How to find the Euler Lagrange Equation of this energy function?
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I come across this problem: Find the Euler Lagrange Equation of this energy function
begin{equation*}
J(u) = int_{0}^{1} c(x) left(frac{mathrm{d}u(x)}{mathrm{d}x}right)^{2} mathrm{d}x
end{equation*}
So $frac{mathrm{d}J(u + tv)}{mathrm{d}t}$ at $t = 0$ would be $2 int_{0}^{1} c(x) left( frac{mathrm{d}u(x)}{mathrm{d}x} right)left( frac{mathrm{d}v(x)}{mathrm{d}x}right) mathrm{d}x$. However, without the differentiability of $c(x)$, I cannot do the integration by part, choose $v(x)$ that is compactly supported inside $(0,1)$ and obtain the differential equation. Do you think that it is just the problem forgot to mention the differentiability of $c(x)$, or it is in fact not necessary?
calculus
asked Jan 9 at 13:03
BM YoonBM Yoon
232
$endgroup$
add a comment |
$begingroup$
I come across this problem: Find the Euler Lagrange Equation of this energy function
begin{equation*}
J(u) = int_{0}^{1} c(x) left(frac{mathrm{d}u(x)}{mathrm{d}x}right)^{2} mathrm{d}x
end{equation*}
So $frac{mathrm{d}J(u + tv)}{mathrm{d}t}$ at $t = 0$ would be $2 int_{0}^{1} c(x) left( frac{mathrm{d}u(x)}{mathrm{d}x} right)left( frac{mathrm{d}v(x)}{mathrm{d}x}right) mathrm{d}x$. However, without the differentiability of $c(x)$, I cannot do the integration by part, choose $v(x)$ that is compactly supported inside $(0,1)$ and obtain the differential equation. Do you think that it is just the problem forgot to mention the differentiability of $c(x)$, or it is in fact not necessary?
calculus
asked Jan 9 at 13:03
BM YoonBM Yoon
232
$endgroup$
$begingroup$
Calling $L = c u^2'$ we have $L_u-(L_{u'})_x = -2(c'u'+c u'')=0$
$endgroup$
– Cesareo
Jan 9 at 13:26
add a comment |
$begingroup$
I come across this problem: Find the Euler Lagrange Equation of this energy function
begin{equation*}
J(u) = int_{0}^{1} c(x) left(frac{mathrm{d}u(x)}{mathrm{d}x}right)^{2} mathrm{d}x
end{equation*}
So $frac{mathrm{d}J(u + tv)}{mathrm{d}t}$ at $t = 0$ would be $2 int_{0}^{1} c(x) left( frac{mathrm{d}u(x)}{mathrm{d}x} right)left( frac{mathrm{d}v(x)}{mathrm{d}x}right) mathrm{d}x$. However, without the differentiability of $c(x)$, I cannot do the integration by part, choose $v(x)$ that is compactly supported inside $(0,1)$ and obtain the differential equation. Do you think that it is just the problem forgot to mention the differentiability of $c(x)$, or it is in fact not necessary?
calculus
asked Jan 9 at 13:03
BM YoonBM Yoon
232
$endgroup$
I come across this problem: Find the Euler Lagrange Equation of this energy function
begin{equation*}
J(u) = int_{0}^{1} c(x) left(frac{mathrm{d}u(x)}{mathrm{d}x}right)^{2} mathrm{d}x
end{equation*}
So $frac{mathrm{d}J(u + tv)}{mathrm{d}t}$ at $t = 0$ would be $2 int_{0}^{1} c(x) left( frac{mathrm{d}u(x)}{mathrm{d}x} right)left( frac{mathrm{d}v(x)}{mathrm{d}x}right) mathrm{d}x$. However, without the differentiability of $c(x)$, I cannot do the integration by part, choose $v(x)$ that is compactly supported inside $(0,1)$ and obtain the differential equation. Do you think that it is just the problem forgot to mention the differentiability of $c(x)$, or it is in fact not necessary?
