Solution of diffusion equation on infinite region using residue theorem
$begingroup$
The problem is as follows: Solve the partial diffusion equation $frac{partial }{{partial t}}Cleft( {x,t} right) = frac{{{partial ^2}}}{{partial {x^2}}}Cleft( {x,t} right)$ with initial condition $Cleft( {x,0} right) = frac{1}{{pi left( {1 + {{left( {x - 2} right)}^2}} right)}}$, where $mathop {lim }limits_{x to pm infty } Cleft( {x,t} right) = mathop {lim }limits_{x to pm infty } frac{partial }{{partial x}}Cleft( {x,t} right) = 0$.
Now, it is well known that the solution is given by $uleft( {x,t} right) = frac{1}{{2sqrt {pi t} }}int_{ - infty }^{ + infty } {frac{{exp left( { - frac{{{{left( {x - xi } right)}^2}}}{{4t}}} right)}}{{pi left( {1 + {{left( {xi - 2} right)}^2}} right)}}dxi } $.
What I was wondering was if there is any way to evaluate the integral and if there is, what method should I use.
I've tried using the residue theorem and have concluded that for $z = R{e^{itheta }}$ the integrals along parts of contour for $theta in left[ { - frac{pi }{2},frac{pi }{2}} right] cup left[ {frac{{3pi }}{2},frac{{5pi }}{2}} right]$ vanish as $R to + infty $.
So instead, I have decided to use a rectangle in the upper half-plane with vertices $ pm R$ and $ pm R + ai$ as an integration contour, where $a > 0$ is such that it contains the pole $2 + i$. The previous observation guarantees that the integrals on the vertical sides vanish as $R to + infty $, leaving me with $int_{ - infty }^{ + infty } {fleft( z right)dz} - int_{ - infty }^{ + infty } {fleft( {z + ai} right)dz} = 2pi ioperatorname{Res} left( {f,2 + i} right)$.
However, I don't see how to evaluate $int_{ - infty }^{ + infty } {fleft( {z + ai} right)dz} $ for $fleft( z right) = frac{{exp left( { - frac{{{{left( {x - z} right)}^2}}}{{4t}}} right)}}{{pi left( {1 + {{left( {z - 2} right)}^2}} right)}}$.
Can I use this approach to evaluate the integral? Should I choose a different integration contour? Is there another method?
integration pde residue-calculus
asked Dec 6 '18 at 20:17
AlenAlen
1,341620
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add a comment |
$begingroup$
The problem is as follows: Solve the partial diffusion equation $frac{partial }{{partial t}}Cleft( {x,t} right) = frac{{{partial ^2}}}{{partial {x^2}}}Cleft( {x,t} right)$ with initial condition $Cleft( {x,0} right) = frac{1}{{pi left( {1 + {{left( {x - 2} right)}^2}} right)}}$, where $mathop {lim }limits_{x to pm infty } Cleft( {x,t} right) = mathop {lim }limits_{x to pm infty } frac{partial }{{partial x}}Cleft( {x,t} right) = 0$.
Now, it is well known that the solution is given by $uleft( {x,t} right) = frac{1}{{2sqrt {pi t} }}int_{ - infty }^{ + infty } {frac{{exp left( { - frac{{{{left( {x - xi } right)}^2}}}{{4t}}} right)}}{{pi left( {1 + {{left( {xi - 2} right)}^2}} right)}}dxi } $.
What I was wondering was if there is any way to evaluate the integral and if there is, what method should I use.
I've tried using the residue theorem and have concluded that for $z = R{e^{itheta }}$ the integrals along parts of contour for $theta in left[ { - frac{pi }{2},frac{pi }{2}} right] cup left[ {frac{{3pi }}{2},frac{{5pi }}{2}} right]$ vanish as $R to + infty $.
So instead, I have decided to use a rectangle in the upper half-plane with vertices $ pm R$ and $ pm R + ai$ as an integration contour, where $a > 0$ is such that it contains the pole $2 + i$. The previous observation guarantees that the integrals on the vertical sides vanish as $R to + infty $, leaving me with $int_{ - infty }^{ + infty } {fleft( z right)dz} - int_{ - infty }^{ + infty } {fleft( {z + ai} right)dz} = 2pi ioperatorname{Res} left( {f,2 + i} right)$.
However, I don't see how to evaluate $int_{ - infty }^{ + infty } {fleft( {z + ai} right)dz} $ for $fleft( z right) = frac{{exp left( { - frac{{{{left( {x - z} right)}^2}}}{{4t}}} right)}}{{pi left( {1 + {{left( {z - 2} right)}^2}} right)}}$.
Can I use this approach to evaluate the integral? Should I choose a different integration contour? Is there another method?
integration pde residue-calculus
asked Dec 6 '18 at 20:17
AlenAlen
1,341620
$endgroup$
$begingroup$
Some integrals you just simply can’t write out explicitly(perhaps try a CAS). You get much more information out of the convolution integral anyways, and it would be far more elegant than anything else we could come up with.
$endgroup$
– DaveNine
Dec 7 '18 at 19:13
add a comment |
$begingroup$
The problem is as follows: Solve the partial diffusion equation $frac{partial }{{partial t}}Cleft( {x,t} right) = frac{{{partial ^2}}}{{partial {x^2}}}Cleft( {x,t} right)$ with initial condition $Cleft( {x,0} right) = frac{1}{{pi left( {1 + {{left( {x - 2} right)}^2}} right)}}$, where $mathop {lim }limits_{x to pm infty } Cleft( {x,t} right) = mathop {lim }limits_{x to pm infty } frac{partial }{{partial x}}Cleft( {x,t} right) = 0$.
