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First variation
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<h2 id="example">Example</h2> <p>Compute the first variation of </p> J ( y ) = ∫ a b y y ′ d x . {\displaystyle J(y)=\int _{a}^{b}yy'\mathrm {d} x.} <p>From the definition above, </p> δ J ( y , h ) = d d ε J ( y + ε h ) | ε = 0 = d d ε ∫ a b ( y + ε h ) ( y ′ + ε h ′ ) d x | ε = 0 = d d ε ∫ a b ( y y ′ + y ε h ′ + y ′ ε h + ε 2 h h ′ ) d x | ε = 0 = ∫ a b d d ε ( y y ′ + y ε h ′ + y ′ ε h + ε 2 h h ′ ) d x | ε = 0 = ∫ a b ( y h ′ + y ′ h + 2 ε h h ′ ) d x | ε = 0 = ∫ a b ( y h ′ + y ′ h ) d x {\displaystyle {\begin{aligned}\delta J(y,h)&=\left.{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}J(y+\varepsilon h)\right|_{\varepsilon =0}\\&=\left.{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\int _{a}^{b}(y+\varepsilon h)(y^{\prime }+\varepsilon h^{\prime })\ \mathrm {d} x\right|_{\varepsilon =0}\\&=\left.{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\int _{a}^{b}(yy^{\prime }+y\varepsilon h^{\prime }+y^{\prime }\varepsilon h+\varepsilon ^{2}hh^{\prime })\ \mathrm {d} x\right|_{\varepsilon =0}\\&=\left.\int _{a}^{b}{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}(yy^{\prime }+y\varepsilon h^{\prime }+y^{\prime }\varepsilon h+\varepsilon ^{2}hh^{\prime })\ \mathrm {d} x\right|_{\varepsilon =0}\\&=\left.\int _{a}^{b}(yh^{\prime }+y^{\prime }h+2\varepsilon hh^{\prime })\ \mathrm {d} x\right|_{\varepsilon =0}\\&=\int _{a}^{b}(yh^{\prime }+y^{\prime }h)\ \mathrm {d} x\\\end{aligned}}} <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Calculus_of_variations/Ya5Lsma3">Calculus of variations</a></li> <li><a href="/facts/Functional_derivative/urNSIWYc">Functional derivative</a></li></ul>