Find the streamline equation of a velocity field
$begingroup$
Assume the velocity field:
$$
v_x=-Ue^{-at}cos left( frac{pi x}{L} right)sin left( frac{pi y}{L} right)
$$
$$
v_y=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)
$$
The streamline equation is the solution of the differential equation:
$$
ds=frac{dx}{v_x}=frac{dy}{v_y}
$$
$$
iff -Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right) dx=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)dy
$$
$$ Rightarrow tan left( frac{pi x}{L} right)dx=tan left( frac{pi y}{L} right)dy
$$
indefinite integration and adjusting of the constant gives:
$$
left| cos left( frac{pi x}{L} right)right|=c cdot left|cos left( frac{pi y}{L} right)right|
$$
$$
iff cos^2 left( frac{pi x}{L} right)+lambda cdot cos^2 left( frac{pi y}{L} right)=0
$$
Given that we are given no initial point $(x_0,y_0)$ to determine $c$, how can a final result be obtained?
ordinary-differential-equations fluid-dynamics
asked Dec 11 '18 at 18:13
JevautJevaut
1,151112
$endgroup$
add a comment |
$begingroup$
Assume the velocity field:
$$
v_x=-Ue^{-at}cos left( frac{pi x}{L} right)sin left( frac{pi y}{L} right)
$$
$$
v_y=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)
$$
The streamline equation is the solution of the differential equation:
$$
ds=frac{dx}{v_x}=frac{dy}{v_y}
$$
$$
iff -Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right) dx=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)dy
$$
$$ Rightarrow tan left( frac{pi x}{L} right)dx=tan left( frac{pi y}{L} right)dy
$$
indefinite integration and adjusting of the constant gives:
$$
left| cos left( frac{pi x}{L} right)right|=c cdot left|cos left( frac{pi y}{L} right)right|
$$
$$
iff cos^2 left( frac{pi x}{L} right)+lambda cdot cos^2 left( frac{pi y}{L} right)=0
$$
Given that we are given no initial point $(x_0,y_0)$ to determine $c$, how can a final result be obtained?
ordinary-differential-equations fluid-dynamics
asked Dec 11 '18 at 18:13
JevautJevaut
1,151112
$endgroup$
add a comment |
$begingroup$
Assume the velocity field:
$$
v_x=-Ue^{-at}cos left( frac{pi x}{L} right)sin left( frac{pi y}{L} right)
$$
$$
v_y=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)
$$
The streamline equation is the solution of the differential equation:
$$
ds=frac{dx}{v_x}=frac{dy}{v_y}
$$
$$
iff -Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right) dx=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)dy
$$
$$ Rightarrow tan left( frac{pi x}{L} right)dx=tan left( frac{pi y}{L} right)dy
$$
indefinite integration and adjusting of the constant gives:
$$
left| cos left( frac{pi x}{L} right)right|=c cdot left|cos left( frac{pi y}{L} right)right|
$$
$$
iff cos^2 left( frac{pi x}{L} right)+lambda cdot cos^2 left( frac{pi y}{L} right)=0
$$
Given that we are given no initial point $(x_0,y_0)$ to determine $c$, how can a final result be obtained?
ordinary-differential-equations fluid-dynamics
asked Dec 11 '18 at 18:13
JevautJevaut
1,151112
$endgroup$
Assume the velocity field:
$$
v_x=-Ue^{-at}cos left( frac{pi x}{L} right)sin left( frac{pi y}{L} right)
$$
$$
v_y=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)
$$
The streamline equation is the solution of the differential equation:
$$
ds=frac{dx}{v_x}=frac{dy}{v_y}
$$
$$
iff -Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right) dx=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)dy
$$
$$ Rightarrow tan left( frac{pi x}{L} right)dx=tan left( frac{pi y}{L} right)dy
$$
indefinite integration and adjusting of the constant gives:
$$
left| cos left( frac{pi x}{L} right)right|=c cdot left|cos left( frac{pi y}{L} right)right|
$$
$$
iff cos^2 left( frac{pi x}{L} right)+lambda cdot cos^2 left( frac{pi y}{L} right)=0
$$
Given that we are given no initial point $(x_0,y_0)$ to determine $c$, how can a final result be obtained?
