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Determining uniqueness of solution of ODE

0

$begingroup$

We have an IVP:

$$x'=frac2xsqrtln(x)t$$
$$x(t_0)=1$$
and $t_0 >0$ , $x>1$

The question is:

Does this IVP have a unique solution for any value of $t_0$ ?

---

I found the general solution which is $x(t)=e^(ln(t) +C)^2$ but ODEs are quite a new thing to me and the existance and uniqueness issues are still a bit unclear for me, so I would appreciate any explanation.

ordinary-differential-equations

share|cite|improve this question

asked Apr 2 at 17:03

Wojtek MaślakiewiczWojtek Maślakiewicz

585

$endgroup$

add a comment | 

0

$begingroup$

We have an IVP:

$$x'=frac2xsqrtln(x)t$$
$$x(t_0)=1$$
and $t_0 >0$ , $x>1$

The question is:

Does this IVP have a unique solution for any value of $t_0$ ?

---

I found the general solution which is $x(t)=e^(ln(t) +C)^2$ but ODEs are quite a new thing to me and the existance and uniqueness issues are still a bit unclear for me, so I would appreciate any explanation.

ordinary-differential-equations

share|cite|improve this question

asked Apr 2 at 17:03

Wojtek MaślakiewiczWojtek Maślakiewicz

585

$endgroup$

add a comment | 

$begingroup$

We have an IVP:

$$x'=frac2xsqrtln(x)t$$
$$x(t_0)=1$$
and $t_0 >0$ , $x>1$

The question is:

Does this IVP have a unique solution for any value of $t_0$ ?

---

I found the general solution which is $x(t)=e^(ln(t) +C)^2$ but ODEs are quite a new thing to me and the existance and uniqueness issues are still a bit unclear for me, so I would appreciate any explanation.

ordinary-differential-equations

share|cite|improve this question

asked Apr 2 at 17:03

Wojtek MaślakiewiczWojtek Maślakiewicz

585

$endgroup$

share|cite|improve this question

asked Apr 2 at 17:03

Wojtek MaślakiewiczWojtek Maślakiewicz

585

share|cite|improve this question

asked Apr 2 at 17:03

Wojtek MaślakiewiczWojtek Maślakiewicz

585

asked Apr 2 at 17:03

Wojtek MaślakiewiczWojtek Maślakiewicz

585

asked Apr 2 at 17:03

Wojtek MaślakiewiczWojtek Maślakiewicz

585

1 Answer
1

active

oldest

votes

1

$begingroup$

The usual Picard–Lindelöf Existence and Uniqueness Theorem applies when the right side of your equation and its partial derivative with respect to $x$ are continuous in a neighbourhood of the initial condition $(x_0, t_0)$. This is the case for your initial condition
since $x_0 > 1$ (so there is no problem with the square root of the logarithm) and $t_0 > 0$ (so there is no problem with division by $t$). So yes, there is a unique solution satisfying the initial condition defined in some neighbourhood of $t_0$. It may stop existing or stop being unique if it runs off to $pm infty$ or leaves the region $(x > 1, t>0$).

EDITED AGAIN: For example, consider the case $t_0 = 1$, $x_0 = e$. The solution you found is $x = e^(1 + ln t)^2$. However, (assuming as usual we take the positive square root) that is only a solution when $1 + ln t ge 0$, i.e. $t ge 1/e$. On the other hand the constant $x=1$ is a solution. Thus the actual (unique) solution with this initial point is
$$ x(t) = cases1 & if $0 < t le 1/e$cr
e^(1+ln t)^2 & if $t ge 1/e$cr$$

share|cite|improve this answer

edited Apr 2 at 20:15

answered Apr 2 at 17:17

Robert IsraelRobert Israel

332k23222482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So am I correct that if we leave the assumption $x_0>1$ and set the initial value to $x(t_0)=1$ than we can have a constant solution $x=1$ and therefore there would be no uniqueness, right?
  $endgroup$
  – Wojtek Maślakiewicz
  Apr 2 at 17:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, with initial condition $x(t_0) = 1$ we don't have uniqueness because we have $x=1$ and also solutions of the form $$x(t) = cases1 & if $0 < t le t_1$cr e^(ln(t) - ln(t_1))^2 & if $t ge t_1$$$ for any $t_1 > t_0$.
  $endgroup$
  – Robert Israel
  Apr 2 at 20:18
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

The usual Picard–Lindelöf Existence and Uniqueness Theorem applies when the right side of your equation and its partial derivative with respect to $x$ are continuous in a neighbourhood of the initial condition $(x_0, t_0)$. This is the case for your initial condition
since $x_0 > 1$ (so there is no problem with the square root of the logarithm) and $t_0 > 0$ (so there is no problem with division by $t$). So yes, there is a unique solution satisfying the initial condition defined in some neighbourhood of $t_0$. It may stop existing or stop being unique if it runs off to $pm infty$ or leaves the region $(x > 1, t>0$).

