extension of a regular path












0












$begingroup$


Given a smooth path in $gamma: I to mathbb R^n$. such that $gamma(0) = x$, $gamma(1)=y$, $gamma'(t) neq 0, forall t$, $x neq y$. Let $z neq x, z neq y$, Is it always possible to extend $gamma$ to a smooth path connecting $y$ to $z$, such that $gamma'(t)neq 0$ everywhere?



What is the right tool to deal with this question? Maybe should I take piecewise path first then try to smooth it using convolution fixing endpoints?






real-analysis fourier-analysis differential-topology







share|cite|improve this question









$endgroup$












  • $begingroup$
    Couldn't you tack on a Bezier curve at the end going from y to z? Of course this will only give finite smoothness
    $endgroup$
    – Dunham
    Jan 13 at 0:39










  • $begingroup$
    I am sorry, but I do not know what is a Bezier curve.@Dunham
    $endgroup$
    – koch
    Jan 13 at 3:06
















0












$begingroup$


Given a smooth path in $gamma: I to mathbb R^n$. such that $gamma(0) = x$, $gamma(1)=y$, $gamma'(t) neq 0, forall t$, $x neq y$. Let $z neq x, z neq y$, Is it always possible to extend $gamma$ to a smooth path connecting $y$ to $z$, such that $gamma'(t)neq 0$ everywhere?



What is the right tool to deal with this question? Maybe should I take piecewise path first then try to smooth it using convolution fixing endpoints?






real-analysis fourier-analysis differential-topology







share|cite|improve this question









$endgroup$












  • $begingroup$
    Couldn't you tack on a Bezier curve at the end going from y to z? Of course this will only give finite smoothness
    $endgroup$
    – Dunham
    Jan 13 at 0:39










  • $begingroup$
    I am sorry, but I do not know what is a Bezier curve.@Dunham
    $endgroup$
    – koch
    Jan 13 at 3:06














0












0








0





$begingroup$


Given a smooth path in $gamma: I to mathbb R^n$. such that $gamma(0) = x$, $gamma(1)=y$, $gamma'(t) neq 0, forall t$, $x neq y$. Let $z neq x, z neq y$, Is it always possible to extend $gamma$ to a smooth path connecting $y$ to $z$, such that $gamma'(t)neq 0$ everywhere?



What is the right tool to deal with this question? Maybe should I take piecewise path first then try to smooth it using convolution fixing endpoints?






real-analysis fourier-analysis differential-topology







share|cite|improve this question









$endgroup$




Given a smooth path in $gamma: I to mathbb R^n$. such that $gamma(0) = x$, $gamma(1)=y$, $gamma'(t) neq 0, forall t$, $x neq y$. Let $z neq x, z neq y$, Is it always possible to extend $gamma$ to a smooth path connecting $y$ to $z$, such that $gamma'(t)neq 0$ everywhere?



What is the right tool to deal with this question? Maybe should I take piecewise path first then try to smooth it using convolution fixing endpoints?






real-analysis fourier-analysis differential-topology




real-analysis fourier-analysis differential-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 12 at 23:39









kochkoch

20318




20318












  • $begingroup$
    Couldn't you tack on a Bezier curve at the end going from y to z? Of course this will only give finite smoothness
    $endgroup$
    – Dunham
    Jan 13 at 0:39










  • $begingroup$
    I am sorry, but I do not know what is a Bezier curve.@Dunham
    $endgroup$
    – koch
    Jan 13 at 3:06


















  • $begingroup$
    Couldn't you tack on a Bezier curve at the end going from y to z? Of course this will only give finite smoothness
    $endgroup$
    – Dunham
    Jan 13 at 0:39










  • $begingroup$
    I am sorry, but I do not know what is a Bezier curve.@Dunham
    $endgroup$
    – koch
    Jan 13 at 3:06
















$begingroup$
Couldn't you tack on a Bezier curve at the end going from y to z? Of course this will only give finite smoothness
$endgroup$
– Dunham
Jan 13 at 0:39




