       Re: Derivative recurrence

• To: mathgroup at smc.vnet.net
• Subject: [mg72718] Re: Derivative recurrence
• From: dh <dh at metrohm.ch>
• Date: Tue, 16 Jan 2007 03:27:40 -0500 (EST)
• References: <eofikq\$gc1\$1@smc.vnet.net>

```
Hi,

Eij is obvioulsy the sum of a simplier tensors, say eij, and its

transpose. eij can be written as an outer  product. Assume you defined

your functions in funs and the variables in vars

funs={z f1[x,y,t],z f2[x,y,t],z f3[x,y,t]};

vars={x,y,z}

eij=Outer[D[#1,#2]&,funs,vars];

Eij=(1/2)(eij+Transpose[eij])

Daniel

KFUPM wrote:

> Dear All

>

> I have the following

>

> u1(x,x,z,t) = z f1(x,y,t)

> u2(x,y,z,t) = z f2(x,y,t)

> u3(x,y,z,t) = z f3(x,y,t)

>

> which is basically the displacement fied. I want to compute the strain

> which is given in indecial notation as:

>

> Eij = 1/2( dui/dxj + duj/dxj ) where i, j, k go from 1 to 3,,, x1=x, x2

> =y, x3=z,

>

> I want mathematica to calculate the strain Eij autmoatically,

>

> for example

>

> when i = 1 and j= 2

>

> E12 =1/2 (du1/dx2+du2/dx1 )     but x1 =x and x2 =y

>

> therefore

>

> E12 =1/2 (du1/dy + du2/dx) = 1/2 ( z df1/dx + z df2/dy),,,   where f1,

> f2 and f3 are arbitrary functions.

>

> Eij should generate 3 by 3 matrix containing the strain components.

> Frankly, i don't know how to do it using mathematica and i would

> appreciate any help in this regard.

>

>

> Thanks in anticipation

>

```

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