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characteristic method for a pde
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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 solve this equation with the characteristic's method.
If my computations are correct the equations of the characeristics are given by
beginalign
&fracdtds=-1,\
&fracdrds=alpha(r-beta)
endalign
while the equation along the characteristics is
beginalign
fracduds=-alpha u
endalign
and consequently the total solution is given by $u(t,r)=u_0((r-beta)e^alpha t)+beta)e^alpha t$.
My problem is that, under the hypotesis $u_0(r)in[0,1]$ I should have $u(t,r)in[0,1]$ for every $tgeq 0$, but according to the solution that I get, it doesen't seem to be in this way. There is something wrong in my argument?
Could someone help me?
real-analysis analysis pde characteristics
asked Jul 25 at 16:23
user495333
766
add a comment |Â
up vote
0
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 solve this equation with the characteristic's method.
If my computations are correct the equations of the characeristics are given by
beginalign
&fracdtds=-1,\
&fracdrds=alpha(r-beta)
endalign
while the equation along the characteristics is
beginalign
fracduds=-alpha u
endalign
and consequently the total solution is given by $u(t,r)=u_0((r-beta)e^alpha t)+beta)e^alpha t$.
My problem is that, under the hypotesis $u_0(r)in[0,1]$ I should have $u(t,r)in[0,1]$ for every $tgeq 0$, but according to the solution that I get, it doesen't seem to be in this way. There is something wrong in my argument?
Could someone help me?
real-analysis analysis pde characteristics
asked Jul 25 at 16:23
user495333
766
You are totally right, I modified the text! Anyway the solution is $u(s)=u_0e^-alpha s$ but $s=-t$ and this means that the sign in the exponential is positive... Am I wrong?
– user495333
Jul 25 at 17:14
Your result $u(t,r)=u_0(r-beta(1-e^-alpha t))e^alpha t$ is not correct. But you have not edited the intermediate steps. So, it is impossible to say where exactly is the mistake. May be, comparing to my answer will allow you so find what is wrong in your argument. Anyways, it is not recommended to simply write "Consequently the total solution is given by … '' : the missing steps are not so simple.
– JJacquelin
Jul 26 at 6:52
add a comment |Â
up vote
0
down vote
favorite
up vote
0
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 solve this equation with the characteristic's method.
If my computations are correct the equations of the characeristics are given by
beginalign
&fracdtds=-1,\
&fracdrds=alpha(r-beta)
endalign
while the equation along the characteristics is
beginalign
fracduds=-alpha u
endalign
and consequently the total solution is given by $u(t,r)=u_0((r-beta)e^alpha t)+beta)e^alpha t$.
My problem is that, under the hypotesis $u_0(r)in[0,1]$ I should have $u(t,r)in[0,1]$ for every $tgeq 0$, but according to the solution that I get, it doesen't seem to be in this way. There is something wrong in my argument?
Could someone help me?
real-analysis analysis pde characteristics
asked Jul 25 at 16:23
user495333
766
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 solve this equation with the characteristic's method.
If my computations are correct the equations of the characeristics are given by
beginalign
&fracdtds=-1,\
&fracdrds=alpha(r-beta)
endalign
while the equation along the characteristics is
beginalign
fracduds=-alpha u
endalign
and consequently the total solution is given by $u(t,r)=u_0((r-beta)e^alpha t)+beta)e^alpha t$.
My problem is that, under the hypotesis $u_0(r)in[0,1]$ I should have $u(t,r)in[0,1]$ for every $tgeq 0$, but according to the solution that I get, it doesen't seem to be in this way. There is something wrong in my argument?
Could someone help me?
real-analysis analysis pde characteristics
asked Jul 25 at 16:23
user495333
766
edited Jul 26 at 10:16
asked Jul 25 at 16:23
user495333
766
asked Jul 25 at 16:23
user495333
766
asked Jul 25 at 16:23
user495333
766
766
You are totally right, I modified the text! Anyway the solution is $u(s)=u_0e^-alpha s$ but $s=-t$ and this means that the sign in the exponential is positive... Am I wrong?
