Mixing
Partial Differential Equation with Two Dependent Variables
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I've come across a PDE where I feel there should be some clever way of solving it, but I can't seem to find it. It has a form like this:
$$fracpartialpartial tleft(y+czright)=fracpartialpartial xleft(-by+aczright)$$
Here $y$ and $z$ are the dependent variables and a, b, and c are all constants. Ideally, there would exist some other variable $u$ thats a function of $y, z$ such that we can rewrite the equation in an equivalent form like:
$$fracpartialpartial tleft(uright)=fracpartialpartial xleft(uright)$$
which could be solved with method of characteristics, though I can't seem to find such a variable. Assuming there isn't one, can you recommend any other approaches?
Edit: Here's the original two PDE's from which the above expression was obtained:
$$fracpartial ypartial t=-a_1fracpartial ypartial x-b_1left(y-zright)$$
$$fracpartial zpartial t=a_2fracpartial zpartial x+b_2left(y-zright)$$
We took a linear combination of these 2 equations to try eliminate the terms with $left(y-zright)$ in them and then tried to find an invariant variable to solve as I was talking about earlier. $a_1, a_2, b_1, b_2$ are all constants.
pde
asked Jul 18 at 16:50
Leif Ericson
1248
add a comment |Â
up vote
0
down vote
favorite
I've come across a PDE where I feel there should be some clever way of solving it, but I can't seem to find it. It has a form like this:
$$fracpartialpartial tleft(y+czright)=fracpartialpartial xleft(-by+aczright)$$
Here $y$ and $z$ are the dependent variables and a, b, and c are all constants. Ideally, there would exist some other variable $u$ thats a function of $y, z$ such that we can rewrite the equation in an equivalent form like:
$$fracpartialpartial tleft(uright)=fracpartialpartial xleft(uright)$$
which could be solved with method of characteristics, though I can't seem to find such a variable. Assuming there isn't one, can you recommend any other approaches?
Edit: Here's the original two PDE's from which the above expression was obtained:
$$fracpartial ypartial t=-a_1fracpartial ypartial x-b_1left(y-zright)$$
$$fracpartial zpartial t=a_2fracpartial zpartial x+b_2left(y-zright)$$
We took a linear combination of these 2 equations to try eliminate the terms with $left(y-zright)$ in them and then tried to find an invariant variable to solve as I was talking about earlier. $a_1, a_2, b_1, b_2$ are all constants.
pde
asked Jul 18 at 16:50
Leif Ericson
1248
How about Fourier transforming it?
– joriki
Jul 18 at 16:59
1
They are two unknowns $y(x,t)$ and $z(x,t)$. But there is only one equation. The problem is undetermined, or something is missing in the wording of the question, isn't it ?
– JJacquelin
Jul 18 at 17:07
Yeah, this pde actually comes from the linear combination of two other pdes. One for y, and one for z. We ended up doing that combination to eliminate a term that was present in both of them.
– Leif Ericson
Jul 18 at 17:15
1
@Leif Ericson. What are the two PDEs ? Why don't you write the whole problem ? This makes doubtful what you did and what really is the key issue..
– JJacquelin
Jul 18 at 17:51
Sorry about the lateness of the reply, I edited the original post to show the PDE's from which the combined expression came from. Thanks again for your help.
– Leif Ericson
Jul 19 at 16:21
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've come across a PDE where I feel there should be some clever way of solving it, but I can't seem to find it. It has a form like this:
$$fracpartialpartial tleft(y+czright)=fracpartialpartial xleft(-by+aczright)$$
Here $y$ and $z$ are the dependent variables and a, b, and c are all constants. Ideally, there would exist some other variable $u$ thats a function of $y, z$ such that we can rewrite the equation in an equivalent form like:
$$fracpartialpartial tleft(uright)=fracpartialpartial xleft(uright)$$
which could be solved with method of characteristics, though I can't seem to find such a variable. Assuming there isn't one, can you recommend any other approaches?
