Solution verification for finding an extremal under constraints

Question

Find the extremal of $\int_0^1 \left[\dot{x}^2 + 2x\dot{x} + 2x\right] dt$ with $x(0)=0$, and $x(1)=\frac12$ subject to the constraint $\int_0^1 12tx dt=24$

Could anyone verify the anwer to this?

I have obtained $x=-\frac{9t^3}{16}+\frac{t^2}{2}+\frac{9t}{16}$

Thank you.

Answer 1

Define the functionals

$$F(t,x,\dot{x})= \dot{x}^2 + 2x\dot{x} + 2x,\\G(t,x,\dot{x})= xt$$

An extremal satisfies the (constrained) Euler-Lagrange equation

$$F_x - \frac{d}{dt}F_{\dot{x}} + \lambda\left(G_x - \frac{d}{dt}G_{\dot{x}}\right)=0.$$

This reduces to

$$x''= 1 +\frac{\lambda}{2}t,$$

with general solution

$$x = C_1 + C_2t + \frac1{2}t^2 + \frac{\lambda}{12}t^3.$$

Applying the boundary conditions

$$x = \frac{\lambda}{12}(t^3-t) + \frac1{2}t^2$$

Applying the constraint

$$2 = \int_0^1 t\left[\frac{\lambda}{12}(t^3-t) + \frac1{2}t^2\right]dt=\frac{\lambda}{12}\left(\frac1{5}-\frac1{3}\right)+ \frac1{8}$$

and

$$\frac{\lambda}{12} = \frac{-225}{16}.$$

The extremal is

$$x = -\frac{225}{16}(t^3-t) + \frac1{2}t^2$$

Answer 2

$$L = \dot{x}^2+2x\dot{x}+2x + 12tx\lambda$$

$$\frac{\partial L}{\partial x} - \frac{d}{dt}\left[\frac{\partial L}{\partial \dot{x}}\right]=0$$

$$2\dot{x}+2+12t\lambda-\frac{d}{dt}\left[2\dot{x}+2x\right]=0$$ $$2\dot{x}+2+12t\lambda-2\ddot{x} -2\dot{x}=0$$ $$\ddot{x}=1 + 6t\lambda$$ $$\dot{x}=t+3t^2\lambda +C$$ $$x=\frac{t^2}{2}+t^3\lambda + Ct+D$$

$$0=D$$ $$\frac12=\frac12+\lambda+C$$ $$\lambda=-C$$ $$\int_0^1 12tx dt=24$$ $$\int_0^1 12t(\frac{t^2}{2}+t^3\lambda + Ct)dt=24$$ $$\int_0^1\frac{12t^3}{2}+12t^4\lambda + 12Ct^2dt=12$$ $$\left[\frac{3t^4}{2}+\frac{12t^5\lambda}{5} + 4Ct^3\right]_0^1=24$$ $$\frac32+\frac{12\lambda}{5}+4C=24$$ $$\frac32+\frac{12}5\lambda - 4\lambda = 24$$ $$\frac{-8\lambda}{5}=22.5$$ $$\lambda = -\frac{225}{16}$$ $$C=\frac{225}{16}$$

$$x=\frac{t^2}{2}-\frac{225t^3}{16} +\frac{225t}{16}$$