Problem 2
Question
A certain physical problem has the functional $$ \Pi_{p}=\int_{0}^{L}\left(\frac{1}{2} \phi_{x}^{2}-50 \phi\right) d x $$ Determine the Euler equation for this functional. Then solve the Euler equation with the essential boundary conditions \(\phi=0\) at \(x=0\) and \(\phi=20\) at \(x=L\) to determine \(\phi\) as a function of \(x\) and \(L\).
Step-by-Step Solution
Verified Answer
The solution of the Euler equation with the given boundary conditions is \( \phi(x) = -25x^{2} + (50L - 25L^{2})x \)
1Step 1: Formulate Euler's equation
From calculus of variations, we know the Euler's equation for a functional of this form is: \[ \frac{d}{dx}\left(\frac{\partial F}{\partial \phi_{x}}\right) - \frac{\partial F}{\partial \phi} = 0 \] Here the integrand \(F\) in the functional is \( \phi_{x}^{2}/2 - 50 \phi \), so the Euler's equation for this functional is: \[ \frac{d}{dx}\phi_{x} + 50 = 0\]
2Step 2: Solving the differential equation
We can solve the ordinary differential equation from Step 1 using standard methods. Integrating both sides with respect to \(x\) gives \( \phi_{x} = -50x + C \), where \(C\) is the constant of integration. We can then integrate once more to find \( \phi \). This gives \( \phi = -25x^{2} + Cx + D \).
3Step 3: Applying the boundary conditions
With the general solution for \( \phi \), we can now use the boundary conditions to find the constants \(C\) and \(D\). Using the condition \( \phi(0) = 0 \), we have \(D = 0\). Now use the second boundary condition \( \phi(L) = 20 \) to solve for \(C\) which leads to \(C = 50L - 25L^{2}\).
4Step 4: Final solution
Substituting the values of \(C\) and \(D\) back into the expression for \( \phi \) gives the final solution: \( \phi(x) = -25x^{2} + (50L - 25L^{2})x \)
Key Concepts
Euler's EquationFunctional AnalysisBoundary Conditions
Euler's Equation
Euler's Equation is a cornerstone in the calculus of variations, serving as a tool to find functions that optimize or extremize functionals. In this context, a functional is an expression involving an integral, which in this case is \( \Pi_{p}=\int_{0}^{L}\left(\frac{1}{2} \phi_{x}^{2}-50 \phi\right) d x \). To find the function \( \phi \) that makes this integral stationary (neither maximizing nor minimizing), we apply Euler's equation.
Specifically, Euler's equation for a functional \( \int F(\phi, \phi_x, x) \, dx \) is given by:
Specifically, Euler's equation for a functional \( \int F(\phi, \phi_x, x) \, dx \) is given by:
- \( \frac{d}{dx}\left(\frac{\partial F}{\partial \phi_{x}}\right) - \frac{\partial F}{\partial \phi} = 0\)
- \( \frac{d}{dx}\phi_{x} + 50 = 0 \)
Functional Analysis
Functional analysis provides a framework to study spaces of functions and their applications. In the context of calculus of variations, it allows us to treat functions in terms of their properties and the operations we can perform on them. This is particularly relevant when we deal with functionals, which are mappings from a space of functions to a set of real numbers.
Our problem involves the functional \( \Pi_{p} \), which is a key focus of functional analysis. In this space:
Our problem involves the functional \( \Pi_{p} \), which is a key focus of functional analysis. In this space:
- The goal is to understand how changes in functions like \( \phi \) affect the value of the functional \( \Pi_{p} \).
- Results such as Euler's equation arise from such analysis, helping specify the conditions under which functionals remain stationary.
Boundary Conditions
Boundary conditions are constraints necessary to find specific solutions to differential equations arising from functionals. They describe how the function \( \phi(x) \) behaves at the boundaries of the interval they are defined on. In our case, these conditions are essential to ensure we select the correct solution among possible general solutions.
The problem specifies two boundary conditions:
The problem specifies two boundary conditions:
- \( \phi(0) = 0 \)
- \( \phi(L) = 20 \)
- From \( \phi(0) = 0 \), it follows \( D = 0 \).
- From \( \phi(L) = 20 \), we solve for \( C \), resulting in \( C = 50L - 25L^{2} \).
Other exercises in this chapter
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