Mixing
Heat equation inequality
$begingroup$
Let $u in C^{1}(0,infty) times C^{1}[0,1]$ be a solution to heat equation (1) with inital boundaries (2),(3)
(1) $partial_tu(t,x)-partial_{xx}u(t,x)=0$, for all $(t,x) in[0,infty)times[0,1]$
(2) $u(t,0)=u(t,1)=0,$ for $t in[0,infty]$
(3) $u(0,x)=u_{0}(x),$ for $xin[0,1]$
Show that there exists constants $C_{1},C_{2}$ such that
$$
int_{0}^{1}u^2(t,x)dx leq C_{1}e^{-tC_2};text{ for all }tgeq0
$$
Thoughts
I want to get this into a form where I can apply Wirtinger's inequality for functions: I tried by multiplying the through by $u(t,x)$ and intergrating but I couldn't get it out that way, so I'm not sure how to solve this.
functional-analysis pde heat-equation
edited Jan 20 at 20:57
Daniele Tampieri
2,3372922
asked Jan 20 at 20:31
RogerRoger
847
$endgroup$
add a comment |
$begingroup$
Let $u in C^{1}(0,infty) times C^{1}[0,1]$ be a solution to heat equation (1) with inital boundaries (2),(3)
(1) $partial_tu(t,x)-partial_{xx}u(t,x)=0$, for all $(t,x) in[0,infty)times[0,1]$
(2) $u(t,0)=u(t,1)=0,$ for $t in[0,infty]$
(3) $u(0,x)=u_{0}(x),$ for $xin[0,1]$
Show that there exists constants $C_{1},C_{2}$ such that
$$
int_{0}^{1}u^2(t,x)dx leq C_{1}e^{-tC_2};text{ for all }tgeq0
$$
Thoughts
I want to get this into a form where I can apply Wirtinger's inequality for functions: I tried by multiplying the through by $u(t,x)$ and intergrating but I couldn't get it out that way, so I'm not sure how to solve this.
functional-analysis pde heat-equation
edited Jan 20 at 20:57
Daniele Tampieri
2,3372922
asked Jan 20 at 20:31
RogerRoger
847
$endgroup$
$begingroup$
I'm having some problems deciphering you statement$u(t, x) in C^1(0, infty) times C^1[0, 1]$; for one thing, the elements of $C^1(0, infty) times C^1[0, 1]$ are ordered pairs of functions; but $u$ doesn't seem to exhibit that form. Can you clarify? Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 21:11
1
$begingroup$
Also, seems that the spatial derivative must be $C^{2}$.
$endgroup$
– alexp9
Jan 20 at 21:13
1
$begingroup$
@RobertLewis, you're right but I just checked the problem again and it's written down the same way as I have put it here, so I'm not really sure what to make of that the question might be inaccurate.
$endgroup$
– Roger
Jan 20 at 21:42
$begingroup$
@Roger: OK thanks for the update!
$endgroup$
– Robert Lewis
Jan 20 at 21:44
add a comment |
$begingroup$
Let $u in C^{1}(0,infty) times C^{1}[0,1]$ be a solution to heat equation (1) with inital boundaries (2),(3)
(1) $partial_tu(t,x)-partial_{xx}u(t,x)=0$, for all $(t,x) in[0,infty)times[0,1]$
(2) $u(t,0)=u(t,1)=0,$ for $t in[0,infty]$
(3) $u(0,x)=u_{0}(x),$ for $xin[0,1]$
Show that there exists constants $C_{1},C_{2}$ such that
$$
int_{0}^{1}u^2(t,x)dx leq C_{1}e^{-tC_2};text{ for all }tgeq0
$$
Thoughts
I want to get this into a form where I can apply Wirtinger's inequality for functions: I tried by multiplying the through by $u(t,x)$ and intergrating but I couldn't get it out that way, so I'm not sure how to solve this.
functional-analysis pde heat-equation
edited Jan 20 at 20:57
Daniele Tampieri
2,3372922
asked Jan 20 at 20:31
RogerRoger
847
$endgroup$
Let $u in C^{1}(0,infty) times C^{1}[0,1]$ be a solution to heat equation (1) with inital boundaries (2),(3)
(1) $partial_tu(t,x)-partial_{xx}u(t,x)=0$, for all $(t,x) in[0,infty)times[0,1]$
(2) $u(t,0)=u(t,1)=0,$ for $t in[0,infty]$
(3) $u(0,x)=u_{0}(x),$ for $xin[0,1]$
Show that there exists constants $C_{1},C_{2}$ such that
$$
int_{0}^{1}u^2(t,x)dx leq C_{1}e^{-tC_2};text{ for all }tgeq0
$$
Thoughts
I want to get this into a form where I can apply Wirtinger's inequality for functions: I tried by multiplying the through by $u(t,x)$ and intergrating but I couldn't get it out that way, so I'm not sure how to solve this.
