Coupled partial differential equations The Next CEO of Stack OverflowFinding the general solution of a second order PDEWhat exactly are partial differential equations?Techniques for solving coupled differential equationsnumerical simulation of 4 coupled nonlinear PDEsFinding general solution to Partial Differential EquationsHow to find the general solution of this PDENumerical solutions of partial differential equationsSolving a system of semilinear coupled partial differential equationsSolve partial differential equations (PDEs) analyticallySolving Coupled Second Order Differential EquationHow do I proceed from here solving partial differential equation with boundary conditions
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Coupled partial differential equations
The Next CEO of Stack OverflowFinding the general solution of a second order PDEWhat exactly are partial differential equations?Techniques for solving coupled differential equationsnumerical simulation of 4 coupled nonlinear PDEsFinding general solution to Partial Differential EquationsHow to find the general solution of this PDENumerical solutions of partial differential equationsSolving a system of semilinear coupled partial differential equationsSolve partial differential equations (PDEs) analyticallySolving Coupled Second Order Differential EquationHow do I proceed from here solving partial differential equation with boundary conditions
$begingroup$
I'm having trouble solving these coupled partial differential equations:
$$fracpartialpartial tf(x,t)-cfracpartialpartial xf(x,t)-Ap(x,t)=0,$$
$$fracpartialpartial tp(x,t)+cfracpartialpartial xp(x,t)+Af(x,t)=0,$$
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
5,49911439
asked Mar 11 '16 at 15:58
new guynew guy
433
$endgroup$
add a comment |
$begingroup$
I'm having trouble solving these coupled partial differential equations:
$$fracpartialpartial tf(x,t)-cfracpartialpartial xf(x,t)-Ap(x,t)=0,$$
$$fracpartialpartial tp(x,t)+cfracpartialpartial xp(x,t)+Af(x,t)=0,$$
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
5,49911439
asked Mar 11 '16 at 15:58
new guynew guy
433
$endgroup$
$begingroup$
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
$begingroup$
@Mattos What do you mean by rearranging $partial_x p$?
$endgroup$
– new guy
Mar 11 '16 at 16:05
$begingroup$
From the first equation $$partial_x p = frac1A partial_x bigg( partial_t f - c partial_x f bigg)$$ 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:
$$fracpartialpartial tf(x,t)-cfracpartialpartial xf(x,t)-Ap(x,t)=0,$$
$$fracpartialpartial tp(x,t)+cfracpartialpartial xp(x,t)+Af(x,t)=0,$$
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
5,49911439
asked Mar 11 '16 at 15:58
new guynew guy
433
$endgroup$
I'm having trouble solving these coupled partial differential equations:
$$fracpartialpartial tf(x,t)-cfracpartialpartial xf(x,t)-Ap(x,t)=0,$$
$$fracpartialpartial tp(x,t)+cfracpartialpartial xp(x,t)+Af(x,t)=0,$$
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
5,49911439
asked Mar 11 '16 at 15:58
new guynew guy
433
edited Feb 25 at 11:36
Harry Peter
5,49911439
asked Mar 11 '16 at 15:58
new guynew guy
433
edited Feb 25 at 11:36
Harry Peter
5,49911439
edited Feb 25 at 11:36
Harry Peter
5,49911439
edited Feb 25 at 11:36
Harry Peter
5,49911439
5,49911439
asked Mar 11 '16 at 15:58
new guynew guy
433
asked Mar 11 '16 at 15:58
new guynew guy
433
asked Mar 11 '16 at 15:58
new guynew guy
433
433
$begingroup$
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 $$partial_x p = frac1A partial_x bigg( partial_t f - c partial_x f bigg)$$ 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 $$partial_x p = frac1A partial_x bigg( partial_t f - c partial_x f bigg)$$ 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 $$partial_x p = frac1A partial_x bigg( partial_t f - c partial_x f bigg)$$ You can get something similar for $partial_t p$.
