Limit behaviour for solutions of $y'=-(1/x) y-(1/x^3)cos(1/x)$
$begingroup$
The ODE $$y'=-frac1x y-frac1{x^3} cosleft(frac1xright)$$
has as its set of solutions defined on $(-infty,0)$ the functions
$$y_C:(-infty,0)tomathbb{R},~~~y_C(x):=frac1xleft(C+sinleft(frac1xright)right).$$ If $Cnot=0$ we see by choosing $x_k:=-frac1{pi k}$ that $lim_{ktoinfty} y_c(x_k)=-text{sign}(C)cdot infty$. For $C=0$ we may choose $x_k'=-frac1{(2k+1/2)pi}$ to get $lim_{ktoinfty}y_0(x_k') =lim_{ktoinfty}(-frac1{x_k})=infty$.
In this case ($C=0$) much more can be shown:
For each real $b$ there is a sequence $(b_n)$ of negative numbers tending to $0$ such that $lim_{ntoinfty} y_0(b_n)=b$.
To this aim let $z_n:=-frac1{(n+1/2)pi}$. Then $z_n<z_{n+1}$ and $lim_{ntoinfty} z_n=0$. Morover let $f(x):=sin(1/x)-b x$. Obviously $f(x_{2n})=-1-b x_{2n}$ and $f(x_{2n+1})=1-b x_{2n+1}$. For large $n$ ($vert bcdot x_{2n}vert<1$) this implies $f(x_{2n})<0$ and
$f(x_{2n+1})>0$. Thus there is some $b_nin(x_{2n},x_{2n+1})$ such that $f(b_n)=0$, i.e., $y_0(b_n)=b$. This trivialy implies $lim_{ntoinfty }y_0(b_n)=b$
For $vert Cvert>1$ there is no sequence of $z_n<0$ converging to $0$ such that $y_C(z_n)$ converges to some real number since otherwise we would have $lim_{ntoinfty} (C+sin(1/z_n))=0$, implying $Cin[-1,1]$.
Question: Given $vert Cvertleq 1$ what are the points $b$ with $lim_{ntoinfty } y_C(z_n)=b$ for some sequence of negative reals $z_n$ converging to $0$?
ordinary-differential-equations
edited Dec 12 '18 at 13:47
LutzL
58.5k42054
asked Dec 12 '18 at 13:05
Jens SchwaigerJens Schwaiger
1,566128
$endgroup$
add a comment |
$begingroup$
The ODE $$y'=-frac1x y-frac1{x^3} cosleft(frac1xright)$$
has as its set of solutions defined on $(-infty,0)$ the functions
$$y_C:(-infty,0)tomathbb{R},~~~y_C(x):=frac1xleft(C+sinleft(frac1xright)right).$$ If $Cnot=0$ we see by choosing $x_k:=-frac1{pi k}$ that $lim_{ktoinfty} y_c(x_k)=-text{sign}(C)cdot infty$. For $C=0$ we may choose $x_k'=-frac1{(2k+1/2)pi}$ to get $lim_{ktoinfty}y_0(x_k') =lim_{ktoinfty}(-frac1{x_k})=infty$.
In this case ($C=0$) much more can be shown:
For each real $b$ there is a sequence $(b_n)$ of negative numbers tending to $0$ such that $lim_{ntoinfty} y_0(b_n)=b$.
To this aim let $z_n:=-frac1{(n+1/2)pi}$. Then $z_n<z_{n+1}$ and $lim_{ntoinfty} z_n=0$. Morover let $f(x):=sin(1/x)-b x$. Obviously $f(x_{2n})=-1-b x_{2n}$ and $f(x_{2n+1})=1-b x_{2n+1}$. For large $n$ ($vert bcdot x_{2n}vert<1$) this implies $f(x_{2n})<0$ and
$f(x_{2n+1})>0$. Thus there is some $b_nin(x_{2n},x_{2n+1})$ such that $f(b_n)=0$, i.e., $y_0(b_n)=b$. This trivialy implies $lim_{ntoinfty }y_0(b_n)=b$
For $vert Cvert>1$ there is no sequence of $z_n<0$ converging to $0$ such that $y_C(z_n)$ converges to some real number since otherwise we would have $lim_{ntoinfty} (C+sin(1/z_n))=0$, implying $Cin[-1,1]$.