functional-analysis pde heat-equation
functional-analysis pde heat-equation
edited Jan 20 at 20:57
Daniele Tampieri
2,3372922
asked Jan 20 at 20:31
RogerRoger
847
edited Jan 20 at 20:57
Daniele Tampieri
2,3372922
asked Jan 20 at 20:31
RogerRoger
847
edited Jan 20 at 20:57
Daniele Tampieri
2,3372922
edited Jan 20 at 20:57
Daniele Tampieri
2,3372922
edited Jan 20 at 20:57
Daniele Tampieri
2,3372922
2,3372922
asked Jan 20 at 20:31
RogerRoger
847
asked Jan 20 at 20:31
RogerRoger
847
asked Jan 20 at 20:31
RogerRoger
847
847
$begingroup$
I'm having some problems deciphering you statement$u(t, x) in C^1(0, infty) times C^1[0, 1]$; for one thing, the elements of $C^1(0, infty) times C^1[0, 1]$ are ordered pairs of functions; but $u$ doesn't seem to exhibit that form. Can you clarify? Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 21:11
1
$begingroup$
Also, seems that the spatial derivative must be $C^{2}$.
$endgroup$
– alexp9
Jan 20 at 21:13
1
$begingroup$
@RobertLewis, you're right but I just checked the problem again and it's written down the same way as I have put it here, so I'm not really sure what to make of that the question might be inaccurate.
$endgroup$
– Roger
Jan 20 at 21:42
$begingroup$
@Roger: OK thanks for the update!
$endgroup$
– Robert Lewis
Jan 20 at 21:44
add a comment |
$begingroup$
I'm having some problems deciphering you statement$u(t, x) in C^1(0, infty) times C^1[0, 1]$; for one thing, the elements of $C^1(0, infty) times C^1[0, 1]$ are ordered pairs of functions; but $u$ doesn't seem to exhibit that form. Can you clarify? Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 21:11
1
$begingroup$
Also, seems that the spatial derivative must be $C^{2}$.
$endgroup$
– alexp9
Jan 20 at 21:13
1
$begingroup$
@RobertLewis, you're right but I just checked the problem again and it's written down the same way as I have put it here, so I'm not really sure what to make of that the question might be inaccurate.
$endgroup$
– Roger
Jan 20 at 21:42
$begingroup$
@Roger: OK thanks for the update!
$endgroup$
– Robert Lewis
Jan 20 at 21:44
$begingroup$
I'm having some problems deciphering you statement$u(t, x) in C^1(0, infty) times C^1[0, 1]$; for one thing, the elements of $C^1(0, infty) times C^1[0, 1]$ are ordered pairs of functions; but $u$ doesn't seem to exhibit that form. Can you clarify? Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 21:11
$begingroup$
I'm having some problems deciphering you statement$u(t, x) in C^1(0, infty) times C^1[0, 1]$; for one thing, the elements of $C^1(0, infty) times C^1[0, 1]$ are ordered pairs of functions; but $u$ doesn't seem to exhibit that form. Can you clarify? Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 21:11
1
1
$begingroup$
Also, seems that the spatial derivative must be $C^{2}$.
$endgroup$
– alexp9
Jan 20 at 21:13
$begingroup$
Also, seems that the spatial derivative must be $C^{2}$.
$endgroup$
– alexp9
Jan 20 at 21:13
1
1
$begingroup$
@RobertLewis, you're right but I just checked the problem again and it's written down the same way as I have put it here, so I'm not really sure what to make of that the question might be inaccurate.
$endgroup$
– Roger
Jan 20 at 21:42
$begingroup$
@RobertLewis, you're right but I just checked the problem again and it's written down the same way as I have put it here, so I'm not really sure what to make of that the question might be inaccurate.
$endgroup$
– Roger
Jan 20 at 21:42
$begingroup$
@Roger: OK thanks for the update!
$endgroup$
– Robert Lewis
Jan 20 at 21:44
$begingroup$
@Roger: OK thanks for the update!
