Mixing
Semi-infinite Wave Equation with piewise initial conditions
$begingroup$
We have the following for a semi-infinite interval
$$ u_{tt}-c^2u_{xx} = 0 quad quad 0 < c < infty $$
Initial conditions are piecewise defined.
$$ u(x,0) = begin{cases} 0, & 0<x<2 \ 1, & 2<x<3 \ 0,& x>3
end{cases} $$
$$ u_t(x,0) = 0 $$
We have boundary conditions:
$$ u_x(0,t) = 0 $$
$textbf{My attempt}$ at this is as follows:
We use D'Alembert solution:
$$u(x,t) = F(x-ct) + G(x+ct) $$
We note that we have a problem with zero initial velocity. This means that if $u(x,0) = f(x)$ then we have that:
$$ u(x,t) = frac{1}{2}f(x-ct) + frac{1}{2} f(x+ct)$$
We note $f(x)$ is given piecewise but it only takes on positive arguments. $f(x-ct)$ has negative arguments though. We summarize this as follows
$$
begin{array}{|c|c|c|c|}
hline
mbox{Interval} & F(x) = frac{1}{2}f(x-ct) & G(x) = frac{1}{2}f(x+ct)&u(x,t)\
hline
0<x<ct & ? & 0 & ? + 0 \
hline
ct<x < 2-ct & 0 & 0 & 0\
hline
2-ct<x<2+ct & 0 & frac{1}{2}(1)&frac{1}{2}\
hline
2+ct <x<3-ct & frac{1}{2}(1) & frac{1}{2}(1)&1\
hline
3-ct < x< 3+ct & frac{1}{2}(1) & 0&frac{1}{2} \
hline x > 3+ct & 0 & 0&0 \
hline
end{array}
$$
So solving for $x in (ct, infty)$ is not a problem. One will simply ass up the row to get $u(x,t)$.
$textbf{The Problem}$ arises there where the $?$ has been placed as $f(x)$ takes a negative argument. We try and solve for it using our boundary conditions.
$$ u(x,t)_x = frac{dF(x-ct)}{d(x-ct)} + frac{G(x+ct)}{d(x+ct)} $$
$$ therefore u(0,t)_x = frac{dF(ct)}{d(ct)} + frac{G(ct)}{d(ct)} = 0 $$
$$ -frac{1}{c}frac{dF(-ct)}{d(t)} + frac{1}{c}frac{G(ct)}{d(t)} = 0 $$
$$therefore frac{dF(-ct)}{d(t)} = frac{G(ct)}{d(t)} $$
$$ therefore F(-ct) = G(ct) + k$$
Let $z$ denote a negative argument. Then we have
$$ F(z) = G(-z) + k $$
$$therefore F(x-ct) = G(ct-x) + k$$
Now from the table, we have that $G(ct-x) = 0, quad 0<x<ct$. So should we assume $F(x-ct) = k$ for this interval. There is no way of determining $k$ right? I think so. Thus the first interval has solution $u(x,t) = k$ where $k$ is an undetermined constant.
$textbf{Question:}$ please help me undertand whether what I have done is indeed correct. If it is wrong, please help me to correct it. If it is correct and you feel I should be using a better method, please help me understand that method please. Thank you very much for your time.
P.S. this question is from here
pde wave-equation
asked Jan 15 at 13:39
YesYes
887
$endgroup$
add a comment |
$begingroup$
We have the following for a semi-infinite interval
$$ u_{tt}-c^2u_{xx} = 0 quad quad 0 < c < infty $$
Initial conditions are piecewise defined.
$$ u(x,0) = begin{cases} 0, & 0<x<2 \ 1, & 2<x<3 \ 0,& x>3
end{cases} $$
$$ u_t(x,0) = 0 $$
We have boundary conditions:
$$ u_x(0,t) = 0 $$
$textbf{My attempt}$ at this is as follows:
We use D'Alembert solution:
$$u(x,t) = F(x-ct) + G(x+ct) $$
We note that we have a problem with zero initial velocity. This means that if $u(x,0) = f(x)$ then we have that:
$$ u(x,t) = frac{1}{2}f(x-ct) + frac{1}{2} f(x+ct)$$
We note $f(x)$ is given piecewise but it only takes on positive arguments. $f(x-ct)$ has negative arguments though. We summarize this as follows
$$
begin{array}{|c|c|c|c|}
hline
mbox{Interval} & F(x) = frac{1}{2}f(x-ct) & G(x) = frac{1}{2}f(x+ct)&u(x,t)\
hline
0<x<ct & ? & 0 & ? + 0 \
hline
ct<x < 2-ct & 0 & 0 & 0\
hline
2-ct<x<2+ct & 0 & frac{1}{2}(1)&frac{1}{2}\
hline
2+ct <x<3-ct & frac{1}{2}(1) & frac{1}{2}(1)&1\
hline
3-ct < x< 3+ct & frac{1}{2}(1) & 0&frac{1}{2} \
hline x > 3+ct & 0 & 0&0 \
hline
end{array}
$$
So solving for $x in (ct, infty)$ is not a problem. One will simply ass up the row to get $u(x,t)$.
