Calculating the Hopf-Lax formula
The Hopf-Lax function I have is
$$u(t,x) = inf_{yin mathbb{R}} left{ C|y| + frac{(x-y)^2}{2t} right},$$
where $C>0$ is constant. By evaluating stationary and non-differentiable points I have found two possibilities for the infimum:
$$u_1(t,x) = frac{x^2}{2t},quad u_2(t,x) = C|x||1-Ct| + frac{(x-|1-Ct|x)^2}{2t}.$$
I know the initial condition for the related Cauchy problem is given by $u(0,x) = C|x|$ which leads me to believe that the function takes the form of $u_2$ for at least small $t$. I also know that these two functions ($u_1$ and $u_2$) intersect at $t=frac{1}{C}$, however I believe there is another intersection which is where I believe the piecewise Hopf-Lax function begins to take values according to $u_1$. I'm struggling to find a "nice" form for this point, if it exists. What is the explicit expression for the above infimum?
pde supremum-and-infimum hamilton-jacobi-equation
edited Dec 11 '18 at 13:26
Harry49
5,99121031
add a comment |
The Hopf-Lax function I have is
$$u(t,x) = inf_{yin mathbb{R}} left{ C|y| + frac{(x-y)^2}{2t} right},$$
where $C>0$ is constant. By evaluating stationary and non-differentiable points I have found two possibilities for the infimum:
$$u_1(t,x) = frac{x^2}{2t},quad u_2(t,x) = C|x||1-Ct| + frac{(x-|1-Ct|x)^2}{2t}.$$
I know the initial condition for the related Cauchy problem is given by $u(0,x) = C|x|$ which leads me to believe that the function takes the form of $u_2$ for at least small $t$. I also know that these two functions ($u_1$ and $u_2$) intersect at $t=frac{1}{C}$, however I believe there is another intersection which is where I believe the piecewise Hopf-Lax function begins to take values according to $u_1$. I'm struggling to find a "nice" form for this point, if it exists. What is the explicit expression for the above infimum?
pde supremum-and-infimum hamilton-jacobi-equation
edited Dec 11 '18 at 13:26
Harry49
5,99121031
add a comment |
The Hopf-Lax function I have is
$$u(t,x) = inf_{yin mathbb{R}} left{ C|y| + frac{(x-y)^2}{2t} right},$$
where $C>0$ is constant. By evaluating stationary and non-differentiable points I have found two possibilities for the infimum:
$$u_1(t,x) = frac{x^2}{2t},quad u_2(t,x) = C|x||1-Ct| + frac{(x-|1-Ct|x)^2}{2t}.$$
I know the initial condition for the related Cauchy problem is given by $u(0,x) = C|x|$ which leads me to believe that the function takes the form of $u_2$ for at least small $t$. I also know that these two functions ($u_1$ and $u_2$) intersect at $t=frac{1}{C}$, however I believe there is another intersection which is where I believe the piecewise Hopf-Lax function begins to take values according to $u_1$. I'm struggling to find a "nice" form for this point, if it exists. What is the explicit expression for the above infimum?
pde supremum-and-infimum hamilton-jacobi-equation
edited Dec 11 '18 at 13:26
Harry49
5,99121031
The Hopf-Lax function I have is
$$u(t,x) = inf_{yin mathbb{R}} left{ C|y| + frac{(x-y)^2}{2t} right},$$
where $C>0$ is constant. By evaluating stationary and non-differentiable points I have found two possibilities for the infimum:
$$u_1(t,x) = frac{x^2}{2t},quad u_2(t,x) = C|x||1-Ct| + frac{(x-|1-Ct|x)^2}{2t}.$$
I know the initial condition for the related Cauchy problem is given by $u(0,x) = C|x|$ which leads me to believe that the function takes the form of $u_2$ for at least small $t$. I also know that these two functions ($u_1$ and $u_2$) intersect at $t=frac{1}{C}$, however I believe there is another intersection which is where I believe the piecewise Hopf-Lax function begins to take values according to $u_1$. I'm struggling to find a "nice" form for this point, if it exists. What is the explicit expression for the above infimum?
pde supremum-and-infimum hamilton-jacobi-equation
pde supremum-and-infimum hamilton-jacobi-equation
edited Dec 11 '18 at 13:26
Harry49
5,99121031
edited Dec 11 '18 at 13:26
Harry49
5,99121031
edited Dec 11 '18 at 13:26
Harry49
5,99121031
edited Dec 11 '18 at 13:26
Harry49
5,99121031
edited Dec 11 '18 at 13:26
Harry49
5,99121031
5,99121031
asked May 5 '17 at 16:38
user383264
add a comment |
add a comment |
1 Answer
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For fixed $(t,x)in (0,+infty)timesmathbb{R}$, the function $varphi(y) := C|y| + frac{(x-y)^2}{2t}$ is strictly convex and superlinear, hence it has a unique minimum point $y=y(t,x)$ which is characterized by the condition $0in [varphi'_-(y), varphi'_+(y)]$.