$endgroup$
– Robert Lewis
Jan 20 at 21:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As mentioned in the comnents, I don't feel totally clear on the differentiability conditions stipulated for $u(x, t)$, so I assume $u(x, t) in C^2([0, 1] times [0, infty))$; this seems to suffice for the present application.
We first develop a couple of useful identities; as usual
$u_x = dfrac{partial u}{partial x}, tag 1$
and so forth.
First,
$(u^2)_{xx} = (2uu_x)_x = 2(uu_x)_x = 2(u_x^2 + uu_{xx}) = 2u_x^2 + 2uu_{xx}; tag 2$
second, by virtue of $u_t = u_{xx}$,
$(u^2)_t = 2uu_t = 2uu_{xx} = (u^2)_{xx} - 2u_x^2 = 2(uu_x)_x - 2u_x^2; tag 3$
we have
$dfrac{d}{dt} displaystyle int_0^1 u^2(x, t) ; dx = int_0^1 (u^2)_t ; dx$
$= displaystyle int_0^1 (2(uu_x)_x - 2u_x^2) ; dx = 2int_0^1 (uu_x)_x ; dx - 2 int_0^1 u_x^2 ; dx; tag 4$
we next take note of the fact that
$displaystyle int_0^1 (uu_x)_x ; dx = u(1, t)u_x(1, t) - u(1, t) u_x(0, t) = 0, tag 5$
which follows from boundary condition (2); thus,
$dfrac{d}{dt} displaystyle int_0^1 u^2(x, t) ; dx = displaystyle int_0^1 (u^2)_t ; dx = -2 int_0^1 u_x^2 ; dx. tag 6$
We are now in position to apply the Wirtinger inequality to the integral on the right of this equation and find
$displaystyle pi^2 int_0^1 u^2 ; dx le int_0^1 u_x^2 ; dx, tag 7$
which we multiply by $-2$:
$-2 displaystyle int_0^1 u_x^2 ; dx le -2pi^2 int_0^1 u^2 ; dx; tag 8$
in combination with (6) this yields
$dfrac{d}{dt} displaystyle int_0^1 u^2; dx le -2pi^2 int_0^1 u^2 ; dx, tag 9$
to which we may apply Gronwall:
$displaystyle int_0^1 u^2(x, t) ; dx le left ( int_0^1 u^2(x, 0) ; dx right ) e^{-2pi^2 t} = left ( int_0^1 u_0^2 ; dx right ) e^{-2pi^2 t} , tag{10}$
and arrive at the desired result. $OEDelta$.
answered Jan 21 at 0:40
Robert LewisRobert Lewis
47.6k23067
$endgroup$
1
$begingroup$
A power-like estimate follows easily from my previous argument, which provides a uniform estimate on [0,1]. If you need an exponential decay for the $L_2$ norm, your argument is the right one of course. I just went with the easiest way to get what was needed.
$endgroup$
– GReyes
Jan 21 at 0:49
$begingroup$
@GReyes: just flying blind, dead-stick by the seat of my pants, old chap! Cheers!
$endgroup$
– Robert Lewis
Jan 21 at 0:52
$begingroup$
@GReyes: I'll have to take a closer look at your answer!
$endgroup$
– Robert Lewis
Jan 21 at 0:53
1
$begingroup$
well... point-wise comparison arguments like this has been my bread and butter for many years... anyways, your estimate is the sharpest one, no doubt.
$endgroup$
– GReyes
Jan 21 at 0:59
1
$begingroup$
BTW did you see my comment to your conjecture about the inequalities $|e^z-1|le|z|$ and similar ones?
$endgroup$
– GReyes
Jan 21 at 1:45
|
show 2 more comments
$begingroup$
Let's assume your solution has the regularity required by the equation. A way to get an estimate of this form is as follows: Consider a shifted multiple of the fundamental solution $G$, $w(x,t)=MG(x,t+1)$. For large enough $M$, $w(x,0)=MG(x,1)ge u_0$. These shifted and rescaled Gaussians are solutions to your equation in all of $[0,infty)timesmathbb{R}$. Therefore they are solutions on $[0,infty)times[0,1]$ (with boundary data equal $w(t,0)>0$ and $w(t,1)>0$). By the parabolic comparison principle, the solution to your equation will be below $w$ at all times, $u(x,t)le w(x,t)$ and, for $xin[0,1]$, $w(x,t)le Ct^{-1/2}$ where $C=M/{sqrt{4pi}}$.
