Can a differential equation have non unique solutions?
up vote
6
down vote
favorite
There are theorems of existence and uniqueness of differential equations.
I was wondering if it is possible that a differential equations has a solution but it is not unique.
differential-equations
asked Nov 17 '12 at 15:30
mariosangiorgio
13315
add a comment |
up vote
6
down vote
favorite
There are theorems of existence and uniqueness of differential equations.
I was wondering if it is possible that a differential equations has a solution but it is not unique.
differential-equations
asked Nov 17 '12 at 15:30
mariosangiorgio
13315
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
There are theorems of existence and uniqueness of differential equations.
I was wondering if it is possible that a differential equations has a solution but it is not unique.
differential-equations
asked Nov 17 '12 at 15:30
mariosangiorgio
13315
There are theorems of existence and uniqueness of differential equations.
I was wondering if it is possible that a differential equations has a solution but it is not unique.
differential-equations
differential-equations
asked Nov 17 '12 at 15:30
mariosangiorgio
13315
asked Nov 17 '12 at 15:30
mariosangiorgio
13315
asked Nov 17 '12 at 15:30
mariosangiorgio
13315
asked Nov 17 '12 at 15:30
mariosangiorgio
13315
asked Nov 17 '12 at 15:30
mariosangiorgio
13315
13315
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
12
down vote
accepted
Consider for example the equation $x' = 2sqrt{|x|}$. For every $a$, the function
$$ x_a(t) = begin{cases} 0 & x le a\
(t-a)^2 & x ge a
end{cases} $$
is a solution. Note that for $a ge 0$ all $x_a$ have $x_a(0) = 0$, so they are all solutions to the IVP $x' = 2sqrt{|x|}, x(0) = 0$ and you usually discuss uniqueness for inital value problems, as otherwise uniqueness will almost never hold ($x' = 0$ has all constants as solutions).
answered Nov 17 '12 at 15:36
martini
70.1k45990
Thanks. You mentioned initial value problems. Do they have an unique solution or no solution at all? Or even in that case we may have no solution, an unique solution or multiple solutions?
– mariosangiorgio
Nov 17 '12 at 15:44
2
@mariosangiorgio In my above answer, I gave an example of an non-unique solvable IVP.
– martini
Nov 17 '12 at 16:38
The definition of the solution on piecewise form should depend on the value of t.
– Jorge E. Cardona
May 1 '16 at 21:57
add a comment |
up vote
4
down vote
Let your ODE be $y'-xsqrt{y}=0, ; y(0)=0$. It is not difficult finding its solution on $mathbb R$. It has at least two solutions as $y=0$ and $y=frac{x^4}{16}$ passing through the origin. Can you see why the ODE has no unique solution?
answered Nov 17 '12 at 15:36
mrs
1
Is it because the two solutions shares only the point (0,0) but they do not intersect?
– mariosangiorgio
Nov 17 '12 at 15:55
3
@mariosangiorgio: Actually no. If we consider the equation as $y'=f(x,y)$ wherein $f(x,y)=xy^{frac{1}{2}}$, then $frac{partial f}{partial y}=frac{x}{2y^{frac{1}{2}}}$ which are both defined just for $y>0$. In fact Picard's theorem doesn't let the equation to have a unique solution in a rectangular region around the origin.
– mrs
Nov 17 '12 at 16:05
1
Thank you for the explanation
– mariosangiorgio
Nov 17 '12 at 16:22
Nice explanation, indeed! ;-)
– amWhy
Apr 5 '13 at 1:33
add a comment |
up vote
0
down vote
Since you didn't specify an IVP, it is trivial to find an ODE with non unique solution, such as $y'(x)=0 implies y(x)=C$, where $Cinmathbb R$.
answered Nov 17 at 10:58
jinawee
1,21511333
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
Consider for example the equation $x' = 2sqrt{|x|}$. For every $a$, the function
$$ x_a(t) = begin{cases} 0 & x le a\
(t-a)^2 & x ge a
end{cases} $$
is a solution. Note that for $a ge 0$ all $x_a$ have $x_a(0) = 0$, so they are all solutions to the IVP $x' = 2sqrt{|x|}, x(0) = 0$ and you usually discuss uniqueness for inital value problems, as otherwise uniqueness will almost never hold ($x' = 0$ has all constants as solutions).
answered Nov 17 '12 at 15:36
martini
70.1k45990
Thanks. You mentioned initial value problems. Do they have an unique solution or no solution at all? Or even in that case we may have no solution, an unique solution or multiple solutions?