$endgroup$
– Mattos
Mar 11 '16 at 16:06
$begingroup$
From the first equation $$partial_x p = frac1A partial_x bigg( partial_t f - c partial_x f bigg)$$ 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
$begingroup$
$$begincases
fracpartialpartial tf(x,t)-cfracpartialpartial xf(x,t)-Ap(x,t)=0 \
fracpartialpartial tp(x,t)+cfracpartialpartial xp(x,t)+Af(x,t)=0
endcases$$
Regularised form with $begincases T=At \ X=fracAcx endcases quadtoquad
begincases
fracpartialpartial Tf(X,T)-fracpartialpartial Xf(X,T)-p(X,T)=0 \
fracpartialpartial Tp(X,T)+fracpartialpartial Xp(X,T)+f(X,T)=0
endcases$
$$begincases
f_T-f_X-p=0 \
p_T+p_X+f=0
endcases$$
$p=f_T-f_X quadtoquad (f_TT-f_XT)+(f_XT-f_XX)+f=0$
$$fracpartial^2 fpartial T^2-fracpartial^2fpartial X^2+f(X,T)=0$$
Solving this hyperbolic PDE leads to $f(X,T)=fleft(At:,:fracAcx right)$
Then $quad p(X,T)=fracpartial fpartial T-fracpartial fpartial X=pleft(At:,:fracAcx right)$
For example of solving see : Finding the general solution of a second order PDE
This method leads to the integral form of solution :
$$f(X,T)=int c(s)e^sqrtalpha(s)-frac12:X +sqrtalpha(s)+frac12 :T ds$$
$c(s)$ 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
45.2k21856
$endgroup$
add a comment |
$begingroup$
Hint.
We have
$$
mathcalD_1 f = A p\
mathcalD_2 p = -A f
$$
then
$$
mathcalD_2mathcalD_1 f = AmathcalD_2 p = -A^2 f\
mathcalD_1mathcalD_2 p = -A mathcalD_1f = -A^2p
$$
answered Jan 11 at 13:35
CesareoCesareo
9,5473517
$endgroup$
add a comment |
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2 Answers
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2 Answers
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votes
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oldest
votes
$begingroup$
$$begincases
fracpartialpartial tf(x,t)-cfracpartialpartial xf(x,t)-Ap(x,t)=0 \
fracpartialpartial tp(x,t)+cfracpartialpartial xp(x,t)+Af(x,t)=0
endcases$$
Regularised form with $begincases T=At \ X=fracAcx endcases quadtoquad
begincases
fracpartialpartial Tf(X,T)-fracpartialpartial Xf(X,T)-p(X,T)=0 \
fracpartialpartial Tp(X,T)+fracpartialpartial Xp(X,T)+f(X,T)=0
endcases$
$$begincases
f_T-f_X-p=0 \
p_T+p_X+f=0
endcases$$
$p=f_T-f_X quadtoquad (f_TT-f_XT)+(f_XT-f_XX)+f=0$
$$fracpartial^2 fpartial T^2-fracpartial^2fpartial X^2+f(X,T)=0$$
Solving this hyperbolic PDE leads to $f(X,T)=fleft(At:,:fracAcx right)$
Then $quad p(X,T)=fracpartial fpartial T-fracpartial fpartial X=pleft(At:,:fracAcx right)$
For example of solving see : Finding the general solution of a second order PDE
This method leads to the integral form of solution :
$$f(X,T)=int c(s)e^sqrtalpha(s)-frac12:X +sqrtalpha(s)+frac12 :T ds$$
$c(s)$ 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
45.2k21856
$endgroup$
add a comment |
$begingroup$
$$begincases
fracpartialpartial tf(x,t)-cfracpartialpartial xf(x,t)-Ap(x,t)=0 \
fracpartialpartial tp(x,t)+cfracpartialpartial xp(x,t)+Af(x,t)=0
endcases$$
Regularised form with $begincases T=At \ X=fracAcx endcases quadtoquad
begincases
fracpartialpartial Tf(X,T)-fracpartialpartial Xf(X,T)-p(X,T)=0 \
fracpartialpartial Tp(X,T)+fracpartialpartial Xp(X,T)+f(X,T)=0
endcases$
$$begincases
f_T-f_X-p=0 \
p_T+p_X+f=0
endcases$$
$p=f_T-f_X quadtoquad (f_TT-f_XT)+(f_XT-f_XX)+f=0$
$$fracpartial^2 fpartial T^2-fracpartial^2fpartial X^2+f(X,T)=0$$
Solving this hyperbolic PDE leads to $f(X,T)=fleft(At:,:fracAcx right)$
Then $quad p(X,T)=fracpartial fpartial T-fracpartial fpartial X=pleft(At:,:fracAcx right)$
For example of solving see : Finding the general solution of a second order PDE
This method leads to the integral form of solution :
$$f(X,T)=int c(s)e^sqrtalpha(s)-frac12:X +sqrtalpha(s)+frac12 :T ds$$
