Mixing
A specific partial differential equation using Fourier Transform
$begingroup$
I have the following PDE problem which I think sounds like a job for the Fourier transform:
$ u_t + 2u_x = u_{xx} space space space -infty < x < infty space space space t>0 $
$ u(x,0) = left{
begin{array}{ll}
0 & mbox{if } x > 0 \
e^x & mbox{if } x < 0
end{array}
right. $
I thought about using an integral transform to solve this and since the added condition is with a specific $t $ I thought it would be natural to transform with respect to $x$ and since $x$ can be any real number , I figured the Fourier Transform is the right choice so transforming the equation gives an ODE int terms of $t$ and transforming the initial condition is feasible but the problem comes with the solution which at first is of course stated in terms of the Fourier transform with respect to $x$ the function at the end is simply awful, so awful that I cannot even write the solution in terms of a convolution integral. Here is what I got
$ U(omega,t) = frac{1}{iomega + 1} e^{-t(omega^2+2iomega)} $
Perhaps my solution was incorrect somewhere or the Fourier transform is not the way to go but I would like to ask for help on this in terms of how to solve it properly, but of course there is the matter of different but equivalent definitions of the Fourier transform (I use the one with no normalization constant and positive complex exponent). All help appreciated thanks.
pde fourier-analysis integral-transforms
edited Jan 22 at 11:19
dmtri
1,5642521
asked Sep 29 '15 at 6:25
kronerkroner
1,3043926
$endgroup$
add a comment |
$begingroup$
I have the following PDE problem which I think sounds like a job for the Fourier transform:
$ u_t + 2u_x = u_{xx} space space space -infty < x < infty space space space t>0 $
$ u(x,0) = left{
begin{array}{ll}
0 & mbox{if } x > 0 \
e^x & mbox{if } x < 0
end{array}
right. $
I thought about using an integral transform to solve this and since the added condition is with a specific $t $ I thought it would be natural to transform with respect to $x$ and since $x$ can be any real number , I figured the Fourier Transform is the right choice so transforming the equation gives an ODE int terms of $t$ and transforming the initial condition is feasible but the problem comes with the solution which at first is of course stated in terms of the Fourier transform with respect to $x$ the function at the end is simply awful, so awful that I cannot even write the solution in terms of a convolution integral. Here is what I got
$ U(omega,t) = frac{1}{iomega + 1} e^{-t(omega^2+2iomega)} $
Perhaps my solution was incorrect somewhere or the Fourier transform is not the way to go but I would like to ask for help on this in terms of how to solve it properly, but of course there is the matter of different but equivalent definitions of the Fourier transform (I use the one with no normalization constant and positive complex exponent). All help appreciated thanks.
pde fourier-analysis integral-transforms
edited Jan 22 at 11:19
dmtri
1,5642521
asked Sep 29 '15 at 6:25
kronerkroner
1,3043926
$endgroup$
3
$begingroup$
A convenient strategy for this type of problem is to use the complex Laplace transform for the time variable and, as you already did, the Fourier transform for the space variable $x$. Tell me if you need further help.
$endgroup$
– Urgje
Sep 29 '15 at 13:36
$begingroup$
@Urgje : That certainly helped thank you it worked great
$endgroup$
– kroner
Sep 30 '15 at 3:26
$begingroup$
@Urgje: if you could post an answer with some more detail it would be appreciated. It is interesting.
$endgroup$
– Giuseppe Negro
Jan 22 at 11:24
add a comment |
$begingroup$
I have the following PDE problem which I think sounds like a job for the Fourier transform:
$ u_t + 2u_x = u_{xx} space space space -infty < x < infty space space space t>0 $
$ u(x,0) = left{
begin{array}{ll}
0 & mbox{if } x > 0 \
e^x & mbox{if } x < 0
end{array}
right. $
I thought about using an integral transform to solve this and since the added condition is with a specific $t $ I thought it would be natural to transform with respect to $x$ and since $x$ can be any real number , I figured the Fourier Transform is the right choice so transforming the equation gives an ODE int terms of $t$ and transforming the initial condition is feasible but the problem comes with the solution which at first is of course stated in terms of the Fourier transform with respect to $x$ the function at the end is simply awful, so awful that I cannot even write the solution in terms of a convolution integral. Here is what I got
$ U(omega,t) = frac{1}{iomega + 1} e^{-t(omega^2+2iomega)} $
Perhaps my solution was incorrect somewhere or the Fourier transform is not the way to go but I would like to ask for help on this in terms of how to solve it properly, but of course there is the matter of different but equivalent definitions of the Fourier transform (I use the one with no normalization constant and positive complex exponent). All help appreciated thanks.
