Solving a Cauchy Problem using Method of Characteristics
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I came across a partial differential equation (IVP PDE) that I would like to solve:
$$ u(x,0)=0$$
This should be a quasilinear PDE, and is in the format of a Cauchy Problem, in the form of:
$$au_x+bu_y=c$$
Such that a, b, and c are constants.
In my particular case, I have:
$$
left{
beginarrayc
a=cos(ky) \
b=1 \
c=ax^2
endarray
right.
$$
Using the Lagrange-Carpit Equations: $$fracdxcos(ky)=fracdy1=fracduax^2$$
Rearranging gives: $$fracdxdy=cos(ky)$$
$$dy=fracduax^2$$
I then canclulated $dx-dy$:
$$du=(ax^2cos(ky)+1)dy-dx$$
Integrate to obtain $u(x,y,a,k)$:
$$u(x,y,a,k)=frac1kax^2sin(ky)+y-x+C_1$$
I applied the initial condition of $u(x,0)=0$ and made $x=0$, which gave $C_1=0$.
The question is thus finalised.
This however does not look correct. Am I doing the right thing?
I would also like to see what the characteristic curves look like, what are these? Are they essentially the terms $a, b, c$ I obtained above? Or rather, I simply differentiate those three variables in terms of $s$. Essentially calculating $fracdyds$, $fracdxds$ and $fracdzds$.
pde initial-value-problems cauchy-problem
add a comment |Â
up vote
2
down vote
favorite
I came across a partial differential equation (IVP PDE) that I would like to solve:
$$ u(x,0)=0$$
This should be a quasilinear PDE, and is in the format of a Cauchy Problem, in the form of:
$$au_x+bu_y=c$$
Such that a, b, and c are constants.
In my particular case, I have:
$$
left{
beginarrayc
a=cos(ky) \
b=1 \
c=ax^2
endarray
right.
$$
Using the Lagrange-Carpit Equations: $$fracdxcos(ky)=fracdy1=fracduax^2$$
Rearranging gives: $$fracdxdy=cos(ky)$$
$$dy=fracduax^2$$
I then canclulated $dx-dy$:
$$du=(ax^2cos(ky)+1)dy-dx$$
Integrate to obtain $u(x,y,a,k)$:
$$u(x,y,a,k)=frac1kax^2sin(ky)+y-x+C_1$$
I applied the initial condition of $u(x,0)=0$ and made $x=0$, which gave $C_1=0$.
The question is thus finalised.
This however does not look correct. Am I doing the right thing?
I would also like to see what the characteristic curves look like, what are these? Are they essentially the terms $a, b, c$ I obtained above? Or rather, I simply differentiate those three variables in terms of $s$. Essentially calculating $fracdyds$, $fracdxds$ and $fracdzds$.
pde initial-value-problems cauchy-problem
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I came across a partial differential equation (IVP PDE) that I would like to solve:
$$ u(x,0)=0$$
This should be a quasilinear PDE, and is in the format of a Cauchy Problem, in the form of:
$$au_x+bu_y=c$$
Such that a, b, and c are constants.
In my particular case, I have:
$$
left{
beginarrayc
a=cos(ky) \
b=1 \
c=ax^2
endarray
right.
$$
Using the Lagrange-Carpit Equations: $$fracdxcos(ky)=fracdy1=fracduax^2$$
Rearranging gives: $$fracdxdy=cos(ky)$$
$$dy=fracduax^2$$
I then canclulated $dx-dy$:
$$du=(ax^2cos(ky)+1)dy-dx$$
Integrate to obtain $u(x,y,a,k)$:
$$u(x,y,a,k)=frac1kax^2sin(ky)+y-x+C_1$$
I applied the initial condition of $u(x,0)=0$ and made $x=0$, which gave $C_1=0$.
The question is thus finalised.
This however does not look correct. Am I doing the right thing?
I would also like to see what the characteristic curves look like, what are these? Are they essentially the terms $a, b, c$ I obtained above? Or rather, I simply differentiate those three variables in terms of $s$. Essentially calculating $fracdyds$, $fracdxds$ and $fracdzds$.
pde initial-value-problems cauchy-problem
I came across a partial differential equation (IVP PDE) that I would like to solve:
$$ u(x,0)=0$$
This should be a quasilinear PDE, and is in the format of a Cauchy Problem, in the form of:
$$au_x+bu_y=c$$
Such that a, b, and c are constants.
In my particular case, I have:
$$
left{
beginarrayc
a=cos(ky) \
b=1 \
c=ax^2
endarray
right.
$$
Using the Lagrange-Carpit Equations: $$fracdxcos(ky)=fracdy1=fracduax^2$$
Rearranging gives: $$fracdxdy=cos(ky)$$
$$dy=fracduax^2$$
I then canclulated $dx-dy$:
$$du=(ax^2cos(ky)+1)dy-dx$$
Integrate to obtain $u(x,y,a,k)$:
$$u(x,y,a,k)=frac1kax^2sin(ky)+y-x+C_1$$
I applied the initial condition of $u(x,0)=0$ and made $x=0$, which gave $C_1=0$.
The question is thus finalised.
This however does not look correct. Am I doing the right thing?
I would also like to see what the characteristic curves look like, what are these? Are they essentially the terms $a, b, c$ I obtained above? Or rather, I simply differentiate those three variables in terms of $s$. Essentially calculating $fracdyds$, $fracdxds$ and $fracdzds$.
pde initial-value-problems cauchy-problem
asked Aug 29 at 10:26
Hypergeometry
607
607
add a comment |Â
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f2898211%2fsolving-a-cauchy-problem-using-method-of-characteristics%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