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Given problem is

$J[y] =\int_{0}^{x_1}y'^2dx$ with $y(0)=0$ and $y(x_1)=-x_1-1$.

After solving Euler Lagrange equation I got $y=Ax+B$ .

And using first boundry conditon I got $y=Ax$

We have transversatity condition $[F+(\phi'-y')F_{y'}]=0 $ at $x=x_1$.

But solving this I am getting $A=0$ but the solution is given as $y = - 2 x \,, x_1= 1.$

Any help will be appreciated.

zafran
  • 904

1 Answers1

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Here $\phi=-x-1$

So transversatity condition will give us $A^2+[-1-A ]2A=0$

$\Longrightarrow$ $A^2-2A-2A^2=0$

$\Longrightarrow$ Either $A=0 $ or $A=-2$

But we also have $y(x_1)=Ax_1$

$\Longrightarrow$ $-x_1-1=Ax_1$

$\Longrightarrow$ $A=-{1\over x_1}-1$

$\Longrightarrow$ $x_1=-{1\over A+1}$

$\Longrightarrow$ $x_1=1$

Therefore $y=-2x$ and $x_1=1$

I had made a mistake therefore I was getting wrong answer.

zafran
  • 904