Calculus of variation with discontinuous solutions
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I'm thinking of the following question:
Consider a function $f: Urightarrowmathbb{R}$ where $U=[0,L_1)cup(L_1,L]$, and an energy functional $$F=int_{U}Big (frac{mathrm{d}f}{mathrm{d}x}Big)^2mathrm{d}x.$$ The boundary condition is $f(0)=f(L)=0$. Also, assume $f$ has a jump discontinuity at $x=L_1$, i.e., $f(L_1^-)=-f(L_1^+)=-a$, where $a$ is a constant, and $L_1$ is a variable.
The question is: what is the minimum energy?
Simple answer: the Euler-Lagrange equation on $[0,L_1)$ and $(L_1,L]$ is $$frac{mathrm{d}^2 f}{mathrm{d}x^2}=0,$$ Therefore we have the following solution: $$f(x) =
begin{cases}
-frac{a}{L_1}x, & text{x$in [0,L_1)$} \
-frac{a}{L-L_1}x+frac{L}{L-L_1}a, & text{x$in (L_1,L]$}
end{cases}$$ Then the energy function $F(L_1)=frac{a^2}{L_1}+frac{a^2}{L-L_1}$. Therefore we have the minimum energy $4a^2/L$ when $L_1=L/2$.
Then I tried Fourier series. The Fourier series for the discontinuous solutions are $$ f(x)=sum_{k=1}^{infty}Big[frac{4a}{kpi}cosfrac{kpi L_1}{L}+frac{2aL}{(kpi)^2}Big(frac{1}{L-L_1}-frac{1}{L_1}Big)sinfrac{kpi L_1}{L}Big]sinBig(frac{kpi x}{L}Big),$$ When we substitute this representation into the energy functional, we get infinite energy. In particular,begin{align}
F(L_1=L/2) & =sum_{k=1}^{infty}Big(4acosBig(frac{kpi}{2}Big)Big)^2cdotfrac{1}{2L}\
& =sum_{k=1}^{infty}frac{4a^2}{L}(1+cos kpi)~mathrm{e}^{-lambda_1 k} & text{$lambda_1to0^+$}\
& tofrac{4a^2}{L}cdotfrac{1}{lambda_1}\
end{align}
begin{align}
F(L_1=L/4) & =sum_{k=1}^{infty}Big(4acosBig(frac{kpi}{4}Big)-frac{16}{3}frac{a}{kpi}sinBig(frac{kpi}{4}Big)Big)^2cdotfrac{1}{2L}\
& =sum_{k=1}^{infty}frac{4a^2}{L}Big(1+cos Big(frac{kpi}{2}Big)Big)~mathrm{e}^{-lambda_2 k}\& +sum_{k=1}^{infty}Big(frac{128a^2}{9L}frac{1}{(kpi)^2}sin^2Big(frac{kpi}{4}Big)-frac{64a^2}{3L}frac{1}{kpi}sinBig(frac{kpi}{4}Big)cosBig(frac{kpi}{4}Big)Big)& text{$lambda_2to0^+$}\
& tofrac{4a^2}{L}cdotfrac{1}{lambda_2}-frac{4}{3}cdotfrac{a^2}{L}\
end{align}
It is not surprising that the energies are infinite due to the Gibbs phenomenon. However, it is surprising when we take a look at the energy difference: $$F(L_1=L/4)-F(L_1=L/2)=Big(frac{4a^2}{L}cdotfrac{1}{lambda_2}-frac{4a^2}{L}cdotfrac{1}{lambda_1}Big)-frac{4}{3}cdotfrac{a^2}{L}$$
While when we use $F(L_1)=frac{a^2}{L_1}+frac{a^2}{L-L_1}$, we get $F(L_1=L/4)-F(L_1=L/2)=frac{4}{3}cdotfrac{a^2}{L}$. My observation is that when we "ignore" the divergent term, the Fourier series give us the correct absolute value for the energy difference but wrong sign (I have checked for all values of $L_1$). I don't feel like it is a coincidence.
