*diff(w(x,t), t$2)+C[em,m]*v(t) = 0;

And:

C[p]*diff(v(t), t)+1/R[l]*v(t) = C[em,e]*(D[1,2](w)(0,t)-D[1,2](w)(ell,t));
 

where "w(x,t)" stands for the beam's vibration and "v(t)" means the electric voltage, which is constant throught the beam. I would like to numerically solve both DE simultaneosly, but maple will not let me do it. I would like to know why. I am getting the following error:

Error, (in pdsolve/numeric/process_PDEs) number of dependent variables and number of PDE must be the same
 

I suppose it is because "w(x,t)" depends on "x" and "t", while "v(t)" depends solely on time, but I am not sure. Could someone help me out? Here is my current code:

restart:
with(PDEtools):
declare(w(x,t), v(t)):

YI*diff(w(x,t), x$4)-N[0]*cos(2*omega*t)*diff(w(x,t), x$2)+c*diff(w(x,t), t)+`ρA`*diff(w(x,t), t$2)+C[em,m]*v(t) = 0;
pde1:= subs([YI = 1e4, N[0] = 5e3, c = 300, omega = 3.2233993, C[em,m] = 1], %):
ibc1:= w(0,t) = 0, D[1,1](w)(0,t) = 0, w(ell,t) = 0, D[1,1](w)(ell,t) = 0, D[2](w)(x,0) = 0, w(x,0) = sin(Pi*x/ell):

C[p]*diff(v(t), t)+1/R[l]*v(t) = C[em,e]*(D[1,2](w)(0,t)-D[1,2](w)(ell,t));
pde2:= subs([C[p] = 10, R[l] = 1000, C[em,e] = 1, ell = 5], %):
ibc2:= v(0) = 0:

pdsolve({pde1, pde2}, {ibc1, ibc2}, numeric);

Thanks.

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