Mixing
continuity of $f(t,u(t))$ with respect to $t$
1
$begingroup$
My question relates to definition of continuity from calculus.
Let $y=f(t,u(t))$. What does it mean $f$ is continuous with respect to $t$ at $t_{0}$?
calculus
asked Jan 15 at 18:50
Zviad KhukhunashviliZviad Khukhunashvili
1208
$endgroup$
add a comment |
1
$begingroup$
My question relates to definition of continuity from calculus.
Let $y=f(t,u(t))$. What does it mean $f$ is continuous with respect to $t$ at $t_{0}$?
calculus
asked Jan 15 at 18:50
Zviad KhukhunashviliZviad Khukhunashvili
1208
$endgroup$
add a comment |
1
1
1
$begingroup$
My question relates to definition of continuity from calculus.
Let $y=f(t,u(t))$. What does it mean $f$ is continuous with respect to $t$ at $t_{0}$?
calculus
asked Jan 15 at 18:50
Zviad KhukhunashviliZviad Khukhunashvili
1208
$endgroup$
My question relates to definition of continuity from calculus.
Let $y=f(t,u(t))$. What does it mean $f$ is continuous with respect to $t$ at $t_{0}$?
calculus
calculus
asked Jan 15 at 18:50
Zviad KhukhunashviliZviad Khukhunashvili
1208
asked Jan 15 at 18:50
Zviad KhukhunashviliZviad Khukhunashvili
1208
asked Jan 15 at 18:50
Zviad KhukhunashviliZviad Khukhunashvili
1208
asked Jan 15 at 18:50
Zviad KhukhunashviliZviad Khukhunashvili
1208
asked Jan 15 at 18:50
Zviad KhukhunashviliZviad Khukhunashvili
1208
1208
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
It seems that the definition $y=f(t,u(t))$ is not related to your question. And it seems that $t$ is the first parameter of the function $f$.
The function $f$ is continuous with respect to $t$ at $t_0$ if
$$lim_{tto t_0} f(t,x) = f(t_0, x)$$
for any $x$ such that there is a neighborhood of $(t_0,x)$ where $f$ is defined.
answered Jan 15 at 19:03
I like SerenaI like Serena
4,2221722
$endgroup$
$begingroup$
shouldn't the fact that $u$ is a function of $t$ be somehow reflected in this definition?
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:23
$begingroup$
$u$ is not part of the definition of $f$ is it @ZviadKhukhunashvili? It would be different if we want to know about the continuity of $y$ with respect to $t$. Then $u$ would be involved.
$endgroup$
– I like Serena
Jan 15 at 19:31
$begingroup$
what do you mean by "not part of the definition of $f$"? $u$ is unknown function, $f$ is known.
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:44
$begingroup$
@ZviadKhukhunashvili, we could have for instance $f(t,x)=t+x$ and $u(t)=sqrt t$. Note that the definition of $f$ does not contain any reference to $u$, nor does its continuity depend on it. However, we have $y(t)=f(t,u(t))=t+sqrt t$, so the continuity of $y$ does depend on $u$.
$endgroup$
– I like Serena
Jan 15 at 19:50
$begingroup$
then I say $f(t,u(t))$ is the right hand side of an ODE with unknown $u(t)$ and I need to check whether it satisfies conditions of the existence and uniqueness theorem, in particular, continuity of $f$ in $t$. Should I go with the first way or the second way?
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:57
|
show 3 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
1
$begingroup$
It seems that the definition $y=f(t,u(t))$ is not related to your question. And it seems that $t$ is the first parameter of the function $f$.
The function $f$ is continuous with respect to $t$ at $t_0$ if
$$lim_{tto t_0} f(t,x) = f(t_0, x)$$
for any $x$ such that there is a neighborhood of $(t_0,x)$ where $f$ is defined.
answered Jan 15 at 19:03
I like SerenaI like Serena
4,2221722
$endgroup$
$begingroup$
shouldn't the fact that $u$ is a function of $t$ be somehow reflected in this definition?
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:23
$begingroup$
$u$ is not part of the definition of $f$ is it @ZviadKhukhunashvili? It would be different if we want to know about the continuity of $y$ with respect to $t$. Then $u$ would be involved.
