Items tagged with diff diff Tagged Items Feed

f:=x+y+y;

diff(f,x);
diff(f,y);
diff(f,z);

What I hope to get is a vector with i-th entry being the dervative of f, differentiated w.r.s.t the i-th parameter, like this

Vector([1,1,1]);

 

Is there a more efficient (built-in) command to do this?

 

VectorCalculus[diff] does not do what I want.

 

Thanks,

 

casper

 


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

Hi everyone,

I have a question regarding the derivation of tensors/matrices.
Let's assume for simplicity, that I have a vector (6x1) s and a matrix A (6x6)defining Transpose(s)*Inverse(A)*s. From this function I want to calculate the derivative w.r.t. s. My approach would be

restartwith(Physics):
with(LinearAlgebra):
Define(s,A)

Diff(
Transpose(s)*Inverse(A)*s, s)

As a result I get

though I'd rather expect something like Inverse(A)*s + Transpose(s)*Inverse(A)

Now as I'm pretty new to Maple, I can imagine that my approach is wrong, but I don't know any better and can't seem to get any information out of the help documents.

Thanks in advance for any of your suggestions!

Dear all,

It's very convenient to define a DE or PDE through Differential Operator D, for example,

((D[1, 1]+D[1, 2]+D[2, 2])(z))(x, y) = exp(x)*sin(y)

Is it possible to realize Inverse Operator Method of Operator D? How to solve the following equation if we rewrite the pde through inverse operator method?

(z)(x, y)=((D[1, 1]+D[1, 2]+D[2, 2])^(-1))exp(x)*sin(y)

 

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

 

hi.i encountered this erroe  [Error, (in dsolve/numeric/bvp/convertsys) unable to convert to an explicit first-order system] with solving set of differential equation.please help me.thanks a lot  

dsys3 := {`1`*h1(theta)+`1`*(diff(h1(theta), theta, theta))+`1`*(diff(h2(theta), theta))+`1`*(diff(h2(theta), theta, theta, theta))+`1`*h3(theta)+`1`*(diff(h3(theta), theta, theta))+`1`*(diff(h1(theta), theta, theta, theta, theta)) = 0, `1`*h2(theta)+`1`*(diff(h2(theta), theta, theta, theta, theta))+`1`*(diff(h2(theta), theta, theta))+`1`*(diff(h1(theta), theta))+`1`*(diff(h1(theta), theta, theta, theta))+`1`*(diff(h3(theta), theta))+`1`*(diff(h3(theta), theta, theta, theta)) = 0, h3(theta)^5*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+(diff(h3(theta), theta, theta))*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+(diff(h3(theta), theta, theta, theta, theta))*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+h1(theta)*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+(diff(h1(theta), theta, theta))*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+(diff(h2(theta), theta))*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+(diff(h2(theta), theta, theta, theta))*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+h3(theta)^4*(diff(h2(theta), theta, theta, theta, theta, theta, theta))*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)-beta*h3(theta)^3*`1`-chi*ln(h3(theta))^2*`1`/kappa-chi*`1`/kappa-2*chi*ln(h3(theta))*`1`/kappa = 0, h1(0) = 0, h1(1) = 0, h2(0) = 0, h2(1) = 0, h3(0) = 1, h3(1) = 1, ((D@@1)(h1))(0) = 0, ((D@@1)(h1))(1) = 0, ((D@@1)(h2))(0) = 0, ((D@@1)(h2))(1) = 0, ((D@@1)(h3))(0) = 0, ((D@@1)(h3))(1) = 0, ((D@@2)(h3))(0) = 0, ((D@@2)(h3))(1) = 0}; dsol5 := dsolve(dsys3, 'maxmesh' = 600, numeric, output = listprocedure);
%;
Error, (in dsolve/numeric/bvp/convertsys) unable to convert to an explicit first-order system

 

Hi,

I have a system of diff equations (see below). I am trying to obtain analytical solution. when I assume that z=wN, I receive such solution. Do anybody have idea if I know that z>wN, does this system has an analytical solution?

diff(K(t), t) = -(1/2)*(Q(t)^2*alpha^2*eta*upsilon-2*eta*alpha*(N*upsilon*w*C[max]-z*alpha*K(t))*Q(t)+N*w*(-2*C[max]*z*eta*alpha*K(t)+upsilon*((-N*w+z)*alpha+N*C[max]^2*w*eta)))*K(t)/((C[max]*w*N-alpha*Q(t))*upsilon*N*w)

diff(Q(t), t) = (1/2)*(-z*(Q(t)^2*alpha^2*eta-2*N*Q(t)*alpha*eta*w*C[max]+w*(w*(eta*C[max]^2-alpha)*N+z*alpha)*N)*K(t)-2*N*upsilon*w*(N*w-z)*(C[max]*w*N-alpha*Q(t)))/((C[max]*w*N-alpha*Q(t))*upsilon*N*w)

K(0) = K0, Q(0) = Q0

Thanks,

Dmitry

Hello fellow maple users,im new to the software,im trying to solve a differential system but it dosent work

 

This is the system :

DE1 := diff(Y(t), t) = 5*Y(t)*ln(b(t)/Y(t))-5*Y(t)

DE2 := diff(b(t), t) = 5*b(t)*Y(t)^(3/2)-5*Y(t)

 

Thank you for your help !

