Let y = y(x) ∈ C4 ([0, 1]) be an extremizing function for the functional \(I(y)=\int^{1}_0\left[\left(\frac{d^2y}{dx^2} \right)^2 -2y\right] dx,\) satisfying y(0) = 0 = y(1). Then an extremal y(x), satisfying the given conditions at 0 and 1 together with the natural boundary conditions, is given by
Extremal Function for Variational Problem
The problem asks us to find an extremal function \(y(x)\) for a given functional \(I(y)\) defined by an integral. The functional is given by:
\[I(y)=\int^{1}_0\left[\left(\frac{d^2y}{dx^2} \right)^2 -2y\right] dx\]This is a problem in Calculus of Variations. The integrand \(F\) depends on \(x\), \(y\), and the second derivative \(y''\). The form of the functional is \(I(y) = \int_a^b F(x, y, y', y'') dx\), where \(a=0\) and \(b=1\), and \(F(x, y, y', y'') = (y'')^2 - 2y\).
An extremal function must satisfy the Euler-Lagrange equation for higher-order derivatives. The general form of the Euler-Lagrange equation for this type of functional is:
\[\frac{\partial F}{\partial y} - \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) + \frac{d^2}{dx^2}\left(\frac{\partial F}{\partial y''}\right) = 0\]Euler-Lagrange Equation Application
Let's find the required partial derivatives of \(F = (y'')^2 - 2y\):
- \(\frac{\partial F}{\partial y} = \frac{\partial}{\partial y}( (y'')^2 - 2y ) = -2\)
- \(\frac{\partial F}{\partial y'} = \frac{\partial}{\partial y'}( (y'')^2 - 2y ) = 0\)
- \(\frac{\partial F}{\partial y''} = \frac{\partial}{\partial y''}( (y'')^2 - 2y ) = 2y''\)
Now substitute these into the Euler-Lagrange equation:
\[-2 - \frac{d}{dx}(0) + \frac{d^2}{dx^2}(2y'') = 0\] \[-2 + 2\frac{d^2}{dx^2}(y'') = 0\] \[2y'''' = 2\] \[y'''' = 1\]This is the fourth-order differential equation that the extremal function \(y(x)\) must satisfy.
Solving the Differential Equation
We need to solve the differential equation \(y'''' = 1\). We can do this by integrating four times with respect to \(x\):
- Integrate once: \(y''' = \int 1 \, dx = x + C_1\)
- Integrate twice: \(y'' = \int (x + C_1) \, dx = \frac{x^2}{2} + C_1x + C_2\)
- Integrate thrice: \(y' = \int \left(\frac{x^2}{2} + C_1x + C_2\right) \, dx = \frac{x^3}{6} + C_1\frac{x^2}{2} + C_2x + C_3\)
- Integrate four times: \(y = \int \left(\frac{x^3}{6} + C_1\frac{x^2}{2} + C_2x + C_3\right) \, dx = \frac{x^4}{24} + C_1\frac{x^3}{6} + C_2\frac{x^2}{2} + C_3x + C_4\)
So, the general solution is \(y(x) = \frac{x^4}{24} + C_1\frac{x^3}{6} + C_2\frac{x^2}{2} + C_3x + C_4\), where \(C_1, C_2, C_3, C_4\) are integration constants.
Boundary Conditions
We are given fixed boundary conditions \(y(0) = 0\) and \(y(1) = 0\). Additionally, the problem states that the extremal satisfies these conditions along with natural boundary conditions. For a functional involving \(y''\), the natural boundary conditions at endpoints \(a\) and \(b\) where \(y'\) is not fixed are \(\left[\frac{\partial F}{\partial y''}\right]_{x=a} = 0\) and \(\left[\frac{\partial F}{\partial y''}\right]_{x=b} = 0\).
In our case, \(\frac{\partial F}{\partial y''} = 2y''\). The natural boundary conditions are:
- At \(x=0\): \(2y''(0) = 0 \implies y''(0) = 0\)
- At \(x=1\): \(2y''(1) = 0 \implies y''(1) = 0\)
So we have a total of four boundary conditions:
- \(y(0) = 0\)
- \(y(1) = 0\)
- \(y''(0) = 0\)
- \(y''(1) = 0\)
Finding Constants
Now we use these boundary conditions to find the values of \(C_1, C_2, C_3, C_4\).
We have \(y(x) = \frac{x^4}{24} + C_1\frac{x^3}{6} + C_2\frac{x^2}{2} + C_3x + C_4\) and \(y''(x) = \frac{x^2}{2} + C_1x + C_2\).
