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Asymptotics Neumann Solution Stefan problem
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Consider the one-phase Stefan problem, defined on the moving domain $[0,s(t)]$ where the temperature inside the domain is determined by the heat equation $$T_t=T_xx,qquad 0<x<s(t),$$ subject to the boundary conditions $$T(0,t)=-1,quad T(s(t),t)=0,$$ and the moving interface is determined by the Stefan condition $$betadot s=T_x|_x=s(t),$$ where $beta$ is the Stefan number. This is one of the rare configurations where the problem has an analytical solution, known as the Neumann solution:$$T(x,t)=fractexterf(x/2sqrtt)texterf(lambda)-1,qquad s(t)=2lambdasqrtt,$$ where $lambda$ satisfies the transcendental equation
beginequation
betasqrtpilambda e^lambda^2texterf(lambda)=1.
endequation
For large Stefan numbers we have $lambdasimbeta^-1$, but what happens in the case where $betall1$? The Stefan problem itself is much more complicated to study in this case, but I am only interested in seeing what happens to the solution of the transcendental equation above. Any hint on how to start?
pde asymptotics roots
edited Jul 24 at 18:30
Antonio Vargas
20.2k244110
asked Jul 24 at 11:01
Marc
42139
add a comment |Â
up vote
3
down vote
favorite
Consider the one-phase Stefan problem, defined on the moving domain $[0,s(t)]$ where the temperature inside the domain is determined by the heat equation $$T_t=T_xx,qquad 0<x<s(t),$$ subject to the boundary conditions $$T(0,t)=-1,quad T(s(t),t)=0,$$ and the moving interface is determined by the Stefan condition $$betadot s=T_x|_x=s(t),$$ where $beta$ is the Stefan number. This is one of the rare configurations where the problem has an analytical solution, known as the Neumann solution:$$T(x,t)=fractexterf(x/2sqrtt)texterf(lambda)-1,qquad s(t)=2lambdasqrtt,$$ where $lambda$ satisfies the transcendental equation
beginequation
betasqrtpilambda e^lambda^2texterf(lambda)=1.
endequation
For large Stefan numbers we have $lambdasimbeta^-1$, but what happens in the case where $betall1$? The Stefan problem itself is much more complicated to study in this case, but I am only interested in seeing what happens to the solution of the transcendental equation above. Any hint on how to start?
pde asymptotics roots
edited Jul 24 at 18:30
Antonio Vargas
20.2k244110
asked Jul 24 at 11:01
Marc
42139
1
I think I've got it. For small Stefan numbers we expect $lambdagg1$. By expanding the error function an rearranging terms in the end I get $lambdasimbeta^-1/2$.
– Marc
Jul 24 at 11:12
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Consider the one-phase Stefan problem, defined on the moving domain $[0,s(t)]$ where the temperature inside the domain is determined by the heat equation $$T_t=T_xx,qquad 0<x<s(t),$$ subject to the boundary conditions $$T(0,t)=-1,quad T(s(t),t)=0,$$ and the moving interface is determined by the Stefan condition $$betadot s=T_x|_x=s(t),$$ where $beta$ is the Stefan number. This is one of the rare configurations where the problem has an analytical solution, known as the Neumann solution:$$T(x,t)=fractexterf(x/2sqrtt)texterf(lambda)-1,qquad s(t)=2lambdasqrtt,$$ where $lambda$ satisfies the transcendental equation
beginequation
betasqrtpilambda e^lambda^2texterf(lambda)=1.
endequation
For large Stefan numbers we have $lambdasimbeta^-1$, but what happens in the case where $betall1$? The Stefan problem itself is much more complicated to study in this case, but I am only interested in seeing what happens to the solution of the transcendental equation above. Any hint on how to start?
pde asymptotics roots
edited Jul 24 at 18:30
Antonio Vargas
20.2k244110
asked Jul 24 at 11:01
Marc
42139
Consider the one-phase Stefan problem, defined on the moving domain $[0,s(t)]$ where the temperature inside the domain is determined by the heat equation $$T_t=T_xx,qquad 0<x<s(t),$$ subject to the boundary conditions $$T(0,t)=-1,quad T(s(t),t)=0,$$ and the moving interface is determined by the Stefan condition $$betadot s=T_x|_x=s(t),$$ where $beta$ is the Stefan number. This is one of the rare configurations where the problem has an analytical solution, known as the Neumann solution:$$T(x,t)=fractexterf(x/2sqrtt)texterf(lambda)-1,qquad s(t)=2lambdasqrtt,$$ where $lambda$ satisfies the transcendental equation
beginequation
betasqrtpilambda e^lambda^2texterf(lambda)=1.
endequation
For large Stefan numbers we have $lambdasimbeta^-1$, but what happens in the case where $betall1$? The Stefan problem itself is much more complicated to study in this case, but I am only interested in seeing what happens to the solution of the transcendental equation above. Any hint on how to start?