– mariosangiorgio
Nov 17 '12 at 15:44
2
@mariosangiorgio In my above answer, I gave an example of an non-unique solvable IVP.
– martini
Nov 17 '12 at 16:38
The definition of the solution on piecewise form should depend on the value of t.
– Jorge E. Cardona
May 1 '16 at 21:57
add a comment |
up vote
12
down vote
accepted
Consider for example the equation $x' = 2sqrt{|x|}$. For every $a$, the function
$$ x_a(t) = begin{cases} 0 & x le a\
(t-a)^2 & x ge a
end{cases} $$
is a solution. Note that for $a ge 0$ all $x_a$ have $x_a(0) = 0$, so they are all solutions to the IVP $x' = 2sqrt{|x|}, x(0) = 0$ and you usually discuss uniqueness for inital value problems, as otherwise uniqueness will almost never hold ($x' = 0$ has all constants as solutions).
answered Nov 17 '12 at 15:36
martini
70.1k45990
Thanks. You mentioned initial value problems. Do they have an unique solution or no solution at all? Or even in that case we may have no solution, an unique solution or multiple solutions?
– mariosangiorgio
Nov 17 '12 at 15:44
2
@mariosangiorgio In my above answer, I gave an example of an non-unique solvable IVP.
– martini
Nov 17 '12 at 16:38
The definition of the solution on piecewise form should depend on the value of t.
– Jorge E. Cardona
May 1 '16 at 21:57
add a comment |
up vote
12
down vote
accepted
up vote
12
down vote
accepted
Consider for example the equation $x' = 2sqrt{|x|}$. For every $a$, the function
$$ x_a(t) = begin{cases} 0 & x le a\
(t-a)^2 & x ge a
end{cases} $$
is a solution. Note that for $a ge 0$ all $x_a$ have $x_a(0) = 0$, so they are all solutions to the IVP $x' = 2sqrt{|x|}, x(0) = 0$ and you usually discuss uniqueness for inital value problems, as otherwise uniqueness will almost never hold ($x' = 0$ has all constants as solutions).
answered Nov 17 '12 at 15:36
martini
70.1k45990
Consider for example the equation $x' = 2sqrt{|x|}$. For every $a$, the function
$$ x_a(t) = begin{cases} 0 & x le a\
(t-a)^2 & x ge a
end{cases} $$
is a solution. Note that for $a ge 0$ all $x_a$ have $x_a(0) = 0$, so they are all solutions to the IVP $x' = 2sqrt{|x|}, x(0) = 0$ and you usually discuss uniqueness for inital value problems, as otherwise uniqueness will almost never hold ($x' = 0$ has all constants as solutions).
answered Nov 17 '12 at 15:36
martini
70.1k45990
answered Nov 17 '12 at 15:36
martini
70.1k45990
answered Nov 17 '12 at 15:36
martini
70.1k45990
answered Nov 17 '12 at 15:36
martini
70.1k45990
70.1k45990
Thanks. You mentioned initial value problems. Do they have an unique solution or no solution at all? Or even in that case we may have no solution, an unique solution or multiple solutions?
– mariosangiorgio
Nov 17 '12 at 15:44
2
@mariosangiorgio In my above answer, I gave an example of an non-unique solvable IVP.
– martini
Nov 17 '12 at 16:38
The definition of the solution on piecewise form should depend on the value of t.
– Jorge E. Cardona
May 1 '16 at 21:57
add a comment |
Thanks. You mentioned initial value problems. Do they have an unique solution or no solution at all? Or even in that case we may have no solution, an unique solution or multiple solutions?
– mariosangiorgio
Nov 17 '12 at 15:44
2
@mariosangiorgio In my above answer, I gave an example of an non-unique solvable IVP.
– martini
Nov 17 '12 at 16:38
The definition of the solution on piecewise form should depend on the value of t.
