$fracdxdt = p, fracdydt = q$: Solution of these ODE imply the solution is constant along characteristics of the form $qx − py = constant$.
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My lecture notes state the following:
When we were dealing with first order equations we saw that a differential operator of the form,
$$pfracpartialpartialx + qfracpartialpartialy$$
Led to the characteristic equations
$$fracdxdt = p, fracdydt = q$$
The solution of these ODE in turn implied that the solution would be constant along characteristics of the form $qx − py = constant$.
Can someone please demonstrate this?
differential-equations pde differential-operators characteristics
asked Aug 11 at 4:51
handler's handle
1508
add a comment |Â
up vote
4
down vote
favorite
My lecture notes state the following:
When we were dealing with first order equations we saw that a differential operator of the form,
$$pfracpartialpartialx + qfracpartialpartialy$$
Led to the characteristic equations
$$fracdxdt = p, fracdydt = q$$
The solution of these ODE in turn implied that the solution would be constant along characteristics of the form $qx − py = constant$.
Can someone please demonstrate this?
differential-equations pde differential-operators characteristics
asked Aug 11 at 4:51
handler's handle
1508
2
Are $p,q$ constants? Usually in the treatment of first order PDE one has $p=frac∂u∂x$, $q=frac∂u∂y$.
– LutzL
Aug 11 at 6:49
@LutzL Yes, $p, q$ are constants. Sorry for not clarifying.
– handler's handle
Aug 11 at 11:47
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
My lecture notes state the following:
When we were dealing with first order equations we saw that a differential operator of the form,
$$pfracpartialpartialx + qfracpartialpartialy$$
Led to the characteristic equations
$$fracdxdt = p, fracdydt = q$$
The solution of these ODE in turn implied that the solution would be constant along characteristics of the form $qx − py = constant$.
Can someone please demonstrate this?
differential-equations pde differential-operators characteristics
asked Aug 11 at 4:51
handler's handle
1508
My lecture notes state the following:
When we were dealing with first order equations we saw that a differential operator of the form,
$$pfracpartialpartialx + qfracpartialpartialy$$
Led to the characteristic equations
$$fracdxdt = p, fracdydt = q$$
The solution of these ODE in turn implied that the solution would be constant along characteristics of the form $qx − py = constant$.
Can someone please demonstrate this?
differential-equations pde differential-operators characteristics
asked Aug 11 at 4:51
handler's handle
1508
asked Aug 11 at 4:51
handler's handle
1508
asked Aug 11 at 4:51
handler's handle
1508
asked Aug 11 at 4:51
handler's handle
1508
1508
2
Are $p,q$ constants? Usually in the treatment of first order PDE one has $p=frac∂u∂x$, $q=frac∂u∂y$.
– LutzL
Aug 11 at 6:49
@LutzL Yes, $p, q$ are constants. Sorry for not clarifying.
– handler's handle
Aug 11 at 11:47
add a comment |Â
2
Are $p,q$ constants? Usually in the treatment of first order PDE one has $p=frac∂u∂x$, $q=frac∂u∂y$.
– LutzL
Aug 11 at 6:49
@LutzL Yes, $p, q$ are constants. Sorry for not clarifying.
– handler's handle
Aug 11 at 11:47
2
2
Are $p,q$ constants? Usually in the treatment of first order PDE one has $p=frac∂u∂x$, $q=frac∂u∂y$.
– LutzL
Aug 11 at 6:49
Are $p,q$ constants? Usually in the treatment of first order PDE one has $p=frac∂u∂x$, $q=frac∂u∂y$.
– LutzL
Aug 11 at 6:49
@LutzL Yes, $p, q$ are constants. Sorry for not clarifying.
– handler's handle
Aug 11 at 11:47
@LutzL Yes, $p, q$ are constants. Sorry for not clarifying.
