Math 2.0

Calculus of variations and optimal control; optimization Definition

Euler–Poisson equation for higher derivatives

Builds on Integral functional with a Lagrangian densityEuler–Lagrange equationPartial derivative

Definition

Let k≥1k \ge 1 and a<ba < b. Let L ⁣:[a,b]×ℝk+1→ℝL \colon [a, b] \times \R^{k+1} \to \R, written L(x,z0,z1,…,zk)L(x, z_0, z_1, \dots, z_k), be a function of class Ck+1C^{k+1}, and write LzjL_{z_j} for its partial derivative in zjz_j. Consider the functional J(y)=∫abL(x,y(x),y′(x),…,y(k)(x)) dx.J(y) = \int_a^b L\big(x, y(x), y'(x), \dots, y^{(k)}(x)\big) \, dx. The Euler–Poisson equation of JJ is the equation ∑j=0k(−1)jdjdxj(Lzj(x,y(x),y′(x),…,y(k)(x)))=0,a≤x≤b,\sum_{j=0}^{k} (-1)^j \frac{d^j}{dx^j} \Big( L_{z_j}\big(x, y(x), y'(x), \dots, y^{(k)}(x)\big) \Big) = 0, \qquad a \le x \le b, for an unknown y∈C2k([a,b])y \in C^{2k}([a, b]).

Remarks

For k=1k = 1 it is the Euler–Lagrange equation Lz0−ddxLz1=0L_{z_0} - \frac{d}{dx} L_{z_1} = 0 of an integral functional. A minimiser y∈C2k([a,b])y \in C^{2k}([a, b]) of JJ among functions with prescribed y(j)(a)y^{(j)}(a) and y(j)(b)y^{(j)}(b) for 0≤j≤k−10 \le j \le k - 1 solves it. To see this, take variations η\eta with η(j)(a)=η(j)(b)=0\eta^{(j)}(a) = \eta^{(j)}(b) = 0 for j<kj < k, integrate the term with η(j)\eta^{(j)} by parts jj times, and apply the fundamental lemma of the calculus of variations.

Example: for k=2k = 2 and L=12z22L = \frac{1}{2} z_2^2, one has Lz0=Lz1=0L_{z_0} = L_{z_1} = 0 and Lz2=z2L_{z_2} = z_2, so the equation is y′′′′=0y'''' = 0. Every cubic polynomial solves it. Non-example: y(x)=x4y(x) = x^4 does not solve this equation, since y′′′′=24y'''' = 24.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation
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Kinddefinition
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MSC49K05, 34B15
DefinesEuler–Poisson equation for higher derivatives
UsesIntegral functional with a Lagrangian density, Euler–Lagrange equation, Partial derivative

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