$endgroup$
– I like Serena
Jan 15 at 19:31
$begingroup$
what do you mean by "not part of the definition of $f$"? $u$ is unknown function, $f$ is known.
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:44
$begingroup$
@ZviadKhukhunashvili, we could have for instance $f(t,x)=t+x$ and $u(t)=sqrt t$. Note that the definition of $f$ does not contain any reference to $u$, nor does its continuity depend on it. However, we have $y(t)=f(t,u(t))=t+sqrt t$, so the continuity of $y$ does depend on $u$.
$endgroup$
– I like Serena
Jan 15 at 19:50
$begingroup$
then I say $f(t,u(t))$ is the right hand side of an ODE with unknown $u(t)$ and I need to check whether it satisfies conditions of the existence and uniqueness theorem, in particular, continuity of $f$ in $t$. Should I go with the first way or the second way?
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:57
|
show 3 more comments
1
$begingroup$
It seems that the definition $y=f(t,u(t))$ is not related to your question. And it seems that $t$ is the first parameter of the function $f$.
The function $f$ is continuous with respect to $t$ at $t_0$ if
$$lim_{tto t_0} f(t,x) = f(t_0, x)$$
for any $x$ such that there is a neighborhood of $(t_0,x)$ where $f$ is defined.
answered Jan 15 at 19:03
I like SerenaI like Serena
4,2221722
$endgroup$
$begingroup$
shouldn't the fact that $u$ is a function of $t$ be somehow reflected in this definition?
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:23
$begingroup$
$u$ is not part of the definition of $f$ is it @ZviadKhukhunashvili? It would be different if we want to know about the continuity of $y$ with respect to $t$. Then $u$ would be involved.
$endgroup$
– I like Serena
Jan 15 at 19:31
$begingroup$
what do you mean by "not part of the definition of $f$"? $u$ is unknown function, $f$ is known.
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:44
$begingroup$
@ZviadKhukhunashvili, we could have for instance $f(t,x)=t+x$ and $u(t)=sqrt t$. Note that the definition of $f$ does not contain any reference to $u$, nor does its continuity depend on it. However, we have $y(t)=f(t,u(t))=t+sqrt t$, so the continuity of $y$ does depend on $u$.
$endgroup$
– I like Serena
Jan 15 at 19:50
$begingroup$
then I say $f(t,u(t))$ is the right hand side of an ODE with unknown $u(t)$ and I need to check whether it satisfies conditions of the existence and uniqueness theorem, in particular, continuity of $f$ in $t$. Should I go with the first way or the second way?
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:57
|
show 3 more comments
1
1
1
$begingroup$
It seems that the definition $y=f(t,u(t))$ is not related to your question. And it seems that $t$ is the first parameter of the function $f$.
The function $f$ is continuous with respect to $t$ at $t_0$ if
$$lim_{tto t_0} f(t,x) = f(t_0, x)$$
for any $x$ such that there is a neighborhood of $(t_0,x)$ where $f$ is defined.
answered Jan 15 at 19:03
I like SerenaI like Serena
4,2221722
$endgroup$
It seems that the definition $y=f(t,u(t))$ is not related to your question. And it seems that $t$ is the first parameter of the function $f$.
The function $f$ is continuous with respect to $t$ at $t_0$ if
$$lim_{tto t_0} f(t,x) = f(t_0, x)$$
for any $x$ such that there is a neighborhood of $(t_0,x)$ where $f$ is defined.
answered Jan 15 at 19:03
I like SerenaI like Serena
4,2221722
answered Jan 15 at 19:03
I like SerenaI like Serena
4,2221722
answered Jan 15 at 19:03
I like SerenaI like Serena
4,2221722
answered Jan 15 at 19:03
I like SerenaI like Serena
4,2221722
4,2221722
$begingroup$
shouldn't the fact that $u$ is a function of $t$ be somehow reflected in this definition?
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:23
$begingroup$
$u$ is not part of the definition of $f$ is it @ZviadKhukhunashvili? It would be different if we want to know about the continuity of $y$ with respect to $t$. Then $u$ would be involved.
$endgroup$
– I like Serena
Jan 15 at 19:31
$begingroup$
what do you mean by "not part of the definition of $f$"? $u$ is unknown function, $f$ is known.
