MHB Existence and uniqueness of solution

[mathmari]

Hey!

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)

 

[I like Serena]

Hey!

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)

 

[I like Serena]

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)

 

[HOI]

Hey!

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?

 

[I like Serena]

Where does it say that f(x) is not 0 on this interval?

$f:\mathbb R^2→(0,∞)$