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How to solve $dot{x}=|x|$
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How to solve this differential equation $dot{x}=|x|$?
differential-equations analysis
asked 20 hours ago
winston
395116
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up vote
1
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How to solve this differential equation $dot{x}=|x|$?
differential-equations analysis
asked 20 hours ago
winston
395116
@Rahul because initial condition is not given.
– bellcircle
20 hours ago
add a comment |
up vote
1
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favorite
up vote
1
down vote
favorite
How to solve this differential equation $dot{x}=|x|$?
differential-equations analysis
asked 20 hours ago
winston
395116
How to solve this differential equation $dot{x}=|x|$?
differential-equations analysis
differential-equations analysis
asked 20 hours ago
winston
395116
asked 20 hours ago
winston
395116
edited 19 hours ago
asked 20 hours ago
winston
395116
asked 20 hours ago
winston
395116
asked 20 hours ago
winston
395116
395116
@Rahul because initial condition is not given.
– bellcircle
20 hours ago
add a comment |
@Rahul because initial condition is not given.
– bellcircle
20 hours ago
@Rahul because initial condition is not given.
– bellcircle
20 hours ago
@Rahul because initial condition is not given.
– bellcircle
20 hours ago
add a comment |
2 Answers
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up vote
4
down vote
One observes that $x=0$ is the trivial solution of the given ODE.
Now find nontrivial solutions of the equation.
Using separation of variables, one gets $$frac{dot{x}}{|x|}=1.$$
Integrating both sides gives
begin{align}text{sgn}(x)log|x|=t+Cquadcdotsquadtext{(a)}end{align} for some constant $C$, where $text{sgn}(x)=begin{cases}1,&x> 0\-1,&x<0.end{cases}$
Therefore, $x(t)=begin{cases}C_1e^t,&x>0\C_2e^{-t},&x<0.end{cases}$
Now find the solution of explicit form.
From the equation, $dot{x}$ is always nonnegative, so $x$ is a continuously differentiable and nondecreasing function of $t$. This shows that $C_1$ must be positive and $C_2$ must be negative.
Since $x$ is a continuous function of $t$ and a nontrivial solution of $x$ cannot take 0 by the above observation, it follows that $x(t)=C_1e^t,C_1>0$ for all $t$ or $x(t)=C_2e^{-t},C_2<0$ for all $t$.
edited 17 hours ago
Shashi
7,0221525
answered 19 hours ago
bellcircle
1,204411
You first divided both sides by $|x|$ to get the following results. But how can you guarantee that you are not dividing by $0$. In other words, if you cannot guarantee that once the solution starts from a non-zero value, it will not reach $0$, then you cannot divide it by $|x|$.
– winston
18 hours ago
@winston Good question. If $dot{x}=0$, then $x=0$ at such point. Now it is obvious that there is no way to combine the distinct (local) solutions $C_1e^t,C_2e^{-t},0$ to make a continuous global solution because the first is always positive, the second always negative, the last always zero. Therefore the global solution must be one of the three.
– bellcircle
18 hours ago
add a comment |
up vote
1
down vote
bellcircle has provided a solution to the problem. However, I would like to express the solution in a more compact form and also provide an answer using a proposition-proof style.
$(Dx)(t) = |x(t)|$ (1)
Proposition. If $I$ is an open interval in $mathbb{R}$ and $C in mathbb{R}$, then $x(t)=Ce^{mathsf{sgn}(C) t}$ is a solution of the differential equation (1) on $I$. If $x(t)$ is a solution of the differential equation (1) on an open interval $I$ in $mathbb{R}$, then there exists $C in mathbb{R}$ such that $x(t)=Ce^{mathsf{sgn}(C) t}$ for all $t in I$.