– user495333
Jul 25 at 17:14
Your result $u(t,r)=u_0(r-beta(1-e^-alpha t))e^alpha t$ is not correct. But you have not edited the intermediate steps. So, it is impossible to say where exactly is the mistake. May be, comparing to my answer will allow you so find what is wrong in your argument. Anyways, it is not recommended to simply write "Consequently the total solution is given by … '' : the missing steps are not so simple.
– JJacquelin
Jul 26 at 6:52
add a comment |Â
You are totally right, I modified the text! Anyway the solution is $u(s)=u_0e^-alpha s$ but $s=-t$ and this means that the sign in the exponential is positive... Am I wrong?
– user495333
Jul 25 at 17:14
Your result $u(t,r)=u_0(r-beta(1-e^-alpha t))e^alpha t$ is not correct. But you have not edited the intermediate steps. So, it is impossible to say where exactly is the mistake. May be, comparing to my answer will allow you so find what is wrong in your argument. Anyways, it is not recommended to simply write "Consequently the total solution is given by … '' : the missing steps are not so simple.
– JJacquelin
Jul 26 at 6:52
You are totally right, I modified the text! Anyway the solution is $u(s)=u_0e^-alpha s$ but $s=-t$ and this means that the sign in the exponential is positive... Am I wrong?
– user495333
Jul 25 at 17:14
You are totally right, I modified the text! Anyway the solution is $u(s)=u_0e^-alpha s$ but $s=-t$ and this means that the sign in the exponential is positive... Am I wrong?
– user495333
Jul 25 at 17:14
Your result $u(t,r)=u_0(r-beta(1-e^-alpha t))e^alpha t$ is not correct. But you have not edited the intermediate steps. So, it is impossible to say where exactly is the mistake. May be, comparing to my answer will allow you so find what is wrong in your argument. Anyways, it is not recommended to simply write "Consequently the total solution is given by … '' : the missing steps are not so simple.
– JJacquelin
Jul 26 at 6:52
Your result $u(t,r)=u_0(r-beta(1-e^-alpha t))e^alpha t$ is not correct. But you have not edited the intermediate steps. So, it is impossible to say where exactly is the mistake. May be, comparing to my answer will allow you so find what is wrong in your argument. Anyways, it is not recommended to simply write "Consequently the total solution is given by … '' : the missing steps are not so simple.
– JJacquelin
Jul 26 at 6:52
add a comment |Â
1 Answer
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up vote
1
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$$fracddtu(t,r)-alpha(r-beta)fracddru(t,r)=alpha u(t,r)$$
$$fracdt1=fracdralpha(beta-r)=fracdualpha u=-ds$$
It doesn't mater the coefficient of the parameter $s$. I put $-1$ to be consistent with the meaning of $s$ in your notations. So, your characteristic ODEs are consistent with mine. Your mistake is later. Compare your further calculus with the following :
First characteristic curves, from $quad fracdt1=fracdualpha u$
$$u e^-alpha t=c_1$$
Second characteristic curves, from $quad fracdt1=fracdralpha(beta-r)$
$$(r-beta)e^alpha t=c_2$$
The general solution on implicit form is $quad u e^-alpha t=Fleft((r-beta)e^alpha t right)quad$ where $F$ is an arbitrary function.
The general solution on explicit form is :
$$u(r,t)=e^alpha tFleft((r-beta)e^alpha t right)$$
Condition :
$u(0,r)=u_0(r)=e^alpha 0Fleft((r-beta)e^alpha 0 right)=Fleft((r-beta) right)$
Let $x=r-beta$. $quad r=x+betaquad;quad u_0(r)=u_0(x+beta)=F(x).$
$$F(x)=u_0(x+beta)$$
Now, the function $F(x)$ is determined. We put it into the above general solution where $x=(r-beta)e^alpha tquad$ thus $quad F(x)=u_0(x+beta)=u_0left((r-beta)e^alpha t+beta right)$.
$$u(r,t)=e^alpha tu_0left((r-beta)e^alpha t+beta right)$$
answered Jul 25 at 17:50
JJacquelin
39.9k21649
Thank you very much for your answer. I checked my computation and I agree that the final solution is the one that you wrote : $u(r,t)=e^alpha tu_0((r-beta)e^alpha t+beta)$.