pde asymptotics roots
edited Jul 24 at 18:30
Antonio Vargas
20.2k244110
asked Jul 24 at 11:01
Marc
42139
edited Jul 24 at 18:30
Antonio Vargas
20.2k244110
edited Jul 24 at 18:30
Antonio Vargas
20.2k244110
edited Jul 24 at 18:30
Antonio Vargas
20.2k244110
20.2k244110
asked Jul 24 at 11:01
Marc
42139
asked Jul 24 at 11:01
Marc
42139
asked Jul 24 at 11:01
Marc
42139
42139
1
I think I've got it. For small Stefan numbers we expect $lambdagg1$. By expanding the error function an rearranging terms in the end I get $lambdasimbeta^-1/2$.
– Marc
Jul 24 at 11:12
add a comment |Â
1
I think I've got it. For small Stefan numbers we expect $lambdagg1$. By expanding the error function an rearranging terms in the end I get $lambdasimbeta^-1/2$.
– Marc
Jul 24 at 11:12
1
1
I think I've got it. For small Stefan numbers we expect $lambdagg1$. By expanding the error function an rearranging terms in the end I get $lambdasimbeta^-1/2$.
– Marc
Jul 24 at 11:12
I think I've got it. For small Stefan numbers we expect $lambdagg1$. By expanding the error function an rearranging terms in the end I get $lambdasimbeta^-1/2$.
– Marc
Jul 24 at 11:12
add a comment |Â
1 Answer
1
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oldest
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up vote
0
down vote
Let's rearrange the equation to
$$
sqrtpilambda e^lambda^2operatornameerf(lambda) = frac1beta.
$$
Note that the function $f(lambda) = lambda e^lambda^2 operatornameerf(lambda)$ is strictly increasing for $lambda geq 0$ with $f(0) = 0$ and $f(infty) = infty$, so this equation has a single positive solution $lambda$ for any $beta > 0$. Further,
$$
lim_beta to infty lambda = 0 qquad textand qquad lim_beta to 0^+ lambda = infty.
$$
So, when $beta$ is large $lambda$ is small, and for small $lambda$ we have the leading order approximation
$$
frac1beta = sqrtpilambda cdot e^lambda^2 cdot operatornameerf(lambda) sim sqrtpi lambda cdot 1 cdot frac2lambdasqrtpi = 2lambda^2,
$$
which imples that
$$
lambda sim frac1sqrt2beta qquad textas beta to infty. tag$*$
$$
When $beta$ is small $lambda$ is large, leading to the approximation
$$
frac1beta = sqrtpilambda e^lambda^2 cdot operatornameerf(lambda) sim sqrtpilambda e^lambda^2 cdot 1 = sqrtpilambda e^lambda^2.
$$
Taking logarithms, this yields
$$
log frac1beta = lambda^2 + log sqrtpilambda + o(1) sim lambda^2,
$$
and thus
$$
lambda sim sqrtlog frac1beta qquad textas beta to 0^+. tag$**$
$$
answered Jul 24 at 18:29
Antonio Vargas
20.2k244110
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let's rearrange the equation to
$$
sqrtpilambda e^lambda^2operatornameerf(lambda) = frac1beta.
$$
Note that the function $f(lambda) = lambda e^lambda^2 operatornameerf(lambda)$ is strictly increasing for $lambda geq 0$ with $f(0) = 0$ and $f(infty) = infty$, so this equation has a single positive solution $lambda$ for any $beta > 0$. Further,
$$
lim_beta to infty lambda = 0 qquad textand qquad lim_beta to 0^+ lambda = infty.
$$
So, when $beta$ is large $lambda$ is small, and for small $lambda$ we have the leading order approximation
$$
frac1beta = sqrtpilambda cdot e^lambda^2 cdot operatornameerf(lambda) sim sqrtpi lambda cdot 1 cdot frac2lambdasqrtpi = 2lambda^2,
$$
which imples that
$$
lambda sim frac1sqrt2beta qquad textas beta to infty. tag$*$
$$
When $beta$ is small $lambda$ is large, leading to the approximation
$$
frac1beta = sqrtpilambda e^lambda^2 cdot operatornameerf(lambda) sim sqrtpilambda e^lambda^2 cdot 1 = sqrtpilambda e^lambda^2.
$$
Taking logarithms, this yields
$$
log frac1beta = lambda^2 + log sqrtpilambda + o(1) sim lambda^2,
$$
and thus
$$
lambda sim sqrtlog frac1beta qquad textas beta to 0^+. tag$**$
$$
answered Jul 24 at 18:29
Antonio Vargas
20.2k244110
add a comment |Â
up vote
0
down vote
Let's rearrange the equation to
$$
sqrtpilambda e^lambda^2operatornameerf(lambda) = frac1beta.