To get the correct value of energy difference, I'm guessing that maybe when we use Fourier series, we introduce a vanishing length scale for the jump discontinuity, and this length scale should be the same whatever the value of $L_1$ is, and somehow $lambda_1$, $lambda_2$ are related to it so that $frac{4a^2}{L}cdotfrac{1}{lambda_2}-frac{4a^2}{L}cdotfrac{1}{lambda_1}$ may be some finite number.
My question is whether we can use Fourier series to get the correct value of energy difference.
sequences-and-series fourier-series calculus-of-variations divergent-series
asked Jun 30 '18 at 16:36
user573229user573229
5115
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add a comment |
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I'm thinking of the following question:
Consider a function $f: Urightarrowmathbb{R}$ where $U=[0,L_1)cup(L_1,L]$, and an energy functional $$F=int_{U}Big (frac{mathrm{d}f}{mathrm{d}x}Big)^2mathrm{d}x.$$ The boundary condition is $f(0)=f(L)=0$. Also, assume $f$ has a jump discontinuity at $x=L_1$, i.e., $f(L_1^-)=-f(L_1^+)=-a$, where $a$ is a constant, and $L_1$ is a variable.
The question is: what is the minimum energy?
Simple answer: the Euler-Lagrange equation on $[0,L_1)$ and $(L_1,L]$ is $$frac{mathrm{d}^2 f}{mathrm{d}x^2}=0,$$ Therefore we have the following solution: $$f(x) =
begin{cases}
-frac{a}{L_1}x, & text{x$in [0,L_1)$} \
-frac{a}{L-L_1}x+frac{L}{L-L_1}a, & text{x$in (L_1,L]$}
end{cases}$$ Then the energy function $F(L_1)=frac{a^2}{L_1}+frac{a^2}{L-L_1}$. Therefore we have the minimum energy $4a^2/L$ when $L_1=L/2$.
Then I tried Fourier series. The Fourier series for the discontinuous solutions are $$ f(x)=sum_{k=1}^{infty}Big[frac{4a}{kpi}cosfrac{kpi L_1}{L}+frac{2aL}{(kpi)^2}Big(frac{1}{L-L_1}-frac{1}{L_1}Big)sinfrac{kpi L_1}{L}Big]sinBig(frac{kpi x}{L}Big),$$ When we substitute this representation into the energy functional, we get infinite energy. In particular,begin{align}
F(L_1=L/2) & =sum_{k=1}^{infty}Big(4acosBig(frac{kpi}{2}Big)Big)^2cdotfrac{1}{2L}\
& =sum_{k=1}^{infty}frac{4a^2}{L}(1+cos kpi)~mathrm{e}^{-lambda_1 k} & text{$lambda_1to0^+$}\
& tofrac{4a^2}{L}cdotfrac{1}{lambda_1}\
end{align}
begin{align}
F(L_1=L/4) & =sum_{k=1}^{infty}Big(4acosBig(frac{kpi}{4}Big)-frac{16}{3}frac{a}{kpi}sinBig(frac{kpi}{4}Big)Big)^2cdotfrac{1}{2L}\
& =sum_{k=1}^{infty}frac{4a^2}{L}Big(1+cos Big(frac{kpi}{2}Big)Big)~mathrm{e}^{-lambda_2 k}\& +sum_{k=1}^{infty}Big(frac{128a^2}{9L}frac{1}{(kpi)^2}sin^2Big(frac{kpi}{4}Big)-frac{64a^2}{3L}frac{1}{kpi}sinBig(frac{kpi}{4}Big)cosBig(frac{kpi}{4}Big)Big)& text{$lambda_2to0^+$}\
& tofrac{4a^2}{L}cdotfrac{1}{lambda_2}-frac{4}{3}cdotfrac{a^2}{L}\
end{align}
It is not surprising that the energies are infinite due to the Gibbs phenomenon. However, it is surprising when we take a look at the energy difference: $$F(L_1=L/4)-F(L_1=L/2)=Big(frac{4a^2}{L}cdotfrac{1}{lambda_2}-frac{4a^2}{L}cdotfrac{1}{lambda_1}Big)-frac{4}{3}cdotfrac{a^2}{L}$$
While when we use $F(L_1)=frac{a^2}{L_1}+frac{a^2}{L-L_1}$, we get $F(L_1=L/4)-F(L_1=L/2)=frac{4}{3}cdotfrac{a^2}{L}$. My observation is that when we "ignore" the divergent term, the Fourier series give us the correct absolute value for the energy difference but wrong sign (I have checked for all values of $L_1$). I don't feel like it is a coincidence.