– handler's handle
Aug 11 at 11:47
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The wording of the question is doubtful about the notations $p$ and $q$ which are not conventional or ambiguously used. So, the answer below might be made obsolete if the OP rewrite the question on another form. Nevertheless, according to the present form of the question, my answer is :
When the differential operator
$quad pfracpartialpartialx + qfracpartialpartialyquad$ is applied to an unknown function $quad u(x,y)quad$ this function is solution of the PDE :
$$left(pfracpartial partialx + qfracpartial partialyright)u(x,y)=0$$
$$pfracpartial upartialx + qfracpartial upartialy=0$$
The differential of $u(x,y)$ is :
$$du=fracpartial upartialxdx + fracpartial upartialydy$$
Consider a curve (C) parametrically defined with parameter $t$ $$quadbegincases x=x(t)\y=y(t)endcases quad ; quad quadbegincases dx=fracdxdtdt\dy=fracdydtdtendcases$$
along this curve :
$$du=fracpartial upartialxfracdxdtdt + fracpartial upartialyfracdydtdt$$
$$fracdudt=fracpartial upartialxfracdxdt + fracpartial upartialyfracdydt$$
Suppose now that (C) is no longer any curve, but a characteristic curve which equation is :
$$qx-py=textconstant$$
where $p$ and $q$ are constant, or equivalently on parametric form :
$$begincases x=pt+x_0\y=qt+y_0endcasesquad textthusquadbegincases fracdxdt=p\ fracdydt=qendcases$$
Putting this into $fracdudt=fracpartial upartialxfracdxdt + fracpartial upartialyfracdydt$
$$fracdudt=pfracpartial upartialx + qfracpartial upartialy$$
Comparing to the PDE : $pfracpartial upartialx + qfracpartial upartialy=0$ gives :
$$fracdudt=0$$
$$u(x,y)=textconstant$$
This proves that the function $u(x,y)$ solution of the PDE is constant along the characteristic curve $qx-py=$constant .
answered Aug 11 at 9:45
JJacquelin
40.4k21650
Thanks for the clear answer! Can you please clarify this part: Comparing to the PDE: $pfracpartial upartialx + qfracpartial upartialy=0$. What is meant by "compare" here? What operation was done? Did you subtract $pfracpartial upartialx + qfracpartial upartialy=0$ from $fracdudt=pfracpartial upartialx + qfracpartial upartialy$?
– handler's handle
Aug 11 at 11:49
1
It is straightforward : On one hand we have $$fracdudt=pfracpartial upartialx + qfracpartial upartialy$$ and on the other hand : $$pfracpartial upartialx + qfracpartial upartialy=0$$ thus : $$fracdudt=0$$
– JJacquelin
Aug 11 at 12:27
Ahh, of course. Thank you!
– handler's handle
Aug 11 at 12:37
My apologies, I noticed something else when studying your answer: how did you get $du=fracpartial upartialxdx + fracpartial upartialydy$ as the differential of $u(x,y)$? It looks a bit weird.
– handler's handle
Aug 12 at 5:08
1
This is basic in differential calculus : $$df(x_1,x_2,...,x_k,...,x_n)=sum_k=1^nfracpartial fpartial x_kdx_k$$ See Eq.(*) in en.wikipedia.org/wiki/Differential_form#Differential_calculus . Note that the indices doesn't denote differentials, but are indices to distinguish different variables. In case of two variables $x_1=x:,: x_2=y:,: f(x_1,x_2)=u(x,y)$ corresponds to the formula that you are asking for.
– JJacquelin
Aug 12 at 7:18
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The wording of the question is doubtful about the notations $p$ and $q$ which are not conventional or ambiguously used. So, the answer below might be made obsolete if the OP rewrite the question on another form. Nevertheless, according to the present form of the question, my answer is :
When the differential operator
$quad pfracpartialpartialx + qfracpartialpartialyquad$ is applied to an unknown function $quad u(x,y)quad$ this function is solution of the PDE :
$$left(pfracpartial partialx + qfracpartial partialyright)u(x,y)=0$$
$$pfracpartial upartialx + qfracpartial upartialy=0$$
The differential of $u(x,y)$ is :
$$du=fracpartial upartialxdx + fracpartial upartialydy$$
Consider a curve (C) parametrically defined with parameter $t$ $$quadbegincases x=x(t)\y=y(t)endcases quad ; quad quadbegincases dx=fracdxdtdt\dy=fracdydtdtendcases$$
along this curve :
$$du=fracpartial upartialxfracdxdtdt + fracpartial upartialyfracdydtdt$$
$$fracdudt=fracpartial upartialxfracdxdt + fracpartial upartialyfracdydt$$
Suppose now that (C) is no longer any curve, but a characteristic curve which equation is :
$$qx-py=textconstant$$
where $p$ and $q$ are constant, or equivalently on parametric form :
$$begincases x=pt+x_0\y=qt+y_0endcasesquad textthusquadbegincases fracdxdt=p\ fracdydt=qendcases$$
Putting this into $fracdudt=fracpartial upartialxfracdxdt + fracpartial upartialyfracdydt$
$$fracdudt=pfracpartial upartialx + qfracpartial upartialy$$
Comparing to the PDE : $pfracpartial upartialx + qfracpartial upartialy=0$ gives :
$$fracdudt=0$$
$$u(x,y)=textconstant$$
This proves that the function $u(x,y)$ solution of the PDE is constant along the characteristic curve $qx-py=$constant .
