mehdibgh

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7 years, 104 days

MaplePrimes Activity


These are replies submitted by mehdibgh

@tomleslie My question is very simple. In math when we see A[i]*B[i,j] expression, we know that i is dummy and j is free and we can write it in open form as below, immediately

I am looking how ask from Maple to write indicial expressions in open form, as above?

As you see I have so many indicial expressions like below that have multiple dummy and free indices, and I need the open form of them. But dont know how to get them in Maple??? As you see this a very simple question.


 

``

restart

Error, invalid input: with expects its 1st argument, pname, to be of type {`module`, package}, but received shareman

 

C__1j := 2*A[1, 1]*Rr[i, m]*(-1)^m*U[i, j]/a+2*A[1, 2]*Rr[l, j]*(-1)^i*V[i, l]/b+A[1, 6]*(2*Rr[l, j]*(-1)^i*U[i, l]/b+2*Rr[i, m]*(-1)^m*V[i, j]/a)+2*B[1, 1]*Rr[i, m]*(-1)^m*Phi[i, j]/a+2*B[1, 2]*Rr[l, j]*(-1)^i*Xi[i, l]/b+B[1, 6]*(2*Rr[l, j]*(-1)^i*Phi[i, l]/b+2*Rr[i, m]*(-1)^m*Xi[i, j]/a)+E[3, 1]*(-1)^i*Omega[i, j]-K__ux0*(-1)^i*U[i, j] = 0

C__11i := 2*A[1, 2]*Rr[k, i]*(-1)^j*U[k, j]/a+2*A[2, 2]*Rr[j, n]*(-1)^n*V[i, j]/b+A[2, 6]*(2*Rr[j, n]*(-1)^n*U[i, j]/b+2*Rr[k, i]*(-1)^m*V[k, j]/a)+2*B[1, 2]*Rr[k, i]*(-1)^j*Phi[k, j]/a+2*B[2, 2]*Rr[j, n]*(-1)^i*Xi[i, l]/b+B[2, 6]*(2*Rr[j, n]*(-1)^n*Phi[i, j]/b+2*Rr[k, i]*(-1)^j*Xi[i, j]/a)+E[3, 2]*(-1)^j*Omega[i, j]-K__vy0*(-1)^j*V[i, j] = 0

C__21j := 2*E[1, 4]*(-1)^i*W[i, j]*Rr[l, j]/b+E[1, 4]*Xi[i, j]*(-1)^i+2*E[1, 5]*(-1)^m*Rr[i, m]*W[i, j]/a+E[1, 5]*Phi[i, j]*(-1)^i-2*`R__1,1`*(-1)^m*Rr[i, m]*Theta[i, j]/a+2*`#mover(mi("z"),mo("&uminus0;"))`*R[1, 1]*(-1)^m*Rr[i, m]*Omega[i, j]/a-2*R[1, 2]*(-1)^i*Theta[i, l]*R[l, j]/b+2*`#mover(mi("z"),mo("&uminus0;"))`*R[1, 2]*(-1)^i*Omega[i, l]*R[l, j]/b-2*S[1, 1]*(-1)^m*Rr[i, m]*Omega[i, j]/a-2*S[1, 2]*(-1)^i*Omega[i, l]*Rr[l, j]/b = 0

C__27i := 2*E[2, 4]*1^m*W[i, l]*Rr[j, m]/b+E[2, 4]*Xi[i, j]*1^j+2*E[2, 5]*1^j*Rr[l, i]*W[l, j]/a+E[2, 5]*Phi[i, j]*1^j-2*`R__1,1`*(-1)^j*Rr[l, i]*Theta[l, j]/a+2*`#mover(mi("z"),mo("&uminus0;"))`*R[1, 1]*1^j*Rr[l, i]*Omega[l, j]/a-2*R[2, 2]*1^m*Theta[i, j]*R[j, m]/b+2*`#mover(mi("z"),mo("&uminus0;"))`*R[2, 2]*1^m*Omega[i, j]*R[j, m]/b-2*S[1, 1]*1^j*Rr[l, i]*Omega[l, j]/a-2*S[2, 2]*1^m*Omega[i, j]*Rr[j, m]/b = 0

``

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``

``


 

Download soal.mw

 

 

@mehdibaghaee Let me ask my explicit question;

How to produce rhs of following equation by lhs in maple???

I this way how to assign I,J,M???

@tomleslie hi

I couldnt expand A[i]*B[i,j] using commands you mentioned. Could you please expand it with maple?

what about if i=1..4 and j=1..6?

 

@taro Dear taro, thanks to your kind reply,

@mehdibaghaee any comment?

@mehdibaghaee Any comment????!!!!!!

@mehdibaghaee Let apply your method to my problem, but I still get error:
 

``

restart

Error, invalid input: with expects its 1st argument, pname, to be of type {`module`, package}, but received shareman

 

with(IntegrationTools):

