Compute principal invariants for an extension of amount $\lambda$ in direction $\boldsymbol{e}$

Question

I'm trying to solve an exercise taken from Gurtin's book (Ex. 3 pg 52)

Let $f$ be an extension of amount $\lambda$ in direction $e$: $$f(p)=f(q)+U(p-q)$$ where $$U = I + (\lambda -1) e \otimes e$$ with $\lambda >0$ and $p,q$ are material points.

Compute $C$ and the list of principal invariants $i_1(C),i_2(C),i_3(C)$, where $C$ is the right Cauchy Green strain tensor defined as $C=U^2$

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First I compute $U^2$ explicitely:
$$U^2 = (I + (\lambda -1) e \otimes e ) (I + (\lambda -1) e \otimes e ) = I + 2(\lambda -1) e \otimes e + (\lambda -1)^2 |e|^2 e \otimes e$$ I don't think I can simplify more, right?

To compute the invariants, I should use the fact that $C=U^2$ is symmetric and positive definite and hence the principal invariants are easy to compute because we only need the eigenvalues, but here I don't really know how I could move, because $e$ may be any direction, so it seems a really weird question.

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EDIT $$i_1(C)=\text{tr}(C)=\lambda^2$$

$$i_2(C)=\frac{1}{2}[\text{tr}(C)^2 - \text{tr(C^2)}]$$

Now $C^2 = U^2 U^2 = I+(\lambda^4 -1) e \otimes e$ and hence $$\text{tr}(C^2)=\text{tr}(U^4) = \lambda^4$$ so $$i_2(C)=\frac{1}{2}\lambda^2 (1 - \lambda^2 )$$

Finally, $$i_3(C)=\det(C)=\det(I + (\lambda -1) e \otimes e)$$

Now, $\det$ is the product of the eigenvalues, and we can see immediately that

• $\lambda$ is an eigenvalue: $$(I-(\lambda -1) e \otimes e)e = \lambda e$$

• all the other eigenvalues are $1$: take $v$ eigenvector, which is orthogonal to $e$ because $U$ is symmetric $$(I-(\lambda -1)e \otimes e)v = v - (\lambda-1) (e\otimes e) v=v$$ because $(e \otimes e )v = 0$

Therefore $$i_3(C) = \lambda 1= \lambda$$

Answer 1

We have $U^2 = I + (\lambda^2-1)e\otimes e$. So the characteristic polynomial is $$\begin{align} \det(tI - U^2) &= \det((t-1)I - (\lambda^2-1)e\otimes e) \\ &= (t-1)^3 \det\left(I - \frac{\lambda^2-1}{t-1}e\otimes e\right) \\ &= (t-1)^3\left(1 - {\rm tr}\left(\frac{\lambda^2-1}{t-1}e\otimes e\right)\right) \\ &= (t-1)^3 \left(1 - \frac{(\lambda^2-1)}{t-1} \right) \\ &= (t-1)^3 - (\lambda^2-1)(t-1)^2 \\ &{\color{red}{= (t-1)^2(t-\lambda^2)}} \\ &= (t^2-2t+1)(t-\lambda^2) \\ &= t^3 - (\lambda^2+2)t^2 + (2\lambda^2+1)t - (\lambda^2). \end{align}$$It seems that the invariants are $\lambda^2+2$, $2\lambda^2+1$ and $\lambda^2$.