answered Aug 11 at 9:45
JJacquelin
40.4k21650
Thanks for the clear answer! Can you please clarify this part: Comparing to the PDE: $pfracpartial upartialx + qfracpartial upartialy=0$. What is meant by "compare" here? What operation was done? Did you subtract $pfracpartial upartialx + qfracpartial upartialy=0$ from $fracdudt=pfracpartial upartialx + qfracpartial upartialy$?
– handler's handle
Aug 11 at 11:49
1
It is straightforward : On one hand we have $$fracdudt=pfracpartial upartialx + qfracpartial upartialy$$ and on the other hand : $$pfracpartial upartialx + qfracpartial upartialy=0$$ thus : $$fracdudt=0$$
– JJacquelin
Aug 11 at 12:27
Ahh, of course. Thank you!
– handler's handle
Aug 11 at 12:37
My apologies, I noticed something else when studying your answer: how did you get $du=fracpartial upartialxdx + fracpartial upartialydy$ as the differential of $u(x,y)$? It looks a bit weird.
– handler's handle
Aug 12 at 5:08
1
This is basic in differential calculus : $$df(x_1,x_2,...,x_k,...,x_n)=sum_k=1^nfracpartial fpartial x_kdx_k$$ See Eq.(*) in en.wikipedia.org/wiki/Differential_form#Differential_calculus . Note that the indices doesn't denote differentials, but are indices to distinguish different variables. In case of two variables $x_1=x:,: x_2=y:,: f(x_1,x_2)=u(x,y)$ corresponds to the formula that you are asking for.
– JJacquelin
Aug 12 at 7:18
add a comment |Â
up vote
2
down vote
accepted
The wording of the question is doubtful about the notations $p$ and $q$ which are not conventional or ambiguously used. So, the answer below might be made obsolete if the OP rewrite the question on another form. Nevertheless, according to the present form of the question, my answer is :
When the differential operator
$quad pfracpartialpartialx + qfracpartialpartialyquad$ is applied to an unknown function $quad u(x,y)quad$ this function is solution of the PDE :
$$left(pfracpartial partialx + qfracpartial partialyright)u(x,y)=0$$
$$pfracpartial upartialx + qfracpartial upartialy=0$$
The differential of $u(x,y)$ is :
$$du=fracpartial upartialxdx + fracpartial upartialydy$$
Consider a curve (C) parametrically defined with parameter $t$ $$quadbegincases x=x(t)\y=y(t)endcases quad ; quad quadbegincases dx=fracdxdtdt\dy=fracdydtdtendcases$$
along this curve :
$$du=fracpartial upartialxfracdxdtdt + fracpartial upartialyfracdydtdt$$
$$fracdudt=fracpartial upartialxfracdxdt + fracpartial upartialyfracdydt$$
Suppose now that (C) is no longer any curve, but a characteristic curve which equation is :
$$qx-py=textconstant$$
where $p$ and $q$ are constant, or equivalently on parametric form :
$$begincases x=pt+x_0\y=qt+y_0endcasesquad textthusquadbegincases fracdxdt=p\ fracdydt=qendcases$$
Putting this into $fracdudt=fracpartial upartialxfracdxdt + fracpartial upartialyfracdydt$
$$fracdudt=pfracpartial upartialx + qfracpartial upartialy$$
Comparing to the PDE : $pfracpartial upartialx + qfracpartial upartialy=0$ gives :
$$fracdudt=0$$
$$u(x,y)=textconstant$$
This proves that the function $u(x,y)$ solution of the PDE is constant along the characteristic curve $qx-py=$constant .
