3

I've gotten the functional $$\int_a^b(y^2+2xyy')dx$$ with Dirichlet boundary conditions. Applying the Euler-Lagrange equation I get:

$$0=\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}= 2y+2xy' -\frac{d}{dx}[2xy] = 2y+2xy'-[2y+2xy']=0$$

So this just gives me $0=0$. Have I done something wrong or does this mean that every curve is a stationary curve? If it's the latter, what's the reason for this functional being path independent (I thought that only happened when the functional is indepedent of $y'$)?

user341126
  • 196
  • 2
    Similar situation: http://math.stackexchange.com/questions/191935/what-if-the-euler-lagrange-equation-yields-a-trivial-answer –  May 21 '16 at 14:45

1 Answers1

2

$$ \frac{d}{dx} \left( xy^2 \right) = y^2 + 2xyy'. $$ Hence $$ \int_a^b \left( y^2 + 2xyy' \right) \, dx = b y(b)^2 - a y(a)^2. $$

Siminore
  • 35,136
  • 1
    Since the resulting integral only depends on $a$, $b$, $y(a)$, and $y(b)$, then does that mean it is "path independent"? @Siminore? – makansij Jan 03 '18 at 04:24