TT := (1/2)*(int(((A[2, 2]+2*A[2, 6]+A[6, 6])*(diff(V0(x, y), x))^2+((2*A[1, 2]+2*A[1, 6])*(diff(U0(x, y), x))+(2*B[1, 2]+2*B[1, 6])*(diff(Phi(x, y), x))+(2*B[2, 6]+2*B[6, 6])*(diff(Phi(x, y), y))+(2*A[2, 6]+2*A[6, 6])*(diff(U0(x, y), y))+(2*B[2, 6]+2*B[6, 6])*(diff(Xi(x, y), x))+(2*B[2, 2]+2*B[2, 6])*(diff(Xi(x, y), y))+Omega(x, y)*(E[3, 2]+E[3, 6]))*(diff(V0(x, y), x))+(diff(U0(x, y), x))^2*A[1, 1]+(2*(diff(Phi(x, y), x))*B[1, 1]+2*(diff(Xi(x, y), y))*B[1, 2]+2*(diff(Xi(x, y), x))*B[1, 6]+2*(diff(Phi(x, y), y))*B[1, 6]+Omega(x, y)*E[3, 1]+2*(diff(U0(x, y), y))*A[1, 6])*(diff(U0(x, y), x))+(diff(Phi(x, y), x))^2*Dd[1, 1]+(2*B[1, 6]*(diff(U0(x, y), y))+F[3, 1]*Omega(x, y)+2*Dd[1, 2]*(diff(Xi(x, y), y))+2*Dd[1, 6]*(diff(Xi(x, y), x))+2*Dd[1, 6]*(diff(Phi(x, y), y)))*(diff(Phi(x, y), x))+(diff(Phi(x, y), y))^2*Dd[6, 6]+(2*B[6, 6]*(diff(U0(x, y), y))+F[3, 6]*Omega(x, y)+2*Dd[2, 6]*(diff(Xi(x, y), y))+2*Dd[6, 6]*(diff(Xi(x, y), x)))*(diff(Phi(x, y), y))+(diff(U0(x, y), y))^2*A[6, 6]+(2*B[2, 6]*(diff(Xi(x, y), y))+2*B[6, 6]*(diff(Xi(x, y), x))+E[3, 6]*Omega(x, y))*(diff(U0(x, y), y))+(diff(Xi(x, y), x))^2*Dd[6, 6]+(F[3, 6]*Omega(x, y)+2*Dd[2, 6]*(diff(Xi(x, y), y)))*(diff(Xi(x, y), x))+(diff(Xi(x, y), y))^2*Dd[2, 2]+(diff(Xi(x, y), y))*Omega(x, y)*F[3, 2]+(diff(W0(x, y), x))^2*A[5, 4]+((A[4, 4]+A[5, 5])*(diff(W0(x, y), y))+2*A[5, 4]*Phi(x, y)+Xi(x, y)*(A[4, 4]+A[5, 5]))*(diff(W0(x, y), x))+(diff(W0(x, y), y))^2*A[4, 5]+((A[4, 4]+A[5, 5])*Phi(x, y)+2*A[4, 5]*Xi(x, y))*(diff(W0(x, y), y))+Phi(x, y)^2*A[5, 4]+Xi(x, y)*(A[4, 4]+A[5, 5])*Phi(x, y)+Xi(x, y)^2*A[4, 5])*exp((2*I)*omega*t)-(diff(U0(x, y), x))*E1__y(x, y, z, t)*exp(I*omega*t)*F[3, 1], [y = 0 .. b, x = 0 .. a])):

``

F := PDEtools:-dchange({x = (1/2)*a*(xi+1), y = (1/2)*b*(eta+1)}, TT, [xi, eta], params = [a, b])

Error, recursive assignment

 

``

``

``

``

NULL

NULL

Where the problem is???
 

Download inttt2-2.mw

@dharr what do you mean: add known=u0 to get this.

explain more

@dharr It seems the output is incorrect because we dont see (1/2)*a*(xi+1), (1/2)*b*(eta+1) instead of x,y in u0(x, y, t).

@dharr Lets apply it to my problem:


 

``

restart

Error, invalid input: with expects its 1st argument, pname, to be of type {`module`, package}, but received shareman

 

with(IntegrationTools):

T := (1/2)*(int(I__0*(diff(u0(x, y, t), t))^2, [y = 0 .. b, x = 0 .. a]))

(1/2)*I__0*(int(int((diff(u0(x, y, t), t))^2, y = 0 .. b), x = 0 .. a))

(1)

with(IntegrationTools):

T := Change(T, {x = (1/2)*a*(xi+1), y = (1/2)*b*(eta+1)}, [xi, eta])

(1/2)*I__0*(int((1/2)*(D[3](u0))((1/2)*a*(xi+1), (1/2)*b*(eta+1), t)^2*b, eta = -1 .. 1))*a

(2)

 

NULL

NULL

It is amasing, it decreased the twin integral to single one. WHyy?

Where the problem is?
 

Download inttt.mw

 

 

@Kitonum I am wondering why Maple do so for 3dimentional matrix. I just wanted to pick some elements of 3-D matrix and set them in 2-D matrix using for loops. But it seems Maple not did so, although Matlab did it with similar for loop.

It seems there are many differences between Maple and Matlab 3D matrices and for loops.

Any one know how I can do it in Maple?

@Kitonum No error. But not logical output for A!!!

>A(1,1)

Vector[column](4, [` 1..3 x 1..3 x 1..5 `*Array, `Data Type: `*anything, `Storage: `*rectangular, `Order: `*Fortran_order])

Why output is so obscure??

What is Vector[column](4, [` 1..3 x 1..3 x 1..5 `*Array, `Data Type: `*anything, `Storage: `*rectangular, `Order: `*Fortran_order])????!!!

@mehdibaghaee Any comments yet?

@Kitonum Lets look at A(1,1)

>A(1,1)

Vector[column](4, [` 1..3 x 1..3 x 1..5 `*Array, `Data Type: `*anything, `Storage: `*rectangular, `Order: `*Fortran_order])

So whats the problem???

 

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