answered Aug 11 at 9:45
JJacquelin
40.4k21650
Thanks for the clear answer! Can you please clarify this part: Comparing to the PDE: $pfracpartial upartialx + qfracpartial upartialy=0$. What is meant by "compare" here? What operation was done? Did you subtract $pfracpartial upartialx + qfracpartial upartialy=0$ from $fracdudt=pfracpartial upartialx + qfracpartial upartialy$?
– handler's handle
Aug 11 at 11:49
1
It is straightforward : On one hand we have $$fracdudt=pfracpartial upartialx + qfracpartial upartialy$$ and on the other hand : $$pfracpartial upartialx + qfracpartial upartialy=0$$ thus : $$fracdudt=0$$
– JJacquelin
Aug 11 at 12:27
Ahh, of course. Thank you!
– handler's handle
Aug 11 at 12:37
My apologies, I noticed something else when studying your answer: how did you get $du=fracpartial upartialxdx + fracpartial upartialydy$ as the differential of $u(x,y)$? It looks a bit weird.
– handler's handle
Aug 12 at 5:08
1
This is basic in differential calculus : $$df(x_1,x_2,...,x_k,...,x_n)=sum_k=1^nfracpartial fpartial x_kdx_k$$ See Eq.(*) in en.wikipedia.org/wiki/Differential_form#Differential_calculus . Note that the indices doesn't denote differentials, but are indices to distinguish different variables. In case of two variables $x_1=x:,: x_2=y:,: f(x_1,x_2)=u(x,y)$ corresponds to the formula that you are asking for.
– JJacquelin
Aug 12 at 7:18
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The wording of the question is doubtful about the notations $p$ and $q$ which are not conventional or ambiguously used. So, the answer below might be made obsolete if the OP rewrite the question on another form. Nevertheless, according to the present form of the question, my answer is :
When the differential operator
$quad pfracpartialpartialx + qfracpartialpartialyquad$ is applied to an unknown function $quad u(x,y)quad$ this function is solution of the PDE :
$$left(pfracpartial partialx + qfracpartial partialyright)u(x,y)=0$$
$$pfracpartial upartialx + qfracpartial upartialy=0$$
The differential of $u(x,y)$ is :
$$du=fracpartial upartialxdx + fracpartial upartialydy$$
Consider a curve (C) parametrically defined with parameter $t$ $$quadbegincases x=x(t)\y=y(t)endcases quad ; quad quadbegincases dx=fracdxdtdt\dy=fracdydtdtendcases$$
along this curve :
$$du=fracpartial upartialxfracdxdtdt + fracpartial upartialyfracdydtdt$$
$$fracdudt=fracpartial upartialxfracdxdt + fracpartial upartialyfracdydt$$
Suppose now that (C) is no longer any curve, but a characteristic curve which equation is :
$$qx-py=textconstant$$
where $p$ and $q$ are constant, or equivalently on parametric form :
$$begincases x=pt+x_0\y=qt+y_0endcasesquad textthusquadbegincases fracdxdt=p\ fracdydt=qendcases$$
Putting this into $fracdudt=fracpartial upartialxfracdxdt + fracpartial upartialyfracdydt$
$$fracdudt=pfracpartial upartialx + qfracpartial upartialy$$
Comparing to the PDE : $pfracpartial upartialx + qfracpartial upartialy=0$ gives :
$$fracdudt=0$$
$$u(x,y)=textconstant$$
This proves that the function $u(x,y)$ solution of the PDE is constant along the characteristic curve $qx-py=$constant .