Suppose $I$ is an open interval in $mathbb{R}$. Suppose $C in mathbb{R}$ and $x$ is such that $x(t)=Ce^{mathsf{sgn}(C) t}$ for all $t in I$. Then $(Dx)(t)=mathsf{sgn}(C)Ce^{mathsf{sgn}(C) t}$. Also, $|x(t)|=mathsf{sgn}(C)Ce^{mathsf{sgn}(C) t}$ for all $t in I$. Therefore, indeed $x(t)=Ce^{mathsf{sgn}(C) t}$ is a solution of (1) on $I$.
Suppose there is a solution $x_2(t)$ of (1) on $I$ such that there is no $C in mathbb{R}$ such that $x_2(t) = Ce^{mathsf{sgn}(C) t}$ for all $t in I$. Then take $T in I$. Define $A=frac{x_2(T)}{e^{mathsf{sgn}(x_2(T)) T}}$. Then $x(t) = A e^{mathsf{sgn}(A)t}$ is a solution of (1) on $I$ with $x(T)=x_2(T)$. The uniqueness of the solution of (1) on $I$ with $x(T)=x_2(T)$ follows from the Lipschitz continuity of $|cdot|$ by the Picard–Lindelöf theorem (link) and the global uniqueness theorem (e.g. link). Therefore $x_2(t) = A e^{mathsf{sgn}(A)t}$ for all $t in I$ and, thus, by contradiction, if $x(t)$ is a solution of the differential equation (1) on an open interval $I$ in $mathbb{R}$, then there exists $C in mathbb{R}$ such that $x(t)=Ce^{mathsf{sgn}(C) t}$ for all $t in I$
answered 18 hours ago
xanonec
1485
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
One observes that $x=0$ is the trivial solution of the given ODE.
Now find nontrivial solutions of the equation.
Using separation of variables, one gets $$frac{dot{x}}{|x|}=1.$$
Integrating both sides gives
begin{align}text{sgn}(x)log|x|=t+Cquadcdotsquadtext{(a)}end{align} for some constant $C$, where $text{sgn}(x)=begin{cases}1,&x> 0\-1,&x<0.end{cases}$
Therefore, $x(t)=begin{cases}C_1e^t,&x>0\C_2e^{-t},&x<0.end{cases}$
Now find the solution of explicit form.
From the equation, $dot{x}$ is always nonnegative, so $x$ is a continuously differentiable and nondecreasing function of $t$. This shows that $C_1$ must be positive and $C_2$ must be negative.
Since $x$ is a continuous function of $t$ and a nontrivial solution of $x$ cannot take 0 by the above observation, it follows that $x(t)=C_1e^t,C_1>0$ for all $t$ or $x(t)=C_2e^{-t},C_2<0$ for all $t$.
edited 17 hours ago
Shashi
7,0221525
answered 19 hours ago
bellcircle
1,204411
You first divided both sides by $|x|$ to get the following results. But how can you guarantee that you are not dividing by $0$. In other words, if you cannot guarantee that once the solution starts from a non-zero value, it will not reach $0$, then you cannot divide it by $|x|$.
– winston
18 hours ago
@winston Good question. If $dot{x}=0$, then $x=0$ at such point. Now it is obvious that there is no way to combine the distinct (local) solutions $C_1e^t,C_2e^{-t},0$ to make a continuous global solution because the first is always positive, the second always negative, the last always zero. Therefore the global solution must be one of the three.
– bellcircle
18 hours ago
add a comment |
up vote
4
down vote
One observes that $x=0$ is the trivial solution of the given ODE.
Now find nontrivial solutions of the equation.
Using separation of variables, one gets $$frac{dot{x}}{|x|}=1.$$
Integrating both sides gives
begin{align}text{sgn}(x)log|x|=t+Cquadcdotsquadtext{(a)}end{align} for some constant $C$, where $text{sgn}(x)=begin{cases}1,&x> 0\-1,&x<0.end{cases}$
Therefore, $x(t)=begin{cases}C_1e^t,&x>0\C_2e^{-t},&x<0.end{cases}$
Now find the solution of explicit form.
From the equation, $dot{x}$ is always nonnegative, so $x$ is a continuously differentiable and nondecreasing function of $t$. This shows that $C_1$ must be positive and $C_2$ must be negative.
