Asymptotic solution
$begingroup$
I am looking for asymptotic solutions to the equation
$$alpha^{-1}x+sqrt{pi}frac{sqrt{x}}{2}text{erf}left(frac{sqrt{x}}{2}right)=beta^{-1}e^{-x/4},qquad alphall1,betagg1.$$ When $alpha$ is large and the first term is negligible, this is easy to do, but I don't know how to proceed with the opposite case.
What I've tried for now is the following: For $alpha,beta^{-1}=0$, which is the limiting case, I get $x=0$, hence I have to introduce a scaling $x=epsilonhat x$, where $epsilon=epsilon(alpha,beta)ll1$. Introducing this into the equation above allows me to simplify terms and reduce the equation (if I'm not wrong) to
$$epsilon(1color{red}{+}alpha/2)hat x=alphabeta^{-1},$$ therefore I can balance the equation by choosing $epsilon=alphabeta^{-1}$ and finally $$hat xapproxcolor{red}{2/(2+alpha)}qquadRightarrowqquad xapproxalphabeta^{-1}.$$
Is this correct? Any hints or help on this?
Thanks in advance!
$color{red}{text{Edit: The leading order term had a mistake, I have corrected it.}}$
asymptotics
asked Dec 21 '18 at 10:20
MarcMarc
44139
$endgroup$
add a comment |
$begingroup$
I am looking for asymptotic solutions to the equation
$$alpha^{-1}x+sqrt{pi}frac{sqrt{x}}{2}text{erf}left(frac{sqrt{x}}{2}right)=beta^{-1}e^{-x/4},qquad alphall1,betagg1.$$ When $alpha$ is large and the first term is negligible, this is easy to do, but I don't know how to proceed with the opposite case.
What I've tried for now is the following: For $alpha,beta^{-1}=0$, which is the limiting case, I get $x=0$, hence I have to introduce a scaling $x=epsilonhat x$, where $epsilon=epsilon(alpha,beta)ll1$. Introducing this into the equation above allows me to simplify terms and reduce the equation (if I'm not wrong) to
$$epsilon(1color{red}{+}alpha/2)hat x=alphabeta^{-1},$$ therefore I can balance the equation by choosing $epsilon=alphabeta^{-1}$ and finally $$hat xapproxcolor{red}{2/(2+alpha)}qquadRightarrowqquad xapproxalphabeta^{-1}.$$
Is this correct? Any hints or help on this?
Thanks in advance!
$color{red}{text{Edit: The leading order term had a mistake, I have corrected it.}}$
asymptotics
asked Dec 21 '18 at 10:20
MarcMarc
44139
$endgroup$
add a comment |
$begingroup$
I am looking for asymptotic solutions to the equation
$$alpha^{-1}x+sqrt{pi}frac{sqrt{x}}{2}text{erf}left(frac{sqrt{x}}{2}right)=beta^{-1}e^{-x/4},qquad alphall1,betagg1.$$ When $alpha$ is large and the first term is negligible, this is easy to do, but I don't know how to proceed with the opposite case.
What I've tried for now is the following: For $alpha,beta^{-1}=0$, which is the limiting case, I get $x=0$, hence I have to introduce a scaling $x=epsilonhat x$, where $epsilon=epsilon(alpha,beta)ll1$. Introducing this into the equation above allows me to simplify terms and reduce the equation (if I'm not wrong) to
$$epsilon(1color{red}{+}alpha/2)hat x=alphabeta^{-1},$$ therefore I can balance the equation by choosing $epsilon=alphabeta^{-1}$ and finally $$hat xapproxcolor{red}{2/(2+alpha)}qquadRightarrowqquad xapproxalphabeta^{-1}.$$
Is this correct? Any hints or help on this?
Thanks in advance!
$color{red}{text{Edit: The leading order term had a mistake, I have corrected it.}}$
asymptotics
asked Dec 21 '18 at 10:20
MarcMarc
44139
$endgroup$
I am looking for asymptotic solutions to the equation
$$alpha^{-1}x+sqrt{pi}frac{sqrt{x}}{2}text{erf}left(frac{sqrt{x}}{2}right)=beta^{-1}e^{-x/4},qquad alphall1,betagg1.$$ When $alpha$ is large and the first term is negligible, this is easy to do, but I don't know how to proceed with the opposite case.