Now, it is well known that the solution is given by $uleft( {x,t} right) = frac{1}{{2sqrt {pi t} }}int_{ - infty }^{ + infty } {frac{{exp left( { - frac{{{{left( {x - xi } right)}^2}}}{{4t}}} right)}}{{pi left( {1 + {{left( {xi - 2} right)}^2}} right)}}dxi } $.
What I was wondering was if there is any way to evaluate the integral and if there is, what method should I use.
I've tried using the residue theorem and have concluded that for $z = R{e^{itheta }}$ the integrals along parts of contour for $theta in left[ { - frac{pi }{2},frac{pi }{2}} right] cup left[ {frac{{3pi }}{2},frac{{5pi }}{2}} right]$ vanish as $R to + infty $.
So instead, I have decided to use a rectangle in the upper half-plane with vertices $ pm R$ and $ pm R + ai$ as an integration contour, where $a > 0$ is such that it contains the pole $2 + i$. The previous observation guarantees that the integrals on the vertical sides vanish as $R to + infty $, leaving me with $int_{ - infty }^{ + infty } {fleft( z right)dz} - int_{ - infty }^{ + infty } {fleft( {z + ai} right)dz} = 2pi ioperatorname{Res} left( {f,2 + i} right)$.
However, I don't see how to evaluate $int_{ - infty }^{ + infty } {fleft( {z + ai} right)dz} $ for $fleft( z right) = frac{{exp left( { - frac{{{{left( {x - z} right)}^2}}}{{4t}}} right)}}{{pi left( {1 + {{left( {z - 2} right)}^2}} right)}}$.
Can I use this approach to evaluate the integral? Should I choose a different integration contour? Is there another method?
integration pde residue-calculus
asked Dec 6 '18 at 20:17
AlenAlen
1,341620
$endgroup$
The problem is as follows: Solve the partial diffusion equation $frac{partial }{{partial t}}Cleft( {x,t} right) = frac{{{partial ^2}}}{{partial {x^2}}}Cleft( {x,t} right)$ with initial condition $Cleft( {x,0} right) = frac{1}{{pi left( {1 + {{left( {x - 2} right)}^2}} right)}}$, where $mathop {lim }limits_{x to pm infty } Cleft( {x,t} right) = mathop {lim }limits_{x to pm infty } frac{partial }{{partial x}}Cleft( {x,t} right) = 0$.
Now, it is well known that the solution is given by $uleft( {x,t} right) = frac{1}{{2sqrt {pi t} }}int_{ - infty }^{ + infty } {frac{{exp left( { - frac{{{{left( {x - xi } right)}^2}}}{{4t}}} right)}}{{pi left( {1 + {{left( {xi - 2} right)}^2}} right)}}dxi } $.
What I was wondering was if there is any way to evaluate the integral and if there is, what method should I use.
I've tried using the residue theorem and have concluded that for $z = R{e^{itheta }}$ the integrals along parts of contour for $theta in left[ { - frac{pi }{2},frac{pi }{2}} right] cup left[ {frac{{3pi }}{2},frac{{5pi }}{2}} right]$ vanish as $R to + infty $.
So instead, I have decided to use a rectangle in the upper half-plane with vertices $ pm R$ and $ pm R + ai$ as an integration contour, where $a > 0$ is such that it contains the pole $2 + i$. The previous observation guarantees that the integrals on the vertical sides vanish as $R to + infty $, leaving me with $int_{ - infty }^{ + infty } {fleft( z right)dz} - int_{ - infty }^{ + infty } {fleft( {z + ai} right)dz} = 2pi ioperatorname{Res} left( {f,2 + i} right)$.
However, I don't see how to evaluate $int_{ - infty }^{ + infty } {fleft( {z + ai} right)dz} $ for $fleft( z right) = frac{{exp left( { - frac{{{{left( {x - z} right)}^2}}}{{4t}}} right)}}{{pi left( {1 + {{left( {z - 2} right)}^2}} right)}}$.
Can I use this approach to evaluate the integral? Should I choose a different integration contour? Is there another method?
integration pde residue-calculus
integration pde residue-calculus
asked Dec 6 '18 at 20:17
AlenAlen
1,341620
asked Dec 6 '18 at 20:17
AlenAlen
1,341620
edited Dec 6 '18 at 22:14
Alen
asked Dec 6 '18 at 20:17
AlenAlen
1,341620
asked Dec 6 '18 at 20:17
AlenAlen
1,341620
asked Dec 6 '18 at 20:17
AlenAlen
1,341620
1,341620
$begingroup$
Some integrals you just simply can’t write out explicitly(perhaps try a CAS). You get much more information out of the convolution integral anyways, and it would be far more elegant than anything else we could come up with.
$endgroup$
– DaveNine
Dec 7 '18 at 19:13
add a comment |
$begingroup$
Some integrals you just simply can’t write out explicitly(perhaps try a CAS). You get much more information out of the convolution integral anyways, and it would be far more elegant than anything else we could come up with.
$endgroup$
– DaveNine
Dec 7 '18 at 19:13
$begingroup$
Some integrals you just simply can’t write out explicitly(perhaps try a CAS). You get much more information out of the convolution integral anyways, and it would be far more elegant than anything else we could come up with.
$endgroup$
– DaveNine
Dec 7 '18 at 19:13
$begingroup$
Some integrals you just simply can’t write out explicitly(perhaps try a CAS). You get much more information out of the convolution integral anyways, and it would be far more elegant than anything else we could come up with.
$endgroup$
– DaveNine
Dec 7 '18 at 19:13
add a comment |
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Some integrals you just simply can’t write out explicitly(perhaps try a CAS). You get much more information out of the convolution integral anyways, and it would be far more elegant than anything else we could come up with.
$endgroup$
– DaveNine
Dec 7 '18 at 19:13