Question: Given $vert Cvertleq 1$ what are the points $b$ with $lim_{ntoinfty } y_C(z_n)=b$ for some sequence of negative reals $z_n$ converging to $0$?
ordinary-differential-equations
edited Dec 12 '18 at 13:47
LutzL
58.5k42054
asked Dec 12 '18 at 13:05
Jens SchwaigerJens Schwaiger
1,566128
$endgroup$
$begingroup$
DERIVE says ∂(1/x·(c + SIN(1/x)), x)=- COS(1/x)/x^3 - SIN(1/x)/x^2 - c/x^2 and - 1/x·(1/x·(c + SIN(1/x))) - 1/x^3·COS(1/x)=- COS(1/x)/x^3 - SIN(1/x)/x^2 - c/x^2, i.e. it is a solution!
$endgroup$
– Jens Schwaiger
Dec 12 '18 at 13:39
$begingroup$
Applying the integrating factor $x$ gives $$(xy(x))'=-x^{-2}cos(x^{-1}).$$ Now $(sin(x^{-1}))'=cos(x^{-1})(-x^{-2})$ so that indeed $xy(x)=C+sin(x^{-1})$, which gives exactly the expression for $y_C$.
$endgroup$
– LutzL
Dec 12 '18 at 13:52
add a comment |
$begingroup$
The ODE $$y'=-frac1x y-frac1{x^3} cosleft(frac1xright)$$
has as its set of solutions defined on $(-infty,0)$ the functions
$$y_C:(-infty,0)tomathbb{R},~~~y_C(x):=frac1xleft(C+sinleft(frac1xright)right).$$ If $Cnot=0$ we see by choosing $x_k:=-frac1{pi k}$ that $lim_{ktoinfty} y_c(x_k)=-text{sign}(C)cdot infty$. For $C=0$ we may choose $x_k'=-frac1{(2k+1/2)pi}$ to get $lim_{ktoinfty}y_0(x_k') =lim_{ktoinfty}(-frac1{x_k})=infty$.
In this case ($C=0$) much more can be shown:
For each real $b$ there is a sequence $(b_n)$ of negative numbers tending to $0$ such that $lim_{ntoinfty} y_0(b_n)=b$.
To this aim let $z_n:=-frac1{(n+1/2)pi}$. Then $z_n<z_{n+1}$ and $lim_{ntoinfty} z_n=0$. Morover let $f(x):=sin(1/x)-b x$. Obviously $f(x_{2n})=-1-b x_{2n}$ and $f(x_{2n+1})=1-b x_{2n+1}$. For large $n$ ($vert bcdot x_{2n}vert<1$) this implies $f(x_{2n})<0$ and
$f(x_{2n+1})>0$. Thus there is some $b_nin(x_{2n},x_{2n+1})$ such that $f(b_n)=0$, i.e., $y_0(b_n)=b$. This trivialy implies $lim_{ntoinfty }y_0(b_n)=b$
For $vert Cvert>1$ there is no sequence of $z_n<0$ converging to $0$ such that $y_C(z_n)$ converges to some real number since otherwise we would have $lim_{ntoinfty} (C+sin(1/z_n))=0$, implying $Cin[-1,1]$.
