Coupled partial differential equations
$begingroup$
I'm having trouble solving these coupled partial differential equations:
$$frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0,$$
$$frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0,$$
with $A,c$ real constants.
What is the "trick" to solve these? I haven't tried a lot since I wouldn't know where to start looking.
pde systems-of-equations partial-derivative
edited Feb 25 at 11:36
Harry Peter
5,47911439
asked Mar 11 '16 at 15:58
new guynew guy
382
$endgroup$
add a comment |
$begingroup$
I'm having trouble solving these coupled partial differential equations:
$$frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0,$$
$$frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0,$$
with $A,c$ real constants.
What is the "trick" to solve these? I haven't tried a lot since I wouldn't know where to start looking.
pde systems-of-equations partial-derivative
edited Feb 25 at 11:36
Harry Peter
5,47911439
asked Mar 11 '16 at 15:58
new guynew guy
382
$endgroup$
$begingroup$
I suppose you could take $partial_{x}$ (and $partial_{t}$) of the first equation, rearrange for $partial_{x} p$ and $partial_{t} p$, then substitute those results into your second equation which should give you everything in $f$.
$endgroup$
– Mattos
Mar 11 '16 at 16:03
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@Mattos What do you mean by rearranging $partial_x p$?
$endgroup$
– new guy
Mar 11 '16 at 16:05
$begingroup$
From the first equation $$partial_{x} p = frac{1}{A} partial_{x} bigg( partial_{t} f - c partial_{x} f bigg)$$ You can get something similar for $partial_{t} p$.
$endgroup$
– Mattos
Mar 11 '16 at 16:06
add a comment |
$begingroup$
I'm having trouble solving these coupled partial differential equations:
$$frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0,$$
$$frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0,$$
with $A,c$ real constants.
What is the "trick" to solve these? I haven't tried a lot since I wouldn't know where to start looking.
pde systems-of-equations partial-derivative
edited Feb 25 at 11:36
Harry Peter
5,47911439
asked Mar 11 '16 at 15:58
new guynew guy
382
$endgroup$
I'm having trouble solving these coupled partial differential equations:
$$frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0,$$
$$frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0,$$
with $A,c$ real constants.
What is the "trick" to solve these? I haven't tried a lot since I wouldn't know where to start looking.
pde systems-of-equations partial-derivative
pde systems-of-equations partial-derivative
edited Feb 25 at 11:36
Harry Peter
5,47911439
asked Mar 11 '16 at 15:58
new guynew guy
382
edited Feb 25 at 11:36
Harry Peter
5,47911439
asked Mar 11 '16 at 15:58
new guynew guy
382
edited Feb 25 at 11:36
Harry Peter
5,47911439
edited Feb 25 at 11:36
Harry Peter
5,47911439
edited Feb 25 at 11:36
Harry Peter
5,47911439
5,47911439
asked Mar 11 '16 at 15:58
new guynew guy
382
asked Mar 11 '16 at 15:58
new guynew guy
382
asked Mar 11 '16 at 15:58
new guynew guy
382
382
$begingroup$
I suppose you could take $partial_{x}$ (and $partial_{t}$) of the first equation, rearrange for $partial_{x} p$ and $partial_{t} p$, then substitute those results into your second equation which should give you everything in $f$.
$endgroup$
– Mattos
Mar 11 '16 at 16:03
$begingroup$
@Mattos What do you mean by rearranging $partial_x p$?
$endgroup$
– new guy
Mar 11 '16 at 16:05
$begingroup$
From the first equation $$partial_{x} p = frac{1}{A} partial_{x} bigg( partial_{t} f - c partial_{x} f bigg)$$ You can get something similar for $partial_{t} p$.
$endgroup$
– Mattos
Mar 11 '16 at 16:06
add a comment |
$begingroup$
I suppose you could take $partial_{x}$ (and $partial_{t}$) of the first equation, rearrange for $partial_{x} p$ and $partial_{t} p$, then substitute those results into your second equation which should give you everything in $f$.
$endgroup$
– Mattos
Mar 11 '16 at 16:03
$begingroup$
@Mattos What do you mean by rearranging $partial_x p$?
