\section*{Problem 1}
We are studying the one-dimentional buckling beam, which can be described by the equation
\begin{align*}
    \gamma \frac{d^{2} u(x)}{dx^{2}} &= -F u(x) &  \rightarrow & & \frac{d^{2} u(x)}{dx^{2}} &= - \frac{F}{\gamma} u(x) \\
\end{align*}
where $\gamma$ is a constant determined by the material of the beam. We want to scale the equation, that is we want to scale by the x-value of the beams endpoint $x=L$. 
Scaling will result in a dimensionless variable $\hat{x} = \frac{1}{L}$. 
%
\begin{align*}
        \frac{d^{2}}{dx^{2}} &= \frac{d}{dx} \frac{d}{dx} = \bigg( \frac{d\hat{x}}{dx} \frac{d}{d\hat{x}} \bigg) \bigg( \frac{d\hat{x}}{dx} \frac{d}{d\hat{x}} \bigg) & \text{where we have used } \frac{d\hat{x}}{dx} \frac{d}{d\hat{x}} = \frac{d}{dx} \frac{d\hat{x}}{d\hat{x}} \\
        &= \bigg( \frac{1}{L} \frac{d}{d\hat{x}} \bigg) \bigg( \frac{1}{L} \frac{d}{d\hat{x}} \bigg) = \frac{1}{L^{2}} \frac{d}{d\hat{x}^{2}} & \text{where } \hat{x} \equiv \frac{x}{L} \text{ and } \frac{d\hat{x}}{dx} = \frac{1}{L} \\
\end{align*}
Now we insert the expression into the original equation
\begin{align*}
    \frac{d u(\hat{x})}{d\hat{x}^{2}} &= - \frac{F L^{2}}{\gamma} u(\hat{x}) \\
\end{align*}