# Finding an explicit expression for a minimizer

Suppose $f$ is a continuous function on the interval (0,1). We consider the energy functional

$F(u) = \int^1_0\frac{1}{2}((u')^2+u^2)\,dx - \int^1_0fu\,dx$

which is well defined for continuously differentiable functions $u$ on $(0,1)$. Suppouse that $u_0$ is a local minimizer of $F$ in the class of $C^1$ functions satisfying $u(0)=a, u(1)=b$ for some fixed $a,b \in \mathbb{R}$.

This question consists of a few parts but im stuck on one in particular,

Suppose that $a=1,b=e^2,f(x)=-3e^{2x}$. Find an explicit expression for $u_0$.

I have found the Euler-Lagrange equations, $(u_0 -f)-\frac{d}{dx} (u_0'') (u_0') = 0$ but I'm not clear as to how to solve the stated question using this.

Any help would be appreciated,

Thanks

• Not to be picky, but I think you mean 'minimizer' (and if 'minizer' is actually a term I have not come across, apologies). – gnometorule Mar 2 '13 at 16:57
• @gnometorule Thats true lol – bobdylan Mar 2 '13 at 17:10
• I fixed obvious typos, but there may be one in your Euler-Lagrange equation too. Should there be an arithmetical sign between two parentheses? – user53153 Mar 2 '13 at 17:11
• @5pm maybe my calculations were in correct, but isnt it just the product of $u_0'$ and $u_0''$? – bobdylan Mar 2 '13 at 17:26
• That's what I'm saying: your calculations were incorrect. The Euler-Lagrange equation for a quadratic functional is always linear, and your equation is not. Once you fix the equation, you'll find it much easier to solve using the standard approach to second-order linear ODE with constant coefficients. – user53153 Mar 2 '13 at 17:37

The stated Euler-Lagrange equation is incorrect. (It should have been linear, since the functional is quadratic). This is the correct derivation of the equations:

1. Write $z=u$ and $p=u'$ in order to abstract away the function $u$.
2. Find $\frac{\partial F}{\partial p}$ and $\frac{\partial F}{\partial z}$
3. The equation is $-\frac{d}{dx} \frac{\partial F}{\partial p}+\frac{\partial F}{\partial z}=0$.
4. Return to the notation $u,u'$.
5. Take the derivative $\frac{d}{dx}$.

In the present case $\frac{\partial F}{\partial p}=p$ and $\frac{\partial F}{\partial z}=z-f$. Therefore, the Euler-Lagrange equation is $$-\frac{d}{dx} (u') +(u-f)=0 \tag1$$ which is $$-u''+(u-f)=0 \tag2$$ As promised, this is a linear equation, to which standard ODE methods apply.

That is, the general solution is $u=c_1u_1+c_2u_2+u_p$ where $u_1,u_2$ solve the homogeneous equation (and are found from the characteristic equation), and $u_p$ is a particular solution, found by the method of undetermined coefficients. The last step is to find $c_1,c_2$ that satisfy the boundary conditions.

The Euler-Lagrange equations are differential equations, since you know all the terms involved in your expression and boundary conditions on $u$ this is a straightforward computation. There are plenty of programs that will solve a diffie for you symbolically or numerically, or if you're feeling perverse you can get a book on ODEs.