Solving for integrand from integrated quantities.












0












$begingroup$


Given equations of the form:



$A(r) = int_{t_{1}}^{t_{2}}F(r,t)dt$



$B(t) = int_a^b F(r,t)r^2dr$



where $A(r)$, $B(t)$, and all of the limits on the integrals are known, is there enough information to solve for F(r,t)? If so how would one do this?



For more context this is a scenario where there is a quantity, $F(r,t)$, that varies in space and time but is measured as only a function of time and a function of space separately. I am trying to figure out if the full space and time dependence can be reconstructed from these two measurements alone.



EDIT:
Perhaps a better way of stating the question.



Is $F(r,t)$ uniquely constrained given $A(r)$ and $B(t)$?






integral-equations constraints







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you have the actual function $F$? Looks like something to do with the second moment of mass/area...
    $endgroup$
    – Karn Watcharasupat
    Dec 8 '18 at 16:58










  • $begingroup$
    I do not have the function F. It is actually an x-ray flux of a plasma with unknown temperature and density distributions (which is what would set F).
    $endgroup$
    – JJR4
    Dec 8 '18 at 17:01
















0












$begingroup$


Given equations of the form:



$A(r) = int_{t_{1}}^{t_{2}}F(r,t)dt$



$B(t) = int_a^b F(r,t)r^2dr$



where $A(r)$, $B(t)$, and all of the limits on the integrals are known, is there enough information to solve for F(r,t)? If so how would one do this?



For more context this is a scenario where there is a quantity, $F(r,t)$, that varies in space and time but is measured as only a function of time and a function of space separately. I am trying to figure out if the full space and time dependence can be reconstructed from these two measurements alone.



EDIT:
Perhaps a better way of stating the question.



Is $F(r,t)$ uniquely constrained given $A(r)$ and $B(t)$?






integral-equations constraints







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you have the actual function $F$? Looks like something to do with the second moment of mass/area...
    $endgroup$
    – Karn Watcharasupat
    Dec 8 '18 at 16:58










  • $begingroup$
    I do not have the function F. It is actually an x-ray flux of a plasma with unknown temperature and density distributions (which is what would set F).
    $endgroup$
    – JJR4
    Dec 8 '18 at 17:01














0












0








0





$begingroup$


Given equations of the form:



$A(r) = int_{t_{1}}^{t_{2}}F(r,t)dt$



$B(t) = int_a^b F(r,t)r^2dr$



where $A(r)$, $B(t)$, and all of the limits on the integrals are known, is there enough information to solve for F(r,t)? If so how would one do this?



For more context this is a scenario where there is a quantity, $F(r,t)$, that varies in space and time but is measured as only a function of time and a function of space separately. I am trying to figure out if the full space and time dependence can be reconstructed from these two measurements alone.



EDIT:
Perhaps a better way of stating the question.



Is $F(r,t)$ uniquely constrained given $A(r)$ and $B(t)$?






integral-equations constraints







share|cite|improve this question











$endgroup$




Given equations of the form:



$A(r) = int_{t_{1}}^{t_{2}}F(r,t)dt$



$B(t) = int_a^b F(r,t)r^2dr$



where $A(r)$, $B(t)$, and all of the limits on the integrals are known, is there enough information to solve for F(r,t)? If so how would one do this?



For more context this is a scenario where there is a quantity, $F(r,t)$, that varies in space and time but is measured as only a function of time and a function of space separately. I am trying to figure out if the full space and time dependence can be reconstructed from these two measurements alone.



EDIT:
Perhaps a better way of stating the question.



Is $F(r,t)$ uniquely constrained given $A(r)$ and $B(t)$?






integral-equations constraints




integral-equations constraints






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 15:40







JJR4

















asked Dec 8 '18 at 16:47









JJR4JJR4

12




12












  • $begingroup$
    Do you have the actual function $F$? Looks like something to do with the second moment of mass/area...
    $endgroup$
    – Karn Watcharasupat
    Dec 8 '18 at 16:58










  • $begingroup$
    I do not have the function F. It is actually an x-ray flux of a plasma with unknown temperature and density distributions (which is what would set F).
    $endgroup$
    – JJR4
    Dec 8 '18 at 17:01


















  • $begingroup$
    Do you have the actual function $F$? Looks like something to do with the second moment of mass/area...
    $endgroup$
    – Karn Watcharasupat
    Dec 8 '18 at 16:58










  • $begingroup$
    I do not have the function F. It is actually an x-ray flux of a plasma with unknown temperature and density distributions (which is what would set F).
    $endgroup$
    – JJR4
    Dec 8 '18 at 17:01
















$begingroup$
Do you have the actual function $F$? Looks like something to do with the second moment of mass/area...
$endgroup$
– Karn Watcharasupat
Dec 8 '18 at 16:58




$begingroup$
Do you have the actual function $F$? Looks like something to do with the second moment of mass/area...
$endgroup$
– Karn Watcharasupat
Dec 8 '18 at 16:58












