Mixing
Enforcing amplitude profile on complex solution of PDE
$begingroup$
Consider the following PDE.
$$psi_z = ipsi_{xx}$$
Suppose that I can generate the function $F=|psi|^2$ using some other PDE. At this point I am considering this PDE to be
$$a(x,z)F_x + b(x,z)F_z=0$$
where by choosing $a$, $b$ and an initial condition $F(x,z=0)$, I generate the desired $F$.
Once this is done, I want to use the initial PDE as follows. At $z=0$ I want to compute the phase so that the resulting solution will satisfy $|psi|^2=F$. Is there such a method?
What I have done so far is as follows:
Suppose that I want a solution such that $F(x,z) = F(x-frac{z^n}{n!})$, for some $nin mathbb{N}$, and as a starting point, consider the case $n = 1$. For this particular case, the choice on $a$ and $b$ is $a = b = 1$.
The solution of the initial PDE can be considered of this form $psi(x,z) = A(x - z)cdot exp(iphi(x,z))$, and by doing so two differential equations result. I have used the property that $A_z = -A_x$ since $A$ depends on $(x-z)$.
$$a(x,z) A_x = 2 phi_x A_x + phi_{xx}A$$
$$Aphi_z = A_{xx} - (phi_x)^2A$$
In this particular case I mention that those equations can be derived directly by assuming the solution of the above mentioned form ($psi(x,z) = A(x - z)cdot exp(iphi(x,z))$) and then by plugging it in the first PDE, but for more general scenarios I assume that this approach is more general. Coming back to the result, I consider that at $z=0$ I have $a=a_0$. For that value of $a$ I can compute numerically the phase $phi$ and it works. In my particular case that value is $1$.
So what did I get out of this? The solution given by computing $F$ for the particular case moves in the $(x,z)$ space at an angle $theta$ that is given by $tan(theta) = a/b = 1$. By changing the value of $a$ I also change the angle $theta$. What I don't get is the spread of the solution when using the first PDE to compute the solution for an initial condition that has the amplitude $A$ and phase $phi$ that I have considered/computed, but I have the angle.
Now, assuming I take $a=z$ and following the same reasoning, when evaluating the two differential equations above at some point $z=0$ in order to compute the phase, the resulting solution of PDE 1 with that phase has a linear trajectory, as in the first case, although the solution given by computing $F$ has a parabolic trajectory. By changing the value of $z$ for some arbitrary value $z_0$, the angle still changes but the trajectory is still linear.
At this point I considered differentiating $a(x,z)F_x + b(x,z)F_z=0$ one time and then to evaluate at $z=0$ but the computation becomes way to complicated (terms like $A_{xxxx}$ start to appear).
So coming back to the main question, do you know other approaches that do something similar to what I'm trying to do? Some terminology for googling the issue is welcomed.
Related question, also asked by me: here
pde
asked Jan 15 at 13:59
Victor PaleaVictor Palea
301312
$endgroup$
add a comment |
$begingroup$
Consider the following PDE.
$$psi_z = ipsi_{xx}$$
Suppose that I can generate the function $F=|psi|^2$ using some other PDE. At this point I am considering this PDE to be
$$a(x,z)F_x + b(x,z)F_z=0$$
where by choosing $a$, $b$ and an initial condition $F(x,z=0)$, I generate the desired $F$.
Once this is done, I want to use the initial PDE as follows. At $z=0$ I want to compute the phase so that the resulting solution will satisfy $|psi|^2=F$. Is there such a method?
What I have done so far is as follows:
Suppose that I want a solution such that $F(x,z) = F(x-frac{z^n}{n!})$, for some $nin mathbb{N}$, and as a starting point, consider the case $n = 1$. For this particular case, the choice on $a$ and $b$ is $a = b = 1$.