– user495333
Jul 26 at 9:43
I made a mistake in my computations. Anyway what bother me is the fact that $u(r,t)$ can be larger then $1$ even if the initial datum $u(r)in [0,1]$ and this shouldn't happen (because of the nature of my problem, the solution represent a quantity which is always in [0,1], so I have the doubt that t he initial PDE is wrong). Do you agree that if the solution is the one you found and $alpha$ is a positive parameter, the solution is exploding as $tto +infty$? Right?
– user495333
Jul 26 at 9:43
All depends on the function $u_0(r)$ which is not explicit in the wording of the question. For example, if $u_0(r)=exp(-r^2)$ the solution is far to exploding.
– JJacquelin
Jul 26 at 9:51
No, sorry I didn't modified the text yet.... I will do it now... But I totally agree with the solution that you have written...
– user495333
Jul 26 at 10:14
OK. but exploding or not depends on $u_0(r)$.
– JJacquelin
Jul 26 at 10:23
 |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
$$fracddtu(t,r)-alpha(r-beta)fracddru(t,r)=alpha u(t,r)$$
$$fracdt1=fracdralpha(beta-r)=fracdualpha u=-ds$$
It doesn't mater the coefficient of the parameter $s$. I put $-1$ to be consistent with the meaning of $s$ in your notations. So, your characteristic ODEs are consistent with mine. Your mistake is later. Compare your further calculus with the following :
First characteristic curves, from $quad fracdt1=fracdualpha u$
$$u e^-alpha t=c_1$$
Second characteristic curves, from $quad fracdt1=fracdralpha(beta-r)$
$$(r-beta)e^alpha t=c_2$$
The general solution on implicit form is $quad u e^-alpha t=Fleft((r-beta)e^alpha t right)quad$ where $F$ is an arbitrary function.
The general solution on explicit form is :
$$u(r,t)=e^alpha tFleft((r-beta)e^alpha t right)$$
Condition :
$u(0,r)=u_0(r)=e^alpha 0Fleft((r-beta)e^alpha 0 right)=Fleft((r-beta) right)$
Let $x=r-beta$. $quad r=x+betaquad;quad u_0(r)=u_0(x+beta)=F(x).$
$$F(x)=u_0(x+beta)$$
Now, the function $F(x)$ is determined. We put it into the above general solution where $x=(r-beta)e^alpha tquad$ thus $quad F(x)=u_0(x+beta)=u_0left((r-beta)e^alpha t+beta right)$.
$$u(r,t)=e^alpha tu_0left((r-beta)e^alpha t+beta right)$$
answered Jul 25 at 17:50
JJacquelin
39.9k21649
Thank you very much for your answer. I checked my computation and I agree that the final solution is the one that you wrote : $u(r,t)=e^alpha tu_0((r-beta)e^alpha t+beta)$.
– user495333
Jul 26 at 9:43
I made a mistake in my computations. Anyway what bother me is the fact that $u(r,t)$ can be larger then $1$ even if the initial datum $u(r)in [0,1]$ and this shouldn't happen (because of the nature of my problem, the solution represent a quantity which is always in [0,1], so I have the doubt that t he initial PDE is wrong). Do you agree that if the solution is the one you found and $alpha$ is a positive parameter, the solution is exploding as $tto +infty$? Right?
– user495333
Jul 26 at 9:43
All depends on the function $u_0(r)$ which is not explicit in the wording of the question. For example, if $u_0(r)=exp(-r^2)$ the solution is far to exploding.
– JJacquelin
Jul 26 at 9:51
No, sorry I didn't modified the text yet.... I will do it now... But I totally agree with the solution that you have written...
