Weak solutions to Hamilton Jacobi equation
$begingroup$
Assume $u^1, u^2$ are two weak solutions of the initial value problems.
begin{align*} u_t^i+H(Du^i)&=0 text{in} mathbb{R}^ntimes (0,infty)\
u^i&=g^i text{on} mathbb{R}^ntimes{t=0} (i=1,2)end{align*}
for $H$ as in section 3.3 (that is $H$ is convex and $lim_{|p|rightarrowinfty}H(p)/|p|=+infty$. Prove the $L^infty$ contraction inequality: $$sup_{mathbb{R}}|u^1(cdot,t)-u^2(cdot ,t)|leq sup_{mathbb{R}}|g^1-g^2| (t>0)$$
This is an exercise in Evans. And this is what we have so far
By Hopf lax, we there exists $y,y'in mathbb{R}^n$, we have $$u_1(x,t)=tL(frac{x-y}{t})+g^1(y)$$ and $$u_2(x,t)=tL(frac{x-y'}{t})+g^2(y')$$
So $$|u^1(x-y',t)-u^2(x-y,t)|=|g^1(y)-g^2(y')|leq sup_{mathbb{R}}|g^1-g^2|$$
From here, is it safe to conclude that $|u^1(z,t)-u^2(w,t)|leq sup_{mathbb{R}}|g^1-g^2|$ for all $z,winmathbb{R}^n$? The left hand side still depends on $y,y'$ so I think the answer is negative. Is there any other ways to solve this problem? Hints? Thank you very much
pde hamilton-jacobi-equation
edited Dec 11 '18 at 11:28
Harry49
8,35931344
asked Nov 29 '15 at 3:18
nerdnerd
718411
$endgroup$
add a comment |
$begingroup$
Assume $u^1, u^2$ are two weak solutions of the initial value problems.
begin{align*} u_t^i+H(Du^i)&=0 text{in} mathbb{R}^ntimes (0,infty)\
u^i&=g^i text{on} mathbb{R}^ntimes{t=0} (i=1,2)end{align*}
for $H$ as in section 3.3 (that is $H$ is convex and $lim_{|p|rightarrowinfty}H(p)/|p|=+infty$. Prove the $L^infty$ contraction inequality: $$sup_{mathbb{R}}|u^1(cdot,t)-u^2(cdot ,t)|leq sup_{mathbb{R}}|g^1-g^2| (t>0)$$
This is an exercise in Evans. And this is what we have so far
By Hopf lax, we there exists $y,y'in mathbb{R}^n$, we have $$u_1(x,t)=tL(frac{x-y}{t})+g^1(y)$$ and $$u_2(x,t)=tL(frac{x-y'}{t})+g^2(y')$$
So $$|u^1(x-y',t)-u^2(x-y,t)|=|g^1(y)-g^2(y')|leq sup_{mathbb{R}}|g^1-g^2|$$
From here, is it safe to conclude that $|u^1(z,t)-u^2(w,t)|leq sup_{mathbb{R}}|g^1-g^2|$ for all $z,winmathbb{R}^n$? The left hand side still depends on $y,y'$ so I think the answer is negative. Is there any other ways to solve this problem? Hints? Thank you very much
pde hamilton-jacobi-equation
edited Dec 11 '18 at 11:28
Harry49
8,35931344
asked Nov 29 '15 at 3:18
nerdnerd
718411
$endgroup$
$begingroup$
Using your notation, you can actually show that $u^1(x,t)-u^2(x,t)ge g^1(y)-g^2(y)$, and $u^1(x,t)-u^2(x,t)le g^1(y')-g^2(y')$. This is because Hopf Lax gives us $u^2(x,t)le tL(x-y/t) + g^2(y)$, and similarly with $u^1$.
$endgroup$
– Joey Zou
Nov 29 '15 at 4:01
1
$begingroup$
Right, good point, I left out the easiest inequality I could use out of Hopf Lax. Thanks for the hint, I truly appreciated
$endgroup$
– nerd
Nov 29 '15 at 5:32
add a comment |
$begingroup$
Assume $u^1, u^2$ are two weak solutions of the initial value problems.
begin{align*} u_t^i+H(Du^i)&=0 text{in} mathbb{R}^ntimes (0,infty)\
u^i&=g^i text{on} mathbb{R}^ntimes{t=0} (i=1,2)end{align*}
for $H$ as in section 3.3 (that is $H$ is convex and $lim_{|p|rightarrowinfty}H(p)/|p|=+infty$. Prove the $L^infty$ contraction inequality: $$sup_{mathbb{R}}|u^1(cdot,t)-u^2(cdot ,t)|leq sup_{mathbb{R}}|g^1-g^2| (t>0)$$
This is an exercise in Evans. And this is what we have so far
By Hopf lax, we there exists $y,y'in mathbb{R}^n$, we have $$u_1(x,t)=tL(frac{x-y}{t})+g^1(y)$$ and $$u_2(x,t)=tL(frac{x-y'}{t})+g^2(y')$$
So $$|u^1(x-y',t)-u^2(x-y,t)|=|g^1(y)-g^2(y')|leq sup_{mathbb{R}}|g^1-g^2|$$
From here, is it safe to conclude that $|u^1(z,t)-u^2(w,t)|leq sup_{mathbb{R}}|g^1-g^2|$ for all $z,winmathbb{R}^n$? The left hand side still depends on $y,y'$ so I think the answer is negative. Is there any other ways to solve this problem? Hints? Thank you very much
pde hamilton-jacobi-equation
edited Dec 11 '18 at 11:28
Harry49
8,35931344
asked Nov 29 '15 at 3:18
nerdnerd
718411
$endgroup$
Assume $u^1, u^2$ are two weak solutions of the initial value problems.
