Mixing
Gradient Blowup for a Parabolic (Heat) Equation
Clash Royale CLAN TAG#URR8PPP
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Let $u(x,t)$ be a solution to the following parabolic PDE:
With $alpha in (0,1)$,
beginalign
partial_t u(x,t) &= alpha (1-t)^alpha - 1 partial_x u(x,t) + frac12 partial_xx u(x,t), quad (x,t) in (-infty,1] times [0,1] \
u(x,0) &= f(x),\
u(1,t) &= u(-infty,t) = 0 quad text(boundary condition)
endalign
Here $f in C^2$, $; f geq 0$, $; int f^2(x)dx < infty$, $; f(1) = f(-infty) = 0$.
I'm especially interested in the case $alpha < 1/2$, but $alpha in [1/2,1)$ is also valuable.
Note that the coefficient of $partial_x u$, i.e $alpha (1-t)^alpha -1$ becomes unbounded at $t = 1$.
I'm trying to show that $$-partial_xu(x,t)|_x=1 geq kappa (1-t)^alpha-1 quad text as quad t to 1, quad (kappa > 0 text constant)$$ i.e. the right side boundary $(x = 1)$ gradient blows up at the rate at least $(1 - t)^alpha - 1$ as $t to 1$.
We can try the transformation $(z,r) = (mu x, mu^2 t)$ for some $mu > 0$, and define $xi(x,t) = u(z,r)$ for $(x,t) in (-infty, 1/mu] times [0,1/mu^2].$ Then we see that
beginalign
xi_x(x,t) &= mu u_z(z,r), \
xi_xx(x,t) &= mu^2 u_zz(z,r), \
xi_t(x,t) &= mu^2 u_r(z,r),
endalign
so that $xi$ solves the Dirichlet problem
beginalign
partial_t xi(x,t) &= alpha mu (1-mu^2 t)^alpha - 1 partial_x xi(x,t) + frac12 partial_xx xi(x,t), quad (x,t) in (-infty,1/mu] times [0,1/mu^2] \
xi(x,0) &= f(mu x),\
xi(1/mu,t) &= xi(-infty,t) = 0 quad textfor t in[0,1/mu^2]
endalign
Next, for each fixed $T < 1$, we restrict the solution of the above problem to $t in [0,T/mu^2]$ and set the scaling
$$mu = mu_T = (1 - mu^2 t)^1 - alpha|_t = T/mu^2 = (1 - T)^1 - alpha.$$
This means that on each strip $(x,t) in (-infty,1/mu_T)times [0,T/mu_T^2]$ and with the scaling $mu = mu_T$, the coefficients of the equation are (uniformly) bounded, and it has a solution in the classical sense.
Also, for each $T$, by Hopf's boundary gradient lemma, $partial_x xi(x, T/mu_T^2 )|_x = 1/mu_T = C_T > 0$ (boundary gradient is strictly positive).
This means that
$$partial_x u(1,T) = C_T / mu_T = C_T(1-T)^alpha - 1$$.
This would be complete the proof, but unfortunately, as $T to 1$, the right lateral boundary $x = 1/mu_T$ becomes unbounded and Hopf lemma fails. In other words, the asymptoic factor $C_T$ is not uniformly boubded from below. In fact, I suspect $C_T to 0$ as $T to 1$, and the argument is not enough to complete the proof. I need to show that $C_T > C > 0, forall T$ (probably not true), or revise the estimate by showing that $C_T searrow 0$ slowly (logarithmic etc), or use a completely different tactic.
NB. I already have an asymptotic upper bound on $-partial_x u(x,t)|_x = 1$:
$$-partial_x u(x,t)|_x = 1 leq (1 - t)^1 - alpha text as t to 1,$$ $alpha in (0,1)$.
Any ideas? I'll start a bounty when I'm allowed.
boundary-value-problem heat-equation regularity-theory-of-pdes singularity parabolic-pde
asked Jul 26 at 8:21
zab
35215
 |Â
show 2 more comments
up vote
2
down vote
favorite
Let $u(x,t)$ be a solution to the following parabolic PDE:
With $alpha in (0,1)$,
beginalign
partial_t u(x,t) &= alpha (1-t)^alpha - 1 partial_x u(x,t) + frac12 partial_xx u(x,t), quad (x,t) in (-infty,1] times [0,1] \
u(x,0) &= f(x),\
u(1,t) &= u(-infty,t) = 0 quad text(boundary condition)
endalign
Here $f in C^2$, $; f geq 0$, $; int f^2(x)dx < infty$, $; f(1) = f(-infty) = 0$.