$begingroup$
Couldn't you tack on a Bezier curve at the end going from y to z? Of course this will only give finite smoothness
$endgroup$
– Dunham
Jan 13 at 0:39












$begingroup$
I am sorry, but I do not know what is a Bezier curve.@Dunham
$endgroup$
– koch
Jan 13 at 3:06




$begingroup$
I am sorry, but I do not know what is a Bezier curve.@Dunham
$endgroup$
– koch
Jan 13 at 3:06










1 Answer
1






active

oldest

votes


















0












$begingroup$

@koch,



It is definitely possible. You need a family of curves rich enough. For example, take
$$
y(s)={mathbf A}+{mathbf B}s +{mathbf C}s^k,qquad sin[0,1]
$$
where ${mathbf A}$, ${mathbf B}$ and ${mathbf C}$ are vectors in $mathbb{R}^n$ and $k>1$ are to be determined. Imposing $y(0)=gamma(1)$ and $y'(0)=gamma'(1)$ (to guarantee continuity of the tangent vector) you get ${mathbf A}=gamma(1)$ and ${mathbf B}=gamma'(1)$. Next, you choose ${mathbf C}$ in such a way that $y(1)=z$. You can choose then $k>1$ to guarantee $y'(s)neq 0$ for all $sin[0,1]$. Finally, you can easily paste both parametrizations together, namely define $x(t)=y(t-1)$ for $tin[1,2]$.



It seems clear to me that a similar procedure can be used to produce an extension with any desired regularity. Ssy, if you want class $C^2$ you can add a quadratic term, etc.
Hope this helps.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Then how do you ensure the resulting extension is smooth?
    $endgroup$
    – koch
    Jan 13 at 4:30










  • $begingroup$
    How much smoothness do you need? C^infty?
    $endgroup$
    – GReyes
    Jan 19 at 7:30











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

@koch,



It is definitely possible. You need a family of curves rich enough. For example, take
$$
y(s)={mathbf A}+{mathbf B}s +{mathbf C}s^k,qquad sin[0,1]
$$
where ${mathbf A}$, ${mathbf B}$ and ${mathbf C}$ are vectors in $mathbb{R}^n$ and $k>1$ are to be determined. Imposing $y(0)=gamma(1)$ and $y'(0)=gamma'(1)$ (to guarantee continuity of the tangent vector) you get ${mathbf A}=gamma(1)$ and ${mathbf B}=gamma'(1)$. Next, you choose ${mathbf C}$ in such a way that $y(1)=z$. You can choose then $k>1$ to guarantee $y'(s)neq 0$ for all $sin[0,1]$. Finally, you can easily paste both parametrizations together, namely define $x(t)=y(t-1)$ for $tin[1,2]$.



It seems clear to me that a similar procedure can be used to produce an extension with any desired regularity. Ssy, if you want class $C^2$ you can add a quadratic term, etc.
Hope this helps.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Then how do you ensure the resulting extension is smooth?
    $endgroup$
    – koch
    Jan 13 at 4:30










  • $begingroup$
    How much smoothness do you need? C^infty?
    $endgroup$
    – GReyes
    Jan 19 at 7:30
















0












$begingroup$

@koch,



It is definitely possible. You need a family of curves rich enough. For example, take
$$
y(s)={mathbf A}+{mathbf B}s +{mathbf C}s^k,qquad sin[0,1]
$$
where ${mathbf A}$, ${mathbf B}$ and ${mathbf C}$ are vectors in $mathbb{R}^n$ and $k>1$ are to be determined. Imposing $y(0)=gamma(1)$ and $y'(0)=gamma'(1)$ (to guarantee continuity of the tangent vector) you get ${mathbf A}=gamma(1)$ and ${mathbf B}=gamma'(1)$. Next, you choose ${mathbf C}$ in such a way that $y(1)=z$. You can choose then $k>1$ to guarantee $y'(s)neq 0$ for all $sin[0,1]$. Finally, you can easily paste both parametrizations together, namely define $x(t)=y(t-1)$ for $tin[1,2]$.