As a consequence,
$$
intlimits_0^1 u^2(x,t), dxle C^2t^{-1}
$$
answered Jan 20 at 23:15
GReyesGReyes
1,73015
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
As mentioned in the comnents, I don't feel totally clear on the differentiability conditions stipulated for $u(x, t)$, so I assume $u(x, t) in C^2([0, 1] times [0, infty))$; this seems to suffice for the present application.
We first develop a couple of useful identities; as usual
$u_x = dfrac{partial u}{partial x}, tag 1$
and so forth.
First,
$(u^2)_{xx} = (2uu_x)_x = 2(uu_x)_x = 2(u_x^2 + uu_{xx}) = 2u_x^2 + 2uu_{xx}; tag 2$
second, by virtue of $u_t = u_{xx}$,
$(u^2)_t = 2uu_t = 2uu_{xx} = (u^2)_{xx} - 2u_x^2 = 2(uu_x)_x - 2u_x^2; tag 3$
we have
$dfrac{d}{dt} displaystyle int_0^1 u^2(x, t) ; dx = int_0^1 (u^2)_t ; dx$
$= displaystyle int_0^1 (2(uu_x)_x - 2u_x^2) ; dx = 2int_0^1 (uu_x)_x ; dx - 2 int_0^1 u_x^2 ; dx; tag 4$
we next take note of the fact that
$displaystyle int_0^1 (uu_x)_x ; dx = u(1, t)u_x(1, t) - u(1, t) u_x(0, t) = 0, tag 5$
which follows from boundary condition (2); thus,
$dfrac{d}{dt} displaystyle int_0^1 u^2(x, t) ; dx = displaystyle int_0^1 (u^2)_t ; dx = -2 int_0^1 u_x^2 ; dx. tag 6$
We are now in position to apply the Wirtinger inequality to the integral on the right of this equation and find
$displaystyle pi^2 int_0^1 u^2 ; dx le int_0^1 u_x^2 ; dx, tag 7$
which we multiply by $-2$:
$-2 displaystyle int_0^1 u_x^2 ; dx le -2pi^2 int_0^1 u^2 ; dx; tag 8$
in combination with (6) this yields
$dfrac{d}{dt} displaystyle int_0^1 u^2; dx le -2pi^2 int_0^1 u^2 ; dx, tag 9$
to which we may apply Gronwall:
$displaystyle int_0^1 u^2(x, t) ; dx le left ( int_0^1 u^2(x, 0) ; dx right ) e^{-2pi^2 t} = left ( int_0^1 u_0^2 ; dx right ) e^{-2pi^2 t} , tag{10}$
and arrive at the desired result. $OEDelta$.
answered Jan 21 at 0:40
Robert LewisRobert Lewis
47.6k23067
$endgroup$
1
$begingroup$
A power-like estimate follows easily from my previous argument, which provides a uniform estimate on [0,1]. If you need an exponential decay for the $L_2$ norm, your argument is the right one of course. I just went with the easiest way to get what was needed.
$endgroup$
– GReyes
Jan 21 at 0:49
$begingroup$
@GReyes: just flying blind, dead-stick by the seat of my pants, old chap! Cheers!
$endgroup$
– Robert Lewis
Jan 21 at 0:52
$begingroup$
@GReyes: I'll have to take a closer look at your answer!
$endgroup$
– Robert Lewis
Jan 21 at 0:53
1
$begingroup$
well... point-wise comparison arguments like this has been my bread and butter for many years... anyways, your estimate is the sharpest one, no doubt.
$endgroup$
– GReyes
Jan 21 at 0:59
1
$begingroup$
BTW did you see my comment to your conjecture about the inequalities $|e^z-1|le|z|$ and similar ones?
$endgroup$
– GReyes
Jan 21 at 1:45
|
show 2 more comments
$begingroup$
As mentioned in the comnents, I don't feel totally clear on the differentiability conditions stipulated for $u(x, t)$, so I assume $u(x, t) in C^2([0, 1] times [0, infty))$; this seems to suffice for the present application.
We first develop a couple of useful identities; as usual
$u_x = dfrac{partial u}{partial x}, tag 1$
and so forth.