It is not difficult to check that
$$
y =
begin{cases}
0, & text{if} |x| leq Ct,\
x-Ct, & text{if} x > Ct,\
x+Ct, & text{if} x < -Ct,
end{cases}
$$
hence
$$
u(t,x) =
begin{cases}
x^2/(2t), & text{if} |x| leq Ct,\
C|x| - C^2t/2, & text{if} |x| > Ct.
end{cases}
$$
answered May 6 '17 at 13:05
Rigel
10.9k11320
Very nice reasoning - thanks!
– user383264
May 6 '17 at 20:43
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
For fixed $(t,x)in (0,+infty)timesmathbb{R}$, the function $varphi(y) := C|y| + frac{(x-y)^2}{2t}$ is strictly convex and superlinear, hence it has a unique minimum point $y=y(t,x)$ which is characterized by the condition $0in [varphi'_-(y), varphi'_+(y)]$.
It is not difficult to check that
$$
y =
begin{cases}
0, & text{if} |x| leq Ct,\
x-Ct, & text{if} x > Ct,\
x+Ct, & text{if} x < -Ct,
end{cases}
$$
hence
$$
u(t,x) =
begin{cases}
x^2/(2t), & text{if} |x| leq Ct,\
C|x| - C^2t/2, & text{if} |x| > Ct.
end{cases}
$$
answered May 6 '17 at 13:05
Rigel
10.9k11320
Very nice reasoning - thanks!
– user383264
May 6 '17 at 20:43
add a comment |
For fixed $(t,x)in (0,+infty)timesmathbb{R}$, the function $varphi(y) := C|y| + frac{(x-y)^2}{2t}$ is strictly convex and superlinear, hence it has a unique minimum point $y=y(t,x)$ which is characterized by the condition $0in [varphi'_-(y), varphi'_+(y)]$.
It is not difficult to check that
$$
y =
begin{cases}
0, & text{if} |x| leq Ct,\
x-Ct, & text{if} x > Ct,\
x+Ct, & text{if} x < -Ct,
end{cases}
$$
hence
$$
u(t,x) =
begin{cases}
x^2/(2t), & text{if} |x| leq Ct,\
C|x| - C^2t/2, & text{if} |x| > Ct.
end{cases}
$$
answered May 6 '17 at 13:05
Rigel
10.9k11320
Very nice reasoning - thanks!
– user383264
May 6 '17 at 20:43
add a comment |
For fixed $(t,x)in (0,+infty)timesmathbb{R}$, the function $varphi(y) := C|y| + frac{(x-y)^2}{2t}$ is strictly convex and superlinear, hence it has a unique minimum point $y=y(t,x)$ which is characterized by the condition $0in [varphi'_-(y), varphi'_+(y)]$.
It is not difficult to check that
$$
y =
begin{cases}
0, & text{if} |x| leq Ct,\
x-Ct, & text{if} x > Ct,\
x+Ct, & text{if} x < -Ct,
end{cases}
$$
hence
$$
u(t,x) =
begin{cases}
x^2/(2t), & text{if} |x| leq Ct,\
C|x| - C^2t/2, & text{if} |x| > Ct.
end{cases}
$$
answered May 6 '17 at 13:05
Rigel
10.9k11320
For fixed $(t,x)in (0,+infty)timesmathbb{R}$, the function $varphi(y) := C|y| + frac{(x-y)^2}{2t}$ is strictly convex and superlinear, hence it has a unique minimum point $y=y(t,x)$ which is characterized by the condition $0in [varphi'_-(y), varphi'_+(y)]$.
It is not difficult to check that
$$
y =
begin{cases}
0, & text{if} |x| leq Ct,\
x-Ct, & text{if} x > Ct,\
x+Ct, & text{if} x < -Ct,
end{cases}
$$
hence
$$
u(t,x) =
begin{cases}
x^2/(2t), & text{if} |x| leq Ct,\
C|x| - C^2t/2, & text{if} |x| > Ct.
end{cases}
$$
answered May 6 '17 at 13:05
Rigel
10.9k11320
answered May 6 '17 at 13:05
Rigel
10.9k11320
answered May 6 '17 at 13:05
Rigel
10.9k11320
answered May 6 '17 at 13:05
Rigel
10.9k11320
10.9k11320
Very nice reasoning - thanks!
– user383264
May 6 '17 at 20:43
add a comment |
Very nice reasoning - thanks!
– user383264
May 6 '17 at 20:43
Very nice reasoning - thanks!
– user383264
May 6 '17 at 20:43
Very nice reasoning - thanks!
– user383264
May 6 '17 at 20:43
add a comment |
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