To get the correct value of energy difference, I'm guessing that maybe when we use Fourier series, we introduce a vanishing length scale for the jump discontinuity, and this length scale should be the same whatever the value of $L_1$ is, and somehow $lambda_1$, $lambda_2$ are related to it so that $frac{4a^2}{L}cdotfrac{1}{lambda_2}-frac{4a^2}{L}cdotfrac{1}{lambda_1}$ may be some finite number.
My question is whether we can use Fourier series to get the correct value of energy difference.
sequences-and-series fourier-series calculus-of-variations divergent-series
asked Jun 30 '18 at 16:36
user573229user573229
5115
$endgroup$
$begingroup$
It seems to me that you should get the right result if you consider the mollified problem with $f(0)=0,f(L_1-delta)=-a,f(L_1+delta)=a,f(L)=0$ and then ultimately send $delta to 0$. Euler-Lagrange cannot be derived rigorously on the whole interval in this situation, but it can be in the mollified problem (on each of the three subdomains, of course). The only weird thing here is that you are basically looking at the $L^2$ norm of a non-$L^2$ distribution, which is a somewhat weird thing to do in the first place.
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– Ian
Nov 25 '18 at 19:49
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Do you assume $f$ is continuous on $(L_1-delta, L_1+delta)$? I'm not quite sure of how to do it in this way. But I think this problem can be done in separate regions. Then I'm wondering if there is a way to "glue" them. I agree that it is weird.
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– user573229
Nov 25 '18 at 19:55
$begingroup$
Yes, the point is to loosely take into account the jump discontinuity and how it should affect the $L^2$ norm of the derivative, even though the derivative in the real problem isn't $L^2$.
$endgroup$
– Ian
Nov 25 '18 at 21:01
add a comment |
$begingroup$
I'm thinking of the following question:
Consider a function $f: Urightarrowmathbb{R}$ where $U=[0,L_1)cup(L_1,L]$, and an energy functional $$F=int_{U}Big (frac{mathrm{d}f}{mathrm{d}x}Big)^2mathrm{d}x.$$ The boundary condition is $f(0)=f(L)=0$. Also, assume $f$ has a jump discontinuity at $x=L_1$, i.e., $f(L_1^-)=-f(L_1^+)=-a$, where $a$ is a constant, and $L_1$ is a variable.
The question is: what is the minimum energy?
Simple answer: the Euler-Lagrange equation on $[0,L_1)$ and $(L_1,L]$ is $$frac{mathrm{d}^2 f}{mathrm{d}x^2}=0,$$ Therefore we have the following solution: $$f(x) =
begin{cases}
-frac{a}{L_1}x, & text{x$in [0,L_1)$} \
-frac{a}{L-L_1}x+frac{L}{L-L_1}a, & text{x$in (L_1,L]$}
end{cases}$$ Then the energy function $F(L_1)=frac{a^2}{L_1}+frac{a^2}{L-L_1}$. Therefore we have the minimum energy $4a^2/L$ when $L_1=L/2$.
