Mixing
Modified Heat Equation: k is not a constant
$begingroup$
Given the heat equation,
$$
u_{t} = ku_{xx},
$$
how do we modify the solution below (when $k$ is a constant)
$$
u(x,t) = frac{1}{sqrt{4pi kt}}intlimits_{-infty}^{infty} g(y)e^{frac{-(x-y)^{2}}{4kt}} dy
$$
to solve the equation also for $k = k(t)$? Thank you.
pde heat-equation
edited Jan 23 at 20:41
Daniele Tampieri
2,1371621
asked Jan 7 at 23:15
robertrobert
115
$endgroup$
add a comment |
$begingroup$
Given the heat equation,
$$
u_{t} = ku_{xx},
$$
how do we modify the solution below (when $k$ is a constant)
$$
u(x,t) = frac{1}{sqrt{4pi kt}}intlimits_{-infty}^{infty} g(y)e^{frac{-(x-y)^{2}}{4kt}} dy
$$
to solve the equation also for $k = k(t)$? Thank you.
pde heat-equation
edited Jan 23 at 20:41
Daniele Tampieri
2,1371621
asked Jan 7 at 23:15
robertrobert
115
$endgroup$
$begingroup$
Keep in mind that the heat equation is only well-posed for $k>0$. So, you need to keep in mind that you time-dependent coefficient $k(t)>0$ for all $t$.
$endgroup$
– D.B.
Jan 8 at 2:50
add a comment |
$begingroup$
Given the heat equation,
$$
u_{t} = ku_{xx},
$$
how do we modify the solution below (when $k$ is a constant)
$$
u(x,t) = frac{1}{sqrt{4pi kt}}intlimits_{-infty}^{infty} g(y)e^{frac{-(x-y)^{2}}{4kt}} dy
$$
to solve the equation also for $k = k(t)$? Thank you.
pde heat-equation
edited Jan 23 at 20:41
Daniele Tampieri
2,1371621
asked Jan 7 at 23:15
robertrobert
115
$endgroup$
Given the heat equation,
$$
u_{t} = ku_{xx},
$$
how do we modify the solution below (when $k$ is a constant)
$$
u(x,t) = frac{1}{sqrt{4pi kt}}intlimits_{-infty}^{infty} g(y)e^{frac{-(x-y)^{2}}{4kt}} dy
$$
to solve the equation also for $k = k(t)$? Thank you.
pde heat-equation
pde heat-equation
edited Jan 23 at 20:41
Daniele Tampieri
2,1371621
asked Jan 7 at 23:15
robertrobert
115
edited Jan 23 at 20:41
Daniele Tampieri
2,1371621
asked Jan 7 at 23:15
robertrobert
115
edited Jan 23 at 20:41
Daniele Tampieri
2,1371621
edited Jan 23 at 20:41
Daniele Tampieri
2,1371621
edited Jan 23 at 20:41
Daniele Tampieri
2,1371621
2,1371621
asked Jan 7 at 23:15
robertrobert
115
asked Jan 7 at 23:15
robertrobert
115
asked Jan 7 at 23:15
robertrobert
115
115
$begingroup$
Keep in mind that the heat equation is only well-posed for $k>0$. So, you need to keep in mind that you time-dependent coefficient $k(t)>0$ for all $t$.
$endgroup$
– D.B.
Jan 8 at 2:50
add a comment |
$begingroup$
Keep in mind that the heat equation is only well-posed for $k>0$. So, you need to keep in mind that you time-dependent coefficient $k(t)>0$ for all $t$.
$endgroup$
– D.B.
Jan 8 at 2:50
$begingroup$
Keep in mind that the heat equation is only well-posed for $k>0$. So, you need to keep in mind that you time-dependent coefficient $k(t)>0$ for all $t$.
$endgroup$
– D.B.
Jan 8 at 2:50
$begingroup$
Keep in mind that the heat equation is only well-posed for $k>0$. So, you need to keep in mind that you time-dependent coefficient $k(t)>0$ for all $t$.
$endgroup$
– D.B.
Jan 8 at 2:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The solution does not change much, denote $K(t) = int_0^t k(z),dz$, then the solution is
$$u(x,t) = g(x) ast frac{e^{frac{-x^2}{4K(t)}}}{sqrt{4 pi K(t)}}$$
I omit the details because this is obviously a homework question, but my hint is to notice how the Fourier Transform does not care about $t$.
answered Jan 8 at 6:03
DaveNineDaveNine
1,286914
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The solution does not change much, denote $K(t) = int_0^t k(z),dz$, then the solution is
$$u(x,t) = g(x) ast frac{e^{frac{-x^2}{4K(t)}}}{sqrt{4 pi K(t)}}$$
I omit the details because this is obviously a homework question, but my hint is to notice how the Fourier Transform does not care about $t$.
answered Jan 8 at 6:03
DaveNineDaveNine
1,286914
$endgroup$
add a comment |
$begingroup$
The solution does not change much, denote $K(t) = int_0^t k(z),dz$, then the solution is
$$u(x,t) = g(x) ast frac{e^{frac{-x^2}{4K(t)}}}{sqrt{4 pi K(t)}}$$
I omit the details because this is obviously a homework question, but my hint is to notice how the Fourier Transform does not care about $t$.
answered Jan 8 at 6:03
DaveNineDaveNine
1,286914
$endgroup$
add a comment |
$begingroup$
The solution does not change much, denote $K(t) = int_0^t k(z),dz$, then the solution is
$$u(x,t) = g(x) ast frac{e^{frac{-x^2}{4K(t)}}}{sqrt{4 pi K(t)}}$$
I omit the details because this is obviously a homework question, but my hint is to notice how the Fourier Transform does not care about $t$.
answered Jan 8 at 6:03
DaveNineDaveNine
1,286914
$endgroup$
The solution does not change much, denote $K(t) = int_0^t k(z),dz$, then the solution is
$$u(x,t) = g(x) ast frac{e^{frac{-x^2}{4K(t)}}}{sqrt{4 pi K(t)}}$$
I omit the details because this is obviously a homework question, but my hint is to notice how the Fourier Transform does not care about $t$.
answered Jan 8 at 6:03
DaveNineDaveNine
1,286914
answered Jan 8 at 6:03
DaveNineDaveNine
1,286914
answered Jan 8 at 6:03
DaveNineDaveNine
1,286914
answered Jan 8 at 6:03
DaveNineDaveNine
1,286914
1,286914
add a comment |
add a comment |
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$begingroup$
Keep in mind that the heat equation is only well-posed for $k>0$. So, you need to keep in mind that you time-dependent coefficient $k(t)>0$ for all $t$.
$endgroup$
– D.B.
Jan 8 at 2:50