Mixing
Exercise in Applied Mathematics Logan's Book (functional)
$begingroup$
From David Logan book (Applied Mathematics) page 175.
Show that Euler Equation for the functional
$$J(y)=int_{a}^{b} f(x,y)sqrt{1+(y')^2)} dx$$
has the form
$$f_y-f_xy'-ffrac{y''}{1+(y')^2}=0$$
I have found that:
$$L_y(x,y,y')=f_ysqrt{1+(y')^2}$$
$$L_{y'}(x,y,y')=f(x,y)frac{y'}{sqrt{1+(y')^2}}$$
$$frac{dL_{y'}(x,y,y')}{dx}=f_{x}(x,y)frac{y'}{sqrt{1+(y')^2}}+f(x,y)frac{y''}{(1+(y')^{2})^{frac{3}{2}}}$$
and finally (by Euler Equation) i have found:
$$f_y(1+(y')^2)-f_xy'-ffrac{y''}{1+(y')^2}=0$$
What wrong with that?
calculus
asked Jan 4 at 19:05
Θάνος Κ.Θάνος Κ.
306
$endgroup$
add a comment |
$begingroup$
From David Logan book (Applied Mathematics) page 175.
Show that Euler Equation for the functional
$$J(y)=int_{a}^{b} f(x,y)sqrt{1+(y')^2)} dx$$
has the form
$$f_y-f_xy'-ffrac{y''}{1+(y')^2}=0$$
I have found that:
$$L_y(x,y,y')=f_ysqrt{1+(y')^2}$$
$$L_{y'}(x,y,y')=f(x,y)frac{y'}{sqrt{1+(y')^2}}$$
$$frac{dL_{y'}(x,y,y')}{dx}=f_{x}(x,y)frac{y'}{sqrt{1+(y')^2}}+f(x,y)frac{y''}{(1+(y')^{2})^{frac{3}{2}}}$$
and finally (by Euler Equation) i have found:
$$f_y(1+(y')^2)-f_xy'-ffrac{y''}{1+(y')^2}=0$$
What wrong with that?
calculus
asked Jan 4 at 19:05
Θάνος Κ.Θάνος Κ.
306
$endgroup$
$begingroup$
I think you missed out the third term in the detivative of $L_{y'}$.
$endgroup$
– Botond
Jan 4 at 19:15
$begingroup$
oh my goddddd thank you very much!!!!
$endgroup$
– Θάνος Κ.
Jan 4 at 19:28
add a comment |
$begingroup$
From David Logan book (Applied Mathematics) page 175.
Show that Euler Equation for the functional
$$J(y)=int_{a}^{b} f(x,y)sqrt{1+(y')^2)} dx$$
has the form
$$f_y-f_xy'-ffrac{y''}{1+(y')^2}=0$$
I have found that:
$$L_y(x,y,y')=f_ysqrt{1+(y')^2}$$
$$L_{y'}(x,y,y')=f(x,y)frac{y'}{sqrt{1+(y')^2}}$$
$$frac{dL_{y'}(x,y,y')}{dx}=f_{x}(x,y)frac{y'}{sqrt{1+(y')^2}}+f(x,y)frac{y''}{(1+(y')^{2})^{frac{3}{2}}}$$
and finally (by Euler Equation) i have found:
$$f_y(1+(y')^2)-f_xy'-ffrac{y''}{1+(y')^2}=0$$
What wrong with that?
calculus
asked Jan 4 at 19:05
Θάνος Κ.Θάνος Κ.
306
$endgroup$
From David Logan book (Applied Mathematics) page 175.
Show that Euler Equation for the functional
$$J(y)=int_{a}^{b} f(x,y)sqrt{1+(y')^2)} dx$$
has the form
$$f_y-f_xy'-ffrac{y''}{1+(y')^2}=0$$
I have found that:
$$L_y(x,y,y')=f_ysqrt{1+(y')^2}$$
$$L_{y'}(x,y,y')=f(x,y)frac{y'}{sqrt{1+(y')^2}}$$
$$frac{dL_{y'}(x,y,y')}{dx}=f_{x}(x,y)frac{y'}{sqrt{1+(y')^2}}+f(x,y)frac{y''}{(1+(y')^{2})^{frac{3}{2}}}$$
and finally (by Euler Equation) i have found:
$$f_y(1+(y')^2)-f_xy'-ffrac{y''}{1+(y')^2}=0$$
What wrong with that?
calculus
calculus
asked Jan 4 at 19:05
Θάνος Κ.Θάνος Κ.
306
asked Jan 4 at 19:05
Θάνος Κ.Θάνος Κ.
306
asked Jan 4 at 19:05
Θάνος Κ.Θάνος Κ.
306
asked Jan 4 at 19:05
Θάνος Κ.Θάνος Κ.
306
asked Jan 4 at 19:05
Θάνος Κ.Θάνος Κ.
306
306
$begingroup$
I think you missed out the third term in the detivative of $L_{y'}$.
$endgroup$
– Botond
Jan 4 at 19:15
$begingroup$
oh my goddddd thank you very much!!!!
$endgroup$
– Θάνος Κ.
Jan 4 at 19:28
add a comment |
$begingroup$
I think you missed out the third term in the detivative of $L_{y'}$.
$endgroup$
– Botond
Jan 4 at 19:15
$begingroup$
oh my goddddd thank you very much!!!!
$endgroup$
– Θάνος Κ.