$textbf{The Problem}$ arises there where the $?$ has been placed as $f(x)$ takes a negative argument. We try and solve for it using our boundary conditions.
$$ u(x,t)_x = frac{dF(x-ct)}{d(x-ct)} + frac{G(x+ct)}{d(x+ct)} $$
$$ therefore u(0,t)_x = frac{dF(ct)}{d(ct)} + frac{G(ct)}{d(ct)} = 0 $$
$$ -frac{1}{c}frac{dF(-ct)}{d(t)} + frac{1}{c}frac{G(ct)}{d(t)} = 0 $$
$$therefore frac{dF(-ct)}{d(t)} = frac{G(ct)}{d(t)} $$
$$ therefore F(-ct) = G(ct) + k$$
Let $z$ denote a negative argument. Then we have
$$ F(z) = G(-z) + k $$
$$therefore F(x-ct) = G(ct-x) + k$$
Now from the table, we have that $G(ct-x) = 0, quad 0<x<ct$. So should we assume $F(x-ct) = k$ for this interval. There is no way of determining $k$ right? I think so. Thus the first interval has solution $u(x,t) = k$ where $k$ is an undetermined constant.
$textbf{Question:}$ please help me undertand whether what I have done is indeed correct. If it is wrong, please help me to correct it. If it is correct and you feel I should be using a better method, please help me understand that method please. Thank you very much for your time.
P.S. this question is from here
pde wave-equation
asked Jan 15 at 13:39
YesYes
887
$endgroup$
add a comment |
$begingroup$
We have the following for a semi-infinite interval
$$ u_{tt}-c^2u_{xx} = 0 quad quad 0 < c < infty $$
Initial conditions are piecewise defined.
$$ u(x,0) = begin{cases} 0, & 0<x<2 \ 1, & 2<x<3 \ 0,& x>3
end{cases} $$
$$ u_t(x,0) = 0 $$
We have boundary conditions:
$$ u_x(0,t) = 0 $$
$textbf{My attempt}$ at this is as follows:
We use D'Alembert solution:
$$u(x,t) = F(x-ct) + G(x+ct) $$
We note that we have a problem with zero initial velocity. This means that if $u(x,0) = f(x)$ then we have that:
$$ u(x,t) = frac{1}{2}f(x-ct) + frac{1}{2} f(x+ct)$$
We note $f(x)$ is given piecewise but it only takes on positive arguments. $f(x-ct)$ has negative arguments though. We summarize this as follows
$$
begin{array}{|c|c|c|c|}
hline
mbox{Interval} & F(x) = frac{1}{2}f(x-ct) & G(x) = frac{1}{2}f(x+ct)&u(x,t)\
hline
0<x<ct & ? & 0 & ? + 0 \
hline
ct<x < 2-ct & 0 & 0 & 0\
hline
2-ct<x<2+ct & 0 & frac{1}{2}(1)&frac{1}{2}\
hline
2+ct <x<3-ct & frac{1}{2}(1) & frac{1}{2}(1)&1\
hline
3-ct < x< 3+ct & frac{1}{2}(1) & 0&frac{1}{2} \
hline x > 3+ct & 0 & 0&0 \
hline
end{array}
$$
So solving for $x in (ct, infty)$ is not a problem. One will simply ass up the row to get $u(x,t)$.