$$
Note that the function $f(lambda) = lambda e^lambda^2 operatornameerf(lambda)$ is strictly increasing for $lambda geq 0$ with $f(0) = 0$ and $f(infty) = infty$, so this equation has a single positive solution $lambda$ for any $beta > 0$. Further,
$$
lim_beta to infty lambda = 0 qquad textand qquad lim_beta to 0^+ lambda = infty.
$$
So, when $beta$ is large $lambda$ is small, and for small $lambda$ we have the leading order approximation
$$
frac1beta = sqrtpilambda cdot e^lambda^2 cdot operatornameerf(lambda) sim sqrtpi lambda cdot 1 cdot frac2lambdasqrtpi = 2lambda^2,
$$
which imples that
$$
lambda sim frac1sqrt2beta qquad textas beta to infty. tag$*$
$$
When $beta$ is small $lambda$ is large, leading to the approximation
$$
frac1beta = sqrtpilambda e^lambda^2 cdot operatornameerf(lambda) sim sqrtpilambda e^lambda^2 cdot 1 = sqrtpilambda e^lambda^2.
$$
Taking logarithms, this yields
$$
log frac1beta = lambda^2 + log sqrtpilambda + o(1) sim lambda^2,
$$
and thus
$$
lambda sim sqrtlog frac1beta qquad textas beta to 0^+. tag$**$
$$
answered Jul 24 at 18:29
Antonio Vargas
20.2k244110
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let's rearrange the equation to
$$
sqrtpilambda e^lambda^2operatornameerf(lambda) = frac1beta.
$$
Note that the function $f(lambda) = lambda e^lambda^2 operatornameerf(lambda)$ is strictly increasing for $lambda geq 0$ with $f(0) = 0$ and $f(infty) = infty$, so this equation has a single positive solution $lambda$ for any $beta > 0$. Further,
$$
lim_beta to infty lambda = 0 qquad textand qquad lim_beta to 0^+ lambda = infty.
$$
So, when $beta$ is large $lambda$ is small, and for small $lambda$ we have the leading order approximation
$$
frac1beta = sqrtpilambda cdot e^lambda^2 cdot operatornameerf(lambda) sim sqrtpi lambda cdot 1 cdot frac2lambdasqrtpi = 2lambda^2,
$$
which imples that
$$
lambda sim frac1sqrt2beta qquad textas beta to infty. tag$*$
$$
When $beta$ is small $lambda$ is large, leading to the approximation
$$
frac1beta = sqrtpilambda e^lambda^2 cdot operatornameerf(lambda) sim sqrtpilambda e^lambda^2 cdot 1 = sqrtpilambda e^lambda^2.
$$
Taking logarithms, this yields
$$
log frac1beta = lambda^2 + log sqrtpilambda + o(1) sim lambda^2,
$$
and thus
$$
lambda sim sqrtlog frac1beta qquad textas beta to 0^+. tag$**$
$$
answered Jul 24 at 18:29
Antonio Vargas
20.2k244110
Let's rearrange the equation to
$$
sqrtpilambda e^lambda^2operatornameerf(lambda) = frac1beta.
$$
Note that the function $f(lambda) = lambda e^lambda^2 operatornameerf(lambda)$ is strictly increasing for $lambda geq 0$ with $f(0) = 0$ and $f(infty) = infty$, so this equation has a single positive solution $lambda$ for any $beta > 0$. Further,
$$
lim_beta to infty lambda = 0 qquad textand qquad lim_beta to 0^+ lambda = infty.
$$
So, when $beta$ is large $lambda$ is small, and for small $lambda$ we have the leading order approximation
$$
frac1beta = sqrtpilambda cdot e^lambda^2 cdot operatornameerf(lambda) sim sqrtpi lambda cdot 1 cdot frac2lambdasqrtpi = 2lambda^2,
$$
which imples that
$$
lambda sim frac1sqrt2beta qquad textas beta to infty. tag$*$
$$
When $beta$ is small $lambda$ is large, leading to the approximation
$$
frac1beta = sqrtpilambda e^lambda^2 cdot operatornameerf(lambda) sim sqrtpilambda e^lambda^2 cdot 1 = sqrtpilambda e^lambda^2.
$$
Taking logarithms, this yields
$$
log frac1beta = lambda^2 + log sqrtpilambda + o(1) sim lambda^2,
$$
and thus
$$
lambda sim sqrtlog frac1beta qquad textas beta to 0^+. tag$**$
$$
answered Jul 24 at 18:29
Antonio Vargas
20.2k244110
answered Jul 24 at 18:29
Antonio Vargas
20.2k244110
answered Jul 24 at 18:29
Antonio Vargas
20.2k244110
answered Jul 24 at 18:29
Antonio Vargas
20.2k244110
20.2k244110
add a comment |Â
add a comment |Â
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1
I think I've got it. For small Stefan numbers we expect $lambdagg1$. By expanding the error function an rearranging terms in the end I get $lambdasimbeta^-1/2$.
– Marc
Jul 24 at 11:12