ordinary-differential-equations fluid-dynamics
ordinary-differential-equations fluid-dynamics
asked Dec 11 '18 at 18:13
JevautJevaut
1,151112
asked Dec 11 '18 at 18:13
JevautJevaut
1,151112
edited Dec 12 '18 at 11:30
Jevaut
asked Dec 11 '18 at 18:13
JevautJevaut
1,151112
asked Dec 11 '18 at 18:13
JevautJevaut
1,151112
asked Dec 11 '18 at 18:13
JevautJevaut
1,151112
1,151112
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
You should assume a point on streamline is given which then will be treated a paramter $(x_0,y_0)$ in your equation. Take definite integral from a given point $(x_0,y_0)$ where the streamline passes through, to some arbitrary point $(x,y)$. The integral equation becomes
begin{equation}
int_{x_0}^x tan left( frac{pi x}{L} right) mathrm{d} x = int_{y_0}^y tan left( frac{pi y}{L} right) mathrm{d} y
end{equation}
which gives
begin{equation}
- frac{L}{pi} left[ ln left| cos( frac{pi x}{L} ) right| right]_{x_0}^x = - frac{L}{pi} left[ ln left| cos( frac{pi y}{L} ) right| right]_{y_0}^y
end{equation}
or
begin{equation}
lnleft| frac{cos( frac{pi x}{L} )}{cos( frac{pi x_0}{L} )} right| = lnleft| frac{cos( frac{pi y}{L} )}{cos( frac{pi y_0}{L} )} right|
end{equation}
Thus the streamline curve $f(x,y) = 0$ that passes through a given point $(x_0,y_0)$ is
begin{equation}
f(x,y) = left( frac{ cos( frac{pi x}{L} )}{ cos( frac{pi y}{L} )} right)^2 - underbrace{ left( frac{cos( frac{pi x_0}{L} ) }{cos( frac{pi y_0}{L} )}right)^2}_{c^2} = 0.
end{equation}
answered Dec 11 '18 at 21:35
SiaSia
1064
$endgroup$
$begingroup$
Isn't it more appropriate to square the absolutes (just to get rid of the cases)?
$endgroup$
– Jevaut
Dec 12 '18 at 11:29
$begingroup$
You are right, I'll edit the answer with squares to be similar to your derivation.
$endgroup$
– Sia
Dec 12 '18 at 20:22
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You should assume a point on streamline is given which then will be treated a paramter $(x_0,y_0)$ in your equation. Take definite integral from a given point $(x_0,y_0)$ where the streamline passes through, to some arbitrary point $(x,y)$. The integral equation becomes
begin{equation}
int_{x_0}^x tan left( frac{pi x}{L} right) mathrm{d} x = int_{y_0}^y tan left( frac{pi y}{L} right) mathrm{d} y
end{equation}
which gives
begin{equation}
- frac{L}{pi} left[ ln left| cos( frac{pi x}{L} ) right| right]_{x_0}^x = - frac{L}{pi} left[ ln left| cos( frac{pi y}{L} ) right| right]_{y_0}^y
end{equation}
or
begin{equation}
lnleft| frac{cos( frac{pi x}{L} )}{cos( frac{pi x_0}{L} )} right| = lnleft| frac{cos( frac{pi y}{L} )}{cos( frac{pi y_0}{L} )} right|
end{equation}
Thus the streamline curve $f(x,y) = 0$ that passes through a given point $(x_0,y_0)$ is
begin{equation}
f(x,y) = left( frac{ cos( frac{pi x}{L} )}{ cos( frac{pi y}{L} )} right)^2 - underbrace{ left( frac{cos( frac{pi x_0}{L} ) }{cos( frac{pi y_0}{L} )}right)^2}_{c^2} = 0.
end{equation}
answered Dec 11 '18 at 21:35
SiaSia
1064
$endgroup$
$begingroup$
Isn't it more appropriate to square the absolutes (just to get rid of the cases)?
$endgroup$
– Jevaut
Dec 12 '18 at 11:29
$begingroup$
You are right, I'll edit the answer with squares to be similar to your derivation.
$endgroup$
– Sia
Dec 12 '18 at 20:22
add a comment |
$begingroup$
You should assume a point on streamline is given which then will be treated a paramter $(x_0,y_0)$ in your equation. Take definite integral from a given point $(x_0,y_0)$ where the streamline passes through, to some arbitrary point $(x,y)$. The integral equation becomes
begin{equation}
int_{x_0}^x tan left( frac{pi x}{L} right) mathrm{d} x = int_{y_0}^y tan left( frac{pi y}{L} right) mathrm{d} y
end{equation}
which gives
begin{equation}
- frac{L}{pi} left[ ln left| cos( frac{pi x}{L} ) right| right]_{x_0}^x = - frac{L}{pi} left[ ln left| cos( frac{pi y}{L} ) right| right]_{y_0}^y
end{equation}
or
begin{equation}
lnleft| frac{cos( frac{pi x}{L} )}{cos( frac{pi x_0}{L} )} right| = lnleft| frac{cos( frac{pi y}{L} )}{cos( frac{pi y_0}{L} )} right|
end{equation}
Thus the streamline curve $f(x,y) = 0$ that passes through a given point $(x_0,y_0)$ is
begin{equation}
f(x,y) = left( frac{ cos( frac{pi x}{L} )}{ cos( frac{pi y}{L} )} right)^2 - underbrace{ left( frac{cos( frac{pi x_0}{L} ) }{cos( frac{pi y_0}{L} )}right)^2}_{c^2} = 0.
end{equation}
answered Dec 11 '18 at 21:35
SiaSia
1064
$endgroup$
$begingroup$
Isn't it more appropriate to square the absolutes (just to get rid of the cases)?