$textbf{The Problem}$ arises there where the $?$ has been placed as $f(x)$ takes a negative argument. We try and solve for it using our boundary conditions.
$$ u(x,t)_x = frac{dF(x-ct)}{d(x-ct)} + frac{G(x+ct)}{d(x+ct)} $$
$$ therefore u(0,t)_x = frac{dF(ct)}{d(ct)} + frac{G(ct)}{d(ct)} = 0 $$
$$ -frac{1}{c}frac{dF(-ct)}{d(t)} + frac{1}{c}frac{G(ct)}{d(t)} = 0 $$
$$therefore frac{dF(-ct)}{d(t)} = frac{G(ct)}{d(t)} $$
$$ therefore F(-ct) = G(ct) + k$$
Let $z$ denote a negative argument. Then we have
$$ F(z) = G(-z) + k $$
$$therefore F(x-ct) = G(ct-x) + k$$
Now from the table, we have that $G(ct-x) = 0, quad 0<x<ct$. So should we assume $F(x-ct) = k$ for this interval. There is no way of determining $k$ right? I think so. Thus the first interval has solution $u(x,t) = k$ where $k$ is an undetermined constant.
$textbf{Question:}$ please help me undertand whether what I have done is indeed correct. If it is wrong, please help me to correct it. If it is correct and you feel I should be using a better method, please help me understand that method please. Thank you very much for your time.
P.S. this question is from here
pde wave-equation
asked Jan 15 at 13:39
YesYes
887
$endgroup$
We have the following for a semi-infinite interval
$$ u_{tt}-c^2u_{xx} = 0 quad quad 0 < c < infty $$
Initial conditions are piecewise defined.
$$ u(x,0) = begin{cases} 0, & 0<x<2 \ 1, & 2<x<3 \ 0,& x>3
end{cases} $$
$$ u_t(x,0) = 0 $$
We have boundary conditions:
$$ u_x(0,t) = 0 $$
$textbf{My attempt}$ at this is as follows:
We use D'Alembert solution:
$$u(x,t) = F(x-ct) + G(x+ct) $$
We note that we have a problem with zero initial velocity. This means that if $u(x,0) = f(x)$ then we have that:
$$ u(x,t) = frac{1}{2}f(x-ct) + frac{1}{2} f(x+ct)$$
We note $f(x)$ is given piecewise but it only takes on positive arguments. $f(x-ct)$ has negative arguments though. We summarize this as follows
$$
begin{array}{|c|c|c|c|}
hline
mbox{Interval} & F(x) = frac{1}{2}f(x-ct) & G(x) = frac{1}{2}f(x+ct)&u(x,t)\
hline
0<x<ct & ? & 0 & ? + 0 \
hline
ct<x < 2-ct & 0 & 0 & 0\
hline
2-ct<x<2+ct & 0 & frac{1}{2}(1)&frac{1}{2}\
hline
2+ct <x<3-ct & frac{1}{2}(1) & frac{1}{2}(1)&1\
hline
3-ct < x< 3+ct & frac{1}{2}(1) & 0&frac{1}{2} \
hline x > 3+ct & 0 & 0&0 \
hline
end{array}
$$
So solving for $x in (ct, infty)$ is not a problem. One will simply ass up the row to get $u(x,t)$.
$textbf{The Problem}$ arises there where the $?$ has been placed as $f(x)$ takes a negative argument. We try and solve for it using our boundary conditions.
$$ u(x,t)_x = frac{dF(x-ct)}{d(x-ct)} + frac{G(x+ct)}{d(x+ct)} $$
$$ therefore u(0,t)_x = frac{dF(ct)}{d(ct)} + frac{G(ct)}{d(ct)} = 0 $$
$$ -frac{1}{c}frac{dF(-ct)}{d(t)} + frac{1}{c}frac{G(ct)}{d(t)} = 0 $$
$$therefore frac{dF(-ct)}{d(t)} = frac{G(ct)}{d(t)} $$
$$ therefore F(-ct) = G(ct) + k$$
Let $z$ denote a negative argument. Then we have
$$ F(z) = G(-z) + k $$
$$therefore F(x-ct) = G(ct-x) + k$$
Now from the table, we have that $G(ct-x) = 0, quad 0<x<ct$. So should we assume $F(x-ct) = k$ for this interval. There is no way of determining $k$ right? I think so. Thus the first interval has solution $u(x,t) = k$ where $k$ is an undetermined constant.
$textbf{Question:}$ please help me undertand whether what I have done is indeed correct. If it is wrong, please help me to correct it. If it is correct and you feel I should be using a better method, please help me understand that method please. Thank you very much for your time.
P.S. this question is from here
pde wave-equation
pde wave-equation
asked Jan 15 at 13:39
YesYes
887
asked Jan 15 at 13:39
YesYes
887
asked Jan 15 at 13:39
YesYes
887
asked Jan 15 at 13:39
YesYes
887
asked Jan 15 at 13:39
YesYes
887
887
add a comment |
add a comment |
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