First,
$(u^2)_{xx} = (2uu_x)_x = 2(uu_x)_x = 2(u_x^2 + uu_{xx}) = 2u_x^2 + 2uu_{xx}; tag 2$
second, by virtue of $u_t = u_{xx}$,
$(u^2)_t = 2uu_t = 2uu_{xx} = (u^2)_{xx} - 2u_x^2 = 2(uu_x)_x - 2u_x^2; tag 3$
we have
$dfrac{d}{dt} displaystyle int_0^1 u^2(x, t) ; dx = int_0^1 (u^2)_t ; dx$
$= displaystyle int_0^1 (2(uu_x)_x - 2u_x^2) ; dx = 2int_0^1 (uu_x)_x ; dx - 2 int_0^1 u_x^2 ; dx; tag 4$
we next take note of the fact that
$displaystyle int_0^1 (uu_x)_x ; dx = u(1, t)u_x(1, t) - u(1, t) u_x(0, t) = 0, tag 5$
which follows from boundary condition (2); thus,
$dfrac{d}{dt} displaystyle int_0^1 u^2(x, t) ; dx = displaystyle int_0^1 (u^2)_t ; dx = -2 int_0^1 u_x^2 ; dx. tag 6$
We are now in position to apply the Wirtinger inequality to the integral on the right of this equation and find
$displaystyle pi^2 int_0^1 u^2 ; dx le int_0^1 u_x^2 ; dx, tag 7$
which we multiply by $-2$:
$-2 displaystyle int_0^1 u_x^2 ; dx le -2pi^2 int_0^1 u^2 ; dx; tag 8$
in combination with (6) this yields
$dfrac{d}{dt} displaystyle int_0^1 u^2; dx le -2pi^2 int_0^1 u^2 ; dx, tag 9$
to which we may apply Gronwall:
$displaystyle int_0^1 u^2(x, t) ; dx le left ( int_0^1 u^2(x, 0) ; dx right ) e^{-2pi^2 t} = left ( int_0^1 u_0^2 ; dx right ) e^{-2pi^2 t} , tag{10}$
and arrive at the desired result. $OEDelta$.
answered Jan 21 at 0:40
Robert LewisRobert Lewis
47.6k23067
$endgroup$
1
$begingroup$
A power-like estimate follows easily from my previous argument, which provides a uniform estimate on [0,1]. If you need an exponential decay for the $L_2$ norm, your argument is the right one of course. I just went with the easiest way to get what was needed.
$endgroup$
– GReyes
Jan 21 at 0:49
$begingroup$
@GReyes: just flying blind, dead-stick by the seat of my pants, old chap! Cheers!
$endgroup$
– Robert Lewis
Jan 21 at 0:52
$begingroup$
@GReyes: I'll have to take a closer look at your answer!
$endgroup$
– Robert Lewis
Jan 21 at 0:53
1
$begingroup$
well... point-wise comparison arguments like this has been my bread and butter for many years... anyways, your estimate is the sharpest one, no doubt.
$endgroup$
– GReyes
Jan 21 at 0:59
1
$begingroup$
BTW did you see my comment to your conjecture about the inequalities $|e^z-1|le|z|$ and similar ones?
$endgroup$
– GReyes
Jan 21 at 1:45
|
show 2 more comments
$begingroup$
As mentioned in the comnents, I don't feel totally clear on the differentiability conditions stipulated for $u(x, t)$, so I assume $u(x, t) in C^2([0, 1] times [0, infty))$; this seems to suffice for the present application.
We first develop a couple of useful identities; as usual
$u_x = dfrac{partial u}{partial x}, tag 1$
and so forth.
