Question $1^1$

a Let $\{e_1,e_2\}$ be a basis of a vector space $V$ and let $2$ vectors of $V$ be

$U=2{}_1+4e_2$;$W=5e_1+3e_2$

Further, let $e_{i}ox{e}_j$ be the basis vectors of $E_4=$VoxV.

$($i$)$ Find UoxW

$($ii$)$ Can the following tensor be a tensor product of $2$ vectors of $V?$

Justify your answer

$T=11e_{1}ox{e}_1+8e_{1}ox{e}_2+20e_{2}ox{e}_1+12e_{2}ox{e}_2$

$($iii$)$ Show that $T$ is the sum of UoxW and another tensor $x$ and show

that $x$ is a tensor product of $2$ vectors.

b We consider the following $2$ d flow in an orthonormal frame of ref$-$

erence.$($Later$)$

$x=x_0{e}^{\frac{t}{}}$

$y=Y_0{e}^{-\frac{t}{}}$

$z=0$

Express the velocity and acceleration components respectively, in the

Lagrange variables.

${?}^1$ Question $1$ continues on the following page

c Consider $2$ second order tensors A and $B$ acting on the vectors $$ and

$$ respectively. By definition, the operator defined by the following

tensor product:

AoxB$=C$

acts on $ox$ as follows:

$C(ox)=(AoxB)(ox)=AoxB$

Let $$ be an eigenvalue of A and $$ the associated eigenvector and $$ be

an eigenvalue of $B$ and $$ the associated eigenvector. Show that $ox$

are eigenvectors of AoxI and IoxB, where I is the identity matrix

and find the corresponding eigenvalues.

d The matuix of tensor $U$ is given by

$U=\left[\begin{array}{ccc}f& cos(x_2)& sin(x_3)\\ cos(x_2)& g& 2x_2{x}_3\\ sin(x_3)& 2x_2{x}_3& h\end{array}\right]$

where $f,g$ and $h$ are functions of $x_1,x_2$ and $x_3.$ The tensor is divergent$-$free,

that is$,$ grad$*U=0$ and $f=0,$ at $x_1=0$;$g=0,$ at $x_2=0$ and $h=0,$

at $x_3=0.$ Find $f,g$ and $h.$

For

${}_{ij}=\left[\begin{array}{ccc}{}_{11}& {}_{12}& {}_{13}\\ {}_{21}& {}_{22}& {}_{23}\\ {}_{31}& {}_{32}& {}_{31}\end{array}\right]$

the divergence vector of ${}_{ij}$ is given by:

grad$*=([\frac{\text{d}\text{e}\text{l}{}_{11}}{\text{d}\text{e}\text{l}{}_1}+\frac{\text{d}\text{e}\text{l}{}_{12}}{del{x}_1}+\frac{\text{d}\text{e}\text{l}{}_{13}}{del{x}_3}],[\frac{\text{d}\text{e}\text{l}{}_{21}}{del{x}_1}+\frac{\text{d}\text{e}\text{l}{x}_2}{del{x}_2}+\frac{\text{d}\text{e}\text{l}{}_3}{del{x}_3}],[\frac{\text{d}\text{e}\text{l}{}_{31}}{del{x}_1}+\frac{\text{d}\text{e}\text{l}{}_{32}}{del{x}_2}+\frac{\text{d}\text{e}\text{l}{}_1}{del{x}_3}])$