It seems clear to me that a similar procedure can be used to produce an extension with any desired regularity. Ssy, if you want class $C^2$ you can add a quadratic term, etc.
Hope this helps.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Then how do you ensure the resulting extension is smooth?
    $endgroup$
    – koch
    Jan 13 at 4:30










  • $begingroup$
    How much smoothness do you need? C^infty?
    $endgroup$
    – GReyes
    Jan 19 at 7:30














0












0








0





$begingroup$

@koch,



It is definitely possible. You need a family of curves rich enough. For example, take
$$
y(s)={mathbf A}+{mathbf B}s +{mathbf C}s^k,qquad sin[0,1]
$$
where ${mathbf A}$, ${mathbf B}$ and ${mathbf C}$ are vectors in $mathbb{R}^n$ and $k>1$ are to be determined. Imposing $y(0)=gamma(1)$ and $y'(0)=gamma'(1)$ (to guarantee continuity of the tangent vector) you get ${mathbf A}=gamma(1)$ and ${mathbf B}=gamma'(1)$. Next, you choose ${mathbf C}$ in such a way that $y(1)=z$. You can choose then $k>1$ to guarantee $y'(s)neq 0$ for all $sin[0,1]$. Finally, you can easily paste both parametrizations together, namely define $x(t)=y(t-1)$ for $tin[1,2]$.



It seems clear to me that a similar procedure can be used to produce an extension with any desired regularity. Ssy, if you want class $C^2$ you can add a quadratic term, etc.
Hope this helps.






share|cite|improve this answer









$endgroup$



@koch,



It is definitely possible. You need a family of curves rich enough. For example, take
$$
y(s)={mathbf A}+{mathbf B}s +{mathbf C}s^k,qquad sin[0,1]
$$
where ${mathbf A}$, ${mathbf B}$ and ${mathbf C}$ are vectors in $mathbb{R}^n$ and $k>1$ are to be determined. Imposing $y(0)=gamma(1)$ and $y'(0)=gamma'(1)$ (to guarantee continuity of the tangent vector) you get ${mathbf A}=gamma(1)$ and ${mathbf B}=gamma'(1)$. Next, you choose ${mathbf C}$ in such a way that $y(1)=z$. You can choose then $k>1$ to guarantee $y'(s)neq 0$ for all $sin[0,1]$. Finally, you can easily paste both parametrizations together, namely define $x(t)=y(t-1)$ for $tin[1,2]$.



It seems clear to me that a similar procedure can be used to produce an extension with any desired regularity. Ssy, if you want class $C^2$ you can add a quadratic term, etc.
Hope this helps.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 13 at 4:29









GReyesGReyes

2,04815




2,04815












  • $begingroup$
    Then how do you ensure the resulting extension is smooth?
    $endgroup$
    – koch
    Jan 13 at 4:30










  • $begingroup$
    How much smoothness do you need? C^infty?
    $endgroup$
    – GReyes
    Jan 19 at 7:30


















  • $begingroup$
    Then how do you ensure the resulting extension is smooth?
    $endgroup$
    – koch
    Jan 13 at 4:30










  • $begingroup$
    How much smoothness do you need? C^infty?
    $endgroup$
    – GReyes
    Jan 19 at 7:30
















$begingroup$
Then how do you ensure the resulting extension is smooth?
$endgroup$
– koch
Jan 13 at 4:30




$begingroup$
Then how do you ensure the resulting extension is smooth?
$endgroup$
– koch
Jan 13 at 4:30












$begingroup$
How much smoothness do you need? C^infty?
$endgroup$
– GReyes
Jan 19 at 7:30




$begingroup$
How much smoothness do you need? C^infty?
$endgroup$
– GReyes
Jan 19 at 7:30


















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