I'm especially interested in the case $alpha < 1/2$, but $alpha in [1/2,1)$ is also valuable.
Note that the coefficient of $partial_x u$, i.e $alpha (1-t)^alpha -1$ becomes unbounded at $t = 1$.
I'm trying to show that $$-partial_xu(x,t)|_x=1 geq kappa (1-t)^alpha-1 quad text as quad t to 1, quad (kappa > 0 text constant)$$ i.e. the right side boundary $(x = 1)$ gradient blows up at the rate at least $(1 - t)^alpha - 1$ as $t to 1$.
We can try the transformation $(z,r) = (mu x, mu^2 t)$ for some $mu > 0$, and define $xi(x,t) = u(z,r)$ for $(x,t) in (-infty, 1/mu] times [0,1/mu^2].$ Then we see that
beginalign
xi_x(x,t) &= mu u_z(z,r), \
xi_xx(x,t) &= mu^2 u_zz(z,r), \
xi_t(x,t) &= mu^2 u_r(z,r),
endalign
so that $xi$ solves the Dirichlet problem
beginalign
partial_t xi(x,t) &= alpha mu (1-mu^2 t)^alpha - 1 partial_x xi(x,t) + frac12 partial_xx xi(x,t), quad (x,t) in (-infty,1/mu] times [0,1/mu^2] \
xi(x,0) &= f(mu x),\
xi(1/mu,t) &= xi(-infty,t) = 0 quad textfor t in[0,1/mu^2]
endalign
Next, for each fixed $T < 1$, we restrict the solution of the above problem to $t in [0,T/mu^2]$ and set the scaling
$$mu = mu_T = (1 - mu^2 t)^1 - alpha|_t = T/mu^2 = (1 - T)^1 - alpha.$$
This means that on each strip $(x,t) in (-infty,1/mu_T)times [0,T/mu_T^2]$ and with the scaling $mu = mu_T$, the coefficients of the equation are (uniformly) bounded, and it has a solution in the classical sense.
Also, for each $T$, by Hopf's boundary gradient lemma, $partial_x xi(x, T/mu_T^2 )|_x = 1/mu_T = C_T > 0$ (boundary gradient is strictly positive).
This means that
$$partial_x u(1,T) = C_T / mu_T = C_T(1-T)^alpha - 1$$.
This would be complete the proof, but unfortunately, as $T to 1$, the right lateral boundary $x = 1/mu_T$ becomes unbounded and Hopf lemma fails. In other words, the asymptoic factor $C_T$ is not uniformly boubded from below. In fact, I suspect $C_T to 0$ as $T to 1$, and the argument is not enough to complete the proof. I need to show that $C_T > C > 0, forall T$ (probably not true), or revise the estimate by showing that $C_T searrow 0$ slowly (logarithmic etc), or use a completely different tactic.
NB. I already have an asymptotic upper bound on $-partial_x u(x,t)|_x = 1$:
$$-partial_x u(x,t)|_x = 1 leq (1 - t)^1 - alpha text as t to 1,$$ $alpha in (0,1)$.
Any ideas? I'll start a bounty when I'm allowed.
boundary-value-problem heat-equation regularity-theory-of-pdes singularity parabolic-pde
asked Jul 26 at 8:21
zab
35215
1
Some nitpicking: the case $fequiv0$ excluded, the Zaremba-Giraud (or Hopf who was later) lemma gives in this case $partial_xu(x,t)|_x=1<0$ so, if the estimate is required without absolute value, it's trivially true.
– Andrew
Jul 31 at 19:11
1
@Andrew - yes absolutely true on both counts. Boundary gradient is negative, and I am infact looking for an asymptotic lower bound for $−partial_x u(t,x)|_x=1$ as $t to 1$. Any suggestions? A different scaling perhaps? Or a sub-parabolic function (sub-temperature) whose gradient($times -1$) blows up at the same rate while staying below u?