– Jorge E. Cardona
May 1 '16 at 21:57
Thanks. You mentioned initial value problems. Do they have an unique solution or no solution at all? Or even in that case we may have no solution, an unique solution or multiple solutions?
– mariosangiorgio
Nov 17 '12 at 15:44
Thanks. You mentioned initial value problems. Do they have an unique solution or no solution at all? Or even in that case we may have no solution, an unique solution or multiple solutions?
– mariosangiorgio
Nov 17 '12 at 15:44
2
2
@mariosangiorgio In my above answer, I gave an example of an non-unique solvable IVP.
– martini
Nov 17 '12 at 16:38
@mariosangiorgio In my above answer, I gave an example of an non-unique solvable IVP.
– martini
Nov 17 '12 at 16:38
The definition of the solution on piecewise form should depend on the value of t.
– Jorge E. Cardona
May 1 '16 at 21:57
The definition of the solution on piecewise form should depend on the value of t.
– Jorge E. Cardona
May 1 '16 at 21:57
add a comment |
up vote
4
down vote
Let your ODE be $y'-xsqrt{y}=0, ; y(0)=0$. It is not difficult finding its solution on $mathbb R$. It has at least two solutions as $y=0$ and $y=frac{x^4}{16}$ passing through the origin. Can you see why the ODE has no unique solution?
answered Nov 17 '12 at 15:36
mrs
1
Is it because the two solutions shares only the point (0,0) but they do not intersect?
– mariosangiorgio
Nov 17 '12 at 15:55
3
@mariosangiorgio: Actually no. If we consider the equation as $y'=f(x,y)$ wherein $f(x,y)=xy^{frac{1}{2}}$, then $frac{partial f}{partial y}=frac{x}{2y^{frac{1}{2}}}$ which are both defined just for $y>0$. In fact Picard's theorem doesn't let the equation to have a unique solution in a rectangular region around the origin.
– mrs
Nov 17 '12 at 16:05
1
Thank you for the explanation
– mariosangiorgio
Nov 17 '12 at 16:22
Nice explanation, indeed! ;-)
– amWhy
Apr 5 '13 at 1:33
add a comment |
up vote
4
down vote
Let your ODE be $y'-xsqrt{y}=0, ; y(0)=0$. It is not difficult finding its solution on $mathbb R$. It has at least two solutions as $y=0$ and $y=frac{x^4}{16}$ passing through the origin. Can you see why the ODE has no unique solution?
answered Nov 17 '12 at 15:36
mrs
1
Is it because the two solutions shares only the point (0,0) but they do not intersect?
– mariosangiorgio
Nov 17 '12 at 15:55
3
@mariosangiorgio: Actually no. If we consider the equation as $y'=f(x,y)$ wherein $f(x,y)=xy^{frac{1}{2}}$, then $frac{partial f}{partial y}=frac{x}{2y^{frac{1}{2}}}$ which are both defined just for $y>0$. In fact Picard's theorem doesn't let the equation to have a unique solution in a rectangular region around the origin.
– mrs
Nov 17 '12 at 16:05
1
Thank you for the explanation
– mariosangiorgio
Nov 17 '12 at 16:22
Nice explanation, indeed! ;-)
– amWhy
Apr 5 '13 at 1:33
add a comment |
up vote
4
down vote
up vote
4
down vote
Let your ODE be $y'-xsqrt{y}=0, ; y(0)=0$. It is not difficult finding its solution on $mathbb R$. It has at least two solutions as $y=0$ and $y=frac{x^4}{16}$ passing through the origin. Can you see why the ODE has no unique solution?
answered Nov 17 '12 at 15:36
mrs
1
Let your ODE be $y'-xsqrt{y}=0, ; y(0)=0$. It is not difficult finding its solution on $mathbb R$. It has at least two solutions as $y=0$ and $y=frac{x^4}{16}$ passing through the origin. Can you see why the ODE has no unique solution?
answered Nov 17 '12 at 15:36
mrs
1
answered Nov 17 '12 at 15:36
mrs
1
answered Nov 17 '12 at 15:36
mrs
1
answered Nov 17 '12 at 15:36
mrs
1
1
Is it because the two solutions shares only the point (0,0) but they do not intersect?
