Mixing
Asymptotic behaviour of the solution of a PDE
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I have the following PDE:
beginalign
fracddtu(t,r)&=alpha(r-beta)fracddru(t,r)+alpha u(t,r)\
u(0,r)=&u_0(r)
endalign
where $rin [0,1]$ and $alpha$ and $beta$ are fixed constants in $[0,1]$.
I would like to prove that $lim_tto +inftyu(t, r)equiv beta$.
What I know about the initial condition is that $u_0(0)=u_0(1)=0$, $int_0^1u_0(r)dr=1$ and that $int_0^1 u_0(r)rdr=beta$.
The equation can be solved trough the characteristic method and the solution is given by
$u(t,r)=u_0((r-beta)e^alpha t)+beta)e^alpha t$.
How to prove that $lim_tto +inftyu(t,r)=beta, forall rin (0,1)$? Could someone help me?
real-analysis analysis limits pde asymptotics
asked 2 days ago
user495333
616
 |Â
show 1 more comment
up vote
1
down vote
favorite
I have the following PDE:
beginalign
fracddtu(t,r)&=alpha(r-beta)fracddru(t,r)+alpha u(t,r)\
u(0,r)=&u_0(r)
endalign
where $rin [0,1]$ and $alpha$ and $beta$ are fixed constants in $[0,1]$.
I would like to prove that $lim_tto +inftyu(t, r)equiv beta$.
What I know about the initial condition is that $u_0(0)=u_0(1)=0$, $int_0^1u_0(r)dr=1$ and that $int_0^1 u_0(r)rdr=beta$.
The equation can be solved trough the characteristic method and the solution is given by
$u(t,r)=u_0((r-beta)e^alpha t)+beta)e^alpha t$.
How to prove that $lim_tto +inftyu(t,r)=beta, forall rin (0,1)$? Could someone help me?
real-analysis analysis limits pde asymptotics
asked 2 days ago
user495333
616
Are you sure it's $int u_0(r)rdr$ not $int_0^1u_0(r)rdr$?
– TheSimpliFire
2 days ago
You are right, it was a misprint, I modified the text. Thank you!
– user495333
2 days ago
Why would you expect that limit? After all, $uequivbeta$ does not solve the equation (unless $alphabeta=0$). Also, a look at the characteristics reveals that your solution formula needs data from the entire real line, which is outside the domain of the equation. Accordingly, you really need boundary conditions at $r=0$ and $r=1$.
– Harald Hanche-Olsen
2 days ago
I have the conditions for $r=0$ and $r=1$, I just added to the text. Thank you!
– user495333
2 days ago
I expect that limit because of the nature of my problem and also because my professor said that.
– user495333
2 days ago
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have the following PDE:
beginalign
fracddtu(t,r)&=alpha(r-beta)fracddru(t,r)+alpha u(t,r)\
u(0,r)=&u_0(r)
endalign
where $rin [0,1]$ and $alpha$ and $beta$ are fixed constants in $[0,1]$.
I would like to prove that $lim_tto +inftyu(t, r)equiv beta$.
What I know about the initial condition is that $u_0(0)=u_0(1)=0$, $int_0^1u_0(r)dr=1$ and that $int_0^1 u_0(r)rdr=beta$.
The equation can be solved trough the characteristic method and the solution is given by
$u(t,r)=u_0((r-beta)e^alpha t)+beta)e^alpha t$.
How to prove that $lim_tto +inftyu(t,r)=beta, forall rin (0,1)$? Could someone help me?
real-analysis analysis limits pde asymptotics
asked 2 days ago
user495333
616
I have the following PDE:
beginalign
fracddtu(t,r)&=alpha(r-beta)fracddru(t,r)+alpha u(t,r)\
u(0,r)=&u_0(r)
endalign
where $rin [0,1]$ and $alpha$ and $beta$ are fixed constants in $[0,1]$.
I would like to prove that $lim_tto +inftyu(t, r)equiv beta$.
What I know about the initial condition is that $u_0(0)=u_0(1)=0$, $int_0^1u_0(r)dr=1$ and that $int_0^1 u_0(r)rdr=beta$.
The equation can be solved trough the characteristic method and the solution is given by
$u(t,r)=u_0((r-beta)e^alpha t)+beta)e^alpha t$.