– user495333
Jul 26 at 10:14
OK. but exploding or not depends on $u_0(r)$.
– JJacquelin
Jul 26 at 10:23
 |Â
show 2 more comments
up vote
1
down vote
$$fracddtu(t,r)-alpha(r-beta)fracddru(t,r)=alpha u(t,r)$$
$$fracdt1=fracdralpha(beta-r)=fracdualpha u=-ds$$
It doesn't mater the coefficient of the parameter $s$. I put $-1$ to be consistent with the meaning of $s$ in your notations. So, your characteristic ODEs are consistent with mine. Your mistake is later. Compare your further calculus with the following :
First characteristic curves, from $quad fracdt1=fracdualpha u$
$$u e^-alpha t=c_1$$
Second characteristic curves, from $quad fracdt1=fracdralpha(beta-r)$
$$(r-beta)e^alpha t=c_2$$
The general solution on implicit form is $quad u e^-alpha t=Fleft((r-beta)e^alpha t right)quad$ where $F$ is an arbitrary function.
The general solution on explicit form is :
$$u(r,t)=e^alpha tFleft((r-beta)e^alpha t right)$$
Condition :
$u(0,r)=u_0(r)=e^alpha 0Fleft((r-beta)e^alpha 0 right)=Fleft((r-beta) right)$
Let $x=r-beta$. $quad r=x+betaquad;quad u_0(r)=u_0(x+beta)=F(x).$
$$F(x)=u_0(x+beta)$$
Now, the function $F(x)$ is determined. We put it into the above general solution where $x=(r-beta)e^alpha tquad$ thus $quad F(x)=u_0(x+beta)=u_0left((r-beta)e^alpha t+beta right)$.
$$u(r,t)=e^alpha tu_0left((r-beta)e^alpha t+beta right)$$
answered Jul 25 at 17:50
JJacquelin
39.9k21649
Thank you very much for your answer. I checked my computation and I agree that the final solution is the one that you wrote : $u(r,t)=e^alpha tu_0((r-beta)e^alpha t+beta)$.
– user495333
Jul 26 at 9:43
I made a mistake in my computations. Anyway what bother me is the fact that $u(r,t)$ can be larger then $1$ even if the initial datum $u(r)in [0,1]$ and this shouldn't happen (because of the nature of my problem, the solution represent a quantity which is always in [0,1], so I have the doubt that t he initial PDE is wrong). Do you agree that if the solution is the one you found and $alpha$ is a positive parameter, the solution is exploding as $tto +infty$? Right?
– user495333
Jul 26 at 9:43
All depends on the function $u_0(r)$ which is not explicit in the wording of the question. For example, if $u_0(r)=exp(-r^2)$ the solution is far to exploding.
– JJacquelin
Jul 26 at 9:51
No, sorry I didn't modified the text yet.... I will do it now... But I totally agree with the solution that you have written...
– user495333
Jul 26 at 10:14
OK. but exploding or not depends on $u_0(r)$.
– JJacquelin
Jul 26 at 10:23
 |Â
show 2 more comments
up vote
1
down vote
up vote
1
down vote
$$fracddtu(t,r)-alpha(r-beta)fracddru(t,r)=alpha u(t,r)$$
$$fracdt1=fracdralpha(beta-r)=fracdualpha u=-ds$$
It doesn't mater the coefficient of the parameter $s$. I put $-1$ to be consistent with the meaning of $s$ in your notations. So, your characteristic ODEs are consistent with mine. Your mistake is later. Compare your further calculus with the following :
First characteristic curves, from $quad fracdt1=fracdualpha u$
$$u e^-alpha t=c_1$$
Second characteristic curves, from $quad fracdt1=fracdralpha(beta-r)$
$$(r-beta)e^alpha t=c_2$$
The general solution on implicit form is $quad u e^-alpha t=Fleft((r-beta)e^alpha t right)quad$ where $F$ is an arbitrary function.