– zab
Aug 1 at 17:34
1
Was the equation obtained by a change of variables to straighten the boundary? Btw returning to the heat eqn and curved boundary gets that there will be no growth at least for $alpha>1/2$.
– Andrew
Aug 1 at 21:02
1
Yes exactly - the original problem was the standard heat equation in the domain $(t,x):t∈[0,1], x leq (1−t)^alpha$. I'm trying to find the rate of gradient blowup at the right boundary $(x=(1−t)^alpha)$ as $t to 1$. The solution itself remains bounded by the $int f(x) e^-x^2/(2t)/sqrt2pi tdx$ (obviously) but my intuition is that the gradient at the boundary should go like $(1−t)^alpha−1$ as $t to infty$ as soon as $alpha <1/2$. It is not immediately obvious to me that "there will be no growth for $alpha>1/2$. Can you direct me to a proof? I don't know the pde literature well.
– zab
Aug 2 at 5:36
1
In the original problem the boundary belongs to the Holder class $C^2alpha$ (exponent counted with factor 2 because of anisotropy). For $2alpha=1+beta$ when $0<beta<1$ ($alpha>1/2$) the solution belongs to the parabolic Holder class $C^1+beta(bar Omega)$ in the closure of the domain in question $Omega$. it means $partial_xuin C^beta$, and so continuous and bounded up in $bar Omega$. For n-dimensional case see Baderko, E. A., ‘Boundary value problems for a parabolic equation, and boundary integral equations’, Diff. Eq., 28, 15—20 (1992), there are ref. for the case $n=1$.
– Andrew
Aug 2 at 9:00
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $u(x,t)$ be a solution to the following parabolic PDE:
With $alpha in (0,1)$,
beginalign
partial_t u(x,t) &= alpha (1-t)^alpha - 1 partial_x u(x,t) + frac12 partial_xx u(x,t), quad (x,t) in (-infty,1] times [0,1] \
u(x,0) &= f(x),\
u(1,t) &= u(-infty,t) = 0 quad text(boundary condition)
endalign
Here $f in C^2$, $; f geq 0$, $; int f^2(x)dx < infty$, $; f(1) = f(-infty) = 0$.
I'm especially interested in the case $alpha < 1/2$, but $alpha in [1/2,1)$ is also valuable.
Note that the coefficient of $partial_x u$, i.e $alpha (1-t)^alpha -1$ becomes unbounded at $t = 1$.
I'm trying to show that $$-partial_xu(x,t)|_x=1 geq kappa (1-t)^alpha-1 quad text as quad t to 1, quad (kappa > 0 text constant)$$ i.e. the right side boundary $(x = 1)$ gradient blows up at the rate at least $(1 - t)^alpha - 1$ as $t to 1$.
We can try the transformation $(z,r) = (mu x, mu^2 t)$ for some $mu > 0$, and define $xi(x,t) = u(z,r)$ for $(x,t) in (-infty, 1/mu] times [0,1/mu^2].$ Then we see that
beginalign
xi_x(x,t) &= mu u_z(z,r), \
xi_xx(x,t) &= mu^2 u_zz(z,r), \
xi_t(x,t) &= mu^2 u_r(z,r),
endalign
so that $xi$ solves the Dirichlet problem
beginalign
partial_t xi(x,t) &= alpha mu (1-mu^2 t)^alpha - 1 partial_x xi(x,t) + frac12 partial_xx xi(x,t), quad (x,t) in (-infty,1/mu] times [0,1/mu^2] \
xi(x,0) &= f(mu x),\
xi(1/mu,t) &= xi(-infty,t) = 0 quad textfor t in[0,1/mu^2]
endalign
Next, for each fixed $T < 1$, we restrict the solution of the above problem to $t in [0,T/mu^2]$ and set the scaling
$$mu = mu_T = (1 - mu^2 t)^1 - alpha|_t = T/mu^2 = (1 - T)^1 - alpha.$$
This means that on each strip $(x,t) in (-infty,1/mu_T)times [0,T/mu_T^2]$ and with the scaling $mu = mu_T$, the coefficients of the equation are (uniformly) bounded, and it has a solution in the classical sense.