First,
$(u^2)_{xx} = (2uu_x)_x = 2(uu_x)_x = 2(u_x^2 + uu_{xx}) = 2u_x^2 + 2uu_{xx}; tag 2$
second, by virtue of $u_t = u_{xx}$,
$(u^2)_t = 2uu_t = 2uu_{xx} = (u^2)_{xx} - 2u_x^2 = 2(uu_x)_x - 2u_x^2; tag 3$
we have
$dfrac{d}{dt} displaystyle int_0^1 u^2(x, t) ; dx = int_0^1 (u^2)_t ; dx$
$= displaystyle int_0^1 (2(uu_x)_x - 2u_x^2) ; dx = 2int_0^1 (uu_x)_x ; dx - 2 int_0^1 u_x^2 ; dx; tag 4$
we next take note of the fact that
$displaystyle int_0^1 (uu_x)_x ; dx = u(1, t)u_x(1, t) - u(1, t) u_x(0, t) = 0, tag 5$
which follows from boundary condition (2); thus,
$dfrac{d}{dt} displaystyle int_0^1 u^2(x, t) ; dx = displaystyle int_0^1 (u^2)_t ; dx = -2 int_0^1 u_x^2 ; dx. tag 6$
We are now in position to apply the Wirtinger inequality to the integral on the right of this equation and find
$displaystyle pi^2 int_0^1 u^2 ; dx le int_0^1 u_x^2 ; dx, tag 7$
which we multiply by $-2$:
$-2 displaystyle int_0^1 u_x^2 ; dx le -2pi^2 int_0^1 u^2 ; dx; tag 8$
in combination with (6) this yields
$dfrac{d}{dt} displaystyle int_0^1 u^2; dx le -2pi^2 int_0^1 u^2 ; dx, tag 9$
to which we may apply Gronwall:
$displaystyle int_0^1 u^2(x, t) ; dx le left ( int_0^1 u^2(x, 0) ; dx right ) e^{-2pi^2 t} = left ( int_0^1 u_0^2 ; dx right ) e^{-2pi^2 t} , tag{10}$
and arrive at the desired result. $OEDelta$.
answered Jan 21 at 0:40
Robert LewisRobert Lewis
47.6k23067
$endgroup$
As mentioned in the comnents, I don't feel totally clear on the differentiability conditions stipulated for $u(x, t)$, so I assume $u(x, t) in C^2([0, 1] times [0, infty))$; this seems to suffice for the present application.
We first develop a couple of useful identities; as usual
$u_x = dfrac{partial u}{partial x}, tag 1$
and so forth.
First,
$(u^2)_{xx} = (2uu_x)_x = 2(uu_x)_x = 2(u_x^2 + uu_{xx}) = 2u_x^2 + 2uu_{xx}; tag 2$
second, by virtue of $u_t = u_{xx}$,
$(u^2)_t = 2uu_t = 2uu_{xx} = (u^2)_{xx} - 2u_x^2 = 2(uu_x)_x - 2u_x^2; tag 3$
we have
$dfrac{d}{dt} displaystyle int_0^1 u^2(x, t) ; dx = int_0^1 (u^2)_t ; dx$
$= displaystyle int_0^1 (2(uu_x)_x - 2u_x^2) ; dx = 2int_0^1 (uu_x)_x ; dx - 2 int_0^1 u_x^2 ; dx; tag 4$
we next take note of the fact that
$displaystyle int_0^1 (uu_x)_x ; dx = u(1, t)u_x(1, t) - u(1, t) u_x(0, t) = 0, tag 5$
which follows from boundary condition (2); thus,
$dfrac{d}{dt} displaystyle int_0^1 u^2(x, t) ; dx = displaystyle int_0^1 (u^2)_t ; dx = -2 int_0^1 u_x^2 ; dx. tag 6$
We are now in position to apply the Wirtinger inequality to the integral on the right of this equation and find
$displaystyle pi^2 int_0^1 u^2 ; dx le int_0^1 u_x^2 ; dx, tag 7$
which we multiply by $-2$:
$-2 displaystyle int_0^1 u_x^2 ; dx le -2pi^2 int_0^1 u^2 ; dx; tag 8$
in combination with (6) this yields
$dfrac{d}{dt} displaystyle int_0^1 u^2; dx le -2pi^2 int_0^1 u^2 ; dx, tag 9$
to which we may apply Gronwall:
$displaystyle int_0^1 u^2(x, t) ; dx le left ( int_0^1 u^2(x, 0) ; dx right ) e^{-2pi^2 t} = left ( int_0^1 u_0^2 ; dx right ) e^{-2pi^2 t} , tag{10}$
and arrive at the desired result. $OEDelta$.
answered Jan 21 at 0:40
Robert LewisRobert Lewis
47.6k23067
edited Jan 21 at 1:27
answered Jan 21 at 0:40
Robert LewisRobert Lewis
47.6k23067
answered Jan 21 at 0:40
Robert LewisRobert Lewis
47.6k23067
answered Jan 21 at 0:40
Robert LewisRobert Lewis
47.6k23067
47.6k23067
1
$begingroup$
A power-like estimate follows easily from my previous argument, which provides a uniform estimate on [0,1]. If you need an exponential decay for the $L_2$ norm, your argument is the right one of course. I just went with the easiest way to get what was needed.
$endgroup$
– GReyes
Jan 21 at 0:49
$begingroup$
@GReyes: just flying blind, dead-stick by the seat of my pants, old chap! Cheers!
