Calculate the extreme value of functional

[skrat]

Homework Statement

We have functional $I(y)=\int_{0}^{2}{y}'(2+e^x{y}')dx$ where $y\in C^1(\mathbb{R})$ and $y(0)=0$. Calculate the extreme value.

Homework Equations

The Attempt at a Solution

I am having some troubles here... :/

From Euler-Lagrange equation we get $\frac{\partial L}{\partial y}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial L}{\partial {y}'}=0$ where first part is obviously $\frac{\partial L}{\partial y}=0$ and the second part tells me that $\frac{\partial L}{\partial {y}'}=konst.=C$.

So $(2+e^x{y}')+{y}'e^x=C$

$e^x{y}'=\frac{C}{2}-1=D$, where D is just another constant...

${y}'=De^{-x}$ so

$y=-De^{-x}+E$

Now we know that $y(0)=0$ so $y(0)=-D+E=0$ therefore $E=D$. The second condition says that $\frac{\partial }{\partial {y}'}L(2)=0$.

Than $(2+e^x{y}')+{y}'e^x=0$ so ${y}'=0$.

From $y=-De^{-x}+E$ we get that ${y}'=De^{-x}$, therefore ${y}'(2)=De^{-2}=0$ and this gives me $D=0$ (remember that $E=D$).

So everything together tells that $y$ is actually $y\equiv 0$ and also $I(y)=0$.I seriously doubt that this is right? :/

 

[pasmith]

Homework Statement

We have functional $I(y)=\int_{0}^{2}{y}'(2+e^x{y}')dx$ where $y\in C^1(\mathbb{R})$ and $y(0)=0$. Calculate the extreme value.

Homework Equations

The Attempt at a Solution

I am having some troubles here... :/

From Euler-Lagrange equation we get $\frac{\partial L}{\partial y}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial L}{\partial {y}'}=0$ where first part is obviously $\frac{\partial L}{\partial y}=0$ and the second part tells me that $\frac{\partial L}{\partial {y}'}=konst.=C$.

So $(2+e^x{y}')+{y}'e^x=C$

$e^x{y}'=\frac{C}{2}-1=D$, where D is just another constant...

${y}'=De^{-x}$ so

$y=-De^{-x}+E$

Now we know that $y(0)=0$ so $y(0)=-D+E=0$ therefore $E=D$. The second condition says that $\frac{\partial }{\partial {y}'}L(2)=0$.

Where do you get this condition from? All that's stated in the problem is that $y(0)= 0$ and $y$ is once continuously differentiable. We know that $\dfrac{\partial L}{\partial y'} = 2(1 + D)$ is constant, but without more information we cannot determine that constant (although we can choose it such that $dI/dD = 0$).

Assuming you've omitted that condition from the problem, your error is here:

Than $(2+e^xy′)+y′e^x=0$ so $y′=0$.

If $\left.\dfrac{\partial L}{\partial y'}\right|_{(x=2)} = 2(1 + e^2 y'(2)) = 0$ then $y'(2) = -e^{-2} \neq 0$.

 

[skrat]

Where do you get this condition from? All that's stated in the problem is that $y(0)= 0$ and $y$ is once continuously differentiable. We know that $\dfrac{\partial L}{\partial y'} = 2(1 + D)$ is constant, but without more information we cannot determine that constant.

That's something that should be true in general, at least my professor said so. So if one condition $y(a)=c$ is given the other one is $\frac{\partial }{\partial {y}'}L=0$ calculated for $x=b$. We've probably also proven that.

Assuming you've omitted that condition from the problem, your error is here:


If $\left.\dfrac{\partial L}{\partial y'}\right|_{(x=2)} = 2(1 + e^2 y'(2)) = 0$ then $y'(2) = -e^{-2} \neq 0$.

You are right. My mistake. Taking that into account, $D=E=-1$ so $y(x)=e^{-x}-1$.