Consider the functional $S[y]=\int_{1}^{2}ln(1+x^2y')dx, y(1)=0, y(2)=A$, where $A$ is a constant and $y$ is a continuously differentiable function for $1\leq x\leq 2$. Let $h$ be a continuously differentiable function for $1\leq x\leq 2$, and let $\epsilon$ be a constant. Let $\bigtriangleup=S[y+\epsilon h]-S[y]$. Show that $\bigtriangleup=\epsilon\int_{1}^{2}\frac{x^2h'}{1+x^2y'}dx-\frac{\epsilon^2}{2}\int_{1}^{2}\frac{x^4h'^2}{(1+x^2y')^2}dx+O(\epsilon^3)$.
Here's my work:
Let $\bigtriangleup=S[y+\epsilon h]-S[y]$.
Then $\bigtriangleup=\int_{1}^{2}ln(1+x^2(y'+\epsilon h'))dx-\int_{1}^{2}ln(1+x^2y')dx=\int_{1}^{2}ln(\frac{1+x^2(y'+\epsilon h')}{1+x^2y'})dx$.
From here what should I do in order to complete the proof?
Best Answer
Since $\varepsilon$ has to be interpreted as an infinitesimal quantity, you should use the second-order Taylor expansion of the logarithm, i.e. $$ \ln(1+z) = \displaystyle\sum_{k\ge1} (-1)^{k+1}\frac{z^k}{k} = z - \frac{z^2}{2} + \mathcal{O}(z^3), $$ as follows : $$ \ln\left(\frac{1+x^2(y'+\varepsilon h')}{1+x^2y'}\right) = \ln\left(1+\frac{\varepsilon x^2h'}{1+x^2y'}\right) = \frac{\varepsilon x^2h'}{1+x^2y'} - \frac{1}{2}\left(\frac{\varepsilon x^2h'}{1+x^2y'}\right)^2 + \mathcal{O}(\varepsilon^3) $$ And the job is done !