answered Aug 11 at 9:45
JJacquelin
40.4k21650
The wording of the question is doubtful about the notations $p$ and $q$ which are not conventional or ambiguously used. So, the answer below might be made obsolete if the OP rewrite the question on another form. Nevertheless, according to the present form of the question, my answer is :
When the differential operator
$quad pfracpartialpartialx + qfracpartialpartialyquad$ is applied to an unknown function $quad u(x,y)quad$ this function is solution of the PDE :
$$left(pfracpartial partialx + qfracpartial partialyright)u(x,y)=0$$
$$pfracpartial upartialx + qfracpartial upartialy=0$$
The differential of $u(x,y)$ is :
$$du=fracpartial upartialxdx + fracpartial upartialydy$$
Consider a curve (C) parametrically defined with parameter $t$ $$quadbegincases x=x(t)\y=y(t)endcases quad ; quad quadbegincases dx=fracdxdtdt\dy=fracdydtdtendcases$$
along this curve :
$$du=fracpartial upartialxfracdxdtdt + fracpartial upartialyfracdydtdt$$
$$fracdudt=fracpartial upartialxfracdxdt + fracpartial upartialyfracdydt$$
Suppose now that (C) is no longer any curve, but a characteristic curve which equation is :
$$qx-py=textconstant$$
where $p$ and $q$ are constant, or equivalently on parametric form :
$$begincases x=pt+x_0\y=qt+y_0endcasesquad textthusquadbegincases fracdxdt=p\ fracdydt=qendcases$$
Putting this into $fracdudt=fracpartial upartialxfracdxdt + fracpartial upartialyfracdydt$
$$fracdudt=pfracpartial upartialx + qfracpartial upartialy$$
Comparing to the PDE : $pfracpartial upartialx + qfracpartial upartialy=0$ gives :
$$fracdudt=0$$
$$u(x,y)=textconstant$$
This proves that the function $u(x,y)$ solution of the PDE is constant along the characteristic curve $qx-py=$constant .
answered Aug 11 at 9:45
JJacquelin
40.4k21650
edited Aug 11 at 9:51
answered Aug 11 at 9:45
JJacquelin
40.4k21650
answered Aug 11 at 9:45
JJacquelin
40.4k21650
answered Aug 11 at 9:45
JJacquelin
40.4k21650
40.4k21650
Thanks for the clear answer! Can you please clarify this part: Comparing to the PDE: $pfracpartial upartialx + qfracpartial upartialy=0$. What is meant by "compare" here? What operation was done? Did you subtract $pfracpartial upartialx + qfracpartial upartialy=0$ from $fracdudt=pfracpartial upartialx + qfracpartial upartialy$?
– handler's handle
Aug 11 at 11:49
1
It is straightforward : On one hand we have $$fracdudt=pfracpartial upartialx + qfracpartial upartialy$$ and on the other hand : $$pfracpartial upartialx + qfracpartial upartialy=0$$ thus : $$fracdudt=0$$
– JJacquelin
Aug 11 at 12:27
Ahh, of course. Thank you!
– handler's handle
Aug 11 at 12:37
My apologies, I noticed something else when studying your answer: how did you get $du=fracpartial upartialxdx + fracpartial upartialydy$ as the differential of $u(x,y)$? It looks a bit weird.
– handler's handle
Aug 12 at 5:08
1
This is basic in differential calculus : $$df(x_1,x_2,...,x_k,...,x_n)=sum_k=1^nfracpartial fpartial x_kdx_k$$ See Eq.(*) in en.wikipedia.org/wiki/Differential_form#Differential_calculus . Note that the indices doesn't denote differentials, but are indices to distinguish different variables. In case of two variables $x_1=x:,: x_2=y:,: f(x_1,x_2)=u(x,y)$ corresponds to the formula that you are asking for.
– JJacquelin
Aug 12 at 7:18
add a comment |Â
Thanks for the clear answer! Can you please clarify this part: Comparing to the PDE: $pfracpartial upartialx + qfracpartial upartialy=0$. What is meant by "compare" here? What operation was done? Did you subtract $pfracpartial upartialx + qfracpartial upartialy=0$ from $fracdudt=pfracpartial upartialx + qfracpartial upartialy$?
– handler's handle
Aug 11 at 11:49
1
It is straightforward : On one hand we have $$fracdudt=pfracpartial upartialx + qfracpartial upartialy$$ and on the other hand : $$pfracpartial upartialx + qfracpartial upartialy=0$$ thus : $$fracdudt=0$$
– JJacquelin
Aug 11 at 12:27
Ahh, of course. Thank you!