$c(s)$ 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
45.2k21856
$endgroup$
add a comment |
$begingroup$
$$begincases
fracpartialpartial tf(x,t)-cfracpartialpartial xf(x,t)-Ap(x,t)=0 \
fracpartialpartial tp(x,t)+cfracpartialpartial xp(x,t)+Af(x,t)=0
endcases$$
Regularised form with $begincases T=At \ X=fracAcx endcases quadtoquad
begincases
fracpartialpartial Tf(X,T)-fracpartialpartial Xf(X,T)-p(X,T)=0 \
fracpartialpartial Tp(X,T)+fracpartialpartial Xp(X,T)+f(X,T)=0
endcases$
$$begincases
f_T-f_X-p=0 \
p_T+p_X+f=0
endcases$$
$p=f_T-f_X quadtoquad (f_TT-f_XT)+(f_XT-f_XX)+f=0$
$$fracpartial^2 fpartial T^2-fracpartial^2fpartial X^2+f(X,T)=0$$
Solving this hyperbolic PDE leads to $f(X,T)=fleft(At:,:fracAcx right)$
Then $quad p(X,T)=fracpartial fpartial T-fracpartial fpartial X=pleft(At:,:fracAcx right)$
For example of solving see : Finding the general solution of a second order PDE
This method leads to the integral form of solution :
$$f(X,T)=int c(s)e^sqrtalpha(s)-frac12:X +sqrtalpha(s)+frac12 :T ds$$
$c(s)$ 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
45.2k21856
$endgroup$
$$begincases
fracpartialpartial tf(x,t)-cfracpartialpartial xf(x,t)-Ap(x,t)=0 \
fracpartialpartial tp(x,t)+cfracpartialpartial xp(x,t)+Af(x,t)=0
endcases$$
Regularised form with $begincases T=At \ X=fracAcx endcases quadtoquad
begincases
fracpartialpartial Tf(X,T)-fracpartialpartial Xf(X,T)-p(X,T)=0 \
fracpartialpartial Tp(X,T)+fracpartialpartial Xp(X,T)+f(X,T)=0
endcases$
$$begincases
f_T-f_X-p=0 \
p_T+p_X+f=0
endcases$$
$p=f_T-f_X quadtoquad (f_TT-f_XT)+(f_XT-f_XX)+f=0$
$$fracpartial^2 fpartial T^2-fracpartial^2fpartial X^2+f(X,T)=0$$
Solving this hyperbolic PDE leads to $f(X,T)=fleft(At:,:fracAcx right)$
Then $quad p(X,T)=fracpartial fpartial T-fracpartial fpartial X=pleft(At:,:fracAcx right)$
For example of solving see : Finding the general solution of a second order PDE
This method leads to the integral form of solution :
$$f(X,T)=int c(s)e^sqrtalpha(s)-frac12:X +sqrtalpha(s)+frac12 :T ds$$
$c(s)$ 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
45.2k21856
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
45.2k21856
answered Mar 16 '16 at 8:49
JJacquelinJJacquelin
45.2k21856
answered Mar 16 '16 at 8:49
JJacquelinJJacquelin
45.2k21856
45.2k21856
add a comment |
add a comment |
$begingroup$
Hint.
We have
$$
mathcalD_1 f = A p\
mathcalD_2 p = -A f
$$
then
$$
mathcalD_2mathcalD_1 f = AmathcalD_2 p = -A^2 f\
mathcalD_1mathcalD_2 p = -A mathcalD_1f = -A^2p
$$
answered Jan 11 at 13:35
CesareoCesareo
9,5473517
$endgroup$
add a comment |
$begingroup$
Hint.
We have
$$
mathcalD_1 f = A p\
mathcalD_2 p = -A f
$$
then
$$
mathcalD_2mathcalD_1 f = AmathcalD_2 p = -A^2 f\
mathcalD_1mathcalD_2 p = -A mathcalD_1f = -A^2p
$$
answered Jan 11 at 13:35
CesareoCesareo
9,5473517
$endgroup$
add a comment |
$begingroup$
Hint.
We have
$$
mathcalD_1 f = A p\
mathcalD_2 p = -A f
$$
then
$$
mathcalD_2mathcalD_1 f = AmathcalD_2 p = -A^2 f\
mathcalD_1mathcalD_2 p = -A mathcalD_1f = -A^2p
$$
answered Jan 11 at 13:35
CesareoCesareo
9,5473517
$endgroup$
Hint.
We have
$$
mathcalD_1 f = A p\
mathcalD_2 p = -A f
$$
then
$$
mathcalD_2mathcalD_1 f = AmathcalD_2 p = -A^2 f\
mathcalD_1mathcalD_2 p = -A mathcalD_1f = -A^2p
$$
answered Jan 11 at 13:35
CesareoCesareo
9,5473517
answered Jan 11 at 13:35
CesareoCesareo
9,5473517
answered Jan 11 at 13:35
CesareoCesareo
9,5473517
answered Jan 11 at 13:35
CesareoCesareo
9,5473517
9,5473517
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 $$partial_x p = frac1A partial_x bigg( partial_t f - c partial_x f bigg)$$ You can get something similar for $partial_t p$.
$endgroup$
– Mattos
Mar 11 '16 at 16:06