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 Post subject: Existence of function
PostPosted: Sat Jan 16, 2016 2:04 am 
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Let \(\displaystyle{x\,,y:\left[0,1\right]\longrightarrow \mathbb{R}\,\, t\mapsto x(t)\,,t\mapsto y(t)}\) be \(\displaystyle{\rm{C}^{\infty}}\) functions


with \(\displaystyle{\left(x(t),y(t)\right)\neq \overline{0}\,\,,\forall\,t\in\left[0,1\right]}\) .


Consider the functions \(\displaystyle{a\,,b:\left[0,1\right]\longrightarrow \mathbb{R}}\,,a(t)=\dfrac{x(t)}{\sqrt{x^2(t)+y^2(t)}}\,,b(t)=\dfrac{y(t)}{\sqrt{x^2(t)+y^2(t)}}\)


such that \(\displaystyle{\cos\,\phi_{0}=a(0)\,,\sin\,\phi_{0}=b(0)}\) for some \(\displaystyle{\phi_{0}\in\mathbb{R}}\) .

Prove that there exists a function \(\displaystyle{\phi:\left[0,1\right]\longrightarrow \mathbb{R}}\) such that


\(\displaystyle{\cos\,\phi(t)=a(t)\,,\sin\,\phi(t)=b(t)\,,\forall\,t\in\left[0,1\right]}\) .


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PostPosted: Sat Jan 16, 2016 2:05 am 
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Firstly, observe that
\[ \displaystyle a^2(t) + b^2(t) = 1 , \, \forall t \in [0,1] \; \; (1) \]and therefore \[ \displaystyle a(t)a^{\prime}(t) + b(t)b^{\prime}(t) = 0 , \, \forall t \in [0,1] \; \; (2) \]Now, define \[ \displaystyle \phi : [0,1] \longrightarrow \mathbb{R} \, , \, \phi(t) = \int_{0}^{t} \left[ a(u)b^{\prime}(u)-a^{\prime}(u)b(u) \right] \mathrm{d}u + \phi_{0} \]We have that \[ \displaystyle \phi^{\prime}(t) = a(t)b^{\prime}(t)-a^{\prime}(t)b(t) , \, \forall t \in [0,1] \; \; (3) \]and that
\begin{align*}
\left[ a(t)\cos\phi(t) + b(t)\sin\phi(t) \right]^{\prime} &= a^{\prime}(t)\cos\phi(t) - a(t)\phi^{\prime}(t)\sin\phi(t) + b^{\prime}(t)\sin\phi(t) + b(t)\phi^{\prime}(t)\cos\phi(t) \\
&\overset{(3)}{=} a^{\prime}(t)\cos\phi(t) -a(t)\left[ a(t)b^{\prime}(t)-a^{\prime}(t)b(t) \right]\sin\phi(t) + \\ &+ b^{\prime}(t)\sin\phi(t) + b(t)\left[ a(t)b^{\prime}(t)-a^{\prime}(t)b(t) \right]\cos\phi(t) \\
&\overset{(2)}{=} -b^{\prime}(t)\sin\phi(t)\left[ a^2(t) + b^2(t) \right] -a^{\prime}(t)\cos\phi(t)\left[ a^2(t) + b^2(t) \right] + \\ &+ a^{\prime}(t)\cos\phi(t) + b^{\prime}(t)\sin\phi(t) \\
&\overset{(1)}{=} 0
\end{align*}
Hence \[ \displaystyle a(t)\cos\phi(t) + b(t)\sin\phi(t) = constant , \, \forall t \in [0,1] \]and since \( \displaystyle \cos\phi_{0}=a(0) , \sin\phi_{0}=b(0) \), we conclude, due to (1), that
\[ \displaystyle a(t)\cos\phi(t) + b(t)\sin\phi(t) = 1 , \, \forall t \in [0,1] \; \; (4) \]It is easy to see that
\[ \displaystyle \left[ a(t) - \cos\phi(t) \right]^2 + \left[ b(t) - \sin\phi(t) \right]^2 = 2 - 2 \left[ a(t)\cos\phi(t) + b(t)\sin\phi(t) \right] \overset{(4)}{=} 0 , \, \forall t \in [0,1] \]whence we have the desired result.


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