Then I tried Fourier series. The Fourier series for the discontinuous solutions are $$ f(x)=sum_{k=1}^{infty}Big[frac{4a}{kpi}cosfrac{kpi L_1}{L}+frac{2aL}{(kpi)^2}Big(frac{1}{L-L_1}-frac{1}{L_1}Big)sinfrac{kpi L_1}{L}Big]sinBig(frac{kpi x}{L}Big),$$ When we substitute this representation into the energy functional, we get infinite energy. In particular,begin{align}
F(L_1=L/2) & =sum_{k=1}^{infty}Big(4acosBig(frac{kpi}{2}Big)Big)^2cdotfrac{1}{2L}\
& =sum_{k=1}^{infty}frac{4a^2}{L}(1+cos kpi)~mathrm{e}^{-lambda_1 k} & text{$lambda_1to0^+$}\
& tofrac{4a^2}{L}cdotfrac{1}{lambda_1}\
end{align}
begin{align}
F(L_1=L/4) & =sum_{k=1}^{infty}Big(4acosBig(frac{kpi}{4}Big)-frac{16}{3}frac{a}{kpi}sinBig(frac{kpi}{4}Big)Big)^2cdotfrac{1}{2L}\
& =sum_{k=1}^{infty}frac{4a^2}{L}Big(1+cos Big(frac{kpi}{2}Big)Big)~mathrm{e}^{-lambda_2 k}\& +sum_{k=1}^{infty}Big(frac{128a^2}{9L}frac{1}{(kpi)^2}sin^2Big(frac{kpi}{4}Big)-frac{64a^2}{3L}frac{1}{kpi}sinBig(frac{kpi}{4}Big)cosBig(frac{kpi}{4}Big)Big)& text{$lambda_2to0^+$}\
& tofrac{4a^2}{L}cdotfrac{1}{lambda_2}-frac{4}{3}cdotfrac{a^2}{L}\
end{align}
It is not surprising that the energies are infinite due to the Gibbs phenomenon. However, it is surprising when we take a look at the energy difference: $$F(L_1=L/4)-F(L_1=L/2)=Big(frac{4a^2}{L}cdotfrac{1}{lambda_2}-frac{4a^2}{L}cdotfrac{1}{lambda_1}Big)-frac{4}{3}cdotfrac{a^2}{L}$$
While when we use $F(L_1)=frac{a^2}{L_1}+frac{a^2}{L-L_1}$, we get $F(L_1=L/4)-F(L_1=L/2)=frac{4}{3}cdotfrac{a^2}{L}$. My observation is that when we "ignore" the divergent term, the Fourier series give us the correct absolute value for the energy difference but wrong sign (I have checked for all values of $L_1$). I don't feel like it is a coincidence.
To get the correct value of energy difference, I'm guessing that maybe when we use Fourier series, we introduce a vanishing length scale for the jump discontinuity, and this length scale should be the same whatever the value of $L_1$ is, and somehow $lambda_1$, $lambda_2$ are related to it so that $frac{4a^2}{L}cdotfrac{1}{lambda_2}-frac{4a^2}{L}cdotfrac{1}{lambda_1}$ may be some finite number.
My question is whether we can use Fourier series to get the correct value of energy difference.
sequences-and-series fourier-series calculus-of-variations divergent-series
asked Jun 30 '18 at 16:36
user573229user573229
5115
$endgroup$
I'm thinking of the following question:
Consider a function $f: Urightarrowmathbb{R}$ where $U=[0,L_1)cup(L_1,L]$, and an energy functional $$F=int_{U}Big (frac{mathrm{d}f}{mathrm{d}x}Big)^2mathrm{d}x.$$ The boundary condition is $f(0)=f(L)=0$. Also, assume $f$ has a jump discontinuity at $x=L_1$, i.e., $f(L_1^-)=-f(L_1^+)=-a$, where $a$ is a constant, and $L_1$ is a variable.
The question is: what is the minimum energy?
Simple answer: the Euler-Lagrange equation on $[0,L_1)$ and $(L_1,L]$ is $$frac{mathrm{d}^2 f}{mathrm{d}x^2}=0,$$ Therefore we have the following solution: $$f(x) =
begin{cases}
-frac{a}{L_1}x, & text{x$in [0,L_1)$} \
-frac{a}{L-L_1}x+frac{L}{L-L_1}a, & text{x$in (L_1,L]$}
end{cases}$$ Then the energy function $F(L_1)=frac{a^2}{L_1}+frac{a^2}{L-L_1}$. Therefore we have the minimum energy $4a^2/L$ when $L_1=L/2$.