What I've tried for now is the following: For $alpha,beta^{-1}=0$, which is the limiting case, I get $x=0$, hence I have to introduce a scaling $x=epsilonhat x$, where $epsilon=epsilon(alpha,beta)ll1$. Introducing this into the equation above allows me to simplify terms and reduce the equation (if I'm not wrong) to
$$epsilon(1color{red}{+}alpha/2)hat x=alphabeta^{-1},$$ therefore I can balance the equation by choosing $epsilon=alphabeta^{-1}$ and finally $$hat xapproxcolor{red}{2/(2+alpha)}qquadRightarrowqquad xapproxalphabeta^{-1}.$$
Is this correct? Any hints or help on this?
Thanks in advance!
$color{red}{text{Edit: The leading order term had a mistake, I have corrected it.}}$
asymptotics
asymptotics
asked Dec 21 '18 at 10:20
MarcMarc
44139
asked Dec 21 '18 at 10:20
MarcMarc
44139
edited Feb 1 at 10:34
Marc
asked Dec 21 '18 at 10:20
MarcMarc
44139
asked Dec 21 '18 at 10:20
MarcMarc
44139
asked Dec 21 '18 at 10:20
MarcMarc
44139
44139
add a comment |
add a comment |
1 Answer
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Let $tilde beta = 1/beta$. Multiplying by $alpha$ and getting rid of the square roots, we can rewrite the equation as
$$x - alpha tilde beta e^{-x/4} +
alpha x int_0^{1/2} e^{-x t^2} d t = 0.$$
Now we can look for $x$ in the form $sum c_{i,j} alpha^i tilde beta {}^j$ by substituting the sum into the equation and taking the bivariate Taylor expansion around $alpha = 0, ,tilde beta = 0$.
Taking $x = alpha tilde beta + sum_{i = 0}^3 c_{i, 3 - i} alpha^i tilde beta {}^{3 - i}$ gives
$$c_{3, 0} alpha^3 +
left( c_{2, 1} + frac 1 2 right) alpha^2 tilde beta +
c_{1, 2} alpha tilde beta {}^2 +
c_{0, 3} tilde beta {}^3 = 0,$$
therefore we get one third-order term $-alpha^2 tilde beta/2$.
On the next step we get two fourth-order terms, which gives the approximation
$$x approx frac alpha beta -frac {alpha^2} {2 beta} +
frac {alpha^3} {4 beta} - frac {alpha^2} {4 beta^2}.$$
answered Dec 27 '18 at 22:12
MaximMaxim
6,2731221
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add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Let $tilde beta = 1/beta$. Multiplying by $alpha$ and getting rid of the square roots, we can rewrite the equation as
$$x - alpha tilde beta e^{-x/4} +
alpha x int_0^{1/2} e^{-x t^2} d t = 0.$$
Now we can look for $x$ in the form $sum c_{i,j} alpha^i tilde beta {}^j$ by substituting the sum into the equation and taking the bivariate Taylor expansion around $alpha = 0, ,tilde beta = 0$.
Taking $x = alpha tilde beta + sum_{i = 0}^3 c_{i, 3 - i} alpha^i tilde beta {}^{3 - i}$ gives
$$c_{3, 0} alpha^3 +
left( c_{2, 1} + frac 1 2 right) alpha^2 tilde beta +
c_{1, 2} alpha tilde beta {}^2 +
c_{0, 3} tilde beta {}^3 = 0,$$
therefore we get one third-order term $-alpha^2 tilde beta/2$.
