BVP where $u$ is finite as $xrightarrow 0^+$
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I am trying to solve the boundary value problem
$$x^2u''+2xu'-2u=18x^4 0<x<2,$$
$$u textis finite, as xrightarrow 0^+$$
$$u'-u=0 textat x=2$$
My attempt:
I have shown that the general solution is $$u(x)=u_H(x)+u_P(x)=C_1x^-2+C_2x+x^4.$$
I have also shown that the second condition yields $$C_1+2C_2=32.$$
However, I am unsure how to utilise the first condition. How can $u$ be finite when $C_1x^-2rightarrowinfty$ as $xrightarrow 0^+$? A hint would be very helpful.
differential-equations proof-verification boundary-value-problem
asked Sep 4 at 6:14
Bell
835314
add a comment |Â
up vote
0
down vote
favorite
I am trying to solve the boundary value problem
$$x^2u''+2xu'-2u=18x^4 0<x<2,$$
$$u textis finite, as xrightarrow 0^+$$
$$u'-u=0 textat x=2$$
My attempt:
I have shown that the general solution is $$u(x)=u_H(x)+u_P(x)=C_1x^-2+C_2x+x^4.$$
I have also shown that the second condition yields $$C_1+2C_2=32.$$
However, I am unsure how to utilise the first condition. How can $u$ be finite when $C_1x^-2rightarrowinfty$ as $xrightarrow 0^+$? A hint would be very helpful.
differential-equations proof-verification boundary-value-problem
asked Sep 4 at 6:14
Bell
835314
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to solve the boundary value problem
$$x^2u''+2xu'-2u=18x^4 0<x<2,$$
$$u textis finite, as xrightarrow 0^+$$
$$u'-u=0 textat x=2$$
My attempt:
I have shown that the general solution is $$u(x)=u_H(x)+u_P(x)=C_1x^-2+C_2x+x^4.$$
I have also shown that the second condition yields $$C_1+2C_2=32.$$
However, I am unsure how to utilise the first condition. How can $u$ be finite when $C_1x^-2rightarrowinfty$ as $xrightarrow 0^+$? A hint would be very helpful.
differential-equations proof-verification boundary-value-problem
asked Sep 4 at 6:14
Bell
835314
I am trying to solve the boundary value problem
$$x^2u''+2xu'-2u=18x^4 0<x<2,$$
$$u textis finite, as xrightarrow 0^+$$
$$u'-u=0 textat x=2$$
My attempt:
I have shown that the general solution is $$u(x)=u_H(x)+u_P(x)=C_1x^-2+C_2x+x^4.$$
I have also shown that the second condition yields $$C_1+2C_2=32.$$
However, I am unsure how to utilise the first condition. How can $u$ be finite when $C_1x^-2rightarrowinfty$ as $xrightarrow 0^+$? A hint would be very helpful.
differential-equations proof-verification boundary-value-problem
differential-equations proof-verification boundary-value-problem
asked Sep 4 at 6:14
Bell
835314
asked Sep 4 at 6:14
Bell
835314
edited Sep 4 at 6:19
asked Sep 4 at 6:14
Bell
835314
asked Sep 4 at 6:14
Bell
835314
asked Sep 4 at 6:14
Bell
835314
835314
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
$ lim_x to 0+ u(x)$ finite implies $C_1=0$ and this gives $C_2=16$.
answered Sep 4 at 6:16
Kavi Rama Murthy
26k31437
Why does this imply $C_1=0$?
– Bell
Sep 4 at 6:24
$lim C_2x+x^4=0$. So if $lim u(x)$ exists we can conclude that $lim C_1x^-2$ must exist. Since $lim_xto 0 x^-2=infty $ it follows that $lim C_1x^-2$ can exist only when $C_1=0$. (Consider the cases $C_1>0$ and $C_1<0$ to see that the limit does not exist in these cases).
– Kavi Rama Murthy
Sep 4 at 6:27
I think I understand. $C_1=0$ is necessary to insure $u$ is finite. If $C_1>0$ and $C_1<0$, does $C_1x^-2rightarrow infty$ and $-infty$ respectively? Hence $u$ is not finite and thus $C_1$ must be $0$.
– Bell
Sep 4 at 6:31
Yes, that is exactly the argument.
– Kavi Rama Murthy
Sep 4 at 6:33
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$ lim_x to 0+ u(x)$ finite implies $C_1=0$ and this gives $C_2=16$.
answered Sep 4 at 6:16
Kavi Rama Murthy
26k31437
Why does this imply $C_1=0$?
– Bell
Sep 4 at 6:24
$lim C_2x+x^4=0$. So if $lim u(x)$ exists we can conclude that $lim C_1x^-2$ must exist. Since $lim_xto 0 x^-2=infty $ it follows that $lim C_1x^-2$ can exist only when $C_1=0$. (Consider the cases $C_1>0$ and $C_1<0$ to see that the limit does not exist in these cases).
– Kavi Rama Murthy
Sep 4 at 6:27
I think I understand. $C_1=0$ is necessary to insure $u$ is finite. If $C_1>0$ and $C_1<0$, does $C_1x^-2rightarrow infty$ and $-infty$ respectively? Hence $u$ is not finite and thus $C_1$ must be $0$.
– Bell
Sep 4 at 6:31
Yes, that is exactly the argument.