How to prove that $lim_tto +inftyu(t,r)=beta, forall rin (0,1)$? Could someone help me?
real-analysis analysis limits pde asymptotics
asked 2 days ago
user495333
616
edited 2 days ago
asked 2 days ago
user495333
616
asked 2 days ago
user495333
616
asked 2 days ago
user495333
616
616
Are you sure it's $int u_0(r)rdr$ not $int_0^1u_0(r)rdr$?
– TheSimpliFire
2 days ago
You are right, it was a misprint, I modified the text. Thank you!
– user495333
2 days ago
Why would you expect that limit? After all, $uequivbeta$ does not solve the equation (unless $alphabeta=0$). Also, a look at the characteristics reveals that your solution formula needs data from the entire real line, which is outside the domain of the equation. Accordingly, you really need boundary conditions at $r=0$ and $r=1$.
– Harald Hanche-Olsen
2 days ago
I have the conditions for $r=0$ and $r=1$, I just added to the text. Thank you!
– user495333
2 days ago
I expect that limit because of the nature of my problem and also because my professor said that.
– user495333
2 days ago
 |Â
show 1 more comment
Are you sure it's $int u_0(r)rdr$ not $int_0^1u_0(r)rdr$?
– TheSimpliFire
2 days ago
You are right, it was a misprint, I modified the text. Thank you!
– user495333
2 days ago
Why would you expect that limit? After all, $uequivbeta$ does not solve the equation (unless $alphabeta=0$). Also, a look at the characteristics reveals that your solution formula needs data from the entire real line, which is outside the domain of the equation. Accordingly, you really need boundary conditions at $r=0$ and $r=1$.
– Harald Hanche-Olsen
2 days ago
I have the conditions for $r=0$ and $r=1$, I just added to the text. Thank you!
– user495333
2 days ago
I expect that limit because of the nature of my problem and also because my professor said that.
– user495333
2 days ago
Are you sure it's $int u_0(r)rdr$ not $int_0^1u_0(r)rdr$?
– TheSimpliFire
2 days ago
Are you sure it's $int u_0(r)rdr$ not $int_0^1u_0(r)rdr$?
– TheSimpliFire
2 days ago
You are right, it was a misprint, I modified the text. Thank you!
– user495333
2 days ago
You are right, it was a misprint, I modified the text. Thank you!
– user495333
2 days ago
Why would you expect that limit? After all, $uequivbeta$ does not solve the equation (unless $alphabeta=0$). Also, a look at the characteristics reveals that your solution formula needs data from the entire real line, which is outside the domain of the equation. Accordingly, you really need boundary conditions at $r=0$ and $r=1$.
– Harald Hanche-Olsen
2 days ago
Why would you expect that limit? After all, $uequivbeta$ does not solve the equation (unless $alphabeta=0$). Also, a look at the characteristics reveals that your solution formula needs data from the entire real line, which is outside the domain of the equation. Accordingly, you really need boundary conditions at $r=0$ and $r=1$.
– Harald Hanche-Olsen
2 days ago
I have the conditions for $r=0$ and $r=1$, I just added to the text. Thank you!
– user495333
2 days ago
I have the conditions for $r=0$ and $r=1$, I just added to the text. Thank you!
– user495333
2 days ago
I expect that limit because of the nature of my problem and also because my professor said that.
– user495333
2 days ago
I expect that limit because of the nature of my problem and also because my professor said that.
– user495333
2 days ago
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
The solution by characteristics that you give, is only valid when $(r-beta)e^alpha t+betain[0,1]$, that is, for
$$ beta-beta e^-alpha t le r le beta+(1-beta) e^-alpha t, $$
an interval of width $e^-alpha t$. Outside that interval, the characteristic will hit one of the boundary conditions at $r=0$ or $r=1$, and so $u(t,r)=0$ at those points.
You should find that $int_0^1 u(t,r) ,dr = int_0^1 u_0(r),dr=beta$ (a constant).
Conclusion: The asymptotic limit of the solution as $ttoinfty$ is $betadelta_beta$, where $delta_beta$ is a delta function located at $beta$ (sometimes written $delta_beta(r)=delta(r-beta)$).
Edited to add:
This analysis assumes $0<beta<1$ and $alpha>0$. If $alpha=0$, the solution is of course independent of $t$. And if $beta=0$ or $beta=1$, the analysis needs to be changed a bit, but much the same will still hold
Edit the second:
Here is a picture of the solution. I chose $u_0(r)=4(r-r^2)$, $beta=2/3$, and $alpha>0$. The red graph is the initial condition $u_0$,
blue is the solution for $e^alpha t=2$, and the tallest one (khaki) is for $e^alpha t=4$. The area under each curve is the same in each case.