Then I tried Fourier series. The Fourier series for the discontinuous solutions are $$ f(x)=sum_{k=1}^{infty}Big[frac{4a}{kpi}cosfrac{kpi L_1}{L}+frac{2aL}{(kpi)^2}Big(frac{1}{L-L_1}-frac{1}{L_1}Big)sinfrac{kpi L_1}{L}Big]sinBig(frac{kpi x}{L}Big),$$ When we substitute this representation into the energy functional, we get infinite energy. In particular,begin{align}
F(L_1=L/2) & =sum_{k=1}^{infty}Big(4acosBig(frac{kpi}{2}Big)Big)^2cdotfrac{1}{2L}\
& =sum_{k=1}^{infty}frac{4a^2}{L}(1+cos kpi)~mathrm{e}^{-lambda_1 k} & text{$lambda_1to0^+$}\
& tofrac{4a^2}{L}cdotfrac{1}{lambda_1}\
end{align}
begin{align}
F(L_1=L/4) & =sum_{k=1}^{infty}Big(4acosBig(frac{kpi}{4}Big)-frac{16}{3}frac{a}{kpi}sinBig(frac{kpi}{4}Big)Big)^2cdotfrac{1}{2L}\
& =sum_{k=1}^{infty}frac{4a^2}{L}Big(1+cos Big(frac{kpi}{2}Big)Big)~mathrm{e}^{-lambda_2 k}\& +sum_{k=1}^{infty}Big(frac{128a^2}{9L}frac{1}{(kpi)^2}sin^2Big(frac{kpi}{4}Big)-frac{64a^2}{3L}frac{1}{kpi}sinBig(frac{kpi}{4}Big)cosBig(frac{kpi}{4}Big)Big)& text{$lambda_2to0^+$}\
& tofrac{4a^2}{L}cdotfrac{1}{lambda_2}-frac{4}{3}cdotfrac{a^2}{L}\
end{align}
It is not surprising that the energies are infinite due to the Gibbs phenomenon. However, it is surprising when we take a look at the energy difference: $$F(L_1=L/4)-F(L_1=L/2)=Big(frac{4a^2}{L}cdotfrac{1}{lambda_2}-frac{4a^2}{L}cdotfrac{1}{lambda_1}Big)-frac{4}{3}cdotfrac{a^2}{L}$$
While when we use $F(L_1)=frac{a^2}{L_1}+frac{a^2}{L-L_1}$, we get $F(L_1=L/4)-F(L_1=L/2)=frac{4}{3}cdotfrac{a^2}{L}$. My observation is that when we "ignore" the divergent term, the Fourier series give us the correct absolute value for the energy difference but wrong sign (I have checked for all values of $L_1$). I don't feel like it is a coincidence.
To get the correct value of energy difference, I'm guessing that maybe when we use Fourier series, we introduce a vanishing length scale for the jump discontinuity, and this length scale should be the same whatever the value of $L_1$ is, and somehow $lambda_1$, $lambda_2$ are related to it so that $frac{4a^2}{L}cdotfrac{1}{lambda_2}-frac{4a^2}{L}cdotfrac{1}{lambda_1}$ may be some finite number.
My question is whether we can use Fourier series to get the correct value of energy difference.
sequences-and-series fourier-series calculus-of-variations divergent-series
sequences-and-series fourier-series calculus-of-variations divergent-series
asked Jun 30 '18 at 16:36
user573229user573229
5115
asked Jun 30 '18 at 16:36
user573229user573229
5115
edited Nov 26 '18 at 19:53
user573229
asked Jun 30 '18 at 16:36
user573229user573229
5115
asked Jun 30 '18 at 16:36
user573229user573229
5115
asked Jun 30 '18 at 16:36
user573229user573229
5115
5115
$begingroup$
It seems to me that you should get the right result if you consider the mollified problem with $f(0)=0,f(L_1-delta)=-a,f(L_1+delta)=a,f(L)=0$ and then ultimately send $delta to 0$. Euler-Lagrange cannot be derived rigorously on the whole interval in this situation, but it can be in the mollified problem (on each of the three subdomains, of course). The only weird thing here is that you are basically looking at the $L^2$ norm of a non-$L^2$ distribution, which is a somewhat weird thing to do in the first place.