$endgroup$
– Jevaut
Dec 12 '18 at 11:29
$begingroup$
You are right, I'll edit the answer with squares to be similar to your derivation.
$endgroup$
– Sia
Dec 12 '18 at 20:22
add a comment |
$begingroup$
You should assume a point on streamline is given which then will be treated a paramter $(x_0,y_0)$ in your equation. Take definite integral from a given point $(x_0,y_0)$ where the streamline passes through, to some arbitrary point $(x,y)$. The integral equation becomes
begin{equation}
int_{x_0}^x tan left( frac{pi x}{L} right) mathrm{d} x = int_{y_0}^y tan left( frac{pi y}{L} right) mathrm{d} y
end{equation}
which gives
begin{equation}
- frac{L}{pi} left[ ln left| cos( frac{pi x}{L} ) right| right]_{x_0}^x = - frac{L}{pi} left[ ln left| cos( frac{pi y}{L} ) right| right]_{y_0}^y
end{equation}
or
begin{equation}
lnleft| frac{cos( frac{pi x}{L} )}{cos( frac{pi x_0}{L} )} right| = lnleft| frac{cos( frac{pi y}{L} )}{cos( frac{pi y_0}{L} )} right|
end{equation}
Thus the streamline curve $f(x,y) = 0$ that passes through a given point $(x_0,y_0)$ is
begin{equation}
f(x,y) = left( frac{ cos( frac{pi x}{L} )}{ cos( frac{pi y}{L} )} right)^2 - underbrace{ left( frac{cos( frac{pi x_0}{L} ) }{cos( frac{pi y_0}{L} )}right)^2}_{c^2} = 0.
end{equation}
answered Dec 11 '18 at 21:35
SiaSia
1064
$endgroup$
You should assume a point on streamline is given which then will be treated a paramter $(x_0,y_0)$ in your equation. Take definite integral from a given point $(x_0,y_0)$ where the streamline passes through, to some arbitrary point $(x,y)$. The integral equation becomes
begin{equation}
int_{x_0}^x tan left( frac{pi x}{L} right) mathrm{d} x = int_{y_0}^y tan left( frac{pi y}{L} right) mathrm{d} y
end{equation}
which gives
begin{equation}
- frac{L}{pi} left[ ln left| cos( frac{pi x}{L} ) right| right]_{x_0}^x = - frac{L}{pi} left[ ln left| cos( frac{pi y}{L} ) right| right]_{y_0}^y
end{equation}
or
begin{equation}
lnleft| frac{cos( frac{pi x}{L} )}{cos( frac{pi x_0}{L} )} right| = lnleft| frac{cos( frac{pi y}{L} )}{cos( frac{pi y_0}{L} )} right|
end{equation}
Thus the streamline curve $f(x,y) = 0$ that passes through a given point $(x_0,y_0)$ is
begin{equation}
f(x,y) = left( frac{ cos( frac{pi x}{L} )}{ cos( frac{pi y}{L} )} right)^2 - underbrace{ left( frac{cos( frac{pi x_0}{L} ) }{cos( frac{pi y_0}{L} )}right)^2}_{c^2} = 0.
end{equation}
answered Dec 11 '18 at 21:35
SiaSia
1064
edited Dec 13 '18 at 1:47
answered Dec 11 '18 at 21:35
SiaSia
1064
answered Dec 11 '18 at 21:35
SiaSia
1064
answered Dec 11 '18 at 21:35
SiaSia
1064
1064
$begingroup$
Isn't it more appropriate to square the absolutes (just to get rid of the cases)?
$endgroup$
– Jevaut
Dec 12 '18 at 11:29
$begingroup$
You are right, I'll edit the answer with squares to be similar to your derivation.
$endgroup$
– Sia
Dec 12 '18 at 20:22
add a comment |
$begingroup$
Isn't it more appropriate to square the absolutes (just to get rid of the cases)?
$endgroup$
– Jevaut
Dec 12 '18 at 11:29
$begingroup$
You are right, I'll edit the answer with squares to be similar to your derivation.
$endgroup$
– Sia
Dec 12 '18 at 20:22
$begingroup$
Isn't it more appropriate to square the absolutes (just to get rid of the cases)?
$endgroup$
– Jevaut
Dec 12 '18 at 11:29
$begingroup$
Isn't it more appropriate to square the absolutes (just to get rid of the cases)?
$endgroup$
– Jevaut
Dec 12 '18 at 11:29
$begingroup$
You are right, I'll edit the answer with squares to be similar to your derivation.
$endgroup$
– Sia
Dec 12 '18 at 20:22
$begingroup$
You are right, I'll edit the answer with squares to be similar to your derivation.
$endgroup$
– Sia
Dec 12 '18 at 20:22
add a comment |
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