$endgroup$
– Robert Lewis
Jan 21 at 0:52
$begingroup$
@GReyes: I'll have to take a closer look at your answer!
$endgroup$
– Robert Lewis
Jan 21 at 0:53
1
$begingroup$
well... point-wise comparison arguments like this has been my bread and butter for many years... anyways, your estimate is the sharpest one, no doubt.
$endgroup$
– GReyes
Jan 21 at 0:59
1
$begingroup$
BTW did you see my comment to your conjecture about the inequalities $|e^z-1|le|z|$ and similar ones?
$endgroup$
– GReyes
Jan 21 at 1:45
|
show 2 more comments
1
$begingroup$
A power-like estimate follows easily from my previous argument, which provides a uniform estimate on [0,1]. If you need an exponential decay for the $L_2$ norm, your argument is the right one of course. I just went with the easiest way to get what was needed.
$endgroup$
– GReyes
Jan 21 at 0:49
$begingroup$
@GReyes: just flying blind, dead-stick by the seat of my pants, old chap! Cheers!
$endgroup$
– Robert Lewis
Jan 21 at 0:52
$begingroup$
@GReyes: I'll have to take a closer look at your answer!
$endgroup$
– Robert Lewis
Jan 21 at 0:53
1
$begingroup$
well... point-wise comparison arguments like this has been my bread and butter for many years... anyways, your estimate is the sharpest one, no doubt.
$endgroup$
– GReyes
Jan 21 at 0:59
1
$begingroup$
BTW did you see my comment to your conjecture about the inequalities $|e^z-1|le|z|$ and similar ones?
$endgroup$
– GReyes
Jan 21 at 1:45
1
1
$begingroup$
A power-like estimate follows easily from my previous argument, which provides a uniform estimate on [0,1]. If you need an exponential decay for the $L_2$ norm, your argument is the right one of course. I just went with the easiest way to get what was needed.
$endgroup$
– GReyes
Jan 21 at 0:49
$begingroup$
A power-like estimate follows easily from my previous argument, which provides a uniform estimate on [0,1]. If you need an exponential decay for the $L_2$ norm, your argument is the right one of course. I just went with the easiest way to get what was needed.
$endgroup$
– GReyes
Jan 21 at 0:49
$begingroup$
@GReyes: just flying blind, dead-stick by the seat of my pants, old chap! Cheers!
$endgroup$
– Robert Lewis
Jan 21 at 0:52
$begingroup$
@GReyes: just flying blind, dead-stick by the seat of my pants, old chap! Cheers!
$endgroup$
– Robert Lewis
Jan 21 at 0:52
$begingroup$
@GReyes: I'll have to take a closer look at your answer!
$endgroup$
– Robert Lewis
Jan 21 at 0:53
$begingroup$
@GReyes: I'll have to take a closer look at your answer!
$endgroup$
– Robert Lewis
Jan 21 at 0:53
1
1
$begingroup$
well... point-wise comparison arguments like this has been my bread and butter for many years... anyways, your estimate is the sharpest one, no doubt.
$endgroup$
– GReyes
Jan 21 at 0:59
$begingroup$
well... point-wise comparison arguments like this has been my bread and butter for many years... anyways, your estimate is the sharpest one, no doubt.
$endgroup$
– GReyes
Jan 21 at 0:59
1
1
$begingroup$
BTW did you see my comment to your conjecture about the inequalities $|e^z-1|le|z|$ and similar ones?
$endgroup$
– GReyes
Jan 21 at 1:45
$begingroup$
BTW did you see my comment to your conjecture about the inequalities $|e^z-1|le|z|$ and similar ones?
$endgroup$
– GReyes
Jan 21 at 1:45
|
show 2 more comments
$begingroup$
Let's assume your solution has the regularity required by the equation. A way to get an estimate of this form is as follows: Consider a shifted multiple of the fundamental solution $G$, $w(x,t)=MG(x,t+1)$. For large enough $M$, $w(x,0)=MG(x,1)ge u_0$. These shifted and rescaled Gaussians are solutions to your equation in all of $[0,infty)timesmathbb{R}$. Therefore they are solutions on $[0,infty)times[0,1]$ (with boundary data equal $w(t,0)>0$ and $w(t,1)>0$). By the parabolic comparison principle, the solution to your equation will be below $w$ at all times, $u(x,t)le w(x,t)$ and, for $xin[0,1]$, $w(x,t)le Ct^{-1/2}$ where $C=M/{sqrt{4pi}}$.