$endgroup$
– Ian
Nov 25 '18 at 19:49
$begingroup$
Do you assume $f$ is continuous on $(L_1-delta, L_1+delta)$? I'm not quite sure of how to do it in this way. But I think this problem can be done in separate regions. Then I'm wondering if there is a way to "glue" them. I agree that it is weird.
$endgroup$
– user573229
Nov 25 '18 at 19:55
$begingroup$
Yes, the point is to loosely take into account the jump discontinuity and how it should affect the $L^2$ norm of the derivative, even though the derivative in the real problem isn't $L^2$.
$endgroup$
– Ian
Nov 25 '18 at 21:01
add a comment |
$begingroup$
It seems to me that you should get the right result if you consider the mollified problem with $f(0)=0,f(L_1-delta)=-a,f(L_1+delta)=a,f(L)=0$ and then ultimately send $delta to 0$. Euler-Lagrange cannot be derived rigorously on the whole interval in this situation, but it can be in the mollified problem (on each of the three subdomains, of course). The only weird thing here is that you are basically looking at the $L^2$ norm of a non-$L^2$ distribution, which is a somewhat weird thing to do in the first place.
$endgroup$
– Ian
Nov 25 '18 at 19:49
$begingroup$
Do you assume $f$ is continuous on $(L_1-delta, L_1+delta)$? I'm not quite sure of how to do it in this way. But I think this problem can be done in separate regions. Then I'm wondering if there is a way to "glue" them. I agree that it is weird.
$endgroup$
– user573229
Nov 25 '18 at 19:55
$begingroup$
Yes, the point is to loosely take into account the jump discontinuity and how it should affect the $L^2$ norm of the derivative, even though the derivative in the real problem isn't $L^2$.
$endgroup$
– Ian
Nov 25 '18 at 21:01
$begingroup$
It seems to me that you should get the right result if you consider the mollified problem with $f(0)=0,f(L_1-delta)=-a,f(L_1+delta)=a,f(L)=0$ and then ultimately send $delta to 0$. Euler-Lagrange cannot be derived rigorously on the whole interval in this situation, but it can be in the mollified problem (on each of the three subdomains, of course). The only weird thing here is that you are basically looking at the $L^2$ norm of a non-$L^2$ distribution, which is a somewhat weird thing to do in the first place.
$endgroup$
– Ian
Nov 25 '18 at 19:49
$begingroup$
It seems to me that you should get the right result if you consider the mollified problem with $f(0)=0,f(L_1-delta)=-a,f(L_1+delta)=a,f(L)=0$ and then ultimately send $delta to 0$. Euler-Lagrange cannot be derived rigorously on the whole interval in this situation, but it can be in the mollified problem (on each of the three subdomains, of course). The only weird thing here is that you are basically looking at the $L^2$ norm of a non-$L^2$ distribution, which is a somewhat weird thing to do in the first place.
$endgroup$
– Ian
Nov 25 '18 at 19:49
$begingroup$
Do you assume $f$ is continuous on $(L_1-delta, L_1+delta)$? I'm not quite sure of how to do it in this way. But I think this problem can be done in separate regions. Then I'm wondering if there is a way to "glue" them. I agree that it is weird.
$endgroup$
– user573229
Nov 25 '18 at 19:55
$begingroup$
Do you assume $f$ is continuous on $(L_1-delta, L_1+delta)$? I'm not quite sure of how to do it in this way. But I think this problem can be done in separate regions. Then I'm wondering if there is a way to "glue" them. I agree that it is weird.
$endgroup$
– user573229
Nov 25 '18 at 19:55
$begingroup$
Yes, the point is to loosely take into account the jump discontinuity and how it should affect the $L^2$ norm of the derivative, even though the derivative in the real problem isn't $L^2$.
$endgroup$
– Ian
Nov 25 '18 at 21:01
$begingroup$
Yes, the point is to loosely take into account the jump discontinuity and how it should affect the $L^2$ norm of the derivative, even though the derivative in the real problem isn't $L^2$.
