- #1
Kashmir
- 466
- 74
I dont Understand how we get the final equations relating ##Q_r## with ##\lambda## given the conditions above?
Graduate Help in understanding this derivation of Lagrange Equations in Non-Holonomic case
- Thread starter Kashmir
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The discussion centers on understanding the derivation of Lagrange equations in the non-holonomic case, specifically how the final equations relate the linear functionals ##Q_r## and ##\lambda##. A theorem is presented, stating that if the intersection of the kernels of several linear functionals is contained within the kernel of another functional, then a linear combination of these functionals can express the latter. This theorem is further generalized to include linear operators between vector spaces, establishing a relationship between their kernels. The participants seek clarity on applying these concepts to derive the necessary equations in the context of Lagrange mechanics. Understanding these relationships is crucial for solving problems in non-holonomic systems.
I dont Understand how we get the final equations relating ##Q_r## with ##\lambda## given the conditions above?
- #2
wrobel
Science Advisor
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There is a nice theorem.
Let ##f,f_1,\ldots, f_n:X\to\mathbb{R}## be linear functionals defined on a vector space ##X##.
Theorem. Assume that $$\bigcap_{k\in\{1,\ldots,n\}}\ker f_k\subset \ker f.$$
Then there are constants ##\lambda_1,\ldots,\lambda_n## such that
$$f=\sum_{k=1}^n\lambda_k f_k.$$
Moreover this theorem is a special case of the following fact. Let ##X,Y,Z## be vector spaces perhaps infinite dimensional. Let
$$A:X\to Y,\quad B:X\to Z$$ be linear operators such that ##\ker A\subset\ker B##. Then there is a linear operator ##\Lambda:Y\to Z## such that ##B=\Lambda A##.
Let ##f,f_1,\ldots, f_n:X\to\mathbb{R}## be linear functionals defined on a vector space ##X##.
Theorem. Assume that $$\bigcap_{k\in\{1,\ldots,n\}}\ker f_k\subset \ker f.$$
Then there are constants ##\lambda_1,\ldots,\lambda_n## such that
$$f=\sum_{k=1}^n\lambda_k f_k.$$
Moreover this theorem is a special case of the following fact. Let ##X,Y,Z## be vector spaces perhaps infinite dimensional. Let
$$A:X\to Y,\quad B:X\to Z$$ be linear operators such that ##\ker A\subset\ker B##. Then there is a linear operator ##\Lambda:Y\to Z## such that ##B=\Lambda A##.
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