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Topic: How to calculate the partial derivative?
Replies: 6   Last Post: Nov 21, 2012 5:17 AM

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Tang Laoya

Posts: 23
Registered: 11/21/12
Re: How to calculate the partial derivative?
Posted: Nov 21, 2012 4:43 AM
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Dear Dr. M. Abbasi,

The following are all the code I want to implement, Besides the DNx, DNy and DNz, I would also replace all "L1^aa*L2^bb*L3^cc" in matrix K by "aa!*bb!*cc!/((aa+bb+cc+3)!)*6*V, how to do?

Thanks,
Tang Laoya




Clear["Global`*"];
eq1 = L1 == (a1 + b1 x + c1 y + d1 z)/6/V;
eq2 = L2 == (a2 + b2 x + c2 y + d2 z)/6/V;
eq3 = L3 == (a3 + b3 x + c3 y + d3 z)/6/V;
eq4 = L4 == (a4 + b4 x + c4 y + d4 z)/6/V;
lhs[eq_] := eq /. (lhs_ == rhs_) -> lhs;
rhs[eq_] := eq /. (lhs_ == rhs_) -> rhs;

nod = 10;
NN = Table[0, {nod}];
NN[[1]] = rhs[eq1] (2 rhs[eq1] - 1);
NN[[2]] = rhs[eq2] (2 rhs[eq2] - 1);
NN[[3]] = rhs[eq3] (2 rhs[eq3] - 1);
NN[[4]] = rhs[eq4] (2 rhs[eq4] - 1);
NN[[5]] = 4 rhs[eq1] rhs[eq2];
NN[[6]] = 4 rhs[eq2] rhs[eq3];
NN[[7]] = 4 rhs[eq3] rhs[eq1];
NN[[8]] = 4 rhs[eq3] rhs[eq4];
NN[[9]] = 4 rhs[eq4] rhs[eq1];
NN[[10]] = 4 rhs[eq2] rhs[eq4];

NNT = Flatten[Table[NN[[i]], {i, 1, nod}]];
Print["Dimensions of NNT:", Dimensions[NNT]];
DNx = D[NNT, x];
DNx = DNx /. {rhs[eq1] -> lhs[eq1], rhs[eq2] -> lhs[eq2], rhs[eq3] -> lhs[eq3], rhs[eq4] -> lhs[eq4]};
DNy = D[NNT, y];
DNy = DNy /. {rhs[eq1] -> lhs[eq1], rhs[eq2] -> lhs[eq2], rhs[eq3] -> lhs[eq3], rhs[eq4] -> lhs[eq4]};
DNz = D[NNT, z];
DNz = DNz /. {rhs[eq1] -> lhs[eq1], rhs[eq2] -> lhs[eq2], rhs[eq3] -> lhs[eq3], rhs[eq4] -> lhs[eq4]};
DNxx = Transpose[{DNx}].{DNx};
DNyy = Transpose[{DNy}].{DNy};
DNzz = Transpose[{DNz}].{DNz};
Print["Dimensions of DNxx:", Dimensions[DNxx]];

K = DNxx + DNyy + DNzz;

(* I would replace all "L1^aa*L2^bb*L3^cc" in matrix K by "aa!*bb!*cc!/((aa+bb+cc+3)!)*6*V, how to do? *)

K = Expand[K];
Print["Simplifying K"];
K = Simplify[K]



On Wednesday, November 21, 2012 4:17:21 PM UTC+8, Nasser M. Abbasi wrote:
> On 11/21/2012 12:15 AM, Tang Laoya wrote:
>

> > Dear all,
>
> >
>
> > I am trying to calculate the partial derivative by mathematica, I have the following commands:
>
> > L1=a1+b1*x+c1*y;
>
> > L2=a2+b2*x+c2*y;
>
> > L3=a3+b3*x+c3*y;
>
> >
>
> > NN=L1*L2;
>
> >
>
> > DNx=D[NN,x];
>
> >
>
> > I got the following result:
>
> > DNex=b2 (a1+b1 x+c1 y)+b1 (a2+b2 x+c2 y)
>
> >
>
> > How to do to have the following result?
>
> >
>
> > DNex=b2*L1 + b1 * L2
>
> >
>
>
>
> The problem is that you did not use equations, but used expresions.
>
> Hence in your case L1 is now 'a1+b1*x+c1*y' and similarly with L2.
>
>
>
> To do this right, I'd write things as equations, and use rules, like
>
> this:
>
>
>
> --------------------------
>
> Clear[L1, L2, a1, b2, x, c1, y, a2, b2, c2, lhs, rhs]
>
> eq1 = L1 == a1 + b1*x + c1*y;
>
> eq2 = L2 == a2 + b2*x + c2*y;
>
> lhs[eq_] := eq /. (lhs_ == rhs_) -> lhs;
>
> rhs[eq_] := eq /. (lhs_ == rhs_) -> rhs;
>
>
>
> NN = rhs[eq1]*rhs[eq2];
>
> DNx = D[NN, x]
>
> ----------------------------
>
>
>
> which gives what you showed
>
>
>
> b2 (a1+b1 x+c1 y)+b1 (a2+b2 x+c2 y)
>
>
>
> Now it is easy to use the rules to get what you want
>
>
>
> ------------------
>
> DNx/.{rhs[eq1]->lhs[eq1],rhs[eq2]->lhs[eq2]}
>
> ---------------------
>
>
>
> which gives
>
>
>
> b2 L1+b1 L2
>
>
>
> --Nasser





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