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Given the following state equation of a mechanical structure:

$K(x) u = F(x)$

where $K$ is the stiffness matrix of size $n \times n$, $x$ is the design variable, $u$ is the state variable (displacement vector) , and $F$ is the force vector.

If $K$ was invertible then $u = K^{-1} ~F$

But, if $K$ is not invertible, then we can find $u$ by solving a system of $n$ equations, right?

user134613
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1 Answers1

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Since the equation is linear, the general solution $u = u_h + u_p$ will be made up of the general solution to the associated homogeneous problem, namely $u_h \in \ker K$, i.e. $Ku_h = 0$, and a particular solution the inhomogeneous equation, i.e. $Ku_p = F$. So, determining the kernel of $K$ (which is non-trivial, since $K$ is non-invertible) will constitute your first task, before finding a particular solution $u_p$ by any means. Finally, it has to be highlighted that no such $u_p$ exists if $F \not\in \mathrm{im\,} K$.

Abezhiko
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  • Thank you. Do you mind please providing more details to your answer (as my Mathematics knowledge is more basic)? – user134613 Mar 01 '24 at 17:35
  • @user134613 It is hard for me to develop my answer without a concrete example or knowing what is unclear for you. – Abezhiko Mar 02 '24 at 08:54
  • Sure. I don't know what is ker, and the reason we're dividing it into homogeneous and inhomogeneous. – user134613 Mar 02 '24 at 15:57
  • @user134613 $\ker K$ represents the kernel of $K$ and it contains the vectors whose image is zero, i.e. $u \in \ker K \Leftrightarrow Ku = 0$. The particular solution solving the original inhomogeneous part can be seen as the "usual" solution in the sense that $u_p = K^{-1}F$ (if it exists, which is not guaranteed, since the map $K$ is not bijective in the present case). – Abezhiko Mar 03 '24 at 09:20
  • The homogeneous solution is usually absent/ignored, since the kernel is reduced to the zero-vector when $K$ is bijective; however, it is needed here, because $Ku = K(u_h + u_p) = Ku_h + Ku_p = 0 + Ku_p = F$. This stuff works exactly as in the case of linear differential equations. – Abezhiko Mar 03 '24 at 09:24
  • It is a pity that this technique is taught only when dealing with ODEs, while it comes from linear algebra initially. The link lies in the fact that derivatives are precisely non-invertible linear maps (acting on functional vector spaces), but it is almost never highlighted. – Abezhiko Mar 03 '24 at 09:27
  • I will have to study further. But thanks a lot for giving these details. I still need to ask please, if $K$ is invetible we say that $u$ is a function of $x$ but if $K$ isn't invertible then we don't say that $u$ is a function of $x$? – user134613 Mar 04 '24 at 22:47
  • @user134613 It depends on the context where your equation arises. Here, $x$ seems to play the role of a parameter only, in such a way that nothing prevents $u$ to be a function of $x$. – Abezhiko Mar 05 '24 at 10:11