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这是一道最优控制作业题，求详细解答 3.拓展欧拉方程应用 (15分) Consider the functional \[ J(x)=\int_{t_{0}}...
Nov 22, 2024
这是一道最优控制作业题，求详细解答
Solution by Steps
step 1
We start with the functional J(x)=∫01[x(t)x˙(t)+x¨2(t)]dt J(x) = \int_{0}^{1} \left[ x(t) \dot{x}(t) + \ddot{x}^2(t) \right] dt . We need to apply the extended Euler equation to find the extremals
step 2
The extended Euler equation is given by ∑k=0r(−1)kdkdtk[∂g∂x(k)]=0 \sum_{k=0}^{r} (-1)^{k} \frac{d^{k}}{dt^{k}} \left[ \frac{\partial g}{\partial x^{(k)}} \right] = 0 . Here, g=x(t)x˙(t)+x¨2(t) g = x(t) \dot{x}(t) + \ddot{x}^2(t) . We will compute the necessary derivatives
step 3
We compute the first and second derivatives of g g with respect to x x and its derivatives. We find ∂g∂x=x˙(t) \frac{\partial g}{\partial x} = \dot{x}(t) and ∂g∂x˙=x(t) \frac{\partial g}{\partial \dot{x}} = x(t) . The second derivative gives ∂g∂x¨=2x¨(t) \frac{\partial g}{\partial \ddot{x}} = 2 \ddot{x}(t)
step 4
Now we substitute these into the extended Euler equation. We have ddt[x˙(t)]−d2dt2[2x¨(t)]=0 \frac{d}{dt} \left[ \dot{x}(t) \right] - \frac{d^2}{dt^2} \left[ 2 \ddot{x}(t) \right] = 0 . This simplifies to x¨(t)−2d2dt2x¨(t)=0 \ddot{x}(t) - 2 \frac{d^2}{dt^2} \ddot{x}(t) = 0
step 5
We solve the resulting differential equation. The general solution is x(t)=A+Bt+Ct2+Dt3 x(t) = A + Bt + Ct^2 + Dt^3 . We apply the boundary conditions x(0)=0,x˙(0)=1,x(1)=2,x˙(1)=4 x(0) = 0, \dot{x}(0) = 1, x(1) = 2, \dot{x}(1) = 4 to find the constants A,B,C,D A, B, C, D
step 6
Applying the boundary conditions, we find A=0,B=1,C=1,D=0 A = 0, B = 1, C = 1, D = 0 . Thus, the extremal function is x(t)=t+t2 x(t) = t + t^2
Answer
The extremal function is x(t)=t+t2 x(t) = t + t^2
Key Concept
The extended Euler equation is a tool used in calculus of variations to find extremals of functionals.
Explanation
By applying the extended Euler equation and using the specified boundary conditions, we derived the extremal function that minimizes the functional J(x) J(x) .
这是一道最优控制作业题，求详细解答
Solution by Steps
step 1
We start with the functional J(x)=∫01[x(t)x˙(t)+x¨2(t)]dt J(x) = \int_{0}^{1} \left[ x(t) \dot{x}(t) + \ddot{x}^2(t) \right] dt . We need to apply the extended Euler equation to find the extremals
step 2
The extended Euler equation is given by ∑k=0r(−1)kdkdtk[∂g∂x(k)]=0 \sum_{k=0}^{r} (-1)^{k} \frac{d^{k}}{dt^{k}} \left[ \frac{\partial g}{\partial x^{(k)}} \right] = 0 . Here, we will identify g g from the functional
step 3
We compute the necessary derivatives and apply the boundary conditions x(0)=0,x˙(0)=1,x(1)=2,x˙(1)=4 x(0) = 0, \dot{x}(0) = 1, x(1) = 2, \dot{x}(1) = 4 to find the extremals
step 4
After solving the equations, we find that the extremal function is x(t)=t+t2 x(t) = t + t^2
Answer
The extremal function is x(t)=t+t2 x(t) = t + t^2
Key Concept
The extended Euler equation is used to find extremals of functionals in optimal control problems.
Explanation
The solution involves applying the extended Euler equation to derive the extremal function that satisfies the given boundary conditions.
这是一道最优控制作业题，求详细解答
Solution by Steps
step 1
We start with the curve defined by y2=y12−4.5 y_{2} = y_{1}^{2} - 4.5 . We need to express the function f(y1,y2)=y12+y22 f(y_{1}, y_{2}) = y_{1}^{2} + y_{2}^{2} in terms of y1 y_{1} only
step 2
Substituting y2 y_{2} into f f , we have f(y1)=y12+(y12−4.5)2 f(y_{1}) = y_{1}^{2} + (y_{1}^{2} - 4.5)^{2}
step 3
To minimize f(y1) f(y_{1}) , we take the derivative f′(y1) f'(y_{1}) and set it to zero: f′(y1)=2y1+2(y12−4.5)(2y1)=0 f'(y_{1}) = 2y_{1} + 2(y_{1}^{2} - 4.5)(2y_{1}) = 0
step 4
Simplifying gives 2y1(1+(y12−4.5))=0 2y_{1}(1 + (y_{1}^{2} - 4.5)) = 0 . This leads to y1=0 y_{1} = 0 or y12−4.5+1=0 y_{1}^{2} - 4.5 + 1 = 0
step 5
Solving y12−3.5=0 y_{1}^{2} - 3.5 = 0 gives y1=±3.5 y_{1} = \pm \sqrt{3.5} . Thus, the critical points are y1=0,3.5,−3.5 y_{1} = 0, \sqrt{3.5}, -\sqrt{3.5}
step 6
We substitute these values back into the curve equation to find corresponding y2 y_{2} values: y2=y12−4.5 y_{2} = y_{1}^{2} - 4.5 . For y1=0 y_{1} = 0 , y2=−4.5 y_{2} = -4.5 ; for y1=3.5 y_{1} = \sqrt{3.5} , y2=−1 y_{2} = -1 ; for y1=−3.5 y_{1} = -\sqrt{3.5} , y2=−1 y_{2} = -1
step 7
The points on the curve that minimize f f are (0,−4.5) (0, -4.5) and (±3.5,−1) (\pm \sqrt{3.5}, -1)
Answer
The points on the curve that minimize the function are (0,−4.5) (0, -4.5) and (±3.5,−1) (\pm \sqrt{3.5}, -1) .
Key Concept
To minimize a function subject to a constraint, we can substitute the constraint into the function and find critical points.
Explanation
By substituting the constraint into the function and finding the critical points, we can determine the points that minimize the function on the given curve.
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