Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

I'm trying to show that there are an infinite number of functions that minimize the integral: $$\int_0^2[(y')^2*(1 + y')^2] dx$$ subject to $y(0) = 1$ and $y(2) = 0$.

(They are continuous functions with piecewise continuous first derivatives.)

share|cite|improve this question
Do you mean $C^1$ functions or piecewise $C^1$ functions? (if you need continous first derivatives I don't think this has minimizers) – Jose27 Dec 2 '12 at 2:08
piecewise C^inf functions – Buddy Holly Dec 2 '12 at 2:10
up vote 1 down vote accepted

Consider the two lines given by the graphs of $$ f(x)=1-x \\ g(x)=2-x $$ Now pick $0<s<1$ and consider $$ h_s(x)= \left\{ \begin{array}{lcl} f(x) & \text{if} & 0\leq x<s\\ 1-s & \text{if} & s\leq x<1+s\\ g(x) & \text{if} & 1+s\leq x \leq 2 \end{array} \right. $$ then it's easy to see that $J(h_s)=0$ ($J$ is the functional in question, and note that trivially $J\geq0$), so that $h_s$ are minimizers for all $0<s<1$.

share|cite|improve this answer
The setup to use is the Euler Lagrange equation. Can you solve it using this? – Buddy Holly Dec 2 '12 at 6:37
@BuddyHolly: Sure, just derive the E-L equations and plug in. You'll get two possible general candidates. One being (arbitary) affine functions and the other some special affine functions. Discard some candidates by evaluating the functional and conclude that the only restriction is that the minimizers have $y'=0,-1$. (By the way you really should put the E-L restriction on the question) – Jose27 Dec 2 '12 at 16:27

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.