Mixing
Substitution into a differential equation
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I have typical 2nd order differential equation:
$nabla^2U + frac1Vfracpartial Upartial z = 0$
Let, $z = r(x,t)$ and $r(x,t) = r_0exp(ikx+omega t)$.
I would like to substitute $r(x,t)$ into the above differential equation.
What I'm doing now, $fracpartial^2 Upartial x^2 = fracpartial partial xfracpartial Upartial x = fracpartial partial r[ fracpartial U partial rfracpartial rpartial x]$. Here I calculate $fracpartial rpartial x$ w.r.t $r(x,t)$.
But, I get stuck at calculating $fracpartial Upartial z$ and its second derivative. Will it be zero ? since $r(x,t)$ is not a function of $z$.
As the substituted equation in an article contains $frac1Vfracpartial Upartial r$ term
differential-equations pde substitution
asked Jul 24 at 13:29
newstudent
62
add a comment |Â
up vote
1
down vote
favorite
I have typical 2nd order differential equation:
$nabla^2U + frac1Vfracpartial Upartial z = 0$
Let, $z = r(x,t)$ and $r(x,t) = r_0exp(ikx+omega t)$.
I would like to substitute $r(x,t)$ into the above differential equation.
What I'm doing now, $fracpartial^2 Upartial x^2 = fracpartial partial xfracpartial Upartial x = fracpartial partial r[ fracpartial U partial rfracpartial rpartial x]$. Here I calculate $fracpartial rpartial x$ w.r.t $r(x,t)$.
But, I get stuck at calculating $fracpartial Upartial z$ and its second derivative. Will it be zero ? since $r(x,t)$ is not a function of $z$.
As the substituted equation in an article contains $frac1Vfracpartial Upartial r$ term
differential-equations pde substitution
asked Jul 24 at 13:29
newstudent
62
You are right saying that $r(x,t)$ is not a function of $z$. Hence you want to substitute $z$ by $r(x,t)$ I guess you also need to substitute $partial z$ in terms of $partial r(x,t)$.
– mrtaurho
Jul 24 at 16:39
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have typical 2nd order differential equation:
$nabla^2U + frac1Vfracpartial Upartial z = 0$
Let, $z = r(x,t)$ and $r(x,t) = r_0exp(ikx+omega t)$.
I would like to substitute $r(x,t)$ into the above differential equation.
What I'm doing now, $fracpartial^2 Upartial x^2 = fracpartial partial xfracpartial Upartial x = fracpartial partial r[ fracpartial U partial rfracpartial rpartial x]$. Here I calculate $fracpartial rpartial x$ w.r.t $r(x,t)$.
But, I get stuck at calculating $fracpartial Upartial z$ and its second derivative. Will it be zero ? since $r(x,t)$ is not a function of $z$.
As the substituted equation in an article contains $frac1Vfracpartial Upartial r$ term
differential-equations pde substitution
asked Jul 24 at 13:29
newstudent
62
I have typical 2nd order differential equation:
$nabla^2U + frac1Vfracpartial Upartial z = 0$
Let, $z = r(x,t)$ and $r(x,t) = r_0exp(ikx+omega t)$.
I would like to substitute $r(x,t)$ into the above differential equation.
What I'm doing now, $fracpartial^2 Upartial x^2 = fracpartial partial xfracpartial Upartial x = fracpartial partial r[ fracpartial U partial rfracpartial rpartial x]$. Here I calculate $fracpartial rpartial x$ w.r.t $r(x,t)$.
But, I get stuck at calculating $fracpartial Upartial z$ and its second derivative. Will it be zero ? since $r(x,t)$ is not a function of $z$.
As the substituted equation in an article contains $frac1Vfracpartial Upartial r$ term
differential-equations pde substitution
asked Jul 24 at 13:29
newstudent
62
asked Jul 24 at 13:29
newstudent
62
asked Jul 24 at 13:29
newstudent
62
asked Jul 24 at 13:29
newstudent
62
62
You are right saying that $r(x,t)$ is not a function of $z$. Hence you want to substitute $z$ by $r(x,t)$ I guess you also need to substitute $partial z$ in terms of $partial r(x,t)$.
– mrtaurho
Jul 24 at 16:39
add a comment |Â
You are right saying that $r(x,t)$ is not a function of $z$. Hence you want to substitute $z$ by $r(x,t)$ I guess you also need to substitute $partial z$ in terms of $partial r(x,t)$.