Since $x$ is a continuous function of $t$ and a nontrivial solution of $x$ cannot take 0 by the above observation, it follows that $x(t)=C_1e^t,C_1>0$ for all $t$ or $x(t)=C_2e^{-t},C_2<0$ for all $t$.
edited 17 hours ago
Shashi
7,0221525
answered 19 hours ago
bellcircle
1,204411
You first divided both sides by $|x|$ to get the following results. But how can you guarantee that you are not dividing by $0$. In other words, if you cannot guarantee that once the solution starts from a non-zero value, it will not reach $0$, then you cannot divide it by $|x|$.
– winston
18 hours ago
@winston Good question. If $dot{x}=0$, then $x=0$ at such point. Now it is obvious that there is no way to combine the distinct (local) solutions $C_1e^t,C_2e^{-t},0$ to make a continuous global solution because the first is always positive, the second always negative, the last always zero. Therefore the global solution must be one of the three.
– bellcircle
18 hours ago
add a comment |
up vote
4
down vote
up vote
4
down vote
One observes that $x=0$ is the trivial solution of the given ODE.
Now find nontrivial solutions of the equation.
Using separation of variables, one gets $$frac{dot{x}}{|x|}=1.$$
Integrating both sides gives
begin{align}text{sgn}(x)log|x|=t+Cquadcdotsquadtext{(a)}end{align} for some constant $C$, where $text{sgn}(x)=begin{cases}1,&x> 0\-1,&x<0.end{cases}$
Therefore, $x(t)=begin{cases}C_1e^t,&x>0\C_2e^{-t},&x<0.end{cases}$
Now find the solution of explicit form.
From the equation, $dot{x}$ is always nonnegative, so $x$ is a continuously differentiable and nondecreasing function of $t$. This shows that $C_1$ must be positive and $C_2$ must be negative.
Since $x$ is a continuous function of $t$ and a nontrivial solution of $x$ cannot take 0 by the above observation, it follows that $x(t)=C_1e^t,C_1>0$ for all $t$ or $x(t)=C_2e^{-t},C_2<0$ for all $t$.
edited 17 hours ago
Shashi
7,0221525
answered 19 hours ago
bellcircle
1,204411
One observes that $x=0$ is the trivial solution of the given ODE.
Now find nontrivial solutions of the equation.
Using separation of variables, one gets $$frac{dot{x}}{|x|}=1.$$
Integrating both sides gives
begin{align}text{sgn}(x)log|x|=t+Cquadcdotsquadtext{(a)}end{align} for some constant $C$, where $text{sgn}(x)=begin{cases}1,&x> 0\-1,&x<0.end{cases}$
Therefore, $x(t)=begin{cases}C_1e^t,&x>0\C_2e^{-t},&x<0.end{cases}$
Now find the solution of explicit form.
From the equation, $dot{x}$ is always nonnegative, so $x$ is a continuously differentiable and nondecreasing function of $t$. This shows that $C_1$ must be positive and $C_2$ must be negative.
Since $x$ is a continuous function of $t$ and a nontrivial solution of $x$ cannot take 0 by the above observation, it follows that $x(t)=C_1e^t,C_1>0$ for all $t$ or $x(t)=C_2e^{-t},C_2<0$ for all $t$.
edited 17 hours ago
Shashi
7,0221525
answered 19 hours ago
bellcircle
1,204411
edited 17 hours ago
Shashi
7,0221525
edited 17 hours ago
Shashi
7,0221525
edited 17 hours ago
Shashi
7,0221525
7,0221525
answered 19 hours ago
bellcircle
1,204411
answered 19 hours ago
bellcircle
1,204411
answered 19 hours ago
bellcircle
1,204411
1,204411
You first divided both sides by $|x|$ to get the following results. But how can you guarantee that you are not dividing by $0$. In other words, if you cannot guarantee that once the solution starts from a non-zero value, it will not reach $0$, then you cannot divide it by $|x|$.