Edit: Here's the original two PDE's from which the above expression was obtained:
$$fracpartial ypartial t=-a_1fracpartial ypartial x-b_1left(y-zright)$$
$$fracpartial zpartial t=a_2fracpartial zpartial x+b_2left(y-zright)$$
We took a linear combination of these 2 equations to try eliminate the terms with $left(y-zright)$ in them and then tried to find an invariant variable to solve as I was talking about earlier. $a_1, a_2, b_1, b_2$ are all constants.
pde
asked Jul 18 at 16:50
Leif Ericson
1248
I've come across a PDE where I feel there should be some clever way of solving it, but I can't seem to find it. It has a form like this:
$$fracpartialpartial tleft(y+czright)=fracpartialpartial xleft(-by+aczright)$$
Here $y$ and $z$ are the dependent variables and a, b, and c are all constants. Ideally, there would exist some other variable $u$ thats a function of $y, z$ such that we can rewrite the equation in an equivalent form like:
$$fracpartialpartial tleft(uright)=fracpartialpartial xleft(uright)$$
which could be solved with method of characteristics, though I can't seem to find such a variable. Assuming there isn't one, can you recommend any other approaches?
Edit: Here's the original two PDE's from which the above expression was obtained:
$$fracpartial ypartial t=-a_1fracpartial ypartial x-b_1left(y-zright)$$
$$fracpartial zpartial t=a_2fracpartial zpartial x+b_2left(y-zright)$$
We took a linear combination of these 2 equations to try eliminate the terms with $left(y-zright)$ in them and then tried to find an invariant variable to solve as I was talking about earlier. $a_1, a_2, b_1, b_2$ are all constants.
pde
asked Jul 18 at 16:50
Leif Ericson
1248
edited Jul 19 at 16:20
asked Jul 18 at 16:50
Leif Ericson
1248
asked Jul 18 at 16:50
Leif Ericson
1248
asked Jul 18 at 16:50
Leif Ericson
1248
1248
How about Fourier transforming it?
– joriki
Jul 18 at 16:59
1
They are two unknowns $y(x,t)$ and $z(x,t)$. But there is only one equation. The problem is undetermined, or something is missing in the wording of the question, isn't it ?
– JJacquelin
Jul 18 at 17:07
Yeah, this pde actually comes from the linear combination of two other pdes. One for y, and one for z. We ended up doing that combination to eliminate a term that was present in both of them.
– Leif Ericson
Jul 18 at 17:15
1
@Leif Ericson. What are the two PDEs ? Why don't you write the whole problem ? This makes doubtful what you did and what really is the key issue..
– JJacquelin
Jul 18 at 17:51
Sorry about the lateness of the reply, I edited the original post to show the PDE's from which the combined expression came from. Thanks again for your help.
– Leif Ericson
Jul 19 at 16:21
add a comment |Â
How about Fourier transforming it?
– joriki
Jul 18 at 16:59
1
They are two unknowns $y(x,t)$ and $z(x,t)$. But there is only one equation. The problem is undetermined, or something is missing in the wording of the question, isn't it ?
– JJacquelin
Jul 18 at 17:07
Yeah, this pde actually comes from the linear combination of two other pdes. One for y, and one for z. We ended up doing that combination to eliminate a term that was present in both of them.
– Leif Ericson
Jul 18 at 17:15
1
@Leif Ericson. What are the two PDEs ? Why don't you write the whole problem ? This makes doubtful what you did and what really is the key issue..
– JJacquelin
Jul 18 at 17:51
Sorry about the lateness of the reply, I edited the original post to show the PDE's from which the combined expression came from. Thanks again for your help.
– Leif Ericson
Jul 19 at 16:21
How about Fourier transforming it?
– joriki
Jul 18 at 16:59
How about Fourier transforming it?
– joriki
Jul 18 at 16:59
1
1
They are two unknowns $y(x,t)$ and $z(x,t)$. But there is only one equation. The problem is undetermined, or something is missing in the wording of the question, isn't it ?
– JJacquelin
Jul 18 at 17:07
They are two unknowns $y(x,t)$ and $z(x,t)$. But there is only one equation. The problem is undetermined, or something is missing in the wording of the question, isn't it ?
– JJacquelin
Jul 18 at 17:07
Yeah, this pde actually comes from the linear combination of two other pdes. One for y, and one for z. We ended up doing that combination to eliminate a term that was present in both of them.
– Leif Ericson
Jul 18 at 17:15
Yeah, this pde actually comes from the linear combination of two other pdes. One for y, and one for z. We ended up doing that combination to eliminate a term that was present in both of them.
– Leif Ericson
Jul 18 at 17:15
1
1
@Leif Ericson. What are the two PDEs ? Why don't you write the whole problem ? This makes doubtful what you did and what really is the key issue..
– JJacquelin
Jul 18 at 17:51
@Leif Ericson. What are the two PDEs ? Why don't you write the whole problem ? This makes doubtful what you did and what really is the key issue..
– JJacquelin
Jul 18 at 17:51
Sorry about the lateness of the reply, I edited the original post to show the PDE's from which the combined expression came from. Thanks again for your help.