– handler's handle
Aug 11 at 12:37
My apologies, I noticed something else when studying your answer: how did you get $du=fracpartial upartialxdx + fracpartial upartialydy$ as the differential of $u(x,y)$? It looks a bit weird.
– handler's handle
Aug 12 at 5:08
1
This is basic in differential calculus : $$df(x_1,x_2,...,x_k,...,x_n)=sum_k=1^nfracpartial fpartial x_kdx_k$$ See Eq.(*) in en.wikipedia.org/wiki/Differential_form#Differential_calculus . Note that the indices doesn't denote differentials, but are indices to distinguish different variables. In case of two variables $x_1=x:,: x_2=y:,: f(x_1,x_2)=u(x,y)$ corresponds to the formula that you are asking for.
– JJacquelin
Aug 12 at 7:18
Thanks for the clear answer! Can you please clarify this part: Comparing to the PDE: $pfracpartial upartialx + qfracpartial upartialy=0$. What is meant by "compare" here? What operation was done? Did you subtract $pfracpartial upartialx + qfracpartial upartialy=0$ from $fracdudt=pfracpartial upartialx + qfracpartial upartialy$?
– handler's handle
Aug 11 at 11:49
Thanks for the clear answer! Can you please clarify this part: Comparing to the PDE: $pfracpartial upartialx + qfracpartial upartialy=0$. What is meant by "compare" here? What operation was done? Did you subtract $pfracpartial upartialx + qfracpartial upartialy=0$ from $fracdudt=pfracpartial upartialx + qfracpartial upartialy$?
– handler's handle
Aug 11 at 11:49
1
1
It is straightforward : On one hand we have $$fracdudt=pfracpartial upartialx + qfracpartial upartialy$$ and on the other hand : $$pfracpartial upartialx + qfracpartial upartialy=0$$ thus : $$fracdudt=0$$
– JJacquelin
Aug 11 at 12:27
It is straightforward : On one hand we have $$fracdudt=pfracpartial upartialx + qfracpartial upartialy$$ and on the other hand : $$pfracpartial upartialx + qfracpartial upartialy=0$$ thus : $$fracdudt=0$$
– JJacquelin
Aug 11 at 12:27
Ahh, of course. Thank you!
– handler's handle
Aug 11 at 12:37
Ahh, of course. Thank you!
– handler's handle
Aug 11 at 12:37
My apologies, I noticed something else when studying your answer: how did you get $du=fracpartial upartialxdx + fracpartial upartialydy$ as the differential of $u(x,y)$? It looks a bit weird.
– handler's handle
Aug 12 at 5:08
My apologies, I noticed something else when studying your answer: how did you get $du=fracpartial upartialxdx + fracpartial upartialydy$ as the differential of $u(x,y)$? It looks a bit weird.
– handler's handle
Aug 12 at 5:08
1
1
This is basic in differential calculus : $$df(x_1,x_2,...,x_k,...,x_n)=sum_k=1^nfracpartial fpartial x_kdx_k$$ See Eq.(*) in en.wikipedia.org/wiki/Differential_form#Differential_calculus . Note that the indices doesn't denote differentials, but are indices to distinguish different variables. In case of two variables $x_1=x:,: x_2=y:,: f(x_1,x_2)=u(x,y)$ corresponds to the formula that you are asking for.
– JJacquelin
Aug 12 at 7:18
This is basic in differential calculus : $$df(x_1,x_2,...,x_k,...,x_n)=sum_k=1^nfracpartial fpartial x_kdx_k$$ See Eq.(*) in en.wikipedia.org/wiki/Differential_form#Differential_calculus . Note that the indices doesn't denote differentials, but are indices to distinguish different variables. In case of two variables $x_1=x:,: x_2=y:,: f(x_1,x_2)=u(x,y)$ corresponds to the formula that you are asking for.
– JJacquelin
Aug 12 at 7:18
add a comment |Â
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2
Are $p,q$ constants? Usually in the treatment of first order PDE one has $p=frac∂u∂x$, $q=frac∂u∂y$.
– LutzL
Aug 11 at 6:49
@LutzL Yes, $p, q$ are constants. Sorry for not clarifying.
– handler's handle
Aug 11 at 11:47