EDITED AGAIN: For example, consider the case $t_0 = 1$, $x_0 = e$. The solution you found is $x = e^(1 + ln t)^2$. However, (assuming as usual we take the positive square root) that is only a solution when $1 + ln t ge 0$, i.e. $t ge 1/e$. On the other hand the constant $x=1$ is a solution. Thus the actual (unique) solution with this initial point is
$$ x(t) = cases1 & if $0 < t le 1/e$cr
e^(1+ln t)^2 & if $t ge 1/e$cr$$

share|cite|improve this answer

edited Apr 2 at 20:15

answered Apr 2 at 17:17

Robert IsraelRobert Israel

332k23222482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So am I correct that if we leave the assumption $x_0>1$ and set the initial value to $x(t_0)=1$ than we can have a constant solution $x=1$ and therefore there would be no uniqueness, right?
  $endgroup$
  – Wojtek Maślakiewicz
  Apr 2 at 17:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, with initial condition $x(t_0) = 1$ we don't have uniqueness because we have $x=1$ and also solutions of the form $$x(t) = cases1 & if $0 < t le t_1$cr e^(ln(t) - ln(t_1))^2 & if $t ge t_1$$$ for any $t_1 > t_0$.
  $endgroup$
  – Robert Israel
  Apr 2 at 20:18
  
  

  
  
  
  

add a comment | 

1

$begingroup$

The usual Picard–Lindelöf Existence and Uniqueness Theorem applies when the right side of your equation and its partial derivative with respect to $x$ are continuous in a neighbourhood of the initial condition $(x_0, t_0)$. This is the case for your initial condition
since $x_0 > 1$ (so there is no problem with the square root of the logarithm) and $t_0 > 0$ (so there is no problem with division by $t$). So yes, there is a unique solution satisfying the initial condition defined in some neighbourhood of $t_0$. It may stop existing or stop being unique if it runs off to $pm infty$ or leaves the region $(x > 1, t>0$).

EDITED AGAIN: For example, consider the case $t_0 = 1$, $x_0 = e$. The solution you found is $x = e^(1 + ln t)^2$. However, (assuming as usual we take the positive square root) that is only a solution when $1 + ln t ge 0$, i.e. $t ge 1/e$. On the other hand the constant $x=1$ is a solution. Thus the actual (unique) solution with this initial point is
$$ x(t) = cases1 & if $0 < t le 1/e$cr
e^(1+ln t)^2 & if $t ge 1/e$cr$$

share|cite|improve this answer

edited Apr 2 at 20:15

answered Apr 2 at 17:17

Robert IsraelRobert Israel

332k23222482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So am I correct that if we leave the assumption $x_0>1$ and set the initial value to $x(t_0)=1$ than we can have a constant solution $x=1$ and therefore there would be no uniqueness, right?
  $endgroup$
  – Wojtek Maślakiewicz
  Apr 2 at 17:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, with initial condition $x(t_0) = 1$ we don't have uniqueness because we have $x=1$ and also solutions of the form $$x(t) = cases1 & if $0 < t le t_1$cr e^(ln(t) - ln(t_1))^2 & if $t ge t_1$$$ for any $t_1 > t_0$.
  $endgroup$
  – Robert Israel
  Apr 2 at 20:18
  
  

  
  
  
  

add a comment | 

$begingroup$

The usual Picard–Lindelöf Existence and Uniqueness Theorem applies when the right side of your equation and its partial derivative with respect to $x$ are continuous in a neighbourhood of the initial condition $(x_0, t_0)$. This is the case for your initial condition
since $x_0 > 1$ (so there is no problem with the square root of the logarithm) and $t_0 > 0$ (so there is no problem with division by $t$). So yes, there is a unique solution satisfying the initial condition defined in some neighbourhood of $t_0$. It may stop existing or stop being unique if it runs off to $pm infty$ or leaves the region $(x > 1, t>0$).

EDITED AGAIN: For example, consider the case $t_0 = 1$, $x_0 = e$. The solution you found is $x = e^(1 + ln t)^2$. However, (assuming as usual we take the positive square root) that is only a solution when $1 + ln t ge 0$, i.e. $t ge 1/e$. On the other hand the constant $x=1$ is a solution. Thus the actual (unique) solution with this initial point is
$$ x(t) = cases1 & if $0 < t le 1/e$cr
e^(1+ln t)^2 & if $t ge 1/e$cr$$

share|cite|improve this answer

edited Apr 2 at 20:15

answered Apr 2 at 17:17

Robert IsraelRobert Israel

332k23222482

$endgroup$

share|cite|improve this answer

edited Apr 2 at 20:15

answered Apr 2 at 17:17

Robert IsraelRobert Israel

332k23222482

answered Apr 2 at 17:17

Robert IsraelRobert Israel

332k23222482

answered Apr 2 at 17:17

Robert IsraelRobert Israel

332k23222482