Solving quasilinear p.d.e. with method of charactersticsMethod of characteristics. Small question about initial conditions.Solving a PDE via method of characteristicsMethod of characteristics - eliminating variablesFirst order PDE with Method of Characterization.Solve the PDE $yu_y - xu_x = 1$ by method of characteristicsMethod of Characteristics(Advection equation with initial and boundary condition)Solving a PDE with Method of Characteristics.Solving PDE using the method of characteristicsHow to Solve This PDE and Deal with Boundary Value Using Method of Characteristics. Work Shown.Solving $u_t + cu_x = k$ by method of characteristics
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Solving quasilinear p.d.e. with method of characterstics
Method of characteristics. Small question about initial conditions.Solving a PDE via method of characteristicsMethod of characteristics - eliminating variablesFirst order PDE with Method of Characterization.Solve the PDE $yu_y - xu_x = 1$ by method of characteristicsMethod of Characteristics(Advection equation with initial and boundary condition)Solving a PDE with Method of Characteristics.Solving PDE using the method of characteristicsHow to Solve This PDE and Deal with Boundary Value Using Method of Characteristics. Work Shown.Solving $u_t + cu_x = k$ by method of characteristics
$begingroup$
I am currently working on solving the p.d.e.
$$
fracpartial upartial x + fracpartial upartial y = 1
$$
with the initial condition
$$
u = 0 text on x + y = 0
$$
using the method of characteristics. Thus, so far, I have obtained a system of characteristic equations
$$
begincases
fracdxdt = 1 implies x = t + x_0\
fracdydt = 1 implies y = t + y_0\
fracdudt = 1 implies u = t + u_0
endcases
$$
However, I'm unsure how to proceed to encorporate the initial conditions or solve for $u(x, y)$ explicitly. Any help is appreciated.
....
pde
edited 2 days ago
Alexandros
9401412
asked 2 days ago
ScriniaryScriniary
163
New contributor
Scriniary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am currently working on solving the p.d.e.
$$
fracpartial upartial x + fracpartial upartial y = 1
$$
with the initial condition
$$
u = 0 text on x + y = 0
$$
using the method of characteristics. Thus, so far, I have obtained a system of characteristic equations
$$
begincases
fracdxdt = 1 implies x = t + x_0\
fracdydt = 1 implies y = t + y_0\
fracdudt = 1 implies u = t + u_0
endcases
$$
However, I'm unsure how to proceed to encorporate the initial conditions or solve for $u(x, y)$ explicitly. Any help is appreciated.
....
pde
edited 2 days ago
Alexandros
9401412
asked 2 days ago
ScriniaryScriniary
163
New contributor
Scriniary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Eliminate $t$ from each of the equations i.e $$x-x_0 = y-y_0 = u-u_0$$ which implies $y = x + c_1$ and $u = x + f(c_1) = x + f(y-x)$. Now apply your boundary data.
$endgroup$
– Mattos
2 days ago
$begingroup$
How do you deduce $u = x + f(c_1)$? All that seems clear to me is $u = x - x_0 + u_0$.
$endgroup$
– Scriniary
2 days ago
$begingroup$
Because the implicit solution to the PDE problem is given by $$phi(c_1, c_2) = 0 implies phi(y-x, u-x) = 0$$ which, in a different form, is $$u-x = f(y-x)$$ i.e $c_2 = f(c_1)$.
$endgroup$
– Mattos
2 days ago
add a comment |
$begingroup$
I am currently working on solving the p.d.e.
$$
fracpartial upartial x + fracpartial upartial y = 1
$$
with the initial condition
$$
u = 0 text on x + y = 0
$$
using the method of characteristics. Thus, so far, I have obtained a system of characteristic equations
$$
begincases
fracdxdt = 1 implies x = t + x_0\
fracdydt = 1 implies y = t + y_0\
fracdudt = 1 implies u = t + u_0
endcases
$$
However, I'm unsure how to proceed to encorporate the initial conditions or solve for $u(x, y)$ explicitly. Any help is appreciated.
....
pde
edited 2 days ago
Alexandros
9401412
asked 2 days ago
ScriniaryScriniary
163
New contributor
Scriniary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am currently working on solving the p.d.e.
$$
fracpartial upartial x + fracpartial upartial y = 1
$$
with the initial condition
$$
u = 0 text on x + y = 0
$$
using the method of characteristics. Thus, so far, I have obtained a system of characteristic equations
$$
begincases
fracdxdt = 1 implies x = t + x_0\
fracdydt = 1 implies y = t + y_0\
fracdudt = 1 implies u = t + u_0
endcases
$$
However, I'm unsure how to proceed to encorporate the initial conditions or solve for $u(x, y)$ explicitly. Any help is appreciated.
