Let's break down the steps to **determine the Mayer equation** for the isoperimetric problem given: ### **Step 1: State the Problem** Given the functional: \[ I[x] = \int_0^a \left[ p x^2 + q \dot{x}^2 \right] dt \] with the **isoperimetric (auxiliary) constraint**: \[ \int_0^a r x^2 \, dt = 1 \] and boundary conditions: \[ x(0) = 0, \quad x(a) = 0 \] where \( p, q, r \) are given \( C^2 \) functions of \( t \). --- ### **Step 2: Formulate the Augmented Functional** Introduce a Lagrange multiplier \( \lambda \) for the constraint. The augmented functional becomes: \[ J[x] = \int_0^a \left[ p x^2 + q \dot{x}^2 + \lambda r x^2 \right] dt - \lambda \] Or usually written: \[ J[x] = \int_0^a \left[ (p + \lambda r) x^2 + q \dot{x}^2 \right] dt \] --- ### **Step 3: Euler–Lagrange Equation** The integrand is: \[ L = (p + \lambda r) x^2 + q \dot{x}^2 \] The Euler–Lagrange equation is: \[ \frac{\partial L}{\partial x} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \right) = 0 \] Compute the derivatives: - \( \frac{\partial L}{\partial x} = 2(p + \lambda r)x \) - \( \frac{\partial L}{\partial \dot{x}} = 2q\dot{x} \) - \( \frac{d}{dt}\left(2q\dot{x}\right) = 2q\ddot{x} + 2q'\dot{x} \) Plug into the equation: \[ 2(p + \lambda r)x - \left[2q\ddot{x} + 2q'\dot{x}\right] = 0 \] Divide by 2: \[ (p + \lambda r)x - q\ddot{x} - q'\dot{x} = 0 \] Or: \[ q\ddot{x} + q'\dot{x} - (p + \lambda r)x = 0 \] --- ### **Step 4: Final Mayer Equation** The **Mayer equation** for this isoperimetric problem is: \[ \boxed{ q(t)\ddot{x} + q'(t)\dot{x} - \left[ p(t) + \lambda r(t) \right] x = 0 } \] with auxiliary condition: \[ \int_0^a r(t)x^2(t)\,dt = 1 \] and boundary conditions: \[ x(0) = 0, \quad x(a) = 0 \] --- ## **Summary** ### The Mayer equation is: \[ \boxed{ q(t)\ddot{x} + q'(t)\dot{x} - (p(t) + \lambda r(t)) x = 0 } \] with: - Boundary: \( x(0) = x(a) = 0 \) - Constraint: \( \int_0^a r(t)x^2(t)\,dt = 1 \) - \( p,q,r \) given \( C^2 \) functions, \( \lambda \) to be determined. --- If you need more details on solving for \( \lambda \), let me know!