Jan 4 at 19:28
$begingroup$
I think you missed out the third term in the detivative of $L_{y'}$.
$endgroup$
– Botond
Jan 4 at 19:15
$begingroup$
I think you missed out the third term in the detivative of $L_{y'}$.
$endgroup$
– Botond
Jan 4 at 19:15
$begingroup$
oh my goddddd thank you very much!!!!
$endgroup$
– Θάνος Κ.
Jan 4 at 19:28
$begingroup$
oh my goddddd thank you very much!!!!
$endgroup$
– Θάνος Κ.
Jan 4 at 19:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Applying the first variation we get
$$
frac{left(y'(x)^2+1right) left(f_y(x,y(x))-y'(x) f_x(x,y(x))right)-y''(x) f(x,y(x))}{left(y'(x)^2+1right)^{3/2}}=0
$$
or
$$
left(y'(x)^2+1right) left(f_y(x,y(x))-y'(x) f_x(x,y(x))right)-y''(x) f(x,y(x)) = 0
$$
or
$$
f_y(x,y(x))-y'(x) f_x(x,y(x))-frac{y''(x) f(x,y(x))}{y'(x)^2+1} = 0
$$
answered Jan 4 at 19:29
CesareoCesareo
8,6243516
$endgroup$
$begingroup$
thanks you very much
$endgroup$
– Θάνος Κ.
Jan 4 at 19:31
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
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$begingroup$
Applying the first variation we get
$$
frac{left(y'(x)^2+1right) left(f_y(x,y(x))-y'(x) f_x(x,y(x))right)-y''(x) f(x,y(x))}{left(y'(x)^2+1right)^{3/2}}=0
$$
or
$$
left(y'(x)^2+1right) left(f_y(x,y(x))-y'(x) f_x(x,y(x))right)-y''(x) f(x,y(x)) = 0
$$
or
$$
f_y(x,y(x))-y'(x) f_x(x,y(x))-frac{y''(x) f(x,y(x))}{y'(x)^2+1} = 0
$$
answered Jan 4 at 19:29
CesareoCesareo
8,6243516
$endgroup$
$begingroup$
thanks you very much
$endgroup$
– Θάνος Κ.
Jan 4 at 19:31
add a comment |
$begingroup$
Applying the first variation we get
$$
frac{left(y'(x)^2+1right) left(f_y(x,y(x))-y'(x) f_x(x,y(x))right)-y''(x) f(x,y(x))}{left(y'(x)^2+1right)^{3/2}}=0
$$
or
$$
left(y'(x)^2+1right) left(f_y(x,y(x))-y'(x) f_x(x,y(x))right)-y''(x) f(x,y(x)) = 0
$$
or
$$
f_y(x,y(x))-y'(x) f_x(x,y(x))-frac{y''(x) f(x,y(x))}{y'(x)^2+1} = 0
$$
answered Jan 4 at 19:29
CesareoCesareo
8,6243516
$endgroup$
$begingroup$
thanks you very much
$endgroup$
– Θάνος Κ.
Jan 4 at 19:31
add a comment |
$begingroup$
Applying the first variation we get
$$
frac{left(y'(x)^2+1right) left(f_y(x,y(x))-y'(x) f_x(x,y(x))right)-y''(x) f(x,y(x))}{left(y'(x)^2+1right)^{3/2}}=0
$$
or
$$
left(y'(x)^2+1right) left(f_y(x,y(x))-y'(x) f_x(x,y(x))right)-y''(x) f(x,y(x)) = 0
$$
or
$$
f_y(x,y(x))-y'(x) f_x(x,y(x))-frac{y''(x) f(x,y(x))}{y'(x)^2+1} = 0
$$
answered Jan 4 at 19:29
CesareoCesareo
8,6243516
$endgroup$
Applying the first variation we get
$$
frac{left(y'(x)^2+1right) left(f_y(x,y(x))-y'(x) f_x(x,y(x))right)-y''(x) f(x,y(x))}{left(y'(x)^2+1right)^{3/2}}=0
$$
or
$$
left(y'(x)^2+1right) left(f_y(x,y(x))-y'(x) f_x(x,y(x))right)-y''(x) f(x,y(x)) = 0
$$
or
$$
f_y(x,y(x))-y'(x) f_x(x,y(x))-frac{y''(x) f(x,y(x))}{y'(x)^2+1} = 0
$$
answered Jan 4 at 19:29
CesareoCesareo
8,6243516
answered Jan 4 at 19:29
CesareoCesareo
8,6243516
answered Jan 4 at 19:29
CesareoCesareo
8,6243516
answered Jan 4 at 19:29
CesareoCesareo
8,6243516
8,6243516
$begingroup$
thanks you very much
$endgroup$
– Θάνος Κ.
Jan 4 at 19:31
add a comment |
$begingroup$
thanks you very much
$endgroup$
– Θάνος Κ.
Jan 4 at 19:31
$begingroup$
thanks you very much
$endgroup$
– Θάνος Κ.
Jan 4 at 19:31
$begingroup$
thanks you very much
$endgroup$
– Θάνος Κ.
Jan 4 at 19:31
add a comment |
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$begingroup$
I think you missed out the third term in the detivative of $L_{y'}$.
$endgroup$
– Botond
Jan 4 at 19:15
$begingroup$
oh my goddddd thank you very much!!!!
$endgroup$
– Θάνος Κ.
Jan 4 at 19:28