Existence and uniqueness of solution

In summary, the conversation discusses the existence and uniqueness of a solution to an initial value problem, given that the function involved is continuous and continuously differentiable. The Picard-Lindelöf theorem is mentioned as a way to prove the existence and uniqueness of the solution. The conversation also discusses whether the Lipschitz condition needs to be used in the proof, and concludes that it is not necessary. Overall, the conversation focuses on proving the existence and uniqueness of a solution to the initial value problem.
  • #1
mathmari
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Hey! :eek:

We have the initital value problem $$\begin{cases}y'(t)=1/f(t, y(t)) \\ y(t_0)=y_0\end{cases} \ \ \ \ \ (1)$$ where the function $f:\mathbb{R}^2\rightarrow (0,\infty)$ is continuous in $\mathbb{R}^2$ and continuously differentiable as for $y$ in a domain that contains the point $(t_0, y_0)$.

Show that there exists $h>0$ such that the following two conditions are satisfied:
  • The problem (1) has a solution $\phi=\phi(t)$ that is defined at least for each $t\in (t_0-h, t_0+h)$.
  • In the interval $(t_0-h, t_0+h)$ there is no other solution of the problem (1). (i.e. if a function $\psi$ is a solution of the problem (1), then $\psi (t)=\phi (t)$, if $t\in (t_0-h, t_0+h)$)
For the first point we use the existence theorem, or not?

We have that $f$ is continuous, then $\frac{1}{f}$ is also continuous, since it doesn't get the value $0$. Is this correct?

We consider a region $R=\left \{(t,y) : |t-t_0|\leq a, \ |y-y_0|\leq b\right \}$ with $a,b>0$.

Do we have to show that $\frac{1}{f}$ is bounded in $R$ ?

Because then the IVP (1) would have at least one solution $\phi = \phi (t)$ defined in the interval $|t − t_0| \leq h$ where $h=\min \left \{a, \frac{b}{K}\right \}$, where $K$ is the maximum value of $\frac{1}{f}$ in $R$, or not?

(Wondering) For the second time we use the uniqueness theorem, or not?

Do we have to use for that the Lipschitz condition?

We have that $$\left |\frac{\frac{1}{f(t_1)}-\frac{1}{f(t_2)}}{t_1-t_2}\right |=\left |\frac{\frac{f(t_2)-f(t_1)}{f(t_1)f(t_2)}}{t_1-t_2}\right |=\left |\frac{f(t_2)-f(t_1)}{(t_1-t_2)f(t_1)f(t_2)}\right |=\frac{|f(t_2)-f(t_1)|}{|t_1-t_2||f(t_1)||f(t_2)|}$$ Since $f$ is continuous we get that $|f(t_2)-f(t_1)|\leq C|t_2-t_1|\Rightarrow \frac{|f(t_2)-f(t_1)|}{|t_2-t_1|}\leq C$. So we get $$\left |\frac{\frac{1}{f(t_1)}-\frac{1}{f(t_2)}}{t_1-t_2}\right |\leq C\cdot \frac{1}{|f(t_1)||f(t_2)|}$$ Is everything correct so far? How could we continue?

(Wondering)
 
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  • #2
mathmari said:
Hey! :eek:

We have the initital value problem $$\begin{cases}y'(t)=1/f(t, y(t)) \\ y(t_0)=y_0\end{cases} \ \ \ \ \ (1)$$ where the function $f:\mathbb{R}^2\rightarrow (0,\infty)$ is continuous in $\mathbb{R}^2$ and continuously differentiable as for $y$ in a domain that contains the point $(t_0, y_0)$.

Show that there exists $h>0$ such that the following two conditions are satisfied:
  • The problem (1) has a solution $\phi=\phi(t)$ that is defined at least for each $t\in (t_0-h, t_0+h)$.
  • In the interval $(t_0-h, t_0+h)$ there is no other solution of the problem (1). (i.e. if a function $\psi$ is a solution of the problem (1), then $\psi (t)=\phi (t)$, if $t\in (t_0-h, t_0+h)$)
For the first point we use the existence theorem, or not?

We have that $f$ is continuous, then $\frac{1}{f}$ is also continuous, since it doesn't get the value $0$. Is this correct?

We consider a region $R=\left \{(t,y) : |t-t_0|\leq a, \ |y-y_0|\leq b\right \}$ with $a,b>0$.

Do we have to show that $\frac{1}{f}$ is bounded in $R$ ?

Because then the IVP (1) would have at least one solution $\phi = \phi (t)$ defined in the interval $|t − t_0| \leq h$ where $h=\min \left \{a, \frac{b}{K}\right \}$, where $K$ is the maximum value of $\frac{1}{f}$ in $R$, or not?

(Wondering) For the second time we use the uniqueness theorem, or not?

Do we have to use for that the Lipschitz condition?

We have that $$\left |\frac{\frac{1}{f(t_1)}-\frac{1}{f(t_2)}}{t_1-t_2}\right |=\left |\frac{\frac{f(t_2)-f(t_1)}{f(t_1)f(t_2)}}{t_1-t_2}\right |=\left |\frac{f(t_2)-f(t_1)}{(t_1-t_2)f(t_1)f(t_2)}\right |=\frac{|f(t_2)-f(t_1)|}{|t_1-t_2||f(t_1)||f(t_2)|}$$ Since $f$ is continuous we get that $|f(t_2)-f(t_1)|\leq C|t_2-t_1|\Rightarrow \frac{|f(t_2)-f(t_1)|}{|t_2-t_1|}\leq C$. So we get $$\left |\frac{\frac{1}{f(t_1)}-\frac{1}{f(t_2)}}{t_1-t_2}\right |\leq C\cdot \frac{1}{|f(t_1)||f(t_2)|}$$ Is everything correct so far? How could we continue?