– Leif Ericson
Jul 19 at 16:21
Sorry about the lateness of the reply, I edited the original post to show the PDE's from which the combined expression came from. Thanks again for your help.
– Leif Ericson
Jul 19 at 16:21
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
A good way to find such solutions is too assume that $$fracpartialpartial tleft(y+czright)=fracpartialpartial xleft(-by+aczright)=f'(t)g'(x)$$therefore $$y+cz=f(t)g'(x)+h_1(x)\-by+acz=f'(t)g(x)+h_2(t)$$therefore$$z=dfrac1ac+bc(bf(t)g'(x)+f'(t)g(x)+bh_1(x)+h_2(t))\y=dfrac1a+b(af(t)g'(x)-f'(t)g(x)+ah_1(x)-h_2(t))$$
answered Jul 18 at 18:22
Mostafa Ayaz
8,6023630
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
A good way to find such solutions is too assume that $$fracpartialpartial tleft(y+czright)=fracpartialpartial xleft(-by+aczright)=f'(t)g'(x)$$therefore $$y+cz=f(t)g'(x)+h_1(x)\-by+acz=f'(t)g(x)+h_2(t)$$therefore$$z=dfrac1ac+bc(bf(t)g'(x)+f'(t)g(x)+bh_1(x)+h_2(t))\y=dfrac1a+b(af(t)g'(x)-f'(t)g(x)+ah_1(x)-h_2(t))$$
answered Jul 18 at 18:22
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
1
down vote
A good way to find such solutions is too assume that $$fracpartialpartial tleft(y+czright)=fracpartialpartial xleft(-by+aczright)=f'(t)g'(x)$$therefore $$y+cz=f(t)g'(x)+h_1(x)\-by+acz=f'(t)g(x)+h_2(t)$$therefore$$z=dfrac1ac+bc(bf(t)g'(x)+f'(t)g(x)+bh_1(x)+h_2(t))\y=dfrac1a+b(af(t)g'(x)-f'(t)g(x)+ah_1(x)-h_2(t))$$
answered Jul 18 at 18:22
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A good way to find such solutions is too assume that $$fracpartialpartial tleft(y+czright)=fracpartialpartial xleft(-by+aczright)=f'(t)g'(x)$$therefore $$y+cz=f(t)g'(x)+h_1(x)\-by+acz=f'(t)g(x)+h_2(t)$$therefore$$z=dfrac1ac+bc(bf(t)g'(x)+f'(t)g(x)+bh_1(x)+h_2(t))\y=dfrac1a+b(af(t)g'(x)-f'(t)g(x)+ah_1(x)-h_2(t))$$
answered Jul 18 at 18:22
Mostafa Ayaz
8,6023630
A good way to find such solutions is too assume that $$fracpartialpartial tleft(y+czright)=fracpartialpartial xleft(-by+aczright)=f'(t)g'(x)$$therefore $$y+cz=f(t)g'(x)+h_1(x)\-by+acz=f'(t)g(x)+h_2(t)$$therefore$$z=dfrac1ac+bc(bf(t)g'(x)+f'(t)g(x)+bh_1(x)+h_2(t))\y=dfrac1a+b(af(t)g'(x)-f'(t)g(x)+ah_1(x)-h_2(t))$$
answered Jul 18 at 18:22
Mostafa Ayaz
8,6023630
answered Jul 18 at 18:22
Mostafa Ayaz
8,6023630
answered Jul 18 at 18:22
Mostafa Ayaz
8,6023630
answered Jul 18 at 18:22
Mostafa Ayaz
8,6023630
8,6023630
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2855764%2fpartial-differential-equation-with-two-dependent-variables%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
How about Fourier transforming it?
– joriki
Jul 18 at 16:59
1
They are two unknowns $y(x,t)$ and $z(x,t)$. But there is only one equation. The problem is undetermined, or something is missing in the wording of the question, isn't it ?
– JJacquelin
Jul 18 at 17:07
Yeah, this pde actually comes from the linear combination of two other pdes. One for y, and one for z. We ended up doing that combination to eliminate a term that was present in both of them.
– Leif Ericson
Jul 18 at 17:15
1
@Leif Ericson. What are the two PDEs ? Why don't you write the whole problem ? This makes doubtful what you did and what really is the key issue..
– JJacquelin
Jul 18 at 17:51
Sorry about the lateness of the reply, I edited the original post to show the PDE's from which the combined expression came from. Thanks again for your help.
– Leif Ericson
Jul 19 at 16:21