Also, I removed my totally misguided “consistency check†from the answer. Sorry about that; not enough caffeine, I suppose.
answered 2 days ago
Harald Hanche-Olsen
27.3k23959
Sorry, you mean that $lim_tto+inftyu(t,r)=1$ if $r=beta$ and $0$ otherwise?
– user495333
2 days ago
No, not all! By your own solution formula, $u(t,beta)=u_0(beta)e^alpha t$, which goes to infinity if $u_0(beta)>0$. For any $t>0$, the graph of $u(t,r)$ for $rin[0,1]$ looks like a copy of the graph of $u_0$, squeezed in the horizontal direction down to size $e^-alpha t$ but expanded in the vertical direction by the factor $e^alpha t$, leaving the integral unchanged – joined by zero values on the left and right. (Assuming $betain(0,1)$ and $alpha>0$, which I forgot to mention.)
– Harald Hanche-Olsen
2 days ago
Ok! Thank you very much for your time!, Now I need to think a bit about your answer.
– user495333
2 days ago
I have a question... Since $int_0^1u_0(r)dr=1$ I should have that $int_0^1u(t,r)dr=1$, but if I write the expression of $u(t, r)$ in terms of $u_0$ and I change variable I get that $int_0^1u(t,r)dr=int_beta-beta e^alpha t^beta-beta e^alpha t+1u_0(r')dr'$ that in general is not $1$ unless $u_0$ is periodic... Is that correct?
– user495333
yesterday
Don't forget that $u(t,r)$ is zero outside a small interval (see the first formula in the answer). So you should integrate only over that interval before you apply the change of variables formula.
– Harald Hanche-Olsen
yesterday
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The solution by characteristics that you give, is only valid when $(r-beta)e^alpha t+betain[0,1]$, that is, for
$$ beta-beta e^-alpha t le r le beta+(1-beta) e^-alpha t, $$
an interval of width $e^-alpha t$. Outside that interval, the characteristic will hit one of the boundary conditions at $r=0$ or $r=1$, and so $u(t,r)=0$ at those points.
You should find that $int_0^1 u(t,r) ,dr = int_0^1 u_0(r),dr=beta$ (a constant).
Conclusion: The asymptotic limit of the solution as $ttoinfty$ is $betadelta_beta$, where $delta_beta$ is a delta function located at $beta$ (sometimes written $delta_beta(r)=delta(r-beta)$).
Edited to add:
This analysis assumes $0<beta<1$ and $alpha>0$. If $alpha=0$, the solution is of course independent of $t$. And if $beta=0$ or $beta=1$, the analysis needs to be changed a bit, but much the same will still hold
Edit the second:
Here is a picture of the solution. I chose $u_0(r)=4(r-r^2)$, $beta=2/3$, and $alpha>0$. The red graph is the initial condition $u_0$,
blue is the solution for $e^alpha t=2$, and the tallest one (khaki) is for $e^alpha t=4$. The area under each curve is the same in each case.
Also, I removed my totally misguided “consistency check†from the answer. Sorry about that; not enough caffeine, I suppose.
answered 2 days ago
Harald Hanche-Olsen
27.3k23959
Sorry, you mean that $lim_tto+inftyu(t,r)=1$ if $r=beta$ and $0$ otherwise?
– user495333
2 days ago
No, not all! By your own solution formula, $u(t,beta)=u_0(beta)e^alpha t$, which goes to infinity if $u_0(beta)>0$. For any $t>0$, the graph of $u(t,r)$ for $rin[0,1]$ looks like a copy of the graph of $u_0$, squeezed in the horizontal direction down to size $e^-alpha t$ but expanded in the vertical direction by the factor $e^alpha t$, leaving the integral unchanged – joined by zero values on the left and right. (Assuming $betain(0,1)$ and $alpha>0$, which I forgot to mention.)
– Harald Hanche-Olsen
2 days ago
Ok! Thank you very much for your time!, Now I need to think a bit about your answer.
– user495333
2 days ago
I have a question... Since $int_0^1u_0(r)dr=1$ I should have that $int_0^1u(t,r)dr=1$, but if I write the expression of $u(t, r)$ in terms of $u_0$ and I change variable I get that $int_0^1u(t,r)dr=int_beta-beta e^alpha t^beta-beta e^alpha t+1u_0(r')dr'$ that in general is not $1$ unless $u_0$ is periodic... Is that correct?