– mariosangiorgio
Nov 17 '12 at 15:55
3
@mariosangiorgio: Actually no. If we consider the equation as $y'=f(x,y)$ wherein $f(x,y)=xy^{frac{1}{2}}$, then $frac{partial f}{partial y}=frac{x}{2y^{frac{1}{2}}}$ which are both defined just for $y>0$. In fact Picard's theorem doesn't let the equation to have a unique solution in a rectangular region around the origin.
– mrs
Nov 17 '12 at 16:05
1
Thank you for the explanation
– mariosangiorgio
Nov 17 '12 at 16:22
Nice explanation, indeed! ;-)
– amWhy
Apr 5 '13 at 1:33
add a comment |
Is it because the two solutions shares only the point (0,0) but they do not intersect?
– mariosangiorgio
Nov 17 '12 at 15:55
3
@mariosangiorgio: Actually no. If we consider the equation as $y'=f(x,y)$ wherein $f(x,y)=xy^{frac{1}{2}}$, then $frac{partial f}{partial y}=frac{x}{2y^{frac{1}{2}}}$ which are both defined just for $y>0$. In fact Picard's theorem doesn't let the equation to have a unique solution in a rectangular region around the origin.
– mrs
Nov 17 '12 at 16:05
1
Thank you for the explanation
– mariosangiorgio
Nov 17 '12 at 16:22
Nice explanation, indeed! ;-)
– amWhy
Apr 5 '13 at 1:33
Is it because the two solutions shares only the point (0,0) but they do not intersect?
– mariosangiorgio
Nov 17 '12 at 15:55
Is it because the two solutions shares only the point (0,0) but they do not intersect?
– mariosangiorgio
Nov 17 '12 at 15:55
3
3
@mariosangiorgio: Actually no. If we consider the equation as $y'=f(x,y)$ wherein $f(x,y)=xy^{frac{1}{2}}$, then $frac{partial f}{partial y}=frac{x}{2y^{frac{1}{2}}}$ which are both defined just for $y>0$. In fact Picard's theorem doesn't let the equation to have a unique solution in a rectangular region around the origin.
– mrs
Nov 17 '12 at 16:05
@mariosangiorgio: Actually no. If we consider the equation as $y'=f(x,y)$ wherein $f(x,y)=xy^{frac{1}{2}}$, then $frac{partial f}{partial y}=frac{x}{2y^{frac{1}{2}}}$ which are both defined just for $y>0$. In fact Picard's theorem doesn't let the equation to have a unique solution in a rectangular region around the origin.
– mrs
Nov 17 '12 at 16:05
1
1
Thank you for the explanation
– mariosangiorgio
Nov 17 '12 at 16:22
Thank you for the explanation
– mariosangiorgio
Nov 17 '12 at 16:22
Nice explanation, indeed! ;-)
– amWhy
Apr 5 '13 at 1:33
Nice explanation, indeed! ;-)
– amWhy
Apr 5 '13 at 1:33
add a comment |
up vote
0
down vote
Since you didn't specify an IVP, it is trivial to find an ODE with non unique solution, such as $y'(x)=0 implies y(x)=C$, where $Cinmathbb R$.
answered Nov 17 at 10:58
jinawee
1,21511333
add a comment |
up vote
0
down vote
Since you didn't specify an IVP, it is trivial to find an ODE with non unique solution, such as $y'(x)=0 implies y(x)=C$, where $Cinmathbb R$.
answered Nov 17 at 10:58
jinawee
1,21511333
add a comment |
up vote
0
down vote
up vote
0
down vote
Since you didn't specify an IVP, it is trivial to find an ODE with non unique solution, such as $y'(x)=0 implies y(x)=C$, where $Cinmathbb R$.
answered Nov 17 at 10:58
jinawee
1,21511333
Since you didn't specify an IVP, it is trivial to find an ODE with non unique solution, such as $y'(x)=0 implies y(x)=C$, where $Cinmathbb R$.
answered Nov 17 at 10:58
jinawee
1,21511333
answered Nov 17 at 10:58
jinawee
1,21511333
answered Nov 17 at 10:58
jinawee
1,21511333
answered Nov 17 at 10:58
jinawee
1,21511333
1,21511333
add a comment |
add a comment |
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