– winston
18 hours ago
@winston Good question. If $dot{x}=0$, then $x=0$ at such point. Now it is obvious that there is no way to combine the distinct (local) solutions $C_1e^t,C_2e^{-t},0$ to make a continuous global solution because the first is always positive, the second always negative, the last always zero. Therefore the global solution must be one of the three.
– bellcircle
18 hours ago
add a comment |
You first divided both sides by $|x|$ to get the following results. But how can you guarantee that you are not dividing by $0$. In other words, if you cannot guarantee that once the solution starts from a non-zero value, it will not reach $0$, then you cannot divide it by $|x|$.
– winston
18 hours ago
@winston Good question. If $dot{x}=0$, then $x=0$ at such point. Now it is obvious that there is no way to combine the distinct (local) solutions $C_1e^t,C_2e^{-t},0$ to make a continuous global solution because the first is always positive, the second always negative, the last always zero. Therefore the global solution must be one of the three.
– bellcircle
18 hours ago
You first divided both sides by $|x|$ to get the following results. But how can you guarantee that you are not dividing by $0$. In other words, if you cannot guarantee that once the solution starts from a non-zero value, it will not reach $0$, then you cannot divide it by $|x|$.
– winston
18 hours ago
You first divided both sides by $|x|$ to get the following results. But how can you guarantee that you are not dividing by $0$. In other words, if you cannot guarantee that once the solution starts from a non-zero value, it will not reach $0$, then you cannot divide it by $|x|$.
– winston
18 hours ago
@winston Good question. If $dot{x}=0$, then $x=0$ at such point. Now it is obvious that there is no way to combine the distinct (local) solutions $C_1e^t,C_2e^{-t},0$ to make a continuous global solution because the first is always positive, the second always negative, the last always zero. Therefore the global solution must be one of the three.
– bellcircle
18 hours ago
@winston Good question. If $dot{x}=0$, then $x=0$ at such point. Now it is obvious that there is no way to combine the distinct (local) solutions $C_1e^t,C_2e^{-t},0$ to make a continuous global solution because the first is always positive, the second always negative, the last always zero. Therefore the global solution must be one of the three.
– bellcircle
18 hours ago
add a comment |
up vote
1
down vote
bellcircle has provided a solution to the problem. However, I would like to express the solution in a more compact form and also provide an answer using a proposition-proof style.
$(Dx)(t) = |x(t)|$ (1)
Proposition. If $I$ is an open interval in $mathbb{R}$ and $C in mathbb{R}$, then $x(t)=Ce^{mathsf{sgn}(C) t}$ is a solution of the differential equation (1) on $I$. If $x(t)$ is a solution of the differential equation (1) on an open interval $I$ in $mathbb{R}$, then there exists $C in mathbb{R}$ such that $x(t)=Ce^{mathsf{sgn}(C) t}$ for all $t in I$.
Suppose $I$ is an open interval in $mathbb{R}$. Suppose $C in mathbb{R}$ and $x$ is such that $x(t)=Ce^{mathsf{sgn}(C) t}$ for all $t in I$. Then $(Dx)(t)=mathsf{sgn}(C)Ce^{mathsf{sgn}(C) t}$. Also, $|x(t)|=mathsf{sgn}(C)Ce^{mathsf{sgn}(C) t}$ for all $t in I$. Therefore, indeed $x(t)=Ce^{mathsf{sgn}(C) t}$ is a solution of (1) on $I$.