The solution of the initial PDE can be considered of this form $psi(x,z) = A(x - z)cdot exp(iphi(x,z))$, and by doing so two differential equations result. I have used the property that $A_z = -A_x$ since $A$ depends on $(x-z)$.
$$a(x,z) A_x = 2 phi_x A_x + phi_{xx}A$$
$$Aphi_z = A_{xx} - (phi_x)^2A$$
In this particular case I mention that those equations can be derived directly by assuming the solution of the above mentioned form ($psi(x,z) = A(x - z)cdot exp(iphi(x,z))$) and then by plugging it in the first PDE, but for more general scenarios I assume that this approach is more general. Coming back to the result, I consider that at $z=0$ I have $a=a_0$. For that value of $a$ I can compute numerically the phase $phi$ and it works. In my particular case that value is $1$.
So what did I get out of this? The solution given by computing $F$ for the particular case moves in the $(x,z)$ space at an angle $theta$ that is given by $tan(theta) = a/b = 1$. By changing the value of $a$ I also change the angle $theta$. What I don't get is the spread of the solution when using the first PDE to compute the solution for an initial condition that has the amplitude $A$ and phase $phi$ that I have considered/computed, but I have the angle.
Now, assuming I take $a=z$ and following the same reasoning, when evaluating the two differential equations above at some point $z=0$ in order to compute the phase, the resulting solution of PDE 1 with that phase has a linear trajectory, as in the first case, although the solution given by computing $F$ has a parabolic trajectory. By changing the value of $z$ for some arbitrary value $z_0$, the angle still changes but the trajectory is still linear.
At this point I considered differentiating $a(x,z)F_x + b(x,z)F_z=0$ one time and then to evaluate at $z=0$ but the computation becomes way to complicated (terms like $A_{xxxx}$ start to appear).
So coming back to the main question, do you know other approaches that do something similar to what I'm trying to do? Some terminology for googling the issue is welcomed.
Related question, also asked by me: here
pde
asked Jan 15 at 13:59
Victor PaleaVictor Palea
301312
$endgroup$
add a comment |
$begingroup$
Consider the following PDE.
$$psi_z = ipsi_{xx}$$
Suppose that I can generate the function $F=|psi|^2$ using some other PDE. At this point I am considering this PDE to be
$$a(x,z)F_x + b(x,z)F_z=0$$
where by choosing $a$, $b$ and an initial condition $F(x,z=0)$, I generate the desired $F$.
Once this is done, I want to use the initial PDE as follows. At $z=0$ I want to compute the phase so that the resulting solution will satisfy $|psi|^2=F$. Is there such a method?
What I have done so far is as follows:
Suppose that I want a solution such that $F(x,z) = F(x-frac{z^n}{n!})$, for some $nin mathbb{N}$, and as a starting point, consider the case $n = 1$. For this particular case, the choice on $a$ and $b$ is $a = b = 1$.
The solution of the initial PDE can be considered of this form $psi(x,z) = A(x - z)cdot exp(iphi(x,z))$, and by doing so two differential equations result. I have used the property that $A_z = -A_x$ since $A$ depends on $(x-z)$.
$$a(x,z) A_x = 2 phi_x A_x + phi_{xx}A$$
$$Aphi_z = A_{xx} - (phi_x)^2A$$
In this particular case I mention that those equations can be derived directly by assuming the solution of the above mentioned form ($psi(x,z) = A(x - z)cdot exp(iphi(x,z))$) and then by plugging it in the first PDE, but for more general scenarios I assume that this approach is more general. Coming back to the result, I consider that at $z=0$ I have $a=a_0$. For that value of $a$ I can compute numerically the phase $phi$ and it works. In my particular case that value is $1$.
So what did I get out of this? The solution given by computing $F$ for the particular case moves in the $(x,z)$ space at an angle $theta$ that is given by $tan(theta) = a/b = 1$. By changing the value of $a$ I also change the angle $theta$. What I don't get is the spread of the solution when using the first PDE to compute the solution for an initial condition that has the amplitude $A$ and phase $phi$ that I have considered/computed, but I have the angle.