$endgroup$
– new guy
Mar 11 '16 at 16:05
$begingroup$
From the first equation $$partial_{x} p = frac{1}{A} partial_{x} bigg( partial_{t} f - c partial_{x} f bigg)$$ You can get something similar for $partial_{t} p$.
$endgroup$
– Mattos
Mar 11 '16 at 16:06
$begingroup$
I suppose you could take $partial_{x}$ (and $partial_{t}$) of the first equation, rearrange for $partial_{x} p$ and $partial_{t} p$, then substitute those results into your second equation which should give you everything in $f$.
$endgroup$
– Mattos
Mar 11 '16 at 16:03
$begingroup$
I suppose you could take $partial_{x}$ (and $partial_{t}$) of the first equation, rearrange for $partial_{x} p$ and $partial_{t} p$, then substitute those results into your second equation which should give you everything in $f$.
$endgroup$
– Mattos
Mar 11 '16 at 16:03
$begingroup$
@Mattos What do you mean by rearranging $partial_x p$?
$endgroup$
– new guy
Mar 11 '16 at 16:05
$begingroup$
@Mattos What do you mean by rearranging $partial_x p$?
$endgroup$
– new guy
Mar 11 '16 at 16:05
$begingroup$
From the first equation $$partial_{x} p = frac{1}{A} partial_{x} bigg( partial_{t} f - c partial_{x} f bigg)$$ You can get something similar for $partial_{t} p$.
$endgroup$
– Mattos
Mar 11 '16 at 16:06
$begingroup$
From the first equation $$partial_{x} p = frac{1}{A} partial_{x} bigg( partial_{t} f - c partial_{x} f bigg)$$ You can get something similar for $partial_{t} p$.
$endgroup$
– Mattos
Mar 11 '16 at 16:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$begin{cases}
frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0 \
frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0
end{cases}$$
Regularised form with $begin{cases} T=At \ X=frac{A}{c}x end{cases} quadtoquad
begin{cases}
frac{partial}{partial T}f(X,T)-frac{partial}{partial X}f(X,T)-p(X,T)=0 \
frac{partial}{partial T}p(X,T)+frac{partial}{partial X}p(X,T)+f(X,T)=0
end{cases}$
$$begin{cases}
f_T-f_X-p=0 \
p_T+p_X+f=0
end{cases}$$
$p=f_T-f_X quadtoquad (f_{TT}-f_{XT})+(f_{XT}-f_{XX})+f=0$
$$frac{partial^2 f}{partial T^2}-frac{partial^2f}{partial X^2}+f(X,T)=0$$
Solving this hyperbolic PDE leads to $f(X,T)=fleft(At:,:frac{A}{c}x right)$
Then $quad p(X,T)=frac{partial f}{partial T}-frac{partial f}{partial X}=pleft(At:,:frac{A}{c}x right)$
For example of solving see : Finding the general solution of a second order PDE
This method leads to the integral form of solution :
$$f(X,T)=int c(s)e^{sqrt{alpha(s)-frac{1}{2}}:X +sqrt{alpha(s)+frac{1}{2}} :T } ds$$
$c(s)$ and $alpha(s)$ are arbitrary real or complex functions.
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 16 '16 at 8:49
JJacquelinJJacquelin
44.5k21854
$endgroup$
add a comment |
$begingroup$
Hint.
We have
$$
mathcal{D}_1 f = A p\
mathcal{D}_2 p = -A f
$$
then
$$
mathcal{D}_2mathcal{D}_1 f = Amathcal{D}_2 p = -A^2 f\
mathcal{D}_1mathcal{D}_2 p = -A mathcal{D}_1f = -A^2p
$$
answered Jan 11 at 13:35
CesareoCesareo
9,2563517
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$$begin{cases}
frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0 \
frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0
end{cases}$$
Regularised form with $begin{cases} T=At \ X=frac{A}{c}x end{cases} quadtoquad
begin{cases}
frac{partial}{partial T}f(X,T)-frac{partial}{partial X}f(X,T)-p(X,T)=0 \
frac{partial}{partial T}p(X,T)+frac{partial}{partial X}p(X,T)+f(X,T)=0
end{cases}$
$$begin{cases}
f_T-f_X-p=0 \
p_T+p_X+f=0
end{cases}$$
$p=f_T-f_X quadtoquad (f_{TT}-f_{XT})+(f_{XT}-f_{XX})+f=0$
$$frac{partial^2 f}{partial T^2}-frac{partial^2f}{partial X^2}+f(X,T)=0$$
Solving this hyperbolic PDE leads to $f(X,T)=fleft(At:,:frac{A}{c}x right)$
Then $quad p(X,T)=frac{partial f}{partial T}-frac{partial f}{partial X}=pleft(At:,:frac{A}{c}x right)$
For example of solving see : Finding the general solution of a second order PDE
This method leads to the integral form of solution :
$$f(X,T)=int c(s)e^{sqrt{alpha(s)-frac{1}{2}}:X +sqrt{alpha(s)+frac{1}{2}} :T } ds$$
$c(s)$ and $alpha(s)$ are arbitrary real or complex functions.