- Using \(y(0) = 0\):
\(y(0) = \frac{0^4}{24} + C_1\frac{0^3}{6} + C_2\frac{0^2}{2} + C_3(0) + C_4 = 0\) \(0 + 0 + 0 + 0 + C_4 = 0 \implies C_4 = 0\) - Using \(y''(0) = 0\):
\(y''(0) = \frac{0^2}{2} + C_1(0) + C_2 = 0\) \(0 + 0 + C_2 = 0 \implies C_2 = 0\) - Using \(y''(1) = 0\):
Now \(y''(x) = \frac{x^2}{2} + C_1x\) (since \(C_2=0\)). \(y''(1) = \frac{1^2}{2} + C_1(1) = 0\) \(\frac{1}{2} + C_1 = 0 \implies C_1 = -\frac{1}{2}\) - Using \(y(1) = 0\):
Now \(y(x) = \frac{x^4}{24} + C_1\frac{x^3}{6} + C_3x\) (since \(C_2=0, C_4=0\)). Substitute \(C_1 = -\frac{1}{2}\): \(y(x) = \frac{x^4}{24} + \left(-\frac{1}{2}\right)\frac{x^3}{6} + C_3x = \frac{x^4}{24} - \frac{x^3}{12} + C_3x\) \(y(1) = \frac{1^4}{24} - \frac{1^3}{12} + C_3(1) = 0\) \(\frac{1}{24} - \frac{1}{12} + C_3 = 0\) \(\frac{1}{24} - \frac{2}{24} + C_3 = 0\) \(-\frac{1}{24} + C_3 = 0 \implies C_3 = \frac{1}{24}\)
Final Solution for y(x)
Substituting the values of the constants back into the general solution for \(y(x)\):
\[y(x) = \frac{x^4}{24} + \left(-\frac{1}{2}\right)\frac{x^3}{6} + (0)\frac{x^2}{2} + \left(\frac{1}{24}\right)x + 0\] \[y(x) = \frac{x^4}{24} - \frac{x^3}{12} + \frac{x}{24}\]To compare this with the given options, let's factor out \(\frac{x}{24}\):
\[y(x) = \frac{1}{24} \left( x^3 - 2x^2 + x \right)\]Hmm, let me recheck the calculation. Ah, I factored \(x\) from \(\frac{x^4}{24} - \frac{x^3}{12} + \frac{x}{24}\). Let's factor \(\frac{1}{24}\) first.
\[y(x) = \frac{1}{24} \left( x^4 - 2x^3 + x \right)\]Now factor out \(x\):
\[y(x) = \frac{x}{24} \left( x^3 - 2x^2 + 1 \right)\]Verification Against Options
Let's compare our derived solution \(y(x) = \frac{x}{24}{(x^3-2x^2+1)}\) with the given options:
- Option 1: \(\frac{x}{24}{(x-1)}^3 = \frac{x}{24}(x^3 - 3x^2 + 3x - 1)\). Does not match.
- Option 2: \(\frac{x^2}{24}{(x-1)}^2 = \frac{x^2}{24}(x^2 - 2x + 1) = \frac{1}{24}(x^4 - 2x^3 + x^2)\). Does not match.
- Option 3: \(\frac{x}{24}{(x^3-2x^2+1)}\). This matches our derived solution.
- Option 4: \(\frac{x}{24}{(x^3+x^2-2)}\). Does not match.
The derived extremal function satisfies the Euler-Lagrange equation, the fixed boundary conditions \(y(0)=0\), \(y(1)=0\), and the natural boundary conditions \(y''(0)=0\), \(y''(1)=0\).
Let's quickly check the boundary conditions for the matching option, \(y(x) = \frac{x}{24}(x^3 - 2x^2 + 1)\).
\[y(0) = \frac{0}{24}(0 - 0 + 1) = 0\] \[y(1) = \frac{1}{24}(1^3 - 2(1)^2 + 1) = \frac{1}{24}(1 - 2 + 1) = \frac{1}{24}(0) = 0\]Now let's check \(y''(x)\). \[y(x) = \frac{1}{24} (x^4 - 2x^3 + x)\] \[y'(x) = \frac{1}{24} (4x^3 - 6x^2 + 1)\] \[y''(x) = \frac{1}{24} (12x^2 - 12x) = \frac{12x(x-1)}{24} = \frac{x(x-1)}{2}\] \[y''(0) = \frac{0(0-1)}{2} = 0\] \[y''(1) = \frac{1(1-1)}{2} = \frac{1(0)}{2} = 0\]
All four boundary conditions are satisfied by the function from Option 3.
The final answer is \(\frac{x}{24}{(x^3-2x^2+1)}\).
Important Questions from Euler-lagrange Equation
The minimum value of the functional
\(I(y)=\int^{\pi}_0\left(\frac{dy}{dx} \right)^2 dx,\)
subject to \(\int^{\pi}_0 y^2(x)dx = 1, y(0) = 0 = у(π)\) is equal to
The extremal of the functional
\(J(y)=\int_0^1\left[2\left(y^{\prime}\right)^2+x y\right] d x,\) y(0) = 0, y(1) = 1, y ∈ C2[0, 1] is
The Euler equations satisfied by the extremals of the functional
\(J(y)=\int_0^5\left[y^2+x^3 y^{\prime}\right] d x\)
define a solution curve in the (x, y) - plane which is
Let B be the unit ball in \(\mathbb{R}\)3 centered at origin. The Euler-Lagrange equation corresponding to the functional \(I(u)=\frac{1}{2} \int_B|\nabla u|^2 d x-\frac{1}{5} \int_B|u|^5 d x\) is
Which of the following is an extremal of the functional \(I\left( y \right) = \int_{ - 1}^1 {\left( {{y^{'2}} - 2xy} \right)} \,dx\) that satisfies the boundary conditions y(-1) = -1 and y(1) = 1?