Now, assuming I take $a=z$ and following the same reasoning, when evaluating the two differential equations above at some point $z=0$ in order to compute the phase, the resulting solution of PDE 1 with that phase has a linear trajectory, as in the first case, although the solution given by computing $F$ has a parabolic trajectory. By changing the value of $z$ for some arbitrary value $z_0$, the angle still changes but the trajectory is still linear.
At this point I considered differentiating $a(x,z)F_x + b(x,z)F_z=0$ one time and then to evaluate at $z=0$ but the computation becomes way to complicated (terms like $A_{xxxx}$ start to appear).
So coming back to the main question, do you know other approaches that do something similar to what I'm trying to do? Some terminology for googling the issue is welcomed.
Related question, also asked by me: here
pde
asked Jan 15 at 13:59
Victor PaleaVictor Palea
301312
$endgroup$
Consider the following PDE.
$$psi_z = ipsi_{xx}$$
Suppose that I can generate the function $F=|psi|^2$ using some other PDE. At this point I am considering this PDE to be
$$a(x,z)F_x + b(x,z)F_z=0$$
where by choosing $a$, $b$ and an initial condition $F(x,z=0)$, I generate the desired $F$.
Once this is done, I want to use the initial PDE as follows. At $z=0$ I want to compute the phase so that the resulting solution will satisfy $|psi|^2=F$. Is there such a method?
What I have done so far is as follows:
Suppose that I want a solution such that $F(x,z) = F(x-frac{z^n}{n!})$, for some $nin mathbb{N}$, and as a starting point, consider the case $n = 1$. For this particular case, the choice on $a$ and $b$ is $a = b = 1$.
The solution of the initial PDE can be considered of this form $psi(x,z) = A(x - z)cdot exp(iphi(x,z))$, and by doing so two differential equations result. I have used the property that $A_z = -A_x$ since $A$ depends on $(x-z)$.
$$a(x,z) A_x = 2 phi_x A_x + phi_{xx}A$$
$$Aphi_z = A_{xx} - (phi_x)^2A$$
In this particular case I mention that those equations can be derived directly by assuming the solution of the above mentioned form ($psi(x,z) = A(x - z)cdot exp(iphi(x,z))$) and then by plugging it in the first PDE, but for more general scenarios I assume that this approach is more general. Coming back to the result, I consider that at $z=0$ I have $a=a_0$. For that value of $a$ I can compute numerically the phase $phi$ and it works. In my particular case that value is $1$.
So what did I get out of this? The solution given by computing $F$ for the particular case moves in the $(x,z)$ space at an angle $theta$ that is given by $tan(theta) = a/b = 1$. By changing the value of $a$ I also change the angle $theta$. What I don't get is the spread of the solution when using the first PDE to compute the solution for an initial condition that has the amplitude $A$ and phase $phi$ that I have considered/computed, but I have the angle.
Now, assuming I take $a=z$ and following the same reasoning, when evaluating the two differential equations above at some point $z=0$ in order to compute the phase, the resulting solution of PDE 1 with that phase has a linear trajectory, as in the first case, although the solution given by computing $F$ has a parabolic trajectory. By changing the value of $z$ for some arbitrary value $z_0$, the angle still changes but the trajectory is still linear.
At this point I considered differentiating $a(x,z)F_x + b(x,z)F_z=0$ one time and then to evaluate at $z=0$ but the computation becomes way to complicated (terms like $A_{xxxx}$ start to appear).
So coming back to the main question, do you know other approaches that do something similar to what I'm trying to do? Some terminology for googling the issue is welcomed.
Related question, also asked by me: here
pde
pde
asked Jan 15 at 13:59
Victor PaleaVictor Palea
301312
asked Jan 15 at 13:59
Victor PaleaVictor Palea
301312
asked Jan 15 at 13:59
Victor PaleaVictor Palea
301312
asked Jan 15 at 13:59
Victor PaleaVictor Palea
301312
asked Jan 15 at 13:59
Victor PaleaVictor Palea
301312
301312
add a comment |
add a comment |
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