On the next step we get two fourth-order terms, which gives the approximation
$$x approx frac alpha beta -frac {alpha^2} {2 beta} +
frac {alpha^3} {4 beta} - frac {alpha^2} {4 beta^2}.$$
answered Dec 27 '18 at 22:12
MaximMaxim
6,2731221
$endgroup$
add a comment |
$begingroup$
Let $tilde beta = 1/beta$. Multiplying by $alpha$ and getting rid of the square roots, we can rewrite the equation as
$$x - alpha tilde beta e^{-x/4} +
alpha x int_0^{1/2} e^{-x t^2} d t = 0.$$
Now we can look for $x$ in the form $sum c_{i,j} alpha^i tilde beta {}^j$ by substituting the sum into the equation and taking the bivariate Taylor expansion around $alpha = 0, ,tilde beta = 0$.
Taking $x = alpha tilde beta + sum_{i = 0}^3 c_{i, 3 - i} alpha^i tilde beta {}^{3 - i}$ gives
$$c_{3, 0} alpha^3 +
left( c_{2, 1} + frac 1 2 right) alpha^2 tilde beta +
c_{1, 2} alpha tilde beta {}^2 +
c_{0, 3} tilde beta {}^3 = 0,$$
therefore we get one third-order term $-alpha^2 tilde beta/2$.
On the next step we get two fourth-order terms, which gives the approximation
$$x approx frac alpha beta -frac {alpha^2} {2 beta} +
frac {alpha^3} {4 beta} - frac {alpha^2} {4 beta^2}.$$
answered Dec 27 '18 at 22:12
MaximMaxim
6,2731221
$endgroup$
add a comment |
$begingroup$
Let $tilde beta = 1/beta$. Multiplying by $alpha$ and getting rid of the square roots, we can rewrite the equation as
$$x - alpha tilde beta e^{-x/4} +
alpha x int_0^{1/2} e^{-x t^2} d t = 0.$$
Now we can look for $x$ in the form $sum c_{i,j} alpha^i tilde beta {}^j$ by substituting the sum into the equation and taking the bivariate Taylor expansion around $alpha = 0, ,tilde beta = 0$.
Taking $x = alpha tilde beta + sum_{i = 0}^3 c_{i, 3 - i} alpha^i tilde beta {}^{3 - i}$ gives
$$c_{3, 0} alpha^3 +
left( c_{2, 1} + frac 1 2 right) alpha^2 tilde beta +
c_{1, 2} alpha tilde beta {}^2 +
c_{0, 3} tilde beta {}^3 = 0,$$
therefore we get one third-order term $-alpha^2 tilde beta/2$.
On the next step we get two fourth-order terms, which gives the approximation
$$x approx frac alpha beta -frac {alpha^2} {2 beta} +
frac {alpha^3} {4 beta} - frac {alpha^2} {4 beta^2}.$$
answered Dec 27 '18 at 22:12
MaximMaxim
6,2731221
$endgroup$
Let $tilde beta = 1/beta$. Multiplying by $alpha$ and getting rid of the square roots, we can rewrite the equation as
$$x - alpha tilde beta e^{-x/4} +
alpha x int_0^{1/2} e^{-x t^2} d t = 0.$$
Now we can look for $x$ in the form $sum c_{i,j} alpha^i tilde beta {}^j$ by substituting the sum into the equation and taking the bivariate Taylor expansion around $alpha = 0, ,tilde beta = 0$.
Taking $x = alpha tilde beta + sum_{i = 0}^3 c_{i, 3 - i} alpha^i tilde beta {}^{3 - i}$ gives
$$c_{3, 0} alpha^3 +
left( c_{2, 1} + frac 1 2 right) alpha^2 tilde beta +
c_{1, 2} alpha tilde beta {}^2 +
c_{0, 3} tilde beta {}^3 = 0,$$
therefore we get one third-order term $-alpha^2 tilde beta/2$.
On the next step we get two fourth-order terms, which gives the approximation
$$x approx frac alpha beta -frac {alpha^2} {2 beta} +
frac {alpha^3} {4 beta} - frac {alpha^2} {4 beta^2}.$$
answered Dec 27 '18 at 22:12
MaximMaxim
6,2731221
edited Jan 6 at 22:35
answered Dec 27 '18 at 22:12
MaximMaxim
6,2731221
answered Dec 27 '18 at 22:12
MaximMaxim
6,2731221
answered Dec 27 '18 at 22:12
MaximMaxim
6,2731221
6,2731221
add a comment |
add a comment |
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