– mrtaurho
Jul 24 at 16:39
You are right saying that $r(x,t)$ is not a function of $z$. Hence you want to substitute $z$ by $r(x,t)$ I guess you also need to substitute $partial z$ in terms of $partial r(x,t)$.
– mrtaurho
Jul 24 at 16:39
You are right saying that $r(x,t)$ is not a function of $z$. Hence you want to substitute $z$ by $r(x,t)$ I guess you also need to substitute $partial z$ in terms of $partial r(x,t)$.
– mrtaurho
Jul 24 at 16:39
add a comment |Â
1 Answer
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If I am not mistaken when you substitute $z=r(x,t)$ you also need to compute the derivation of $z$ with respect to $r(x,t)$.
So you get
$$fracpartial zpartial r(x,t)~=~fracpartial (r(x,t))partial r(x,t)~=~1$$
and therefore
$$partial z ~=~ partial r(x,t) $$
and so the term $frac1Vfracpartial Upartial z$ becomes $frac1Vfracpartial Upartial r$.
answered Jul 24 at 16:46
mrtaurho
740219
Thanks. I guess, now the second derivative will be non existent.
– newstudent
Jul 26 at 9:51
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If I am not mistaken when you substitute $z=r(x,t)$ you also need to compute the derivation of $z$ with respect to $r(x,t)$.
So you get
$$fracpartial zpartial r(x,t)~=~fracpartial (r(x,t))partial r(x,t)~=~1$$
and therefore
$$partial z ~=~ partial r(x,t) $$
and so the term $frac1Vfracpartial Upartial z$ becomes $frac1Vfracpartial Upartial r$.
answered Jul 24 at 16:46
mrtaurho
740219
Thanks. I guess, now the second derivative will be non existent.
– newstudent
Jul 26 at 9:51
add a comment |Â
up vote
0
down vote
If I am not mistaken when you substitute $z=r(x,t)$ you also need to compute the derivation of $z$ with respect to $r(x,t)$.
So you get
$$fracpartial zpartial r(x,t)~=~fracpartial (r(x,t))partial r(x,t)~=~1$$
and therefore
$$partial z ~=~ partial r(x,t) $$
and so the term $frac1Vfracpartial Upartial z$ becomes $frac1Vfracpartial Upartial r$.
answered Jul 24 at 16:46
mrtaurho
740219
Thanks. I guess, now the second derivative will be non existent.
– newstudent
Jul 26 at 9:51
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If I am not mistaken when you substitute $z=r(x,t)$ you also need to compute the derivation of $z$ with respect to $r(x,t)$.
So you get
$$fracpartial zpartial r(x,t)~=~fracpartial (r(x,t))partial r(x,t)~=~1$$
and therefore
$$partial z ~=~ partial r(x,t) $$
and so the term $frac1Vfracpartial Upartial z$ becomes $frac1Vfracpartial Upartial r$.
answered Jul 24 at 16:46
mrtaurho
740219
If I am not mistaken when you substitute $z=r(x,t)$ you also need to compute the derivation of $z$ with respect to $r(x,t)$.
So you get
$$fracpartial zpartial r(x,t)~=~fracpartial (r(x,t))partial r(x,t)~=~1$$
and therefore
$$partial z ~=~ partial r(x,t) $$
and so the term $frac1Vfracpartial Upartial z$ becomes $frac1Vfracpartial Upartial r$.
answered Jul 24 at 16:46
mrtaurho
740219
answered Jul 24 at 16:46
mrtaurho
740219
answered Jul 24 at 16:46
mrtaurho
740219
answered Jul 24 at 16:46
mrtaurho
740219
740219
Thanks. I guess, now the second derivative will be non existent.
– newstudent
Jul 26 at 9:51
add a comment |Â
Thanks. I guess, now the second derivative will be non existent.
– newstudent
Jul 26 at 9:51
Thanks. I guess, now the second derivative will be non existent.
– newstudent
Jul 26 at 9:51
Thanks. I guess, now the second derivative will be non existent.
– newstudent
Jul 26 at 9:51
add a comment |Â
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You are right saying that $r(x,t)$ is not a function of $z$. Hence you want to substitute $z$ by $r(x,t)$ I guess you also need to substitute $partial z$ in terms of $partial r(x,t)$.
– mrtaurho
Jul 24 at 16:39