calculus
calculus
asked Jan 9 at 13:03
BM YoonBM Yoon
232
asked Jan 9 at 13:03
BM YoonBM Yoon
232
asked Jan 9 at 13:03
BM YoonBM Yoon
232
asked Jan 9 at 13:03
BM YoonBM Yoon
232
asked Jan 9 at 13:03
BM YoonBM Yoon
232
232
$begingroup$
Calling $L = c u^2'$ we have $L_u-(L_{u'})_x = -2(c'u'+c u'')=0$
$endgroup$
– Cesareo
Jan 9 at 13:26
add a comment |
$begingroup$
Calling $L = c u^2'$ we have $L_u-(L_{u'})_x = -2(c'u'+c u'')=0$
$endgroup$
– Cesareo
Jan 9 at 13:26
$begingroup$
Calling $L = c u^2'$ we have $L_u-(L_{u'})_x = -2(c'u'+c u'')=0$
$endgroup$
– Cesareo
Jan 9 at 13:26
$begingroup$
Calling $L = c u^2'$ we have $L_u-(L_{u'})_x = -2(c'u'+c u'')=0$
$endgroup$
– Cesareo
Jan 9 at 13:26
add a comment |
1 Answer
1
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oldest
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Here's a way to look at it that'll get you started. The Lagrangian $L=cu^{prime 2}$ gives Euler-Lagrange equations $u^{prime 2}=0,,(2cu^prime)^prime=0$, which simplifies to $u^prime=0$ regardless of $c$. In particular, on-shell $cu^prime$ is differentiable because it's identically $0$.
answered Jan 9 at 13:21
J.G.J.G.
24.2k22539
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's a way to look at it that'll get you started. The Lagrangian $L=cu^{prime 2}$ gives Euler-Lagrange equations $u^{prime 2}=0,,(2cu^prime)^prime=0$, which simplifies to $u^prime=0$ regardless of $c$. In particular, on-shell $cu^prime$ is differentiable because it's identically $0$.
answered Jan 9 at 13:21
J.G.J.G.
24.2k22539
$endgroup$
add a comment |
$begingroup$
Here's a way to look at it that'll get you started. The Lagrangian $L=cu^{prime 2}$ gives Euler-Lagrange equations $u^{prime 2}=0,,(2cu^prime)^prime=0$, which simplifies to $u^prime=0$ regardless of $c$. In particular, on-shell $cu^prime$ is differentiable because it's identically $0$.
answered Jan 9 at 13:21
J.G.J.G.
24.2k22539
$endgroup$
add a comment |
$begingroup$
Here's a way to look at it that'll get you started. The Lagrangian $L=cu^{prime 2}$ gives Euler-Lagrange equations $u^{prime 2}=0,,(2cu^prime)^prime=0$, which simplifies to $u^prime=0$ regardless of $c$. In particular, on-shell $cu^prime$ is differentiable because it's identically $0$.
answered Jan 9 at 13:21
J.G.J.G.
24.2k22539
$endgroup$
Here's a way to look at it that'll get you started. The Lagrangian $L=cu^{prime 2}$ gives Euler-Lagrange equations $u^{prime 2}=0,,(2cu^prime)^prime=0$, which simplifies to $u^prime=0$ regardless of $c$. In particular, on-shell $cu^prime$ is differentiable because it's identically $0$.
answered Jan 9 at 13:21
J.G.J.G.
24.2k22539
answered Jan 9 at 13:21
J.G.J.G.
24.2k22539
answered Jan 9 at 13:21
J.G.J.G.
24.2k22539
answered Jan 9 at 13:21
J.G.J.G.
24.2k22539
24.2k22539
add a comment |
add a comment |
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$begingroup$
Calling $L = c u^2'$ we have $L_u-(L_{u'})_x = -2(c'u'+c u'')=0$
$endgroup$
– Cesareo
Jan 9 at 13:26