Also, for each $T$, by Hopf's boundary gradient lemma, $partial_x xi(x, T/mu_T^2 )|_x = 1/mu_T = C_T > 0$ (boundary gradient is strictly positive).
This means that
$$partial_x u(1,T) = C_T / mu_T = C_T(1-T)^alpha - 1$$.
This would be complete the proof, but unfortunately, as $T to 1$, the right lateral boundary $x = 1/mu_T$ becomes unbounded and Hopf lemma fails. In other words, the asymptoic factor $C_T$ is not uniformly boubded from below. In fact, I suspect $C_T to 0$ as $T to 1$, and the argument is not enough to complete the proof. I need to show that $C_T > C > 0, forall T$ (probably not true), or revise the estimate by showing that $C_T searrow 0$ slowly (logarithmic etc), or use a completely different tactic.
NB. I already have an asymptotic upper bound on $-partial_x u(x,t)|_x = 1$:
$$-partial_x u(x,t)|_x = 1 leq (1 - t)^1 - alpha text as t to 1,$$ $alpha in (0,1)$.
Any ideas? I'll start a bounty when I'm allowed.
boundary-value-problem heat-equation regularity-theory-of-pdes singularity parabolic-pde
asked Jul 26 at 8:21
zab
35215
Let $u(x,t)$ be a solution to the following parabolic PDE:
With $alpha in (0,1)$,
beginalign
partial_t u(x,t) &= alpha (1-t)^alpha - 1 partial_x u(x,t) + frac12 partial_xx u(x,t), quad (x,t) in (-infty,1] times [0,1] \
u(x,0) &= f(x),\
u(1,t) &= u(-infty,t) = 0 quad text(boundary condition)
endalign
Here $f in C^2$, $; f geq 0$, $; int f^2(x)dx < infty$, $; f(1) = f(-infty) = 0$.
I'm especially interested in the case $alpha < 1/2$, but $alpha in [1/2,1)$ is also valuable.
Note that the coefficient of $partial_x u$, i.e $alpha (1-t)^alpha -1$ becomes unbounded at $t = 1$.
I'm trying to show that $$-partial_xu(x,t)|_x=1 geq kappa (1-t)^alpha-1 quad text as quad t to 1, quad (kappa > 0 text constant)$$ i.e. the right side boundary $(x = 1)$ gradient blows up at the rate at least $(1 - t)^alpha - 1$ as $t to 1$.
We can try the transformation $(z,r) = (mu x, mu^2 t)$ for some $mu > 0$, and define $xi(x,t) = u(z,r)$ for $(x,t) in (-infty, 1/mu] times [0,1/mu^2].$ Then we see that
beginalign
xi_x(x,t) &= mu u_z(z,r), \
xi_xx(x,t) &= mu^2 u_zz(z,r), \
xi_t(x,t) &= mu^2 u_r(z,r),
endalign
so that $xi$ solves the Dirichlet problem
beginalign
partial_t xi(x,t) &= alpha mu (1-mu^2 t)^alpha - 1 partial_x xi(x,t) + frac12 partial_xx xi(x,t), quad (x,t) in (-infty,1/mu] times [0,1/mu^2] \
xi(x,0) &= f(mu x),\
xi(1/mu,t) &= xi(-infty,t) = 0 quad textfor t in[0,1/mu^2]
endalign
Next, for each fixed $T < 1$, we restrict the solution of the above problem to $t in [0,T/mu^2]$ and set the scaling
$$mu = mu_T = (1 - mu^2 t)^1 - alpha|_t = T/mu^2 = (1 - T)^1 - alpha.$$
This means that on each strip $(x,t) in (-infty,1/mu_T)times [0,T/mu_T^2]$ and with the scaling $mu = mu_T$, the coefficients of the equation are (uniformly) bounded, and it has a solution in the classical sense.
Also, for each $T$, by Hopf's boundary gradient lemma, $partial_x xi(x, T/mu_T^2 )|_x = 1/mu_T = C_T > 0$ (boundary gradient is strictly positive).