begin{align*} u_t^i+H(Du^i)&=0 text{in} mathbb{R}^ntimes (0,infty)\
u^i&=g^i text{on} mathbb{R}^ntimes{t=0} (i=1,2)end{align*}
for $H$ as in section 3.3 (that is $H$ is convex and $lim_{|p|rightarrowinfty}H(p)/|p|=+infty$. Prove the $L^infty$ contraction inequality: $$sup_{mathbb{R}}|u^1(cdot,t)-u^2(cdot ,t)|leq sup_{mathbb{R}}|g^1-g^2| (t>0)$$
This is an exercise in Evans. And this is what we have so far
By Hopf lax, we there exists $y,y'in mathbb{R}^n$, we have $$u_1(x,t)=tL(frac{x-y}{t})+g^1(y)$$ and $$u_2(x,t)=tL(frac{x-y'}{t})+g^2(y')$$
So $$|u^1(x-y',t)-u^2(x-y,t)|=|g^1(y)-g^2(y')|leq sup_{mathbb{R}}|g^1-g^2|$$
From here, is it safe to conclude that $|u^1(z,t)-u^2(w,t)|leq sup_{mathbb{R}}|g^1-g^2|$ for all $z,winmathbb{R}^n$? The left hand side still depends on $y,y'$ so I think the answer is negative. Is there any other ways to solve this problem? Hints? Thank you very much
pde hamilton-jacobi-equation
pde hamilton-jacobi-equation
edited Dec 11 '18 at 11:28
Harry49
8,35931344
asked Nov 29 '15 at 3:18
nerdnerd
718411
edited Dec 11 '18 at 11:28
Harry49
8,35931344
asked Nov 29 '15 at 3:18
nerdnerd
718411
edited Dec 11 '18 at 11:28
Harry49
8,35931344
edited Dec 11 '18 at 11:28
Harry49
8,35931344
edited Dec 11 '18 at 11:28
Harry49
8,35931344
8,35931344
asked Nov 29 '15 at 3:18
nerdnerd
718411
asked Nov 29 '15 at 3:18
nerdnerd
718411
asked Nov 29 '15 at 3:18
nerdnerd
718411
718411
$begingroup$
Using your notation, you can actually show that $u^1(x,t)-u^2(x,t)ge g^1(y)-g^2(y)$, and $u^1(x,t)-u^2(x,t)le g^1(y')-g^2(y')$. This is because Hopf Lax gives us $u^2(x,t)le tL(x-y/t) + g^2(y)$, and similarly with $u^1$.
$endgroup$
– Joey Zou
Nov 29 '15 at 4:01
1
$begingroup$
Right, good point, I left out the easiest inequality I could use out of Hopf Lax. Thanks for the hint, I truly appreciated
$endgroup$
– nerd
Nov 29 '15 at 5:32
add a comment |
$begingroup$
Using your notation, you can actually show that $u^1(x,t)-u^2(x,t)ge g^1(y)-g^2(y)$, and $u^1(x,t)-u^2(x,t)le g^1(y')-g^2(y')$. This is because Hopf Lax gives us $u^2(x,t)le tL(x-y/t) + g^2(y)$, and similarly with $u^1$.
$endgroup$
– Joey Zou
Nov 29 '15 at 4:01
1
$begingroup$
Right, good point, I left out the easiest inequality I could use out of Hopf Lax. Thanks for the hint, I truly appreciated
$endgroup$
– nerd
Nov 29 '15 at 5:32
$begingroup$
Using your notation, you can actually show that $u^1(x,t)-u^2(x,t)ge g^1(y)-g^2(y)$, and $u^1(x,t)-u^2(x,t)le g^1(y')-g^2(y')$. This is because Hopf Lax gives us $u^2(x,t)le tL(x-y/t) + g^2(y)$, and similarly with $u^1$.
$endgroup$
– Joey Zou
Nov 29 '15 at 4:01
$begingroup$
Using your notation, you can actually show that $u^1(x,t)-u^2(x,t)ge g^1(y)-g^2(y)$, and $u^1(x,t)-u^2(x,t)le g^1(y')-g^2(y')$. This is because Hopf Lax gives us $u^2(x,t)le tL(x-y/t) + g^2(y)$, and similarly with $u^1$.
$endgroup$
– Joey Zou
Nov 29 '15 at 4:01
1
1
$begingroup$
Right, good point, I left out the easiest inequality I could use out of Hopf Lax. Thanks for the hint, I truly appreciated
$endgroup$
– nerd
Nov 29 '15 at 5:32
$begingroup$
Right, good point, I left out the easiest inequality I could use out of Hopf Lax. Thanks for the hint, I truly appreciated
$endgroup$
– nerd
Nov 29 '15 at 5:32
add a comment |
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$begingroup$
Using your notation, you can actually show that $u^1(x,t)-u^2(x,t)ge g^1(y)-g^2(y)$, and $u^1(x,t)-u^2(x,t)le g^1(y')-g^2(y')$. This is because Hopf Lax gives us $u^2(x,t)le tL(x-y/t) + g^2(y)$, and similarly with $u^1$.
$endgroup$
– Joey Zou
Nov 29 '15 at 4:01
1
$begingroup$
Right, good point, I left out the easiest inequality I could use out of Hopf Lax. Thanks for the hint, I truly appreciated
$endgroup$
– nerd
Nov 29 '15 at 5:32