Hello,

I wanted to ask whether it's possible to use the taylor command together with vectors from the physics package, maybe I am just doing something wrong here. I tried the following:

and get an error message that diff cannot handle vectors. Of course I could do the expansion by hand an enter the result in Maple but I think it would be a very nice feature because an expansion of vector fields which vary in space and time is such a common problem e.g. in classical electrodynamics. I think of an expansion with non-projected vectors such as

+ higher order terms.

Thanks a lot!

Peter

Hi. I want to differentiate the following expression using "Diff", not "diff". but I want to apply "Diff" to differentiate each separate term based on the chain rule. How can I do that? Does "Diff" apply chain rule for differentiation?

 

 

  

Hello,
I have a system of first order diff. equations which I would like to solve symbolically. Unfortunately, Maple does not solve the system. Do anybody have suggestions how can I solve this system (please see below):

diff(S(t), t) = -eta*(C[max]*w*N-alpha*Q(t))*K(t)*S(t)/(w*N*(S(t)+K(t))),

diff(K(t), t) = S(t)*((z*eta*alpha*(C[max]*w*N-alpha*Q(t))*S(t)-upsilon*(eta*alpha^2*Q(t)^2-2*C[max]*w*N*eta*alpha*Q(t)+((-N*w+z)*alpha+N*C[max]^2*w*eta)*N*w))*K(t)^2+(2*((1/2)*z*eta*(C[max]*w*N-alpha*Q(t))*S(t)+N*w*upsilon*(N*w-z)))*S(t)*alpha*K(t)+N*S(t)^2*w*alpha*upsilon*(N*w-z))/((K(t)^2*alpha*z+3*S(t)*K(t)*alpha*z+S(t)*(2*S(t)*z*alpha+upsilon*(C[max]*w*N-alpha*Q(t))))*(S(t)+K(t))*N*w),

diff(Q(t), t) = (-alpha*z*(z*eta*(C[max]*w*N-alpha*Q(t))*K(t)+N*w*upsilon*(N*w-z))*S(t)^2+(-z^2*eta*alpha*(C[max]*w*N-alpha*Q(t))*K(t)^2-(eta*alpha^2*Q(t)^2-2*C[max]*w*N*eta*alpha*Q(t)+N*w*((2*N*w-2*z)*alpha+N*C[max]^2*w*eta))*z*upsilon*K(t)-N*w*upsilon^2*(N*w-z)*(C[max]*w*N-alpha*Q(t)))*S(t)-N*w*z*alpha*upsilon*K(t)^2*(N*w-z))/((2*S(t)^2*alpha*z+(3*z*alpha*K(t)+upsilon*(C[max]*w*N-alpha*Q(t)))*S(t)+K(t)^2*alpha*z)*N*w*upsilon)

where initials conditions are:

S(0) = S0, K(0) = K0, Q(0) = Q0

Thanks,

Dmitry

 

 

 

Hello Hello everybody 
   I have to solve the following differential equation numerically 


``

 

restart:with(plots):

mb:=765 : mp:=587 : Ib:=76.3*10^3 : Ip:=7.3*10^3 : l:=0.92 : d:=10: F:=490: omega:=0.43 :

eq1:=(mp+mb)*diff(x(t),t$2)+mp*(d*cos(theta(t))+l*cos(alpha(t)+theta(t)))*diff(theta(t),t$2)+mp*l*cos(alpha(t)+theta(t))*diff(alpha(t),t$2)+mp*(d*diff(theta(t),t)^2*sin(theta(t))+l*(diff(theta(t),t)+diff(alpha(t),t))^2*sin(alpha(t)+theta(t)))-F*sin(omega*t)=0;

1352*(diff(diff(x(t), t), t))+587*(10*cos(theta(t))+.92*cos(alpha(t)+theta(t)))*(diff(diff(theta(t), t), t))+540.04*cos(alpha(t)+theta(t))*(diff(diff(alpha(t), t), t))+5870*(diff(theta(t), t))^2*sin(theta(t))+540.04*(diff(theta(t), t)+diff(alpha(t), t))^2*sin(alpha(t)+theta(t))-490*sin(.43*t) = 0

(1)

eq2:=(mp+mb)*diff(z(t),t$2)-mp*d*(sin(theta(t)+alpha(t))+sin(theta(t)))*diff(theta(t),t$2)-mp*l*sin(alpha(t)+theta(t))*diff(alpha(t),t$2)+mp*(d*diff(theta(t),t)^2*cos(theta(t))+l*(diff(theta(t),t)+diff(alpha(t),t))^2*cos(alpha(t)+theta(t)))+9.81*(mp+mb)-F*sin(omega*t)=0;