The general solution on explicit form is :
$$u(r,t)=e^alpha tFleft((r-beta)e^alpha t right)$$
Condition :
$u(0,r)=u_0(r)=e^alpha 0Fleft((r-beta)e^alpha 0 right)=Fleft((r-beta) right)$
Let $x=r-beta$. $quad r=x+betaquad;quad u_0(r)=u_0(x+beta)=F(x).$
$$F(x)=u_0(x+beta)$$
Now, the function $F(x)$ is determined. We put it into the above general solution where $x=(r-beta)e^alpha tquad$ thus $quad F(x)=u_0(x+beta)=u_0left((r-beta)e^alpha t+beta right)$.
$$u(r,t)=e^alpha tu_0left((r-beta)e^alpha t+beta right)$$
answered Jul 25 at 17:50
JJacquelin
39.9k21649
$$fracddtu(t,r)-alpha(r-beta)fracddru(t,r)=alpha u(t,r)$$
$$fracdt1=fracdralpha(beta-r)=fracdualpha u=-ds$$
It doesn't mater the coefficient of the parameter $s$. I put $-1$ to be consistent with the meaning of $s$ in your notations. So, your characteristic ODEs are consistent with mine. Your mistake is later. Compare your further calculus with the following :
First characteristic curves, from $quad fracdt1=fracdualpha u$
$$u e^-alpha t=c_1$$
Second characteristic curves, from $quad fracdt1=fracdralpha(beta-r)$
$$(r-beta)e^alpha t=c_2$$
The general solution on implicit form is $quad u e^-alpha t=Fleft((r-beta)e^alpha t right)quad$ where $F$ is an arbitrary function.
The general solution on explicit form is :
$$u(r,t)=e^alpha tFleft((r-beta)e^alpha t right)$$
Condition :
$u(0,r)=u_0(r)=e^alpha 0Fleft((r-beta)e^alpha 0 right)=Fleft((r-beta) right)$
Let $x=r-beta$. $quad r=x+betaquad;quad u_0(r)=u_0(x+beta)=F(x).$
$$F(x)=u_0(x+beta)$$
Now, the function $F(x)$ is determined. We put it into the above general solution where $x=(r-beta)e^alpha tquad$ thus $quad F(x)=u_0(x+beta)=u_0left((r-beta)e^alpha t+beta right)$.
$$u(r,t)=e^alpha tu_0left((r-beta)e^alpha t+beta right)$$
answered Jul 25 at 17:50
JJacquelin
39.9k21649
edited Jul 26 at 7:23
answered Jul 25 at 17:50
JJacquelin
39.9k21649
answered Jul 25 at 17:50
JJacquelin
39.9k21649
answered Jul 25 at 17:50
JJacquelin
39.9k21649
39.9k21649
Thank you very much for your answer. I checked my computation and I agree that the final solution is the one that you wrote : $u(r,t)=e^alpha tu_0((r-beta)e^alpha t+beta)$.
– user495333
Jul 26 at 9:43
I made a mistake in my computations. Anyway what bother me is the fact that $u(r,t)$ can be larger then $1$ even if the initial datum $u(r)in [0,1]$ and this shouldn't happen (because of the nature of my problem, the solution represent a quantity which is always in [0,1], so I have the doubt that t he initial PDE is wrong). Do you agree that if the solution is the one you found and $alpha$ is a positive parameter, the solution is exploding as $tto +infty$? Right?
– user495333
Jul 26 at 9:43
All depends on the function $u_0(r)$ which is not explicit in the wording of the question. For example, if $u_0(r)=exp(-r^2)$ the solution is far to exploding.
– JJacquelin
Jul 26 at 9:51
No, sorry I didn't modified the text yet.... I will do it now... But I totally agree with the solution that you have written...
– user495333
Jul 26 at 10:14
OK. but exploding or not depends on $u_0(r)$.
– JJacquelin
Jul 26 at 10:23
 |Â
show 2 more comments
Thank you very much for your answer. I checked my computation and I agree that the final solution is the one that you wrote : $u(r,t)=e^alpha tu_0((r-beta)e^alpha t+beta)$.