pde fourier-analysis integral-transforms
edited Jan 22 at 11:19
dmtri
1,5642521
asked Sep 29 '15 at 6:25
kronerkroner
1,3043926
$endgroup$
I have the following PDE problem which I think sounds like a job for the Fourier transform:
$ u_t + 2u_x = u_{xx} space space space -infty < x < infty space space space t>0 $
$ u(x,0) = left{
begin{array}{ll}
0 & mbox{if } x > 0 \
e^x & mbox{if } x < 0
end{array}
right. $
I thought about using an integral transform to solve this and since the added condition is with a specific $t $ I thought it would be natural to transform with respect to $x$ and since $x$ can be any real number , I figured the Fourier Transform is the right choice so transforming the equation gives an ODE int terms of $t$ and transforming the initial condition is feasible but the problem comes with the solution which at first is of course stated in terms of the Fourier transform with respect to $x$ the function at the end is simply awful, so awful that I cannot even write the solution in terms of a convolution integral. Here is what I got
$ U(omega,t) = frac{1}{iomega + 1} e^{-t(omega^2+2iomega)} $
Perhaps my solution was incorrect somewhere or the Fourier transform is not the way to go but I would like to ask for help on this in terms of how to solve it properly, but of course there is the matter of different but equivalent definitions of the Fourier transform (I use the one with no normalization constant and positive complex exponent). All help appreciated thanks.
pde fourier-analysis integral-transforms
pde fourier-analysis integral-transforms
edited Jan 22 at 11:19
dmtri
1,5642521
asked Sep 29 '15 at 6:25
kronerkroner
1,3043926
edited Jan 22 at 11:19
dmtri
1,5642521
asked Sep 29 '15 at 6:25
kronerkroner
1,3043926
edited Jan 22 at 11:19
dmtri
1,5642521
edited Jan 22 at 11:19
dmtri
1,5642521
edited Jan 22 at 11:19
dmtri
1,5642521
1,5642521
asked Sep 29 '15 at 6:25
kronerkroner
1,3043926
asked Sep 29 '15 at 6:25
kronerkroner
1,3043926
asked Sep 29 '15 at 6:25
kronerkroner
1,3043926
1,3043926
3
$begingroup$
A convenient strategy for this type of problem is to use the complex Laplace transform for the time variable and, as you already did, the Fourier transform for the space variable $x$. Tell me if you need further help.
$endgroup$
– Urgje
Sep 29 '15 at 13:36
$begingroup$
@Urgje : That certainly helped thank you it worked great
$endgroup$
– kroner
Sep 30 '15 at 3:26
$begingroup$
@Urgje: if you could post an answer with some more detail it would be appreciated. It is interesting.
$endgroup$
– Giuseppe Negro
Jan 22 at 11:24
add a comment |
3
$begingroup$
A convenient strategy for this type of problem is to use the complex Laplace transform for the time variable and, as you already did, the Fourier transform for the space variable $x$. Tell me if you need further help.
$endgroup$
– Urgje
Sep 29 '15 at 13:36
$begingroup$
@Urgje : That certainly helped thank you it worked great
$endgroup$
– kroner
Sep 30 '15 at 3:26
$begingroup$
@Urgje: if you could post an answer with some more detail it would be appreciated. It is interesting.
$endgroup$
– Giuseppe Negro
Jan 22 at 11:24
3
3
$begingroup$
A convenient strategy for this type of problem is to use the complex Laplace transform for the time variable and, as you already did, the Fourier transform for the space variable $x$. Tell me if you need further help.
$endgroup$
– Urgje
Sep 29 '15 at 13:36
$begingroup$
A convenient strategy for this type of problem is to use the complex Laplace transform for the time variable and, as you already did, the Fourier transform for the space variable $x$. Tell me if you need further help.
$endgroup$
– Urgje
Sep 29 '15 at 13:36
$begingroup$
@Urgje : That certainly helped thank you it worked great
$endgroup$
– kroner
Sep 30 '15 at 3:26
$begingroup$
@Urgje : That certainly helped thank you it worked great
$endgroup$
– kroner
Sep 30 '15 at 3:26
$begingroup$
@Urgje: if you could post an answer with some more detail it would be appreciated. It is interesting.
$endgroup$
– Giuseppe Negro
Jan 22 at 11:24
$begingroup$
@Urgje: if you could post an answer with some more detail it would be appreciated. It is interesting.
$endgroup$
– Giuseppe Negro
Jan 22 at 11:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Maybe I am wrong here for reasons that I do not understand but is it possible to assume that $u(x,t)=w(x,t)e^{p(x,t)}$? I know that this works for a PDE on a finite interval, but I am not sure about your interval here.
