- #1
evinda
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Hi! (Smile)
Consider the initial value problem
$$\left\{\begin{matrix}
y'(t)=\sqrt{|y|}, 0 \leq t \leq 2\\
y(0)=1
\end{matrix}\right. \tag 1$$
Show that for this problem the assumptions of the following theorem hold:
"Let $c>0$ and $f \in C([a,b] \times [y_0-c, y_0+c])$. If $f$ satisfies at $[a,b] \times [y_0-c, y_0+c])$ the Lipschtiz condition in respect to $y$, uniformly in respect of $t$, that means that:$$\exists L \geq 0: \forall t \in [a,b] \ \forall y_1, y_2 \in [y_0-c,y_0+c]: \\ |f(t,y_1)-f(t,y_2)| \leq L |y_1-y_2|$$
then the initial value problem $\left\{\begin{matrix}
y'(t)=f(t,y(t)) &, a \leq t \leq b \\
y(a)=y_0 &
\end{matrix}\right.$ is solved uniquely at least at the interval $[a,b']$ where with $A=\max_{a \leq t \leq b, y_0-c \leq y \leq y_0+c} |f(t,y)|$ we have that $b'=\min \{ b, a+ \frac{c}{A}\}$. "
for appropriate $c$ and $L$.
Determine the solution $y$ with the method of separation of variables.
I tried the following:
$$f(t)=\sqrt{|y|}$$
$$\frac{\partial{f}}{\partial{y}}(t, y(t))=\frac{1}{2 \sqrt{|y|}} \leq M \Rightarrow \frac{1}{2M} \leq \sqrt{|y|} \Rightarrow y \geq \frac{1}{4M}$$
$$y(0)-c=1-c=\frac{1}{4M} \Rightarrow c=1-\frac{1}{4M}$$
Then $(1)$ is solved uniquely, at least at the interval $[0,b']$ where $A=\max_{0 \leq t \leq 2, \frac{1}{4M} \leq y \leq 2-\frac{1}{4M}} |\sqrt{|y|}|=\sqrt{2-\frac{1}{4M}}$
$$b'=\min \{ 2, \frac{1-\frac{1}{4M}}{\sqrt{2-\frac{1}{4M}}}\}=\frac{1-\frac{1}{4M}}{\sqrt{2-\frac{1}{4M}}}$$$$L=M$$$$\frac{dy}{dt}=\sqrt{|y|} \Rightarrow \frac{dy}{2\sqrt{y}}=\frac{dt}{2} \Rightarrow \sqrt{y}=\frac{1}{2}t+c$$
$$y(0)=1 \Rightarrow c=1$$
So:
$$\sqrt{y}=\frac{1}{2}t+1 \Rightarrow y=\left( \frac{1}{2}t+1\right)^2=\frac{1}{4}t^2+t+1$$
Is it right or have I done something wrong? (Thinking)
Consider the initial value problem
$$\left\{\begin{matrix}
y'(t)=\sqrt{|y|}, 0 \leq t \leq 2\\
y(0)=1
\end{matrix}\right. \tag 1$$
Show that for this problem the assumptions of the following theorem hold:
"Let $c>0$ and $f \in C([a,b] \times [y_0-c, y_0+c])$. If $f$ satisfies at $[a,b] \times [y_0-c, y_0+c])$ the Lipschtiz condition in respect to $y$, uniformly in respect of $t$, that means that:$$\exists L \geq 0: \forall t \in [a,b] \ \forall y_1, y_2 \in [y_0-c,y_0+c]: \\ |f(t,y_1)-f(t,y_2)| \leq L |y_1-y_2|$$
then the initial value problem $\left\{\begin{matrix}
y'(t)=f(t,y(t)) &, a \leq t \leq b \\
y(a)=y_0 &
\end{matrix}\right.$ is solved uniquely at least at the interval $[a,b']$ where with $A=\max_{a \leq t \leq b, y_0-c \leq y \leq y_0+c} |f(t,y)|$ we have that $b'=\min \{ b, a+ \frac{c}{A}\}$. "
for appropriate $c$ and $L$.
Determine the solution $y$ with the method of separation of variables.
I tried the following:
$$f(t)=\sqrt{|y|}$$
$$\frac{\partial{f}}{\partial{y}}(t, y(t))=\frac{1}{2 \sqrt{|y|}} \leq M \Rightarrow \frac{1}{2M} \leq \sqrt{|y|} \Rightarrow y \geq \frac{1}{4M}$$
$$y(0)-c=1-c=\frac{1}{4M} \Rightarrow c=1-\frac{1}{4M}$$
Then $(1)$ is solved uniquely, at least at the interval $[0,b']$ where $A=\max_{0 \leq t \leq 2, \frac{1}{4M} \leq y \leq 2-\frac{1}{4M}} |\sqrt{|y|}|=\sqrt{2-\frac{1}{4M}}$
$$b'=\min \{ 2, \frac{1-\frac{1}{4M}}{\sqrt{2-\frac{1}{4M}}}\}=\frac{1-\frac{1}{4M}}{\sqrt{2-\frac{1}{4M}}}$$$$L=M$$$$\frac{dy}{dt}=\sqrt{|y|} \Rightarrow \frac{dy}{2\sqrt{y}}=\frac{dt}{2} \Rightarrow \sqrt{y}=\frac{1}{2}t+c$$
$$y(0)=1 \Rightarrow c=1$$
So:
$$\sqrt{y}=\frac{1}{2}t+1 \Rightarrow y=\left( \frac{1}{2}t+1\right)^2=\frac{1}{4}t^2+t+1$$
Is it right or have I done something wrong? (Thinking)