Question

Write and solve the Euler equations to make the following integrals stationary. \(\int_{x_{1}}^{x_{2}} e^{x} \sqrt{1+y^{\prime 2}} d x\)

Short answer

Euler-Lagrange yields: \(e^{x} \frac{y'}{\sqrt{1+(y')^{2}}} = C\)

Step-by-step solution

Identify the Functional

The given integral is the functional which needs to be made stationary: \[\[\begin{equation} I[y] = \int_{x_{1}}^{x_{2}} e^{x} \sqrt{1+(y')^{2}} \,dx \end{equation}\] \] The integrand is the function \[\[\begin{equation} F = e^{x} \sqrt{1+(y')^{2}} \end{equation}\] \]

Apply the Euler-Lagrange Equation

The Euler-Lagrange equation for a functional \[\[\begin{equation} I[y] = \int_{x_{1}}^{x_{2}} F(x,y,y') \,dx \end{equation}\] \] is \[\[\begin{equation} \frac{\partial F}{\partial y} - \frac{d}{dx} \left(\frac{\partial F}{\partial y'}\right) = 0 \end{equation}\] \] Here, \[\[\begin{equation} F = e^{x} \sqrt{1+(y')^{2}} \end{equation}\] \]

Compute the Partial Derivatives

First, compute the partial derivative of F with respect to y: \[\[\begin{equation} \frac{\partial F}{\partial y} = 0 \end{equation}\] \] since F does not explicitly depend on y. Next, compute the partial derivative of F with respect to y': \[\[\begin{equation} \frac{\partial F}{\partial y'} = e^{x} \frac{y'}{\sqrt{1+(y')^{2}}} \end{equation}\] \]

Substitute into Euler-Lagrange Equation

Substitute the partial derivatives back into the Euler-Lagrange equation: \[\[\begin{equation} 0 - \frac{d}{dx} \left(e^{x} \frac{y'}{\sqrt{1+(y')^{2}}}\right) = 0 \end{equation}\] \] This simplifies to \[\[\begin{equation} \frac{d}{dx} \left(e^{x} \frac{y'}{\sqrt{1+(y')^{2}}}\right) = 0 \end{equation}\] \]

Integrate the Result

Since the derivative of the term inside is zero, it implies the term inside is a constant: \[\[\begin{equation} e^{x} \frac{y'}{\sqrt{1+(y')^{2}}} = C \end{equation}\] \]

Solve for y'

Isolate y' to solve for it: \[\[\begin{equation} y' = k \, e^{-x}\sqrt{1+(y')^{2}} \end{equation}\] \] where k is a constant. Squaring both sides: \[\[\begin{equation} (y')^{2} = k^{2} e^{-2x} (1 + (y')^{2}) \end{equation}\] \] Solving for y': \[\[\begin{equation} (y')^{2}(1 - k^{2}e^{-2x}) = k^{2}e^{-2x} \end{equation}\] \] Thus: \[\[\begin{equation} (y')^{2} = \frac{k^{2}e^{-2x}}{1 - k^{2}e^{-2x}} \end{equation}\] \]

Integrate for y

Now integrate y' to obtain y: \[\[\begin{equation} y = \int \frac{k e^{-x}}{\sqrt{1 - k^{2}e^{-2x}}} dx + C \end{equation}\] \]

Key concepts

Functional

In the context of calculus of variations, a functional is a mapping from a set of functions to the real numbers. To put it simply, a functional takes a function as input and produces a number as output. In our exercise, the functional is given by the integral: \[I[y] = \int_{x_{1}}^{x_{2}} e^{x} \sqrt{1+(y')^{2}} \, dx\]Here, the functional involves the integral of a function of both the independent variable \(x\) and the dependent variable \(y\), along with its derivative \(y'\). The objective is to find the function \(y(y(x))\) that makes this integral reach a stationary value (usually a minimum or maximum). This is the essence of finding extremals in calculus of variations.

Calculus of Variations

Calculus of Variations is a field of mathematical analysis that deals with optimizing functionals. Instead of optimizing functions as in traditional calculus, here we deal with functions of functions. In our exercise, we seek to make the functional \[I[y] = \int_{x_{1}}^{x_{2}} e^{x} \sqrt{1+(y')^{2}} \, dx\] stationary.
To achieve this, we use the Euler-Lagrange equations. These are differential equations that provide the necessary conditions for a functional to have an extremum. The Euler-Lagrange equation for a functional \[I[y] = \int_{x_{1}}^{x_{2}} F(x,y,y') \,dx\] is:
\( \frac{\partial F}{\partial y} - \frac{d}{dx} \left(\frac{\partial F}{\partial y'}\right) = 0 \)
This is the core tool we use to solve problems in calculus of variations.

Partial Derivatives

Partial derivatives are a fundamental concept used in solving the Euler-Lagrange equations. They represent the rate of change of a function with respect to one of its variables while keeping the other variables constant. In our exercise, the function \(F\) is: \[F = e^{x} \sqrt{1+(y')^{2}}\] We need to find the partial derivatives of \(F\) with respect to \(y\) and \(y'\). First, since \(F\) does not explicitly depend on \(y\), its partial derivative with respect to \(y\) is \(0\):
\( \frac{\partial F}{\partial y} = 0 \)
Next, we compute the partial derivative of \(F\) with respect to \(y'\):
\( \frac{\partial F}{\partial y'} = e^{x} \frac{y'}{\sqrt{1+(y')^{2}}} \)
These partial derivatives are crucial and must be substituted back into the Euler-Lagrange equation to proceed with finding the extremal of the functional.

Integrals

Integrals play a key role in functional calculus. They help to sum up infinitesimal changes over an interval, providing a total change. For the given exercise, solving involves definite integrals, particularly in the Euler-Lagrange framework. We start with the integral form of the functional: \[I[y] = \int_{x_{1}}^{x_{2}} e^{x} \sqrt{1+(y')^{2}} \, dx\] When solving, after using partial derivatives and substituting into the Euler-Lagrange equation, we reach:
\[\frac{d}{dx} \left(e^{x} \frac{y'}{\sqrt{1+(y')^{2}}}\right) = 0\].
The above implies that the expression inside the derivative is a constant. From here, we can solve for \(y'\) and eventually integrate to find \(y\). Performing this gives us the integral: \[y = \int \frac{k e^{-x}}{\sqrt{1 - k^{2}e^{-2x}}} dx + C\] The correct use of integrals helps us move from the differential form back to the function \(y\).