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 16 '16 at 8:49
JJacquelinJJacquelin
44.5k21854
$endgroup$
add a comment |
$begingroup$
$$begin{cases}
frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0 \
frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0
end{cases}$$
Regularised form with $begin{cases} T=At \ X=frac{A}{c}x end{cases} quadtoquad
begin{cases}
frac{partial}{partial T}f(X,T)-frac{partial}{partial X}f(X,T)-p(X,T)=0 \
frac{partial}{partial T}p(X,T)+frac{partial}{partial X}p(X,T)+f(X,T)=0
end{cases}$
$$begin{cases}
f_T-f_X-p=0 \
p_T+p_X+f=0
end{cases}$$
$p=f_T-f_X quadtoquad (f_{TT}-f_{XT})+(f_{XT}-f_{XX})+f=0$
$$frac{partial^2 f}{partial T^2}-frac{partial^2f}{partial X^2}+f(X,T)=0$$
Solving this hyperbolic PDE leads to $f(X,T)=fleft(At:,:frac{A}{c}x right)$
Then $quad p(X,T)=frac{partial f}{partial T}-frac{partial f}{partial X}=pleft(At:,:frac{A}{c}x right)$
For example of solving see : Finding the general solution of a second order PDE
This method leads to the integral form of solution :
$$f(X,T)=int c(s)e^{sqrt{alpha(s)-frac{1}{2}}:X +sqrt{alpha(s)+frac{1}{2}} :T } ds$$
$c(s)$ and $alpha(s)$ are arbitrary real or complex functions.
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 16 '16 at 8:49
JJacquelinJJacquelin
44.5k21854
$endgroup$
add a comment |
$begingroup$
$$begin{cases}
frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0 \
frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0
end{cases}$$
Regularised form with $begin{cases} T=At \ X=frac{A}{c}x end{cases} quadtoquad
begin{cases}
frac{partial}{partial T}f(X,T)-frac{partial}{partial X}f(X,T)-p(X,T)=0 \
frac{partial}{partial T}p(X,T)+frac{partial}{partial X}p(X,T)+f(X,T)=0
end{cases}$
$$begin{cases}
f_T-f_X-p=0 \
p_T+p_X+f=0
end{cases}$$
$p=f_T-f_X quadtoquad (f_{TT}-f_{XT})+(f_{XT}-f_{XX})+f=0$
$$frac{partial^2 f}{partial T^2}-frac{partial^2f}{partial X^2}+f(X,T)=0$$
Solving this hyperbolic PDE leads to $f(X,T)=fleft(At:,:frac{A}{c}x right)$
Then $quad p(X,T)=frac{partial f}{partial T}-frac{partial f}{partial X}=pleft(At:,:frac{A}{c}x right)$
For example of solving see : Finding the general solution of a second order PDE
This method leads to the integral form of solution :
$$f(X,T)=int c(s)e^{sqrt{alpha(s)-frac{1}{2}}:X +sqrt{alpha(s)+frac{1}{2}} :T } ds$$
$c(s)$ and $alpha(s)$ are arbitrary real or complex functions.