If so here's a pointer,
$u_t = w_te^{p}+wp_te^{p}$ and then $u_x = w_xe^{p}+wp_xe^p$. Then $u_{xx} = w_{xx}e^p+w_xp_xe^p+w_xp_xe^p+w(p_{xx}e^p+p_xe^p)$.
Plug all of this in back to PDE,
$[w_te^{p}+wp_te^{p}]+2[w_xe^{p}+wp_xe^p] = w_{xx}e^p+w_xp_xe^p+w_xp_xe^p+wp_{xx}e^p+wp_xe^p$. Impose one further condition, that this PDE should reduce to $w_t+2w_x=w_{xx}$. This means equating the rest of the components to zero and possibly solving an exact differential eq.
Now apply the initial conditions
$u(x,0) = w(x,0)e^{p(x,0)}$. Now two separate cases should be analyzed,
Begin with $x < 0$ and let $gamma$ be a positive value. In this case $u(-gamma, 0) = w(-gamma, 0)e^{p(-gamma, 0)} = e^x$ which essentially means that $w(-gamma,0) = e^{x-p(-gamma, 0)}$
The second case $x>0$, then $u(gamma, 0) = w(gamma, 0)e^{p(gamma, 0)} = 0$. This must mean $w(gamma,0) = 0$ when $x>0$.
These conditions are now ideal for either a Fourier transform or Laplace transform.
answered Feb 23 at 23:23
Jonathan AguileraJonathan Aguilera
466
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Maybe I am wrong here for reasons that I do not understand but is it possible to assume that $u(x,t)=w(x,t)e^{p(x,t)}$? I know that this works for a PDE on a finite interval, but I am not sure about your interval here.
If so here's a pointer,
$u_t = w_te^{p}+wp_te^{p}$ and then $u_x = w_xe^{p}+wp_xe^p$. Then $u_{xx} = w_{xx}e^p+w_xp_xe^p+w_xp_xe^p+w(p_{xx}e^p+p_xe^p)$.
Plug all of this in back to PDE,
$[w_te^{p}+wp_te^{p}]+2[w_xe^{p}+wp_xe^p] = w_{xx}e^p+w_xp_xe^p+w_xp_xe^p+wp_{xx}e^p+wp_xe^p$. Impose one further condition, that this PDE should reduce to $w_t+2w_x=w_{xx}$. This means equating the rest of the components to zero and possibly solving an exact differential eq.
Now apply the initial conditions
$u(x,0) = w(x,0)e^{p(x,0)}$. Now two separate cases should be analyzed,
Begin with $x < 0$ and let $gamma$ be a positive value. In this case $u(-gamma, 0) = w(-gamma, 0)e^{p(-gamma, 0)} = e^x$ which essentially means that $w(-gamma,0) = e^{x-p(-gamma, 0)}$
The second case $x>0$, then $u(gamma, 0) = w(gamma, 0)e^{p(gamma, 0)} = 0$. This must mean $w(gamma,0) = 0$ when $x>0$.
These conditions are now ideal for either a Fourier transform or Laplace transform.
answered Feb 23 at 23:23
Jonathan AguileraJonathan Aguilera
466
$endgroup$
add a comment |
$begingroup$
Maybe I am wrong here for reasons that I do not understand but is it possible to assume that $u(x,t)=w(x,t)e^{p(x,t)}$? I know that this works for a PDE on a finite interval, but I am not sure about your interval here.
If so here's a pointer,
$u_t = w_te^{p}+wp_te^{p}$ and then $u_x = w_xe^{p}+wp_xe^p$. Then $u_{xx} = w_{xx}e^p+w_xp_xe^p+w_xp_xe^p+w(p_{xx}e^p+p_xe^p)$.
Plug all of this in back to PDE,
$[w_te^{p}+wp_te^{p}]+2[w_xe^{p}+wp_xe^p] = w_{xx}e^p+w_xp_xe^p+w_xp_xe^p+wp_{xx}e^p+wp_xe^p$. Impose one further condition, that this PDE should reduce to $w_t+2w_x=w_{xx}$. This means equating the rest of the components to zero and possibly solving an exact differential eq.
Now apply the initial conditions
$u(x,0) = w(x,0)e^{p(x,0)}$. Now two separate cases should be analyzed,
Begin with $x < 0$ and let $gamma$ be a positive value. In this case $u(-gamma, 0) = w(-gamma, 0)e^{p(-gamma, 0)} = e^x$ which essentially means that $w(-gamma,0) = e^{x-p(-gamma, 0)}$
The second case $x>0$, then $u(gamma, 0) = w(gamma, 0)e^{p(gamma, 0)} = 0$. This must mean $w(gamma,0) = 0$ when $x>0$.