As a consequence,
$$
intlimits_0^1 u^2(x,t), dxle C^2t^{-1}
$$
answered Jan 20 at 23:15
GReyesGReyes
1,73015
$endgroup$
add a comment |
$begingroup$
Let's assume your solution has the regularity required by the equation. A way to get an estimate of this form is as follows: Consider a shifted multiple of the fundamental solution $G$, $w(x,t)=MG(x,t+1)$. For large enough $M$, $w(x,0)=MG(x,1)ge u_0$. These shifted and rescaled Gaussians are solutions to your equation in all of $[0,infty)timesmathbb{R}$. Therefore they are solutions on $[0,infty)times[0,1]$ (with boundary data equal $w(t,0)>0$ and $w(t,1)>0$). By the parabolic comparison principle, the solution to your equation will be below $w$ at all times, $u(x,t)le w(x,t)$ and, for $xin[0,1]$, $w(x,t)le Ct^{-1/2}$ where $C=M/{sqrt{4pi}}$.
As a consequence,
$$
intlimits_0^1 u^2(x,t), dxle C^2t^{-1}
$$
answered Jan 20 at 23:15
GReyesGReyes
1,73015
$endgroup$
add a comment |
$begingroup$
Let's assume your solution has the regularity required by the equation. A way to get an estimate of this form is as follows: Consider a shifted multiple of the fundamental solution $G$, $w(x,t)=MG(x,t+1)$. For large enough $M$, $w(x,0)=MG(x,1)ge u_0$. These shifted and rescaled Gaussians are solutions to your equation in all of $[0,infty)timesmathbb{R}$. Therefore they are solutions on $[0,infty)times[0,1]$ (with boundary data equal $w(t,0)>0$ and $w(t,1)>0$). By the parabolic comparison principle, the solution to your equation will be below $w$ at all times, $u(x,t)le w(x,t)$ and, for $xin[0,1]$, $w(x,t)le Ct^{-1/2}$ where $C=M/{sqrt{4pi}}$.
As a consequence,
$$
intlimits_0^1 u^2(x,t), dxle C^2t^{-1}
$$
answered Jan 20 at 23:15
GReyesGReyes
1,73015
$endgroup$
Let's assume your solution has the regularity required by the equation. A way to get an estimate of this form is as follows: Consider a shifted multiple of the fundamental solution $G$, $w(x,t)=MG(x,t+1)$. For large enough $M$, $w(x,0)=MG(x,1)ge u_0$. These shifted and rescaled Gaussians are solutions to your equation in all of $[0,infty)timesmathbb{R}$. Therefore they are solutions on $[0,infty)times[0,1]$ (with boundary data equal $w(t,0)>0$ and $w(t,1)>0$). By the parabolic comparison principle, the solution to your equation will be below $w$ at all times, $u(x,t)le w(x,t)$ and, for $xin[0,1]$, $w(x,t)le Ct^{-1/2}$ where $C=M/{sqrt{4pi}}$.
As a consequence,
$$
intlimits_0^1 u^2(x,t), dxle C^2t^{-1}
$$
answered Jan 20 at 23:15
GReyesGReyes
1,73015
edited Jan 21 at 0:53
answered Jan 20 at 23:15
GReyesGReyes
1,73015
answered Jan 20 at 23:15
GReyesGReyes
1,73015
answered Jan 20 at 23:15
GReyesGReyes
1,73015
1,73015
add a comment |
add a comment |
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$begingroup$
I'm having some problems deciphering you statement$u(t, x) in C^1(0, infty) times C^1[0, 1]$; for one thing, the elements of $C^1(0, infty) times C^1[0, 1]$ are ordered pairs of functions; but $u$ doesn't seem to exhibit that form. Can you clarify? Cheers!
$endgroup$
– Robert Lewis
Jan 20 at 21:11
1
$begingroup$
Also, seems that the spatial derivative must be $C^{2}$.
$endgroup$
– alexp9
Jan 20 at 21:13
1
$begingroup$
@RobertLewis, you're right but I just checked the problem again and it's written down the same way as I have put it here, so I'm not really sure what to make of that the question might be inaccurate.
$endgroup$
– Roger
Jan 20 at 21:42
$begingroup$
@Roger: OK thanks for the update!
$endgroup$
– Robert Lewis
Jan 20 at 21:44