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 16 '16 at 8:49
JJacquelinJJacquelin
44.5k21854
$endgroup$
$$begin{cases}
frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0 \
frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0
end{cases}$$
Regularised form with $begin{cases} T=At \ X=frac{A}{c}x end{cases} quadtoquad
begin{cases}
frac{partial}{partial T}f(X,T)-frac{partial}{partial X}f(X,T)-p(X,T)=0 \
frac{partial}{partial T}p(X,T)+frac{partial}{partial X}p(X,T)+f(X,T)=0
end{cases}$
$$begin{cases}
f_T-f_X-p=0 \
p_T+p_X+f=0
end{cases}$$
$p=f_T-f_X quadtoquad (f_{TT}-f_{XT})+(f_{XT}-f_{XX})+f=0$
$$frac{partial^2 f}{partial T^2}-frac{partial^2f}{partial X^2}+f(X,T)=0$$
Solving this hyperbolic PDE leads to $f(X,T)=fleft(At:,:frac{A}{c}x right)$
Then $quad p(X,T)=frac{partial f}{partial T}-frac{partial f}{partial X}=pleft(At:,:frac{A}{c}x right)$
For example of solving see : Finding the general solution of a second order PDE
This method leads to the integral form of solution :
$$f(X,T)=int c(s)e^{sqrt{alpha(s)-frac{1}{2}}:X +sqrt{alpha(s)+frac{1}{2}} :T } ds$$
$c(s)$ and $alpha(s)$ are arbitrary real or complex functions.
edited Apr 13 '17 at 12:20
Community♦
1
answered Mar 16 '16 at 8:49
JJacquelinJJacquelin
44.5k21854
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Mar 16 '16 at 8:49
JJacquelinJJacquelin
44.5k21854
answered Mar 16 '16 at 8:49
JJacquelinJJacquelin
44.5k21854
answered Mar 16 '16 at 8:49
JJacquelinJJacquelin
44.5k21854
44.5k21854
add a comment |
add a comment |
$begingroup$
Hint.
We have
$$
mathcal{D}_1 f = A p\
mathcal{D}_2 p = -A f
$$
then
$$
mathcal{D}_2mathcal{D}_1 f = Amathcal{D}_2 p = -A^2 f\
mathcal{D}_1mathcal{D}_2 p = -A mathcal{D}_1f = -A^2p
$$
answered Jan 11 at 13:35
CesareoCesareo
9,2563517
$endgroup$
add a comment |
$begingroup$
Hint.
We have
$$
mathcal{D}_1 f = A p\
mathcal{D}_2 p = -A f
$$
then
$$
mathcal{D}_2mathcal{D}_1 f = Amathcal{D}_2 p = -A^2 f\
mathcal{D}_1mathcal{D}_2 p = -A mathcal{D}_1f = -A^2p
$$
answered Jan 11 at 13:35
CesareoCesareo
9,2563517
$endgroup$
add a comment |
$begingroup$
Hint.
We have
$$
mathcal{D}_1 f = A p\
mathcal{D}_2 p = -A f
$$
then
$$
mathcal{D}_2mathcal{D}_1 f = Amathcal{D}_2 p = -A^2 f\
mathcal{D}_1mathcal{D}_2 p = -A mathcal{D}_1f = -A^2p
$$
answered Jan 11 at 13:35
CesareoCesareo
9,2563517
$endgroup$
Hint.
We have
$$
mathcal{D}_1 f = A p\
mathcal{D}_2 p = -A f
$$
then
$$
mathcal{D}_2mathcal{D}_1 f = Amathcal{D}_2 p = -A^2 f\
mathcal{D}_1mathcal{D}_2 p = -A mathcal{D}_1f = -A^2p
$$
answered Jan 11 at 13:35
CesareoCesareo
9,2563517
answered Jan 11 at 13:35
CesareoCesareo
9,2563517
answered Jan 11 at 13:35
CesareoCesareo
9,2563517
answered Jan 11 at 13:35
CesareoCesareo
9,2563517
9,2563517
add a comment |
add a comment |
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$begingroup$
I suppose you could take $partial_{x}$ (and $partial_{t}$) of the first equation, rearrange for $partial_{x} p$ and $partial_{t} p$, then substitute those results into your second equation which should give you everything in $f$.
$endgroup$
– Mattos
Mar 11 '16 at 16:03
$begingroup$
@Mattos What do you mean by rearranging $partial_x p$?
$endgroup$
– new guy
Mar 11 '16 at 16:05
$begingroup$
From the first equation $$partial_{x} p = frac{1}{A} partial_{x} bigg( partial_{t} f - c partial_{x} f bigg)$$ You can get something similar for $partial_{t} p$.
$endgroup$
– Mattos
Mar 11 '16 at 16:06