$begingroup$
I do not have the function F. It is actually an x-ray flux of a plasma with unknown temperature and density distributions (which is what would set F).
$endgroup$
– JJR4
Dec 8 '18 at 17:01




$begingroup$
I do not have the function F. It is actually an x-ray flux of a plasma with unknown temperature and density distributions (which is what would set F).
$endgroup$
– JJR4
Dec 8 '18 at 17:01










1 Answer
1






active

oldest

votes


















1












$begingroup$

What I would do is to find
$$A(r,t) = int_{t_1}^t F(r,t) dt$$
so that
$$A'_t(r,t) = F(r,t)$$
Hence,
$$A'_t(r,t_1)=F(r,t_1), A'_t(r,t_2)=F(r,t_2), A'_t(r,t_3)=F(r,t_3), dots$$
which means you should be able reconstruct numerically the function $F$ by finding $A$ as a function of $r$ and different time $t_i$ endpoints if you have enough data taken and are able to find a time gradient of $A$. We can also apply an analogous trick to $B$ and make sure the two reconstructions match.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Although this is a good suggestion unfortunately the measurement is such that it is not a tenable option.
    $endgroup$
    – JJR4
    Dec 10 '18 at 13:39











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

What I would do is to find
$$A(r,t) = int_{t_1}^t F(r,t) dt$$
so that
$$A'_t(r,t) = F(r,t)$$
Hence,
$$A'_t(r,t_1)=F(r,t_1), A'_t(r,t_2)=F(r,t_2), A'_t(r,t_3)=F(r,t_3), dots$$
which means you should be able reconstruct numerically the function $F$ by finding $A$ as a function of $r$ and different time $t_i$ endpoints if you have enough data taken and are able to find a time gradient of $A$. We can also apply an analogous trick to $B$ and make sure the two reconstructions match.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Although this is a good suggestion unfortunately the measurement is such that it is not a tenable option.
    $endgroup$
    – JJR4
    Dec 10 '18 at 13:39
















1












$begingroup$

What I would do is to find
$$A(r,t) = int_{t_1}^t F(r,t) dt$$
so that
$$A'_t(r,t) = F(r,t)$$
Hence,
$$A'_t(r,t_1)=F(r,t_1), A'_t(r,t_2)=F(r,t_2), A'_t(r,t_3)=F(r,t_3), dots$$
which means you should be able reconstruct numerically the function $F$ by finding $A$ as a function of $r$ and different time $t_i$ endpoints if you have enough data taken and are able to find a time gradient of $A$. We can also apply an analogous trick to $B$ and make sure the two reconstructions match.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Although this is a good suggestion unfortunately the measurement is such that it is not a tenable option.
    $endgroup$
    – JJR4
    Dec 10 '18 at 13:39














1












1








1





$begingroup$

What I would do is to find
$$A(r,t) = int_{t_1}^t F(r,t) dt$$
so that
$$A'_t(r,t) = F(r,t)$$
Hence,
$$A'_t(r,t_1)=F(r,t_1), A'_t(r,t_2)=F(r,t_2), A'_t(r,t_3)=F(r,t_3), dots$$
which means you should be able reconstruct numerically the function $F$ by finding $A$ as a function of $r$ and different time $t_i$ endpoints if you have enough data taken and are able to find a time gradient of $A$. We can also apply an analogous trick to $B$ and make sure the two reconstructions match.






share|cite|improve this answer









$endgroup$



What I would do is to find
$$A(r,t) = int_{t_1}^t F(r,t) dt$$
so that
$$A'_t(r,t) = F(r,t)$$
Hence,
$$A'_t(r,t_1)=F(r,t_1), A'_t(r,t_2)=F(r,t_2), A'_t(r,t_3)=F(r,t_3), dots$$
which means you should be able reconstruct numerically the function $F$ by finding $A$ as a function of $r$ and different time $t_i$ endpoints if you have enough data taken and are able to find a time gradient of $A$. We can also apply an analogous trick to $B$ and make sure the two reconstructions match.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 '18 at 17:09









Karn WatcharasupatKarn Watcharasupat

3,9642526




3,9642526












  • $begingroup$
    Although this is a good suggestion unfortunately the measurement is such that it is not a tenable option.
    $endgroup$
    – JJR4
    Dec 10 '18 at 13:39


















  • $begingroup$
    Although this is a good suggestion unfortunately the measurement is such that it is not a tenable option.
    $endgroup$
    – JJR4
    Dec 10 '18 at 13:39
















$begingroup$
Although this is a good suggestion unfortunately the measurement is such that it is not a tenable option.
$endgroup$
– JJR4
Dec 10 '18 at 13:39




$begingroup$
Although this is a good suggestion unfortunately the measurement is such that it is not a tenable option.
$endgroup$
– JJR4
Dec 10 '18 at 13:39


















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