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How to solve $u_t + u_x =0$ with $u_0(x) = sin(pi x)$ with Characteristics?

Solve PDE using method of characteristicsShock-wave solution for PDE $u_t+(u-1)u_x=2$How to solve 3-variable transport equation $u_t + a u_x + b u_y =0$ by the method of characteristics?$u_t + uu_x=2$ where $u(x,0)=x$. Are there any shock forming with this initial condition?Method of characteristics PDE $u_t +uu_x = x$IVP for nonlinear PDE $u_t + frac13u_x^3 = -cu$Solve IVP of $(u_t )^2 + (u_x )^2 − u^2 = 0$ using method of characteristicsSolve inviscid Burgers' equation with shockSolve initial value problem $u_t+x^2u_x=0$ by method of characteristicsSolution for $u_t+u_x=0$ using characteristics

1

$begingroup$

Consider the initial-boundary value problem (IBVP) for the convection equation
beginarray l u _ t + u _ x = 0 quad x in [ a ( t ) , b ( t ) ] , t in [ 0 , T ] \ u ( x , 0 ) = u _ 0 ( x ) quad x in [ a ( 0 ) , b ( 0 ) ] endarray
with initial conditionsu $u_ 0 ( x ) = sin ( pi x )$
The boundaries $a(t)$ and $b(t)$ are determined by the
characteristics that go through $(x = 0; t = 0)$ and $(x = 2; t = 0)$, respectively.
Let $x _ j = j h$ with
$j = 0,...,20$ and $h = 0.1$ .

pde hyperbolic-equations

share|cite|improve this question

edited Mar 14 at 17:00

Rebellos

15.4k31250

asked Mar 14 at 16:48

Junia ShelomiJunia Shelomi

83

$endgroup$

add a comment | 

1

$begingroup$

Consider the initial-boundary value problem (IBVP) for the convection equation
beginarray l u _ t + u _ x = 0 quad x in [ a ( t ) , b ( t ) ] , t in [ 0 , T ] \ u ( x , 0 ) = u _ 0 ( x ) quad x in [ a ( 0 ) , b ( 0 ) ] endarray
with initial conditionsu $u_ 0 ( x ) = sin ( pi x )$
The boundaries $a(t)$ and $b(t)$ are determined by the
characteristics that go through $(x = 0; t = 0)$ and $(x = 2; t = 0)$, respectively.
Let $x _ j = j h$ with
$j = 0,...,20$ and $h = 0.1$ .

pde hyperbolic-equations

share|cite|improve this question

edited Mar 14 at 17:00

Rebellos

15.4k31250

asked Mar 14 at 16:48

Junia ShelomiJunia Shelomi

83

$endgroup$

add a comment | 

$begingroup$

Consider the initial-boundary value problem (IBVP) for the convection equation
beginarray l u _ t + u _ x = 0 quad x in [ a ( t ) , b ( t ) ] , t in [ 0 , T ] \ u ( x , 0 ) = u _ 0 ( x ) quad x in [ a ( 0 ) , b ( 0 ) ] endarray
with initial conditionsu $u_ 0 ( x ) = sin ( pi x )$
The boundaries $a(t)$ and $b(t)$ are determined by the
characteristics that go through $(x = 0; t = 0)$ and $(x = 2; t = 0)$, respectively.
Let $x _ j = j h$ with
$j = 0,...,20$ and $h = 0.1$ .

pde hyperbolic-equations

share|cite|improve this question

edited Mar 14 at 17:00

Rebellos

15.4k31250

asked Mar 14 at 16:48

Junia ShelomiJunia Shelomi

83

$endgroup$

share|cite|improve this question

edited Mar 14 at 17:00

Rebellos

15.4k31250

asked Mar 14 at 16:48

Junia ShelomiJunia Shelomi

83

share|cite|improve this question

edited Mar 14 at 17:00

Rebellos

15.4k31250

asked Mar 14 at 16:48

Junia ShelomiJunia Shelomi

83

edited Mar 14 at 17:00

Rebellos

15.4k31250

edited Mar 14 at 17:00

Rebellos

15.4k31250

asked Mar 14 at 16:48

Junia ShelomiJunia Shelomi

83

asked Mar 14 at 16:48

Junia ShelomiJunia Shelomi

83

1 Answer
1

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0

$begingroup$

I don't really understand the final sentence of your elaboration. The method of characteristics is a hand-solved method, not an iteration.