Question: Given $vert Cvertleq 1$ what are the points $b$ with $lim_{ntoinfty } y_C(z_n)=b$ for some sequence of negative reals $z_n$ converging to $0$?
ordinary-differential-equations
edited Dec 12 '18 at 13:47
LutzL
58.5k42054
asked Dec 12 '18 at 13:05
Jens SchwaigerJens Schwaiger
1,566128
$endgroup$
The ODE $$y'=-frac1x y-frac1{x^3} cosleft(frac1xright)$$
has as its set of solutions defined on $(-infty,0)$ the functions
$$y_C:(-infty,0)tomathbb{R},~~~y_C(x):=frac1xleft(C+sinleft(frac1xright)right).$$ If $Cnot=0$ we see by choosing $x_k:=-frac1{pi k}$ that $lim_{ktoinfty} y_c(x_k)=-text{sign}(C)cdot infty$. For $C=0$ we may choose $x_k'=-frac1{(2k+1/2)pi}$ to get $lim_{ktoinfty}y_0(x_k') =lim_{ktoinfty}(-frac1{x_k})=infty$.
In this case ($C=0$) much more can be shown:
For each real $b$ there is a sequence $(b_n)$ of negative numbers tending to $0$ such that $lim_{ntoinfty} y_0(b_n)=b$.
To this aim let $z_n:=-frac1{(n+1/2)pi}$. Then $z_n<z_{n+1}$ and $lim_{ntoinfty} z_n=0$. Morover let $f(x):=sin(1/x)-b x$. Obviously $f(x_{2n})=-1-b x_{2n}$ and $f(x_{2n+1})=1-b x_{2n+1}$. For large $n$ ($vert bcdot x_{2n}vert<1$) this implies $f(x_{2n})<0$ and
$f(x_{2n+1})>0$. Thus there is some $b_nin(x_{2n},x_{2n+1})$ such that $f(b_n)=0$, i.e., $y_0(b_n)=b$. This trivialy implies $lim_{ntoinfty }y_0(b_n)=b$
For $vert Cvert>1$ there is no sequence of $z_n<0$ converging to $0$ such that $y_C(z_n)$ converges to some real number since otherwise we would have $lim_{ntoinfty} (C+sin(1/z_n))=0$, implying $Cin[-1,1]$.
Question: Given $vert Cvertleq 1$ what are the points $b$ with $lim_{ntoinfty } y_C(z_n)=b$ for some sequence of negative reals $z_n$ converging to $0$?
ordinary-differential-equations
ordinary-differential-equations
edited Dec 12 '18 at 13:47
LutzL
58.5k42054
asked Dec 12 '18 at 13:05
Jens SchwaigerJens Schwaiger
1,566128
edited Dec 12 '18 at 13:47
LutzL
58.5k42054
asked Dec 12 '18 at 13:05
Jens SchwaigerJens Schwaiger
1,566128
edited Dec 12 '18 at 13:47
LutzL
58.5k42054
edited Dec 12 '18 at 13:47
LutzL
58.5k42054
edited Dec 12 '18 at 13:47
LutzL
58.5k42054
58.5k42054
asked Dec 12 '18 at 13:05
Jens SchwaigerJens Schwaiger
1,566128
asked Dec 12 '18 at 13:05
Jens SchwaigerJens Schwaiger
1,566128
asked Dec 12 '18 at 13:05
Jens SchwaigerJens Schwaiger
1,566128
1,566128
$begingroup$
DERIVE says ∂(1/x·(c + SIN(1/x)), x)=- COS(1/x)/x^3 - SIN(1/x)/x^2 - c/x^2 and - 1/x·(1/x·(c + SIN(1/x))) - 1/x^3·COS(1/x)=- COS(1/x)/x^3 - SIN(1/x)/x^2 - c/x^2, i.e. it is a solution!
$endgroup$
– Jens Schwaiger
Dec 12 '18 at 13:39
$begingroup$
Applying the integrating factor $x$ gives $$(xy(x))'=-x^{-2}cos(x^{-1}).$$ Now $(sin(x^{-1}))'=cos(x^{-1})(-x^{-2})$ so that indeed $xy(x)=C+sin(x^{-1})$, which gives exactly the expression for $y_C$.
