# find the extremal of a function

i need to find the extremal of the functional $\int I(y,y'') dt$. Could anyone tell me the concepts of finding the extremal so that i can go about solving this one?

Update:

So my function was: G= $\int(ay+ $$\frac{1}{2}$$ b$y''^2$)dt$ solving it using using the Euler–Lagrange equation as given in the link below, we get:

a+by''''=0

which on solving gives:

y=-(a(t^4))/(24b).

Am i right in solving this one?

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have you tried en.wikipedia.org/wiki/… (Euler-Lagrange equations - the link is to the case of higher derivatives, since you need it)? – user8268 Apr 10 '11 at 13:50
for some reason my link doesn't go where it's supposed to, so go to the "Single function of single variable with higher derivatives" case – user8268 Apr 10 '11 at 14:00
look up "calculus of variations" and "euler lagrange equations". there's also a cheap, easy to read dover book by gelfand and fomin entitled "calculus of variations" – yoyo Apr 10 '11 at 14:52
@user8286 thank you, the link was very helpful!!! – Siddhartha Apr 10 '11 at 19:40

Principally this should be related to the space on which the functional is defined. This depends on the exact form of $I$. However, if you forget these things, consider for $J(y)=\int I(y,y'')$, the expression $J(u+\tau v)$, where $\tau$ is a parameter, and $u,v$ are functions. Then differentiate the expression which respect to $\tau$, and put $\tau=0$. Rewrite the expression, using partial integrations (this only works if the space you decide to define the functional on is nice enough, for example compactly supported functions on the domain you are working on). You should then find some expression of the form $\int B(u,u'') v$. Since for a critical point this must be zero for all $v$, we conclude $B$ must be identically zero.