Calculating the Extreme Value of a Functional with Given Boundary Conditions

[skrat]

Homework Statement

We have functional $A(y)=\int_{-1}^{1}(4y+({y}')^2)dx$ where $y\in C^1(\mathbb{R})$ and $y(-1)=1$ and $y(1)=3$.
a) Calculate $A(y)$ if graph for $y$ is line segment.
b) Calculate the extreme value of $A(y)$ for that $y$. That does it represent?

Homework Equations

The Attempt at a Solution

a)

Since $y$ is line segment I GUESS it can be written as $y=kx+n$. The functional is therefore calcualted as:

$A(y)=\int_{-1}^{1}(4y+({y}')^2)dx=A(y)=\int_{-1}^{1}(4kx+4n+16k^2)dx=32k^2+8n$

Because $y(-1)=1$ and $y(1)=3$, $k=1$ and $n=2$, so $A(y)=48$

b)

We have a so called Euler–Lagrange equation. Let's say that $L=4y+({y}')^2$.

$\frac{\partial L}{\partial y}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial L}{\partial {y}'}=0$

$4-2{y}''=0$

$y=x^2+Cx+D$

For $y(-1)=1$ we get an equation saying that $D=C$ and from $y(1)=3$ another one saying $C+D=2$ therefore $C=D=1$.

So finally $y(x)=x^2+x+1$ and $A(y)= 0$. $y(x)$ is quadratic function and extreme value of functional represents minimum, i guess?

 

[pasmith]

Homework Statement

We have functional $A(y)=\int_{-1}^{1}(4y+({y}')^2)dx$ where $y\in C^1(\mathbb{R})$ and $y(-1)=1$ and $y(1)=3$.
a) Calculate $A(y)$ if graph for $y$ is line segment.
b) Calculate the extreme value of $A(y)$ for that $y$. That does it represent?

Homework Equations

The Attempt at a Solution

a)

Since $y$ is line segment I GUESS it can be written as $y=kx+n$.The functional is therefore calcualted as:

$A(y)=\int_{-1}^{1}(4y+({y}')^2)dx=A(y)=\int_{-1}^{1}(4kx+4n+16k^2)dx=32k^2+8n$

Because $y(-1)=1$ and $y(1)=3$, $k=1$ and $n=2$, so $A(y)=48$

b)

We have a so called Euler–Lagrange equation. Let's say that $L=4y+({y}')^2$.

$\frac{\partial L}{\partial y}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial L}{\partial {y}'}=0$

$4-2{y}''=0$

$y=x^2+Cx+D$

For $y(-1)=1$ we get an equation saying that $D=C$ and from $y(1)=3$ another one saying $C+D=2$ therefore $C=D=1$.

So finally $y(x)=x^2+x+1$ and $A(y)= 0$.

Are you sure? 4y + (y')^2 = 4x^2 + 4x + 4 + (2x + 1)^2 = 2(2x + 1)^2 + 3 > 0, so A(x^2 + x + 1) > 0.

 

[skrat]

Ahhh, you are right... I made A mistake in my notes.. I used $4y-({y}')^2$...

Thank you!