$endgroup$
– LutzL
Dec 12 '18 at 13:52
add a comment |
$begingroup$
DERIVE says ∂(1/x·(c + SIN(1/x)), x)=- COS(1/x)/x^3 - SIN(1/x)/x^2 - c/x^2 and - 1/x·(1/x·(c + SIN(1/x))) - 1/x^3·COS(1/x)=- COS(1/x)/x^3 - SIN(1/x)/x^2 - c/x^2, i.e. it is a solution!
$endgroup$
– Jens Schwaiger
Dec 12 '18 at 13:39
$begingroup$
Applying the integrating factor $x$ gives $$(xy(x))'=-x^{-2}cos(x^{-1}).$$ Now $(sin(x^{-1}))'=cos(x^{-1})(-x^{-2})$ so that indeed $xy(x)=C+sin(x^{-1})$, which gives exactly the expression for $y_C$.
$endgroup$
– LutzL
Dec 12 '18 at 13:52
$begingroup$
DERIVE says ∂(1/x·(c + SIN(1/x)), x)=- COS(1/x)/x^3 - SIN(1/x)/x^2 - c/x^2 and - 1/x·(1/x·(c + SIN(1/x))) - 1/x^3·COS(1/x)=- COS(1/x)/x^3 - SIN(1/x)/x^2 - c/x^2, i.e. it is a solution!
$endgroup$
– Jens Schwaiger
Dec 12 '18 at 13:39
$begingroup$
DERIVE says ∂(1/x·(c + SIN(1/x)), x)=- COS(1/x)/x^3 - SIN(1/x)/x^2 - c/x^2 and - 1/x·(1/x·(c + SIN(1/x))) - 1/x^3·COS(1/x)=- COS(1/x)/x^3 - SIN(1/x)/x^2 - c/x^2, i.e. it is a solution!
$endgroup$
– Jens Schwaiger
Dec 12 '18 at 13:39
$begingroup$
Applying the integrating factor $x$ gives $$(xy(x))'=-x^{-2}cos(x^{-1}).$$ Now $(sin(x^{-1}))'=cos(x^{-1})(-x^{-2})$ so that indeed $xy(x)=C+sin(x^{-1})$, which gives exactly the expression for $y_C$.
$endgroup$
– LutzL
Dec 12 '18 at 13:52
$begingroup$
Applying the integrating factor $x$ gives $$(xy(x))'=-x^{-2}cos(x^{-1}).$$ Now $(sin(x^{-1}))'=cos(x^{-1})(-x^{-2})$ so that indeed $xy(x)=C+sin(x^{-1})$, which gives exactly the expression for $y_C$.
$endgroup$
– LutzL
Dec 12 '18 at 13:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $|C|<1$ the same applies as for $C=0$, since the minima of $C+sin(1/x)$ are negative and the maxima positive, so each up and down-swing of $y_C$ covers an increasingly large interval around $0$ the closer you get with $x$ to $0$.
For $C=1$ the minima of $C+sin(1/x)$ are exactly $0$, so that the oscillations only reach negative values, for $C=-1$ the oscillations only reach positive values.
answered Dec 12 '18 at 14:00
LutzLLutzL
58.5k42054
$endgroup$
$begingroup$
Very nice. Comment to the case $C=1(-1)$: Since $x<0$ the range of oscillation probably should be reversed.
$endgroup$
– Jens Schwaiger
Dec 12 '18 at 14:26
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For $|C|<1$ the same applies as for $C=0$, since the minima of $C+sin(1/x)$ are negative and the maxima positive, so each up and down-swing of $y_C$ covers an increasingly large interval around $0$ the closer you get with $x$ to $0$.
For $C=1$ the minima of $C+sin(1/x)$ are exactly $0$, so that the oscillations only reach negative values, for $C=-1$ the oscillations only reach positive values.
answered Dec 12 '18 at 14:00
LutzLLutzL
58.5k42054
$endgroup$
$begingroup$
Very nice. Comment to the case $C=1(-1)$: Since $x<0$ the range of oscillation probably should be reversed.