This means that
$$partial_x u(1,T) = C_T / mu_T = C_T(1-T)^alpha - 1$$.
This would be complete the proof, but unfortunately, as $T to 1$, the right lateral boundary $x = 1/mu_T$ becomes unbounded and Hopf lemma fails. In other words, the asymptoic factor $C_T$ is not uniformly boubded from below. In fact, I suspect $C_T to 0$ as $T to 1$, and the argument is not enough to complete the proof. I need to show that $C_T > C > 0, forall T$ (probably not true), or revise the estimate by showing that $C_T searrow 0$ slowly (logarithmic etc), or use a completely different tactic.
NB. I already have an asymptotic upper bound on $-partial_x u(x,t)|_x = 1$:
$$-partial_x u(x,t)|_x = 1 leq (1 - t)^1 - alpha text as t to 1,$$ $alpha in (0,1)$.
Any ideas? I'll start a bounty when I'm allowed.
boundary-value-problem heat-equation regularity-theory-of-pdes singularity parabolic-pde
asked Jul 26 at 8:21
zab
35215
edited Aug 1 at 16:13
asked Jul 26 at 8:21
zab
35215
asked Jul 26 at 8:21
zab
35215
asked Jul 26 at 8:21
zab
35215
35215
1
Some nitpicking: the case $fequiv0$ excluded, the Zaremba-Giraud (or Hopf who was later) lemma gives in this case $partial_xu(x,t)|_x=1<0$ so, if the estimate is required without absolute value, it's trivially true.
– Andrew
Jul 31 at 19:11
1
@Andrew - yes absolutely true on both counts. Boundary gradient is negative, and I am infact looking for an asymptotic lower bound for $−partial_x u(t,x)|_x=1$ as $t to 1$. Any suggestions? A different scaling perhaps? Or a sub-parabolic function (sub-temperature) whose gradient($times -1$) blows up at the same rate while staying below u?
– zab
Aug 1 at 17:34
1
Was the equation obtained by a change of variables to straighten the boundary? Btw returning to the heat eqn and curved boundary gets that there will be no growth at least for $alpha>1/2$.
– Andrew
Aug 1 at 21:02
1
Yes exactly - the original problem was the standard heat equation in the domain $(t,x):t∈[0,1], x leq (1−t)^alpha$. I'm trying to find the rate of gradient blowup at the right boundary $(x=(1−t)^alpha)$ as $t to 1$. The solution itself remains bounded by the $int f(x) e^-x^2/(2t)/sqrt2pi tdx$ (obviously) but my intuition is that the gradient at the boundary should go like $(1−t)^alpha−1$ as $t to infty$ as soon as $alpha <1/2$. It is not immediately obvious to me that "there will be no growth for $alpha>1/2$. Can you direct me to a proof? I don't know the pde literature well.
– zab
Aug 2 at 5:36
1
In the original problem the boundary belongs to the Holder class $C^2alpha$ (exponent counted with factor 2 because of anisotropy). For $2alpha=1+beta$ when $0<beta<1$ ($alpha>1/2$) the solution belongs to the parabolic Holder class $C^1+beta(bar Omega)$ in the closure of the domain in question $Omega$. it means $partial_xuin C^beta$, and so continuous and bounded up in $bar Omega$. For n-dimensional case see Baderko, E. A., ‘Boundary value problems for a parabolic equation, and boundary integral equations’, Diff. Eq., 28, 15—20 (1992), there are ref. for the case $n=1$.
– Andrew
Aug 2 at 9:00
 |Â
show 2 more comments
1
Some nitpicking: the case $fequiv0$ excluded, the Zaremba-Giraud (or Hopf who was later) lemma gives in this case $partial_xu(x,t)|_x=1<0$ so, if the estimate is required without absolute value, it's trivially true.
– Andrew
Jul 31 at 19:11
1
@Andrew - yes absolutely true on both counts. Boundary gradient is negative, and I am infact looking for an asymptotic lower bound for $−partial_x u(t,x)|_x=1$ as $t to 1$. Any suggestions? A different scaling perhaps? Or a sub-parabolic function (sub-temperature) whose gradient($times -1$) blows up at the same rate while staying below u?