The characteristics are :

$$fracmathrmdt1 = fracmathrmdx1 = fracmathrmdu0$$
Taking any of the two first with the last one, yields :
$$fracmathrmdx1 = fracmathrmdu0 implies u_1 = u$$
Taking the first pair, yields :
$$fracmathrmdt1 = fracmathrmdx1 implies u_2 = x-t $$

Since $u_2$ is straight-forwardly expressed by $u$, simply enough, the solution will be given as
$$u_1=F(u_2) Rightarrow u(x,t) = F(x-t)$$
where $F$ is an adequatelly smooth function.

But $u(x,0) = u_0(x) = sin(pi x)$, which means :

$$u(x,0) = F(x) = sin(pi x)$$

Now, letting $x : = x-t$, we yield :
$$u(x,t) = F(x-t) = sin[pi(x-t)]$$

share|cite|improve this answer

answered Mar 14 at 16:59

RebellosRebellos

15.4k31250

$endgroup$

add a comment | 

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0

$begingroup$

I don't really understand the final sentence of your elaboration. The method of characteristics is a hand-solved method, not an iteration.

The characteristics are :

$$fracmathrmdt1 = fracmathrmdx1 = fracmathrmdu0$$
Taking any of the two first with the last one, yields :
$$fracmathrmdx1 = fracmathrmdu0 implies u_1 = u$$
Taking the first pair, yields :
$$fracmathrmdt1 = fracmathrmdx1 implies u_2 = x-t $$

Since $u_2$ is straight-forwardly expressed by $u$, simply enough, the solution will be given as
$$u_1=F(u_2) Rightarrow u(x,t) = F(x-t)$$
where $F$ is an adequatelly smooth function.

But $u(x,0) = u_0(x) = sin(pi x)$, which means :

$$u(x,0) = F(x) = sin(pi x)$$

Now, letting $x : = x-t$, we yield :
$$u(x,t) = F(x-t) = sin[pi(x-t)]$$

share|cite|improve this answer

answered Mar 14 at 16:59

RebellosRebellos

15.4k31250

$endgroup$

add a comment | 

0

$begingroup$

I don't really understand the final sentence of your elaboration. The method of characteristics is a hand-solved method, not an iteration.

The characteristics are :

$$fracmathrmdt1 = fracmathrmdx1 = fracmathrmdu0$$
Taking any of the two first with the last one, yields :
$$fracmathrmdx1 = fracmathrmdu0 implies u_1 = u$$
Taking the first pair, yields :
$$fracmathrmdt1 = fracmathrmdx1 implies u_2 = x-t $$

Since $u_2$ is straight-forwardly expressed by $u$, simply enough, the solution will be given as
$$u_1=F(u_2) Rightarrow u(x,t) = F(x-t)$$
where $F$ is an adequatelly smooth function.

But $u(x,0) = u_0(x) = sin(pi x)$, which means :

$$u(x,0) = F(x) = sin(pi x)$$

Now, letting $x : = x-t$, we yield :
$$u(x,t) = F(x-t) = sin[pi(x-t)]$$

share|cite|improve this answer

answered Mar 14 at 16:59

RebellosRebellos

15.4k31250

$endgroup$

add a comment | 

$begingroup$

I don't really understand the final sentence of your elaboration. The method of characteristics is a hand-solved method, not an iteration.

The characteristics are :

$$fracmathrmdt1 = fracmathrmdx1 = fracmathrmdu0$$
Taking any of the two first with the last one, yields :
$$fracmathrmdx1 = fracmathrmdu0 implies u_1 = u$$
Taking the first pair, yields :
$$fracmathrmdt1 = fracmathrmdx1 implies u_2 = x-t $$

Since $u_2$ is straight-forwardly expressed by $u$, simply enough, the solution will be given as
$$u_1=F(u_2) Rightarrow u(x,t) = F(x-t)$$
where $F$ is an adequatelly smooth function.

But $u(x,0) = u_0(x) = sin(pi x)$, which means :

$$u(x,0) = F(x) = sin(pi x)$$

Now, letting $x : = x-t$, we yield :
$$u(x,t) = F(x-t) = sin[pi(x-t)]$$

share|cite|improve this answer

answered Mar 14 at 16:59

RebellosRebellos

15.4k31250

$endgroup$

share|cite|improve this answer

answered Mar 14 at 16:59

RebellosRebellos

15.4k31250

answered Mar 14 at 16:59

RebellosRebellos

15.4k31250

answered Mar 14 at 16:59

RebellosRebellos

15.4k31250