Solving the diffusion equation with general initial condition $u(x,0)=f(x)$ Announcing the...
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Solving the diffusion equation with general initial condition $u(x,0)=f(x)$
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I know that the solution for the diffusion equation for a general initial condition $u(x,0)=f(x)$ can be given in the form
$$u(x,t)= frac{1}{sqrt{4πDt}} int_{-infty}^{infty}f(y)e^{frac{-(x-y)^2}{4Dt}}dy$$
How do I find an explicit form for this solution when $f(x)=1$ for $x$ between $0$ and $a$, and $0$ otherwise? I was thinking maybe via changing the variables, but didn't reach much further...
It would be helpful if the solution was in terms of the error function $operatorname{erf}(x)$.
pde
asked Mar 23 at 23:41
ISA-ISAISA-ISA
254
$endgroup$
add a comment |
$begingroup$
I know that the solution for the diffusion equation for a general initial condition $u(x,0)=f(x)$ can be given in the form
$$u(x,t)= frac{1}{sqrt{4πDt}} int_{-infty}^{infty}f(y)e^{frac{-(x-y)^2}{4Dt}}dy$$
How do I find an explicit form for this solution when $f(x)=1$ for $x$ between $0$ and $a$, and $0$ otherwise? I was thinking maybe via changing the variables, but didn't reach much further...
It would be helpful if the solution was in terms of the error function $operatorname{erf}(x)$.
pde
asked Mar 23 at 23:41
ISA-ISAISA-ISA
254
$endgroup$
$begingroup$
The solution may be written as a difference of error functions since it involves integrating the Gaussian kernel $frac1{sqrt{2pi}}e^{-u^2/2}$ over some interval which varies in $(x,t)$. So I don't know what you mean by "explicit" because that's necessarily a non-elementary function.
$endgroup$
– Shalop
Mar 24 at 4:45
$begingroup$
@shalop basically what is u(x,t) when the f(x) is as above
$endgroup$
– ISA-ISA
Mar 24 at 12:35
add a comment |
$begingroup$
I know that the solution for the diffusion equation for a general initial condition $u(x,0)=f(x)$ can be given in the form
$$u(x,t)= frac{1}{sqrt{4πDt}} int_{-infty}^{infty}f(y)e^{frac{-(x-y)^2}{4Dt}}dy$$
How do I find an explicit form for this solution when $f(x)=1$ for $x$ between $0$ and $a$, and $0$ otherwise? I was thinking maybe via changing the variables, but didn't reach much further...
It would be helpful if the solution was in terms of the error function $operatorname{erf}(x)$.
pde
asked Mar 23 at 23:41
ISA-ISAISA-ISA
254
$endgroup$
I know that the solution for the diffusion equation for a general initial condition $u(x,0)=f(x)$ can be given in the form
$$u(x,t)= frac{1}{sqrt{4πDt}} int_{-infty}^{infty}f(y)e^{frac{-(x-y)^2}{4Dt}}dy$$
How do I find an explicit form for this solution when $f(x)=1$ for $x$ between $0$ and $a$, and $0$ otherwise? I was thinking maybe via changing the variables, but didn't reach much further...
It would be helpful if the solution was in terms of the error function $operatorname{erf}(x)$.
pde
pde
asked Mar 23 at 23:41
ISA-ISAISA-ISA
254
asked Mar 23 at 23:41
ISA-ISAISA-ISA
254
edited Mar 24 at 12:32
ISA-ISA
asked Mar 23 at 23:41
ISA-ISAISA-ISA
254
asked Mar 23 at 23:41
ISA-ISAISA-ISA
254
asked Mar 23 at 23:41
ISA-ISAISA-ISA
254
254
$begingroup$
The solution may be written as a difference of error functions since it involves integrating the Gaussian kernel $frac1{sqrt{2pi}}e^{-u^2/2}$ over some interval which varies in $(x,t)$. So I don't know what you mean by "explicit" because that's necessarily a non-elementary function.
$endgroup$
– Shalop
Mar 24 at 4:45
$begingroup$
@shalop basically what is u(x,t) when the f(x) is as above
$endgroup$
– ISA-ISA
Mar 24 at 12:35
add a comment |
$begingroup$
The solution may be written as a difference of error functions since it involves integrating the Gaussian kernel $frac1{sqrt{2pi}}e^{-u^2/2}$ over some interval which varies in $(x,t)$. So I don't know what you mean by "explicit" because that's necessarily a non-elementary function.
$endgroup$
– Shalop
Mar 24 at 4:45
$begingroup$
@shalop basically what is u(x,t) when the f(x) is as above
$endgroup$
– ISA-ISA
Mar 24 at 12:35
$begingroup$
The solution may be written as a difference of error functions since it involves integrating the Gaussian kernel $frac1{sqrt{2pi}}e^{-u^2/2}$ over some interval which varies in $(x,t)$. So I don't know what you mean by "explicit" because that's necessarily a non-elementary function.
$endgroup$
– Shalop
Mar 24 at 4:45
$begingroup$
The solution may be written as a difference of error functions since it involves integrating the Gaussian kernel $frac1{sqrt{2pi}}e^{-u^2/2}$ over some interval which varies in $(x,t)$. So I don't know what you mean by "explicit" because that's necessarily a non-elementary function.
$endgroup$
– Shalop
Mar 24 at 4:45
$begingroup$
@shalop basically what is u(x,t) when the f(x) is as above
$endgroup$
– ISA-ISA
Mar 24 at 12:35
$begingroup$
@shalop basically what is u(x,t) when the f(x) is as above
$endgroup$
– ISA-ISA
Mar 24 at 12:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This answer is very close to what you're looking for. See if you can follow through it.
answered Mar 24 at 4:36
AEngineerAEngineer
1,6521318
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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$begingroup$
This answer is very close to what you're looking for. See if you can follow through it.
answered Mar 24 at 4:36
AEngineerAEngineer
1,6521318
$endgroup$
add a comment |
$begingroup$
This answer is very close to what you're looking for. See if you can follow through it.
answered Mar 24 at 4:36
AEngineerAEngineer
1,6521318
$endgroup$
add a comment |
$begingroup$
This answer is very close to what you're looking for. See if you can follow through it.
answered Mar 24 at 4:36
AEngineerAEngineer
1,6521318
$endgroup$
This answer is very close to what you're looking for. See if you can follow through it.
answered Mar 24 at 4:36
AEngineerAEngineer
1,6521318
answered Mar 24 at 4:36
AEngineerAEngineer
1,6521318
answered Mar 24 at 4:36
AEngineerAEngineer
1,6521318
answered Mar 24 at 4:36
AEngineerAEngineer
1,6521318
1,6521318
add a comment |
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$begingroup$
The solution may be written as a difference of error functions since it involves integrating the Gaussian kernel $frac1{sqrt{2pi}}e^{-u^2/2}$ over some interval which varies in $(x,t)$. So I don't know what you mean by "explicit" because that's necessarily a non-elementary function.
$endgroup$
– Shalop
Mar 24 at 4:45
$begingroup$
@shalop basically what is u(x,t) when the f(x) is as above
$endgroup$
– ISA-ISA
Mar 24 at 12:35