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:44
$begingroup$
@ZviadKhukhunashvili, we could have for instance $f(t,x)=t+x$ and $u(t)=sqrt t$. Note that the definition of $f$ does not contain any reference to $u$, nor does its continuity depend on it. However, we have $y(t)=f(t,u(t))=t+sqrt t$, so the continuity of $y$ does depend on $u$.
$endgroup$
– I like Serena
Jan 15 at 19:50
$begingroup$
then I say $f(t,u(t))$ is the right hand side of an ODE with unknown $u(t)$ and I need to check whether it satisfies conditions of the existence and uniqueness theorem, in particular, continuity of $f$ in $t$. Should I go with the first way or the second way?
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:57
|
show 3 more comments
$begingroup$
shouldn't the fact that $u$ is a function of $t$ be somehow reflected in this definition?
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:23
$begingroup$
$u$ is not part of the definition of $f$ is it @ZviadKhukhunashvili? It would be different if we want to know about the continuity of $y$ with respect to $t$. Then $u$ would be involved.
$endgroup$
– I like Serena
Jan 15 at 19:31
$begingroup$
what do you mean by "not part of the definition of $f$"? $u$ is unknown function, $f$ is known.
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:44
$begingroup$
@ZviadKhukhunashvili, we could have for instance $f(t,x)=t+x$ and $u(t)=sqrt t$. Note that the definition of $f$ does not contain any reference to $u$, nor does its continuity depend on it. However, we have $y(t)=f(t,u(t))=t+sqrt t$, so the continuity of $y$ does depend on $u$.
$endgroup$
– I like Serena
Jan 15 at 19:50
$begingroup$
then I say $f(t,u(t))$ is the right hand side of an ODE with unknown $u(t)$ and I need to check whether it satisfies conditions of the existence and uniqueness theorem, in particular, continuity of $f$ in $t$. Should I go with the first way or the second way?
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:57
$begingroup$
shouldn't the fact that $u$ is a function of $t$ be somehow reflected in this definition?
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:23
$begingroup$
shouldn't the fact that $u$ is a function of $t$ be somehow reflected in this definition?
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:23
$begingroup$
$u$ is not part of the definition of $f$ is it @ZviadKhukhunashvili? It would be different if we want to know about the continuity of $y$ with respect to $t$. Then $u$ would be involved.
$endgroup$
– I like Serena
Jan 15 at 19:31
$begingroup$
$u$ is not part of the definition of $f$ is it @ZviadKhukhunashvili? It would be different if we want to know about the continuity of $y$ with respect to $t$. Then $u$ would be involved.
$endgroup$
– I like Serena
Jan 15 at 19:31
$begingroup$
what do you mean by "not part of the definition of $f$"? $u$ is unknown function, $f$ is known.
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:44
$begingroup$
what do you mean by "not part of the definition of $f$"? $u$ is unknown function, $f$ is known.
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:44
$begingroup$
@ZviadKhukhunashvili, we could have for instance $f(t,x)=t+x$ and $u(t)=sqrt t$. Note that the definition of $f$ does not contain any reference to $u$, nor does its continuity depend on it. However, we have $y(t)=f(t,u(t))=t+sqrt t$, so the continuity of $y$ does depend on $u$.
$endgroup$
– I like Serena
Jan 15 at 19:50
$begingroup$
@ZviadKhukhunashvili, we could have for instance $f(t,x)=t+x$ and $u(t)=sqrt t$. Note that the definition of $f$ does not contain any reference to $u$, nor does its continuity depend on it. However, we have $y(t)=f(t,u(t))=t+sqrt t$, so the continuity of $y$ does depend on $u$.
$endgroup$
– I like Serena
Jan 15 at 19:50
$begingroup$
then I say $f(t,u(t))$ is the right hand side of an ODE with unknown $u(t)$ and I need to check whether it satisfies conditions of the existence and uniqueness theorem, in particular, continuity of $f$ in $t$. Should I go with the first way or the second way?
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:57
$begingroup$
then I say $f(t,u(t))$ is the right hand side of an ODE with unknown $u(t)$ and I need to check whether it satisfies conditions of the existence and uniqueness theorem, in particular, continuity of $f$ in $t$. Should I go with the first way or the second way?
$endgroup$
– Zviad Khukhunashvili
Jan 15 at 19:57
|
show 3 more comments
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