$endgroup$
– Jens Schwaiger
Dec 12 '18 at 14:26
add a comment |
$begingroup$
For $|C|<1$ the same applies as for $C=0$, since the minima of $C+sin(1/x)$ are negative and the maxima positive, so each up and down-swing of $y_C$ covers an increasingly large interval around $0$ the closer you get with $x$ to $0$.
For $C=1$ the minima of $C+sin(1/x)$ are exactly $0$, so that the oscillations only reach negative values, for $C=-1$ the oscillations only reach positive values.
answered Dec 12 '18 at 14:00
LutzLLutzL
58.5k42054
$endgroup$
$begingroup$
Very nice. Comment to the case $C=1(-1)$: Since $x<0$ the range of oscillation probably should be reversed.
$endgroup$
– Jens Schwaiger
Dec 12 '18 at 14:26
add a comment |
$begingroup$
For $|C|<1$ the same applies as for $C=0$, since the minima of $C+sin(1/x)$ are negative and the maxima positive, so each up and down-swing of $y_C$ covers an increasingly large interval around $0$ the closer you get with $x$ to $0$.
For $C=1$ the minima of $C+sin(1/x)$ are exactly $0$, so that the oscillations only reach negative values, for $C=-1$ the oscillations only reach positive values.
answered Dec 12 '18 at 14:00
LutzLLutzL
58.5k42054
$endgroup$
For $|C|<1$ the same applies as for $C=0$, since the minima of $C+sin(1/x)$ are negative and the maxima positive, so each up and down-swing of $y_C$ covers an increasingly large interval around $0$ the closer you get with $x$ to $0$.
For $C=1$ the minima of $C+sin(1/x)$ are exactly $0$, so that the oscillations only reach negative values, for $C=-1$ the oscillations only reach positive values.
answered Dec 12 '18 at 14:00
LutzLLutzL
58.5k42054
edited Dec 12 '18 at 14:53
answered Dec 12 '18 at 14:00
LutzLLutzL
58.5k42054
answered Dec 12 '18 at 14:00
LutzLLutzL
58.5k42054
answered Dec 12 '18 at 14:00
LutzLLutzL
58.5k42054
58.5k42054
$begingroup$
Very nice. Comment to the case $C=1(-1)$: Since $x<0$ the range of oscillation probably should be reversed.
$endgroup$
– Jens Schwaiger
Dec 12 '18 at 14:26
add a comment |
$begingroup$
Very nice. Comment to the case $C=1(-1)$: Since $x<0$ the range of oscillation probably should be reversed.
$endgroup$
– Jens Schwaiger
Dec 12 '18 at 14:26
$begingroup$
Very nice. Comment to the case $C=1(-1)$: Since $x<0$ the range of oscillation probably should be reversed.
$endgroup$
– Jens Schwaiger
Dec 12 '18 at 14:26
$begingroup$
Very nice. Comment to the case $C=1(-1)$: Since $x<0$ the range of oscillation probably should be reversed.
$endgroup$
– Jens Schwaiger
Dec 12 '18 at 14:26
add a comment |
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$begingroup$
DERIVE says ∂(1/x·(c + SIN(1/x)), x)=- COS(1/x)/x^3 - SIN(1/x)/x^2 - c/x^2 and - 1/x·(1/x·(c + SIN(1/x))) - 1/x^3·COS(1/x)=- COS(1/x)/x^3 - SIN(1/x)/x^2 - c/x^2, i.e. it is a solution!
$endgroup$
– Jens Schwaiger
Dec 12 '18 at 13:39
$begingroup$
Applying the integrating factor $x$ gives $$(xy(x))'=-x^{-2}cos(x^{-1}).$$ Now $(sin(x^{-1}))'=cos(x^{-1})(-x^{-2})$ so that indeed $xy(x)=C+sin(x^{-1})$, which gives exactly the expression for $y_C$.
$endgroup$
– LutzL
Dec 12 '18 at 13:52