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Show that the first variation \delta J(y_{0},h)\, satisfies the homogeneity condition \delta J(y_{0},\alpha h)=\alpha \delta J(y_{0},h),\alpha \in {\mathbb  {R}}\,.

Define the real-valued function \phi (\varepsilon ):=J(y_{0}+\varepsilon h). Then {\frac  {d}{d\varepsilon }}J(y_{0}+\varepsilon h)=\phi ^{\prime }(\varepsilon ). Consequently, \phi ^{\prime }(0)=\delta J(y_{0},h). So

\delta J(y_{0},\alpha h)\,={\frac  {d}{d\varepsilon }}J(y_{0}+\varepsilon \alpha h)
={\frac  {d}{d\varepsilon }}\phi (\varepsilon \alpha )\left.\right|_{{\varepsilon =0}}
=\phi ^{\prime }(\varepsilon \alpha )\cdot \alpha \left.\right|_{{\varepsilon =0}}
=\alpha \phi ^{\prime }(0)
=\alpha \delta J(y_{0},h)\,

A second method uses the limit definition. Here is an outline of that proof

\delta J(y_{0},\alpha h)=\lim _{{\varepsilon \to 0}}{\frac  {J(y_{0}+\varepsilon \alpha h)-J(y_{0})}{\varepsilon }}
Multiply by 1
\alpha \cdot {\frac  {J(y_{0}+\varepsilon \alpha h)-J(y_{0})}{\alpha \varepsilon }}
With t=\varepsilon \alpha
\delta J(y_{0},\alpha h)=\alpha \cdot \lim _{{t\to 0}}{\frac  {J(y_{0}+th)-J(y_{0})}{t}}
=\alpha \delta J(y_{0},h)\,

Main Page : Calculus of Variations