– user495333
yesterday
Don't forget that $u(t,r)$ is zero outside a small interval (see the first formula in the answer). So you should integrate only over that interval before you apply the change of variables formula.
– Harald Hanche-Olsen
yesterday
 |Â
show 2 more comments
up vote
1
down vote
The solution by characteristics that you give, is only valid when $(r-beta)e^alpha t+betain[0,1]$, that is, for
$$ beta-beta e^-alpha t le r le beta+(1-beta) e^-alpha t, $$
an interval of width $e^-alpha t$. Outside that interval, the characteristic will hit one of the boundary conditions at $r=0$ or $r=1$, and so $u(t,r)=0$ at those points.
You should find that $int_0^1 u(t,r) ,dr = int_0^1 u_0(r),dr=beta$ (a constant).
Conclusion: The asymptotic limit of the solution as $ttoinfty$ is $betadelta_beta$, where $delta_beta$ is a delta function located at $beta$ (sometimes written $delta_beta(r)=delta(r-beta)$).
Edited to add:
This analysis assumes $0<beta<1$ and $alpha>0$. If $alpha=0$, the solution is of course independent of $t$. And if $beta=0$ or $beta=1$, the analysis needs to be changed a bit, but much the same will still hold
Edit the second:
Here is a picture of the solution. I chose $u_0(r)=4(r-r^2)$, $beta=2/3$, and $alpha>0$. The red graph is the initial condition $u_0$,
blue is the solution for $e^alpha t=2$, and the tallest one (khaki) is for $e^alpha t=4$. The area under each curve is the same in each case.
Also, I removed my totally misguided “consistency check†from the answer. Sorry about that; not enough caffeine, I suppose.
answered 2 days ago
Harald Hanche-Olsen
27.3k23959
Sorry, you mean that $lim_tto+inftyu(t,r)=1$ if $r=beta$ and $0$ otherwise?
– user495333
2 days ago
No, not all! By your own solution formula, $u(t,beta)=u_0(beta)e^alpha t$, which goes to infinity if $u_0(beta)>0$. For any $t>0$, the graph of $u(t,r)$ for $rin[0,1]$ looks like a copy of the graph of $u_0$, squeezed in the horizontal direction down to size $e^-alpha t$ but expanded in the vertical direction by the factor $e^alpha t$, leaving the integral unchanged – joined by zero values on the left and right. (Assuming $betain(0,1)$ and $alpha>0$, which I forgot to mention.)
– Harald Hanche-Olsen
2 days ago
Ok! Thank you very much for your time!, Now I need to think a bit about your answer.
– user495333
2 days ago
I have a question... Since $int_0^1u_0(r)dr=1$ I should have that $int_0^1u(t,r)dr=1$, but if I write the expression of $u(t, r)$ in terms of $u_0$ and I change variable I get that $int_0^1u(t,r)dr=int_beta-beta e^alpha t^beta-beta e^alpha t+1u_0(r')dr'$ that in general is not $1$ unless $u_0$ is periodic... Is that correct?
– user495333
yesterday
Don't forget that $u(t,r)$ is zero outside a small interval (see the first formula in the answer). So you should integrate only over that interval before you apply the change of variables formula.
– Harald Hanche-Olsen
yesterday
 |Â
show 2 more comments
up vote
1
down vote
up vote
1
down vote
The solution by characteristics that you give, is only valid when $(r-beta)e^alpha t+betain[0,1]$, that is, for
$$ beta-beta e^-alpha t le r le beta+(1-beta) e^-alpha t, $$
an interval of width $e^-alpha t$. Outside that interval, the characteristic will hit one of the boundary conditions at $r=0$ or $r=1$, and so $u(t,r)=0$ at those points.
You should find that $int_0^1 u(t,r) ,dr = int_0^1 u_0(r),dr=beta$ (a constant).
Conclusion: The asymptotic limit of the solution as $ttoinfty$ is $betadelta_beta$, where $delta_beta$ is a delta function located at $beta$ (sometimes written $delta_beta(r)=delta(r-beta)$).
Edited to add:
This analysis assumes $0<beta<1$ and $alpha>0$. If $alpha=0$, the solution is of course independent of $t$. And if $beta=0$ or $beta=1$, the analysis needs to be changed a bit, but much the same will still hold
Edit the second:
Here is a picture of the solution. I chose $u_0(r)=4(r-r^2)$, $beta=2/3$, and $alpha>0$. The red graph is the initial condition $u_0$,
blue is the solution for $e^alpha t=2$, and the tallest one (khaki) is for $e^alpha t=4$. The area under each curve is the same in each case.