Suppose there is a solution $x_2(t)$ of (1) on $I$ such that there is no $C in mathbb{R}$ such that $x_2(t) = Ce^{mathsf{sgn}(C) t}$ for all $t in I$. Then take $T in I$. Define $A=frac{x_2(T)}{e^{mathsf{sgn}(x_2(T)) T}}$. Then $x(t) = A e^{mathsf{sgn}(A)t}$ is a solution of (1) on $I$ with $x(T)=x_2(T)$. The uniqueness of the solution of (1) on $I$ with $x(T)=x_2(T)$ follows from the Lipschitz continuity of $|cdot|$ by the Picard–Lindelöf theorem (link) and the global uniqueness theorem (e.g. link). Therefore $x_2(t) = A e^{mathsf{sgn}(A)t}$ for all $t in I$ and, thus, by contradiction, if $x(t)$ is a solution of the differential equation (1) on an open interval $I$ in $mathbb{R}$, then there exists $C in mathbb{R}$ such that $x(t)=Ce^{mathsf{sgn}(C) t}$ for all $t in I$
answered 18 hours ago
xanonec
1485
add a comment |
up vote
1
down vote
bellcircle has provided a solution to the problem. However, I would like to express the solution in a more compact form and also provide an answer using a proposition-proof style.
$(Dx)(t) = |x(t)|$ (1)
Proposition. If $I$ is an open interval in $mathbb{R}$ and $C in mathbb{R}$, then $x(t)=Ce^{mathsf{sgn}(C) t}$ is a solution of the differential equation (1) on $I$. If $x(t)$ is a solution of the differential equation (1) on an open interval $I$ in $mathbb{R}$, then there exists $C in mathbb{R}$ such that $x(t)=Ce^{mathsf{sgn}(C) t}$ for all $t in I$.
Suppose $I$ is an open interval in $mathbb{R}$. Suppose $C in mathbb{R}$ and $x$ is such that $x(t)=Ce^{mathsf{sgn}(C) t}$ for all $t in I$. Then $(Dx)(t)=mathsf{sgn}(C)Ce^{mathsf{sgn}(C) t}$. Also, $|x(t)|=mathsf{sgn}(C)Ce^{mathsf{sgn}(C) t}$ for all $t in I$. Therefore, indeed $x(t)=Ce^{mathsf{sgn}(C) t}$ is a solution of (1) on $I$.
Suppose there is a solution $x_2(t)$ of (1) on $I$ such that there is no $C in mathbb{R}$ such that $x_2(t) = Ce^{mathsf{sgn}(C) t}$ for all $t in I$. Then take $T in I$. Define $A=frac{x_2(T)}{e^{mathsf{sgn}(x_2(T)) T}}$. Then $x(t) = A e^{mathsf{sgn}(A)t}$ is a solution of (1) on $I$ with $x(T)=x_2(T)$. The uniqueness of the solution of (1) on $I$ with $x(T)=x_2(T)$ follows from the Lipschitz continuity of $|cdot|$ by the Picard–Lindelöf theorem (link) and the global uniqueness theorem (e.g. link). Therefore $x_2(t) = A e^{mathsf{sgn}(A)t}$ for all $t in I$ and, thus, by contradiction, if $x(t)$ is a solution of the differential equation (1) on an open interval $I$ in $mathbb{R}$, then there exists $C in mathbb{R}$ such that $x(t)=Ce^{mathsf{sgn}(C) t}$ for all $t in I$
answered 18 hours ago
xanonec
1485
add a comment |
up vote
1
down vote
up vote
1
down vote
bellcircle has provided a solution to the problem. However, I would like to express the solution in a more compact form and also provide an answer using a proposition-proof style.
$(Dx)(t) = |x(t)|$ (1)
Proposition. If $I$ is an open interval in $mathbb{R}$ and $C in mathbb{R}$, then $x(t)=Ce^{mathsf{sgn}(C) t}$ is a solution of the differential equation (1) on $I$. If $x(t)$ is a solution of the differential equation (1) on an open interval $I$ in $mathbb{R}$, then there exists $C in mathbb{R}$ such that $x(t)=Ce^{mathsf{sgn}(C) t}$ for all $t in I$.
Suppose $I$ is an open interval in $mathbb{R}$. Suppose $C in mathbb{R}$ and $x$ is such that $x(t)=Ce^{mathsf{sgn}(C) t}$ for all $t in I$. Then $(Dx)(t)=mathsf{sgn}(C)Ce^{mathsf{sgn}(C) t}$. Also, $|x(t)|=mathsf{sgn}(C)Ce^{mathsf{sgn}(C) t}$ for all $t in I$. Therefore, indeed $x(t)=Ce^{mathsf{sgn}(C) t}$ is a solution of (1) on $I$.