1352*(diff(diff(z(t), t), t))-5870*(sin(alpha(t)+theta(t))+sin(theta(t)))*(diff(diff(theta(t), t), t))-540.04*sin(alpha(t)+theta(t))*(diff(diff(alpha(t), t), t))+5870*(diff(theta(t), t))^2*cos(theta(t))+540.04*(diff(theta(t), t)+diff(alpha(t), t))^2*cos(alpha(t)+theta(t))+13263.12-490*sin(.43*t) = 0

(2)

eq3:=mp*(d*cos(theta(t))+l*cos(alpha(t)+theta(t)))*diff(x(t),t$2)-mp*(l*sin(theta(t)+alpha(t))+d*sin(theta(t)))*diff(z(t),t$2)+(Ip+Ib+mp*(d^2+l^2)+2*mp*d*l*cos(alpha(t)))*diff(theta(t),t$2)+[Ip+mp*l^2+mp*d*l*cos(alpha(t))]*diff(alpha(t),t$2)-mp*sin(alpha(t))*(l*d*diff(alpha(t),t)^2-l*d*(diff(alpha(t),t)+diff(theta(t),t))^2)+mp*9.81*l*sin(alpha(t)+theta(t))+mp*9.81*d*sin(theta(t))=0;

587*(10*cos(theta(t))+.92*cos(alpha(t)+theta(t)))*(diff(diff(x(t), t), t))-587*(.92*sin(alpha(t)+theta(t))+10*sin(theta(t)))*(diff(diff(z(t), t), t))+(142796.8368+10800.80*cos(alpha(t)))*(diff(diff(theta(t), t), t))+[7796.8368+5400.40*cos(alpha(t))]*(diff(diff(alpha(t), t), t))-587*sin(alpha(t))*(9.20*(diff(alpha(t), t))^2-9.20*(diff(theta(t), t)+diff(alpha(t), t))^2)+5297.7924*sin(alpha(t)+theta(t))+57584.70*sin(theta(t)) = 0

(3)

eq4:=mp*l*cos(alpha(t)+theta(t))*diff(x(t),t$2)-mp*l*sin(alpha(t)+theta(t))*diff(z(t),t$2)+(Ip+mp*l^2+mp*d*l*cos(alpha(t)))*diff(theta(t),t$2)+(Ip+mp*l^2)*diff(alpha(t),t$2)-mp*9.81*l*sin(alpha(t)+theta(t))+l*d*mp*diff(theta(t),t$1)^2*sin(alpha(t))=0;

540.04*cos(alpha(t)+theta(t))*(diff(diff(x(t), t), t))-540.04*sin(alpha(t)+theta(t))*(diff(diff(z(t), t), t))+(7796.8368+5400.40*cos(alpha(t)))*(diff(diff(theta(t), t), t))+7796.8368*(diff(diff(alpha(t), t), t))-5297.7924*sin(alpha(t)+theta(t))+5400.40*(diff(theta(t), t))^2*sin(alpha(t)) = 0

(4)

CI:= x(0)=0,z(0)=0,theta(0)=0,alpha(0)=0,D(x)(0)=0,D(alpha)(0)=0,D(z)(0)=0,D(theta)(0)=0;

x(0) = 0, z(0) = 0, theta(0) = 0, alpha(0) = 0, (D(x))(0) = 0, (D(alpha))(0) = 0, (D(z))(0) = 0, (D(theta))(0) = 0

(5)

solution:=dsolve([eq1,eq2,eq3,eq4, CI],numeric);

Error, (in f) unable to store '[0.]/(0.17571268341557e16+[-0.25659510610770e15])' when datatype=float[8]

 

 

 

I don't know why it says : Error, (in f) unable to store '[0.]/(0.17571268341557e16+[-0.25659510610770e15])' when datatype=float[8]

 

Help pleaase!

thank you !!!

Download systéme_complet.mw

 

Diff_EQ_sample_questions.pdf

 

This is a link to two sample questions I am trying to learn how to solve using maple. I am using maple student edition of maple. Any help would be great. Thank you.

 

 

can we load part of a package not the package whole !? for example from physics package i only need diff tool not all of the package tools,can i do sth for that !? 

hi,i want to take differential with respect to another differential using physics package,but using D instead of diff,could anyone help me do that ? for example :

restart; with(Physics):
A1 := -(1/24)*1*rho*((diff(phi[1](x, t), t))^2)*(h^3)-(1/2)*1*rho*((diff(u[ref](x, t), t))^2)*h-(1/2)*rho*((diff(w(x, t), t))^2)*h+(1/24)*1*1*((diff(phi[1](x, t), x))^2)*(h^3)+(1/2)*1*(1*((diff(u[ref](x, t), x)+(1/2)*(diff(w(x, t), x))^2)^2)+K*1*((diff(w(x, t), x)+phi[1](x, t))^2))*h-1*q*w(x, t):

A2:=-diff(diff(A1,diff(u[ref](x,t),x)),x);

here i want to compute A2 using D command,not diff and i do not want use convert command ! i just need to calculate A2 directly using D command. tnx for your help.

 

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