Also, I removed my totally misguided “consistency check†from the answer. Sorry about that; not enough caffeine, I suppose.
answered 2 days ago
Harald Hanche-Olsen
27.3k23959
The solution by characteristics that you give, is only valid when $(r-beta)e^alpha t+betain[0,1]$, that is, for
$$ beta-beta e^-alpha t le r le beta+(1-beta) e^-alpha t, $$
an interval of width $e^-alpha t$. Outside that interval, the characteristic will hit one of the boundary conditions at $r=0$ or $r=1$, and so $u(t,r)=0$ at those points.
You should find that $int_0^1 u(t,r) ,dr = int_0^1 u_0(r),dr=beta$ (a constant).
Conclusion: The asymptotic limit of the solution as $ttoinfty$ is $betadelta_beta$, where $delta_beta$ is a delta function located at $beta$ (sometimes written $delta_beta(r)=delta(r-beta)$).
Edited to add:
This analysis assumes $0<beta<1$ and $alpha>0$. If $alpha=0$, the solution is of course independent of $t$. And if $beta=0$ or $beta=1$, the analysis needs to be changed a bit, but much the same will still hold
Edit the second:
Here is a picture of the solution. I chose $u_0(r)=4(r-r^2)$, $beta=2/3$, and $alpha>0$. The red graph is the initial condition $u_0$,
blue is the solution for $e^alpha t=2$, and the tallest one (khaki) is for $e^alpha t=4$. The area under each curve is the same in each case.
Also, I removed my totally misguided “consistency check†from the answer. Sorry about that; not enough caffeine, I suppose.
answered 2 days ago
Harald Hanche-Olsen
27.3k23959
edited 2 days ago
answered 2 days ago
Harald Hanche-Olsen
27.3k23959
answered 2 days ago
Harald Hanche-Olsen
27.3k23959
answered 2 days ago
Harald Hanche-Olsen
27.3k23959
27.3k23959
Sorry, you mean that $lim_tto+inftyu(t,r)=1$ if $r=beta$ and $0$ otherwise?
– user495333
2 days ago
No, not all! By your own solution formula, $u(t,beta)=u_0(beta)e^alpha t$, which goes to infinity if $u_0(beta)>0$. For any $t>0$, the graph of $u(t,r)$ for $rin[0,1]$ looks like a copy of the graph of $u_0$, squeezed in the horizontal direction down to size $e^-alpha t$ but expanded in the vertical direction by the factor $e^alpha t$, leaving the integral unchanged – joined by zero values on the left and right. (Assuming $betain(0,1)$ and $alpha>0$, which I forgot to mention.)
– Harald Hanche-Olsen
2 days ago
Ok! Thank you very much for your time!, Now I need to think a bit about your answer.
– user495333
2 days ago
I have a question... Since $int_0^1u_0(r)dr=1$ I should have that $int_0^1u(t,r)dr=1$, but if I write the expression of $u(t, r)$ in terms of $u_0$ and I change variable I get that $int_0^1u(t,r)dr=int_beta-beta e^alpha t^beta-beta e^alpha t+1u_0(r')dr'$ that in general is not $1$ unless $u_0$ is periodic... Is that correct?
– user495333
yesterday
Don't forget that $u(t,r)$ is zero outside a small interval (see the first formula in the answer). So you should integrate only over that interval before you apply the change of variables formula.
– Harald Hanche-Olsen
yesterday
 |Â
show 2 more comments
Sorry, you mean that $lim_tto+inftyu(t,r)=1$ if $r=beta$ and $0$ otherwise?
– user495333
2 days ago
No, not all! By your own solution formula, $u(t,beta)=u_0(beta)e^alpha t$, which goes to infinity if $u_0(beta)>0$. For any $t>0$, the graph of $u(t,r)$ for $rin[0,1]$ looks like a copy of the graph of $u_0$, squeezed in the horizontal direction down to size $e^-alpha t$ but expanded in the vertical direction by the factor $e^alpha t$, leaving the integral unchanged – joined by zero values on the left and right. (Assuming $betain(0,1)$ and $alpha>0$, which I forgot to mention.)
– Harald Hanche-Olsen
2 days ago
Ok! Thank you very much for your time!, Now I need to think a bit about your answer.