These conditions are now ideal for either a Fourier transform or Laplace transform.
answered Feb 23 at 23:23
Jonathan AguileraJonathan Aguilera
466
$endgroup$
add a comment |
$begingroup$
Maybe I am wrong here for reasons that I do not understand but is it possible to assume that $u(x,t)=w(x,t)e^{p(x,t)}$? I know that this works for a PDE on a finite interval, but I am not sure about your interval here.
If so here's a pointer,
$u_t = w_te^{p}+wp_te^{p}$ and then $u_x = w_xe^{p}+wp_xe^p$. Then $u_{xx} = w_{xx}e^p+w_xp_xe^p+w_xp_xe^p+w(p_{xx}e^p+p_xe^p)$.
Plug all of this in back to PDE,
$[w_te^{p}+wp_te^{p}]+2[w_xe^{p}+wp_xe^p] = w_{xx}e^p+w_xp_xe^p+w_xp_xe^p+wp_{xx}e^p+wp_xe^p$. Impose one further condition, that this PDE should reduce to $w_t+2w_x=w_{xx}$. This means equating the rest of the components to zero and possibly solving an exact differential eq.
Now apply the initial conditions
$u(x,0) = w(x,0)e^{p(x,0)}$. Now two separate cases should be analyzed,
Begin with $x < 0$ and let $gamma$ be a positive value. In this case $u(-gamma, 0) = w(-gamma, 0)e^{p(-gamma, 0)} = e^x$ which essentially means that $w(-gamma,0) = e^{x-p(-gamma, 0)}$
The second case $x>0$, then $u(gamma, 0) = w(gamma, 0)e^{p(gamma, 0)} = 0$. This must mean $w(gamma,0) = 0$ when $x>0$.
These conditions are now ideal for either a Fourier transform or Laplace transform.
answered Feb 23 at 23:23
Jonathan AguileraJonathan Aguilera
466
$endgroup$
Maybe I am wrong here for reasons that I do not understand but is it possible to assume that $u(x,t)=w(x,t)e^{p(x,t)}$? I know that this works for a PDE on a finite interval, but I am not sure about your interval here.
If so here's a pointer,
$u_t = w_te^{p}+wp_te^{p}$ and then $u_x = w_xe^{p}+wp_xe^p$. Then $u_{xx} = w_{xx}e^p+w_xp_xe^p+w_xp_xe^p+w(p_{xx}e^p+p_xe^p)$.
Plug all of this in back to PDE,
$[w_te^{p}+wp_te^{p}]+2[w_xe^{p}+wp_xe^p] = w_{xx}e^p+w_xp_xe^p+w_xp_xe^p+wp_{xx}e^p+wp_xe^p$. Impose one further condition, that this PDE should reduce to $w_t+2w_x=w_{xx}$. This means equating the rest of the components to zero and possibly solving an exact differential eq.
Now apply the initial conditions
$u(x,0) = w(x,0)e^{p(x,0)}$. Now two separate cases should be analyzed,
Begin with $x < 0$ and let $gamma$ be a positive value. In this case $u(-gamma, 0) = w(-gamma, 0)e^{p(-gamma, 0)} = e^x$ which essentially means that $w(-gamma,0) = e^{x-p(-gamma, 0)}$
The second case $x>0$, then $u(gamma, 0) = w(gamma, 0)e^{p(gamma, 0)} = 0$. This must mean $w(gamma,0) = 0$ when $x>0$.
These conditions are now ideal for either a Fourier transform or Laplace transform.
answered Feb 23 at 23:23
Jonathan AguileraJonathan Aguilera
466
answered Feb 23 at 23:23
Jonathan AguileraJonathan Aguilera
466
answered Feb 23 at 23:23
Jonathan AguileraJonathan Aguilera
466
answered Feb 23 at 23:23
Jonathan AguileraJonathan Aguilera
466
466
add a comment |
add a comment |
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$begingroup$
A convenient strategy for this type of problem is to use the complex Laplace transform for the time variable and, as you already did, the Fourier transform for the space variable $x$. Tell me if you need further help.
$endgroup$
– Urgje
Sep 29 '15 at 13:36
$begingroup$
@Urgje : That certainly helped thank you it worked great
$endgroup$
– kroner
Sep 30 '15 at 3:26
$begingroup$
@Urgje: if you could post an answer with some more detail it would be appreciated. It is interesting.
$endgroup$
– Giuseppe Negro
Jan 22 at 11:24