– zab
Aug 1 at 17:34
1
Was the equation obtained by a change of variables to straighten the boundary? Btw returning to the heat eqn and curved boundary gets that there will be no growth at least for $alpha>1/2$.
– Andrew
Aug 1 at 21:02
1
Yes exactly - the original problem was the standard heat equation in the domain $(t,x):t∈[0,1], x leq (1−t)^alpha$. I'm trying to find the rate of gradient blowup at the right boundary $(x=(1−t)^alpha)$ as $t to 1$. The solution itself remains bounded by the $int f(x) e^-x^2/(2t)/sqrt2pi tdx$ (obviously) but my intuition is that the gradient at the boundary should go like $(1−t)^alpha−1$ as $t to infty$ as soon as $alpha <1/2$. It is not immediately obvious to me that "there will be no growth for $alpha>1/2$. Can you direct me to a proof? I don't know the pde literature well.
– zab
Aug 2 at 5:36
1
In the original problem the boundary belongs to the Holder class $C^2alpha$ (exponent counted with factor 2 because of anisotropy). For $2alpha=1+beta$ when $0<beta<1$ ($alpha>1/2$) the solution belongs to the parabolic Holder class $C^1+beta(bar Omega)$ in the closure of the domain in question $Omega$. it means $partial_xuin C^beta$, and so continuous and bounded up in $bar Omega$. For n-dimensional case see Baderko, E. A., ‘Boundary value problems for a parabolic equation, and boundary integral equations’, Diff. Eq., 28, 15—20 (1992), there are ref. for the case $n=1$.
– Andrew
Aug 2 at 9:00
1
1
Some nitpicking: the case $fequiv0$ excluded, the Zaremba-Giraud (or Hopf who was later) lemma gives in this case $partial_xu(x,t)|_x=1<0$ so, if the estimate is required without absolute value, it's trivially true.
– Andrew
Jul 31 at 19:11
Some nitpicking: the case $fequiv0$ excluded, the Zaremba-Giraud (or Hopf who was later) lemma gives in this case $partial_xu(x,t)|_x=1<0$ so, if the estimate is required without absolute value, it's trivially true.
– Andrew
Jul 31 at 19:11
1
1
@Andrew - yes absolutely true on both counts. Boundary gradient is negative, and I am infact looking for an asymptotic lower bound for $−partial_x u(t,x)|_x=1$ as $t to 1$. Any suggestions? A different scaling perhaps? Or a sub-parabolic function (sub-temperature) whose gradient($times -1$) blows up at the same rate while staying below u?
– zab
Aug 1 at 17:34
@Andrew - yes absolutely true on both counts. Boundary gradient is negative, and I am infact looking for an asymptotic lower bound for $−partial_x u(t,x)|_x=1$ as $t to 1$. Any suggestions? A different scaling perhaps? Or a sub-parabolic function (sub-temperature) whose gradient($times -1$) blows up at the same rate while staying below u?
– zab
Aug 1 at 17:34
1
1
Was the equation obtained by a change of variables to straighten the boundary? Btw returning to the heat eqn and curved boundary gets that there will be no growth at least for $alpha>1/2$.
– Andrew
Aug 1 at 21:02
Was the equation obtained by a change of variables to straighten the boundary? Btw returning to the heat eqn and curved boundary gets that there will be no growth at least for $alpha>1/2$.
– Andrew
Aug 1 at 21:02
1
1
Yes exactly - the original problem was the standard heat equation in the domain $(t,x):t∈[0,1], x leq (1−t)^alpha$. I'm trying to find the rate of gradient blowup at the right boundary $(x=(1−t)^alpha)$ as $t to 1$. The solution itself remains bounded by the $int f(x) e^-x^2/(2t)/sqrt2pi tdx$ (obviously) but my intuition is that the gradient at the boundary should go like $(1−t)^alpha−1$ as $t to infty$ as soon as $alpha <1/2$. It is not immediately obvious to me that "there will be no growth for $alpha>1/2$. Can you direct me to a proof? I don't know the pde literature well.