– user495333
Jul 26 at 9:43
I made a mistake in my computations. Anyway what bother me is the fact that $u(r,t)$ can be larger then $1$ even if the initial datum $u(r)in [0,1]$ and this shouldn't happen (because of the nature of my problem, the solution represent a quantity which is always in [0,1], so I have the doubt that t he initial PDE is wrong). Do you agree that if the solution is the one you found and $alpha$ is a positive parameter, the solution is exploding as $tto +infty$? Right?
– user495333
Jul 26 at 9:43
All depends on the function $u_0(r)$ which is not explicit in the wording of the question. For example, if $u_0(r)=exp(-r^2)$ the solution is far to exploding.
– JJacquelin
Jul 26 at 9:51
No, sorry I didn't modified the text yet.... I will do it now... But I totally agree with the solution that you have written...
– user495333
Jul 26 at 10:14
OK. but exploding or not depends on $u_0(r)$.
– JJacquelin
Jul 26 at 10:23
Thank you very much for your answer. I checked my computation and I agree that the final solution is the one that you wrote : $u(r,t)=e^alpha tu_0((r-beta)e^alpha t+beta)$.
– user495333
Jul 26 at 9:43
Thank you very much for your answer. I checked my computation and I agree that the final solution is the one that you wrote : $u(r,t)=e^alpha tu_0((r-beta)e^alpha t+beta)$.
– user495333
Jul 26 at 9:43
I made a mistake in my computations. Anyway what bother me is the fact that $u(r,t)$ can be larger then $1$ even if the initial datum $u(r)in [0,1]$ and this shouldn't happen (because of the nature of my problem, the solution represent a quantity which is always in [0,1], so I have the doubt that t he initial PDE is wrong). Do you agree that if the solution is the one you found and $alpha$ is a positive parameter, the solution is exploding as $tto +infty$? Right?
– user495333
Jul 26 at 9:43
I made a mistake in my computations. Anyway what bother me is the fact that $u(r,t)$ can be larger then $1$ even if the initial datum $u(r)in [0,1]$ and this shouldn't happen (because of the nature of my problem, the solution represent a quantity which is always in [0,1], so I have the doubt that t he initial PDE is wrong). Do you agree that if the solution is the one you found and $alpha$ is a positive parameter, the solution is exploding as $tto +infty$? Right?
– user495333
Jul 26 at 9:43
All depends on the function $u_0(r)$ which is not explicit in the wording of the question. For example, if $u_0(r)=exp(-r^2)$ the solution is far to exploding.
– JJacquelin
Jul 26 at 9:51
All depends on the function $u_0(r)$ which is not explicit in the wording of the question. For example, if $u_0(r)=exp(-r^2)$ the solution is far to exploding.
– JJacquelin
Jul 26 at 9:51
No, sorry I didn't modified the text yet.... I will do it now... But I totally agree with the solution that you have written...
– user495333
Jul 26 at 10:14
No, sorry I didn't modified the text yet.... I will do it now... But I totally agree with the solution that you have written...
– user495333
Jul 26 at 10:14
OK. but exploding or not depends on $u_0(r)$.
– JJacquelin
Jul 26 at 10:23
OK. but exploding or not depends on $u_0(r)$.
– JJacquelin
Jul 26 at 10:23
 |Â
show 2 more comments
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You are totally right, I modified the text! Anyway the solution is $u(s)=u_0e^-alpha s$ but $s=-t$ and this means that the sign in the exponential is positive... Am I wrong?
– user495333
Jul 25 at 17:14
Your result $u(t,r)=u_0(r-beta(1-e^-alpha t))e^alpha t$ is not correct. But you have not edited the intermediate steps. So, it is impossible to say where exactly is the mistake. May be, comparing to my answer will allow you so find what is wrong in your argument. Anyways, it is not recommended to simply write "Consequently the total solution is given by … '' : the missing steps are not so simple.
– JJacquelin
Jul 26 at 6:52