Suppose there is a solution $x_2(t)$ of (1) on $I$ such that there is no $C in mathbb{R}$ such that $x_2(t) = Ce^{mathsf{sgn}(C) t}$ for all $t in I$. Then take $T in I$. Define $A=frac{x_2(T)}{e^{mathsf{sgn}(x_2(T)) T}}$. Then $x(t) = A e^{mathsf{sgn}(A)t}$ is a solution of (1) on $I$ with $x(T)=x_2(T)$. The uniqueness of the solution of (1) on $I$ with $x(T)=x_2(T)$ follows from the Lipschitz continuity of $|cdot|$ by the Picard–Lindelöf theorem (link) and the global uniqueness theorem (e.g. link). Therefore $x_2(t) = A e^{mathsf{sgn}(A)t}$ for all $t in I$ and, thus, by contradiction, if $x(t)$ is a solution of the differential equation (1) on an open interval $I$ in $mathbb{R}$, then there exists $C in mathbb{R}$ such that $x(t)=Ce^{mathsf{sgn}(C) t}$ for all $t in I$
answered 18 hours ago
xanonec
1485
bellcircle has provided a solution to the problem. However, I would like to express the solution in a more compact form and also provide an answer using a proposition-proof style.
$(Dx)(t) = |x(t)|$ (1)
Proposition. If $I$ is an open interval in $mathbb{R}$ and $C in mathbb{R}$, then $x(t)=Ce^{mathsf{sgn}(C) t}$ is a solution of the differential equation (1) on $I$. If $x(t)$ is a solution of the differential equation (1) on an open interval $I$ in $mathbb{R}$, then there exists $C in mathbb{R}$ such that $x(t)=Ce^{mathsf{sgn}(C) t}$ for all $t in I$.
Suppose $I$ is an open interval in $mathbb{R}$. Suppose $C in mathbb{R}$ and $x$ is such that $x(t)=Ce^{mathsf{sgn}(C) t}$ for all $t in I$. Then $(Dx)(t)=mathsf{sgn}(C)Ce^{mathsf{sgn}(C) t}$. Also, $|x(t)|=mathsf{sgn}(C)Ce^{mathsf{sgn}(C) t}$ for all $t in I$. Therefore, indeed $x(t)=Ce^{mathsf{sgn}(C) t}$ is a solution of (1) on $I$.
Suppose there is a solution $x_2(t)$ of (1) on $I$ such that there is no $C in mathbb{R}$ such that $x_2(t) = Ce^{mathsf{sgn}(C) t}$ for all $t in I$. Then take $T in I$. Define $A=frac{x_2(T)}{e^{mathsf{sgn}(x_2(T)) T}}$. Then $x(t) = A e^{mathsf{sgn}(A)t}$ is a solution of (1) on $I$ with $x(T)=x_2(T)$. The uniqueness of the solution of (1) on $I$ with $x(T)=x_2(T)$ follows from the Lipschitz continuity of $|cdot|$ by the Picard–Lindelöf theorem (link) and the global uniqueness theorem (e.g. link). Therefore $x_2(t) = A e^{mathsf{sgn}(A)t}$ for all $t in I$ and, thus, by contradiction, if $x(t)$ is a solution of the differential equation (1) on an open interval $I$ in $mathbb{R}$, then there exists $C in mathbb{R}$ such that $x(t)=Ce^{mathsf{sgn}(C) t}$ for all $t in I$
answered 18 hours ago
xanonec
1485
edited 16 hours ago
answered 18 hours ago
xanonec
1485
answered 18 hours ago
xanonec
1485
answered 18 hours ago
xanonec
1485
1485
add a comment |
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@Rahul because initial condition is not given.
– bellcircle
20 hours ago