....
pde
pde
edited 2 days ago
Alexandros
9401412
asked 2 days ago
ScriniaryScriniary
163
New contributor
Scriniary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Alexandros
9401412
asked 2 days ago
ScriniaryScriniary
163
New contributor
Scriniary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Alexandros
9401412
edited 2 days ago
Alexandros
9401412
edited 2 days ago
Alexandros
9401412
9401412
asked 2 days ago
ScriniaryScriniary
163
New contributor
Scriniary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
ScriniaryScriniary
163
asked 2 days ago
ScriniaryScriniary
163
163
New contributor
Scriniary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Scriniary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Scriniary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Eliminate $t$ from each of the equations i.e $$x-x_0 = y-y_0 = u-u_0$$ which implies $y = x + c_1$ and $u = x + f(c_1) = x + f(y-x)$. Now apply your boundary data.
$endgroup$
– Mattos
2 days ago
$begingroup$
How do you deduce $u = x + f(c_1)$? All that seems clear to me is $u = x - x_0 + u_0$.
$endgroup$
– Scriniary
2 days ago
$begingroup$
Because the implicit solution to the PDE problem is given by $$phi(c_1, c_2) = 0 implies phi(y-x, u-x) = 0$$ which, in a different form, is $$u-x = f(y-x)$$ i.e $c_2 = f(c_1)$.
$endgroup$
– Mattos
2 days ago
add a comment |
$begingroup$
Eliminate $t$ from each of the equations i.e $$x-x_0 = y-y_0 = u-u_0$$ which implies $y = x + c_1$ and $u = x + f(c_1) = x + f(y-x)$. Now apply your boundary data.
$endgroup$
– Mattos
2 days ago
$begingroup$
How do you deduce $u = x + f(c_1)$? All that seems clear to me is $u = x - x_0 + u_0$.
$endgroup$
– Scriniary
2 days ago
$begingroup$
Because the implicit solution to the PDE problem is given by $$phi(c_1, c_2) = 0 implies phi(y-x, u-x) = 0$$ which, in a different form, is $$u-x = f(y-x)$$ i.e $c_2 = f(c_1)$.
$endgroup$
– Mattos
2 days ago
$begingroup$
Eliminate $t$ from each of the equations i.e $$x-x_0 = y-y_0 = u-u_0$$ which implies $y = x + c_1$ and $u = x + f(c_1) = x + f(y-x)$. Now apply your boundary data.
$endgroup$
– Mattos
2 days ago
$begingroup$
Eliminate $t$ from each of the equations i.e $$x-x_0 = y-y_0 = u-u_0$$ which implies $y = x + c_1$ and $u = x + f(c_1) = x + f(y-x)$. Now apply your boundary data.
$endgroup$
– Mattos
2 days ago
$begingroup$
How do you deduce $u = x + f(c_1)$? All that seems clear to me is $u = x - x_0 + u_0$.
$endgroup$
– Scriniary
2 days ago
$begingroup$
How do you deduce $u = x + f(c_1)$? All that seems clear to me is $u = x - x_0 + u_0$.
$endgroup$
– Scriniary
2 days ago
$begingroup$
Because the implicit solution to the PDE problem is given by $$phi(c_1, c_2) = 0 implies phi(y-x, u-x) = 0$$ which, in a different form, is $$u-x = f(y-x)$$ i.e $c_2 = f(c_1)$.
$endgroup$
– Mattos
2 days ago
$begingroup$
Because the implicit solution to the PDE problem is given by $$phi(c_1, c_2) = 0 implies phi(y-x, u-x) = 0$$ which, in a different form, is $$u-x = f(y-x)$$ i.e $c_2 = f(c_1)$.
$endgroup$
– Mattos
2 days ago
add a comment |
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$begingroup$
Eliminate $t$ from each of the equations i.e $$x-x_0 = y-y_0 = u-u_0$$ which implies $y = x + c_1$ and $u = x + f(c_1) = x + f(y-x)$. Now apply your boundary data.
$endgroup$
– Mattos
2 days ago
$begingroup$
How do you deduce $u = x + f(c_1)$? All that seems clear to me is $u = x - x_0 + u_0$.
$endgroup$
– Scriniary
2 days ago
$begingroup$
Because the implicit solution to the PDE problem is given by $$phi(c_1, c_2) = 0 implies phi(y-x, u-x) = 0$$ which, in a different form, is $$u-x = f(y-x)$$ i.e $c_2 = f(c_1)$.
$endgroup$
– Mattos
2 days ago