(Wondering)

Hey mathmari!

The Picard–Lindelöf theorem states:
Consider the initial value problem
$$y'(t)=f(t,y(t)),\qquad y(t_0)=y_0$$
Suppose  f  is uniformly Lipschitz continuous in y (meaning the Lipschitz constant can be taken independent of t) and continuous in t. Then, for some value ε > 0, there exists a unique solution y(t) to the initial value problem on the interval $[t_{0}-\varepsilon ,t_{0}+\varepsilon ]$.

So if we can satisfy those conditions we have the proof for both bullet points don't we? (Wondering)

Since the $f$ of the problem statement is continuously differentiable, it follows that $\frac 1f$ is as well on a sufficiently small interval (inverse function theorem), doesn't it? (Wondering)

And if $\frac 1 f$ is differentiable on a closed interval, it's Lipschitz continuous, isn't it? (Wondering)
 
  • #3
mathmari said:
Do we have to use for that the Lipschitz condition?

We have that $$\left |\frac{\frac{1}{f(t_1)}-\frac{1}{f(t_2)}}{t_1-t_2}\right |=\left |\frac{\frac{f(t_2)-f(t_1)}{f(t_1)f(t_2)}}{t_1-t_2}\right |=\left |\frac{f(t_2)-f(t_1)}{(t_1-t_2)f(t_1)f(t_2)}\right |=\frac{|f(t_2)-f(t_1)|}{|t_1-t_2||f(t_1)||f(t_2)|}$$ Since $f$ is continuous we get that $|f(t_2)-f(t_1)|\leq C|t_2-t_1|\Rightarrow \frac{|f(t_2)-f(t_1)|}{|t_2-t_1|}\leq C$. So we get $$\left |\frac{\frac{1}{f(t_1)}-\frac{1}{f(t_2)}}{t_1-t_2}\right |\leq C\cdot \frac{1}{|f(t_1)||f(t_2)|}$$ Is everything correct so far? How could we continue?

Don't we already have what we need for Lipschitz continuity?
$f$ is known to be a positive function, so $f(t_1)$ is greater than some positive value isn't it? (Wondering)
 
  • #4
mathmari said:
Hey! :eek:

We have the initital value problem $$\begin{cases}y'(t)=1/f(t, y(t)) \\ y(t_0)=y_0\end{cases} \ \ \ \ \ (1)$$ where the function $f:\mathbb{R}^2\rightarrow (0,\infty)$ is continuous in $\mathbb{R}^2$ and continuously differentiable as for $y$ in a domain that contains the point $(t_0, y_0)$.

Show that there exists $h>0$ such that the following two conditions are satisfied:
  • The problem (1) has a solution $\phi=\phi(t)$ that is defined at least for each $t\in (t_0-h, t_0+h)$.
  • In the interval $(t_0-h, t_0+h)$ there is no other solution of the problem (1). (i.e. if a function $\psi$ is a solution of the problem (1), then $\psi (t)=\phi (t)$, if $t\in (t_0-h, t_0+h)$)
For the first point we use the existence theorem, or not?

We have that $f$ is continuous, then $\frac{1}{f}$ is also continuous, since it doesn't get the value $0$. Is this correct?
Where does it say that f(x) is not 0 on this interval?
 
  • #5
Country Boy said:
Where does it say that f(x) is not 0 on this interval?

$f:\mathbb R^2→(0,∞)$
 

FAQ: Existence and uniqueness of solution

1. What is the concept of existence and uniqueness of solution?

The concept of existence and uniqueness of solution refers to the idea that a mathematical equation or problem has a single, specific solution that is valid and can be proven to exist. This means that there is only one possible solution to the problem, and it is the only correct one.

2. Why is the existence and uniqueness of solution important?

The existence and uniqueness of solution is important because it ensures the validity and reliability of mathematical equations and problems. It allows us to confidently use these solutions in various applications, such as in physics, engineering, and other sciences. Without this concept, there would be uncertainty and inconsistency in the solutions we derive.

3. What types of mathematical problems require the concept of existence and uniqueness of solution?

Generally, any problem that involves finding a solution to an equation or set of equations may require the concept of existence and uniqueness of solution. This includes differential equations, optimization problems, and other types of mathematical models used in various fields of science and engineering.

4. How can we prove the existence and uniqueness of solution to a mathematical problem?

The existence and uniqueness of solution can be proven through various mathematical techniques, such as using the method of contradiction, the intermediate value theorem, or the Picard-Lindelöf theorem. These methods involve analyzing the properties and behaviors of the equations and their solutions to determine if they meet the criteria for being unique and valid.

5. Are there any situations where the concept of existence and uniqueness of solution does not apply?

Yes, there are certain mathematical problems, such as chaotic systems or non-linear equations, where the concept of existence and uniqueness of solution may not apply. These types of problems may have multiple solutions or no solutions at all, making it difficult to determine a unique and valid solution. In these cases, other methods and techniques must be used to analyze and solve the problem.

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