– Kavi Rama Murthy
Sep 4 at 6:33
add a comment |Â
up vote
2
down vote
accepted
$ lim_x to 0+ u(x)$ finite implies $C_1=0$ and this gives $C_2=16$.
answered Sep 4 at 6:16
Kavi Rama Murthy
26k31437
Why does this imply $C_1=0$?
– Bell
Sep 4 at 6:24
$lim C_2x+x^4=0$. So if $lim u(x)$ exists we can conclude that $lim C_1x^-2$ must exist. Since $lim_xto 0 x^-2=infty $ it follows that $lim C_1x^-2$ can exist only when $C_1=0$. (Consider the cases $C_1>0$ and $C_1<0$ to see that the limit does not exist in these cases).
– Kavi Rama Murthy
Sep 4 at 6:27
I think I understand. $C_1=0$ is necessary to insure $u$ is finite. If $C_1>0$ and $C_1<0$, does $C_1x^-2rightarrow infty$ and $-infty$ respectively? Hence $u$ is not finite and thus $C_1$ must be $0$.
– Bell
Sep 4 at 6:31
Yes, that is exactly the argument.
– Kavi Rama Murthy
Sep 4 at 6:33
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$ lim_x to 0+ u(x)$ finite implies $C_1=0$ and this gives $C_2=16$.
answered Sep 4 at 6:16
Kavi Rama Murthy
26k31437
$ lim_x to 0+ u(x)$ finite implies $C_1=0$ and this gives $C_2=16$.
answered Sep 4 at 6:16
Kavi Rama Murthy
26k31437
answered Sep 4 at 6:16
Kavi Rama Murthy
26k31437
answered Sep 4 at 6:16
Kavi Rama Murthy
26k31437
answered Sep 4 at 6:16
Kavi Rama Murthy
26k31437
26k31437
Why does this imply $C_1=0$?
– Bell
Sep 4 at 6:24
$lim C_2x+x^4=0$. So if $lim u(x)$ exists we can conclude that $lim C_1x^-2$ must exist. Since $lim_xto 0 x^-2=infty $ it follows that $lim C_1x^-2$ can exist only when $C_1=0$. (Consider the cases $C_1>0$ and $C_1<0$ to see that the limit does not exist in these cases).
– Kavi Rama Murthy
Sep 4 at 6:27
I think I understand. $C_1=0$ is necessary to insure $u$ is finite. If $C_1>0$ and $C_1<0$, does $C_1x^-2rightarrow infty$ and $-infty$ respectively? Hence $u$ is not finite and thus $C_1$ must be $0$.
– Bell
Sep 4 at 6:31
Yes, that is exactly the argument.
– Kavi Rama Murthy
Sep 4 at 6:33
add a comment |Â
Why does this imply $C_1=0$?
– Bell
Sep 4 at 6:24
$lim C_2x+x^4=0$. So if $lim u(x)$ exists we can conclude that $lim C_1x^-2$ must exist. Since $lim_xto 0 x^-2=infty $ it follows that $lim C_1x^-2$ can exist only when $C_1=0$. (Consider the cases $C_1>0$ and $C_1<0$ to see that the limit does not exist in these cases).
– Kavi Rama Murthy
Sep 4 at 6:27
I think I understand. $C_1=0$ is necessary to insure $u$ is finite. If $C_1>0$ and $C_1<0$, does $C_1x^-2rightarrow infty$ and $-infty$ respectively? Hence $u$ is not finite and thus $C_1$ must be $0$.
– Bell
Sep 4 at 6:31
Yes, that is exactly the argument.
– Kavi Rama Murthy
Sep 4 at 6:33
Why does this imply $C_1=0$?
– Bell
Sep 4 at 6:24
Why does this imply $C_1=0$?
– Bell
Sep 4 at 6:24
$lim C_2x+x^4=0$. So if $lim u(x)$ exists we can conclude that $lim C_1x^-2$ must exist. Since $lim_xto 0 x^-2=infty $ it follows that $lim C_1x^-2$ can exist only when $C_1=0$. (Consider the cases $C_1>0$ and $C_1<0$ to see that the limit does not exist in these cases).
– Kavi Rama Murthy
Sep 4 at 6:27
$lim C_2x+x^4=0$. So if $lim u(x)$ exists we can conclude that $lim C_1x^-2$ must exist. Since $lim_xto 0 x^-2=infty $ it follows that $lim C_1x^-2$ can exist only when $C_1=0$. (Consider the cases $C_1>0$ and $C_1<0$ to see that the limit does not exist in these cases).
– Kavi Rama Murthy
Sep 4 at 6:27
I think I understand. $C_1=0$ is necessary to insure $u$ is finite. If $C_1>0$ and $C_1<0$, does $C_1x^-2rightarrow infty$ and $-infty$ respectively? Hence $u$ is not finite and thus $C_1$ must be $0$.
– Bell
Sep 4 at 6:31
I think I understand. $C_1=0$ is necessary to insure $u$ is finite. If $C_1>0$ and $C_1<0$, does $C_1x^-2rightarrow infty$ and $-infty$ respectively? Hence $u$ is not finite and thus $C_1$ must be $0$.
– Bell
Sep 4 at 6:31
Yes, that is exactly the argument.
– Kavi Rama Murthy
Sep 4 at 6:33
Yes, that is exactly the argument.
– Kavi Rama Murthy
Sep 4 at 6:33
add a comment |Â
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