Exercise in Applied Mathematics Logan's Book (functional)












0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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



$$
f_y(x,y(x))-y'(x) f_x(x,y(x))-frac{y''(x) f(x,y(x))}{y'(x)^2+1} = 0
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks you very much
    $endgroup$
    – Θάνος Κ.
    Jan 4 at 19:31











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1 Answer
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1 Answer
1






active

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active

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1












$begingroup$

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



$$
f_y(x,y(x))-y'(x) f_x(x,y(x))-frac{y''(x) f(x,y(x))}{y'(x)^2+1} = 0
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks you very much
    $endgroup$
    – Θάνος Κ.
    Jan 4 at 19:31
















1












$begingroup$

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



$$
f_y(x,y(x))-y'(x) f_x(x,y(x))-frac{y''(x) f(x,y(x))}{y'(x)^2+1} = 0
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks you very much
    $endgroup$
    – Θάνος Κ.
    Jan 4 at 19:31














1












1








1





$begingroup$

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



$$
f_y(x,y(x))-y'(x) f_x(x,y(x))-frac{y''(x) f(x,y(x))}{y'(x)^2+1} = 0
$$






share|cite|improve this answer









$endgroup$



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



$$
f_y(x,y(x))-y'(x) f_x(x,y(x))-frac{y''(x) f(x,y(x))}{y'(x)^2+1} = 0
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 19:29









CesareoCesareo

9,2013517




9,2013517












  • $begingroup$
    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




$begingroup$
thanks you very much
$endgroup$
– Θάνος Κ.
Jan 4 at 19:31


















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