Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I find an extremal for the following functional: $$J[y] = \int_0^1 e^y (y')^2 dx $$

with the boundary conditions $y(0)=1, y(1)=\log(4)$? I have tried computing the Euler-Lagrange equation but it becomes overly complicated.

share|cite|improve this question

The E-L equation is $- e^y (y')^2 - 2 e^y y'' = 0$. Divide by $e^y$ and first solve the first-order equation for $y'$, then integrate to get $y$. Then put in the boundary conditions.

share|cite|improve this answer

Hint: The Lagrangian $$L(y,\dot{y})~=~ e^{y} \dot{y}^2$$ does not depend explicitly on the independent variable. Noether's theorem then states that the energy function $$h(y,\dot{y})~=~\dot{y} \frac{\partial L(y,\dot{y})}{\partial \dot{y}}- L(y,\dot{y})~=~\ldots ~=~L(y,\dot{y})$$ is conserved, and hence provides a first integral $$ e^{y/2} \dot{y}~=~{\rm const}.$$ (This fact is also known in mathematics as the Beltrami identity.)

share|cite|improve this answer

The Euler-Lagrange equation simplifies down to $y''+(y')^{2}=0$. Set $u=y'$ to get a seperable ODE, then integrate to find $y$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.