– user495333
2 days ago
I have a question... Since $int_0^1u_0(r)dr=1$ I should have that $int_0^1u(t,r)dr=1$, but if I write the expression of $u(t, r)$ in terms of $u_0$ and I change variable I get that $int_0^1u(t,r)dr=int_beta-beta e^alpha t^beta-beta e^alpha t+1u_0(r')dr'$ that in general is not $1$ unless $u_0$ is periodic... Is that correct?
– user495333
yesterday
Don't forget that $u(t,r)$ is zero outside a small interval (see the first formula in the answer). So you should integrate only over that interval before you apply the change of variables formula.
– Harald Hanche-Olsen
yesterday
Sorry, you mean that $lim_tto+inftyu(t,r)=1$ if $r=beta$ and $0$ otherwise?
– user495333
2 days ago
Sorry, you mean that $lim_tto+inftyu(t,r)=1$ if $r=beta$ and $0$ otherwise?
– user495333
2 days ago
No, not all! By your own solution formula, $u(t,beta)=u_0(beta)e^alpha t$, which goes to infinity if $u_0(beta)>0$. For any $t>0$, the graph of $u(t,r)$ for $rin[0,1]$ looks like a copy of the graph of $u_0$, squeezed in the horizontal direction down to size $e^-alpha t$ but expanded in the vertical direction by the factor $e^alpha t$, leaving the integral unchanged – joined by zero values on the left and right. (Assuming $betain(0,1)$ and $alpha>0$, which I forgot to mention.)
– Harald Hanche-Olsen
2 days ago
No, not all! By your own solution formula, $u(t,beta)=u_0(beta)e^alpha t$, which goes to infinity if $u_0(beta)>0$. For any $t>0$, the graph of $u(t,r)$ for $rin[0,1]$ looks like a copy of the graph of $u_0$, squeezed in the horizontal direction down to size $e^-alpha t$ but expanded in the vertical direction by the factor $e^alpha t$, leaving the integral unchanged – joined by zero values on the left and right. (Assuming $betain(0,1)$ and $alpha>0$, which I forgot to mention.)
– Harald Hanche-Olsen
2 days ago
Ok! Thank you very much for your time!, Now I need to think a bit about your answer.
– user495333
2 days ago
Ok! Thank you very much for your time!, Now I need to think a bit about your answer.
– user495333
2 days ago
I have a question... Since $int_0^1u_0(r)dr=1$ I should have that $int_0^1u(t,r)dr=1$, but if I write the expression of $u(t, r)$ in terms of $u_0$ and I change variable I get that $int_0^1u(t,r)dr=int_beta-beta e^alpha t^beta-beta e^alpha t+1u_0(r')dr'$ that in general is not $1$ unless $u_0$ is periodic... Is that correct?
– user495333
yesterday
I have a question... Since $int_0^1u_0(r)dr=1$ I should have that $int_0^1u(t,r)dr=1$, but if I write the expression of $u(t, r)$ in terms of $u_0$ and I change variable I get that $int_0^1u(t,r)dr=int_beta-beta e^alpha t^beta-beta e^alpha t+1u_0(r')dr'$ that in general is not $1$ unless $u_0$ is periodic... Is that correct?
– user495333
yesterday
Don't forget that $u(t,r)$ is zero outside a small interval (see the first formula in the answer). So you should integrate only over that interval before you apply the change of variables formula.
– Harald Hanche-Olsen
yesterday
Don't forget that $u(t,r)$ is zero outside a small interval (see the first formula in the answer). So you should integrate only over that interval before you apply the change of variables formula.
– Harald Hanche-Olsen
yesterday
 |Â
show 2 more comments
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871824%2fasymptotic-behaviour-of-the-solution-of-a-pde%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Are you sure it's $int u_0(r)rdr$ not $int_0^1u_0(r)rdr$?
– TheSimpliFire
2 days ago
You are right, it was a misprint, I modified the text. Thank you!
– user495333
2 days ago
Why would you expect that limit? After all, $uequivbeta$ does not solve the equation (unless $alphabeta=0$). Also, a look at the characteristics reveals that your solution formula needs data from the entire real line, which is outside the domain of the equation. Accordingly, you really need boundary conditions at $r=0$ and $r=1$.
– Harald Hanche-Olsen
2 days ago
I have the conditions for $r=0$ and $r=1$, I just added to the text. Thank you!
– user495333
2 days ago
I expect that limit because of the nature of my problem and also because my professor said that.
– user495333
2 days ago