– zab
Aug 2 at 5:36
Yes exactly - the original problem was the standard heat equation in the domain $(t,x):t∈[0,1], x leq (1−t)^alpha$. I'm trying to find the rate of gradient blowup at the right boundary $(x=(1−t)^alpha)$ as $t to 1$. The solution itself remains bounded by the $int f(x) e^-x^2/(2t)/sqrt2pi tdx$ (obviously) but my intuition is that the gradient at the boundary should go like $(1−t)^alpha−1$ as $t to infty$ as soon as $alpha <1/2$. It is not immediately obvious to me that "there will be no growth for $alpha>1/2$. Can you direct me to a proof? I don't know the pde literature well.
– zab
Aug 2 at 5:36
1
1
In the original problem the boundary belongs to the Holder class $C^2alpha$ (exponent counted with factor 2 because of anisotropy). For $2alpha=1+beta$ when $0<beta<1$ ($alpha>1/2$) the solution belongs to the parabolic Holder class $C^1+beta(bar Omega)$ in the closure of the domain in question $Omega$. it means $partial_xuin C^beta$, and so continuous and bounded up in $bar Omega$. For n-dimensional case see Baderko, E. A., ‘Boundary value problems for a parabolic equation, and boundary integral equations’, Diff. Eq., 28, 15—20 (1992), there are ref. for the case $n=1$.
– Andrew
Aug 2 at 9:00
In the original problem the boundary belongs to the Holder class $C^2alpha$ (exponent counted with factor 2 because of anisotropy). For $2alpha=1+beta$ when $0<beta<1$ ($alpha>1/2$) the solution belongs to the parabolic Holder class $C^1+beta(bar Omega)$ in the closure of the domain in question $Omega$. it means $partial_xuin C^beta$, and so continuous and bounded up in $bar Omega$. For n-dimensional case see Baderko, E. A., ‘Boundary value problems for a parabolic equation, and boundary integral equations’, Diff. Eq., 28, 15—20 (1992), there are ref. for the case $n=1$.
– Andrew
Aug 2 at 9:00
 |Â
show 2 more comments
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1
Some nitpicking: the case $fequiv0$ excluded, the Zaremba-Giraud (or Hopf who was later) lemma gives in this case $partial_xu(x,t)|_x=1<0$ so, if the estimate is required without absolute value, it's trivially true.
– Andrew
Jul 31 at 19:11
1
@Andrew - yes absolutely true on both counts. Boundary gradient is negative, and I am infact looking for an asymptotic lower bound for $−partial_x u(t,x)|_x=1$ as $t to 1$. Any suggestions? A different scaling perhaps? Or a sub-parabolic function (sub-temperature) whose gradient($times -1$) blows up at the same rate while staying below u?
– zab
Aug 1 at 17:34
1
Was the equation obtained by a change of variables to straighten the boundary? Btw returning to the heat eqn and curved boundary gets that there will be no growth at least for $alpha>1/2$.
– Andrew
Aug 1 at 21:02
1
Yes exactly - the original problem was the standard heat equation in the domain $(t,x):t∈[0,1], x leq (1−t)^alpha$. I'm trying to find the rate of gradient blowup at the right boundary $(x=(1−t)^alpha)$ as $t to 1$. The solution itself remains bounded by the $int f(x) e^-x^2/(2t)/sqrt2pi tdx$ (obviously) but my intuition is that the gradient at the boundary should go like $(1−t)^alpha−1$ as $t to infty$ as soon as $alpha <1/2$. It is not immediately obvious to me that "there will be no growth for $alpha>1/2$. Can you direct me to a proof? I don't know the pde literature well.
– zab
Aug 2 at 5:36
1
In the original problem the boundary belongs to the Holder class $C^2alpha$ (exponent counted with factor 2 because of anisotropy). For $2alpha=1+beta$ when $0<beta<1$ ($alpha>1/2$) the solution belongs to the parabolic Holder class $C^1+beta(bar Omega)$ in the closure of the domain in question $Omega$. it means $partial_xuin C^beta$, and so continuous and bounded up in $bar Omega$. For n-dimensional case see Baderko, E. A., ‘Boundary value problems for a parabolic equation, and boundary integral equations’, Diff. Eq., 28, 15—20 (1992), there are ref. for the case $n=1$.
– Andrew
Aug 2 at 9:00