$endgroup$
– Ian
Nov 25 '18 at 21:01
add a comment |
1 Answer
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My attempted answer is:
$F(L_1)=sum_{k=1}^{infty}frac{(b_kkpi)^2}{2L}=F_1(L_1)+F_2(L_1)+F_3(L_1)$
where
$F_1(L_1)=frac{1}{2L}cdotsum_{k=1}^{infty}Big(4acosfrac{kpi L_1}{L}Big)^2$
$F_2(L_1)=frac{1}{2L}cdotsum_{k=1}^{infty}Big[frac{2aL}{kpi}Big(frac{1}{L-L_1}-frac{1}{L_1}Big)sinfrac{kpi L_1}{L}Big]^2=frac{a^2}{L_1}+frac{a^2}{L-L_1}-frac{4a^2}{L}$
$F_3(L_1)=frac{1}{L}cdotsum_{k=1}^{infty}Big(4acosfrac{kpi L_1}{L}Big)cdotBig[frac{2aL}{kpi}Big(frac{1}{L-L_1}-frac{1}{L_1}Big)sinfrac{kpi L_1}{L}Big]$
To determine the ground state, we may throw away $F_1(L_1)$ and $F_3(L_1)$, because:
(1) $F_1(L_1)$ contains the fictitious core energy, and we can see this by finding the energy density.
$F(y, L_1)=int_{0}^{y}(frac{mathrm{d}f}{mathrm{d}x})^2mathrm{d}x=F_1(y, L_1)+F_2(y, L_1)+F_3(y, L_1)$,
where $F(L,L_1)=F(L_1)$, $F_1(L,L_1)=F_1(L_1)$, $F_2(L,L_1)=F_2(L_1)$, and $F_3(L,L_1)=F_3(L_1)$.
After some calculations, we have
$F_{1}(y, L_1)
=frac{2a^2y}{L^2}Big(frac{cosfrac{2Npi L_1}{L}-cosfrac{2(N+1)pi L_1}{L}}{1-cosfrac{2pi L_1}{L}}-1Big)-frac{2a^2}{pi L}cdotfrac{sinfrac{pi y}{L}}{cosfrac{pi L_1}{L}-cosfrac{pi y}{L}}+frac{a^2}{pi L}cdotfrac{cosfrac{(2N+1)pi L_1}{L}}{sinfrac{pi L_1}{L}}lnBigg (frac{1-cosfrac{pi (y-L_1)}{L}}{1-cosfrac{pi (y+L_1)}{L}}Bigg)+frac{4a^2 y}{L^2}\ \ +begin{cases}
0+frac{2a^2y}{L^2}Big(1-frac{sinfrac{(2N+1)pi L_1}{L}}{sinfrac{pi L_1}{L}}Big), & text{if } 0<y<L_1\
frac{4(N-1)a^2}{L}+frac{2a^2(L-y)}{L^2}Big(-1+frac{sinfrac{(2N+1)pi L_1}{L}}{sinfrac{pi L_1}{L}}Big), & text{if } L_1<y<L
end{cases}\=-frac{2a^2}{pi L}cdotfrac{sinfrac{pi y}{L}}{cosfrac{pi L_1}{L}-cosfrac{pi y}{L}}+frac{a^2}{pi L}cdotfrac{cosfrac{pi L_1}{L}}{sinfrac{pi L_1}{L}}lnBigg (frac{1-cosfrac{pi (y-L_1)}{L}}{1-cosfrac{pi (y+L_1)}{L}}Bigg)+frac{4a^2 y}{L^2}+begin{cases}
0, & text{if } 0<y<L_1\
frac{4(N-1)a^2}{L}, & text{if } L_1<y<L
end{cases}$
where $Ntoinfty$. And I assume $Ncdotfrac{L_1}{L}in mathbb{Z}$, because that seems to make sure the core size for different $L_1$ and finite $N$ are the same (I don't know a better way to justify it). Now we can see $F_1(L,L_1)=F_1(L_1)$ is infinite but independent of $L_1$. Its energy density profile is
$mathcal{F}_1(y, L_1)=frac{4a^2}{L^2}+frac{a^2}{2L^2}Big(csc ^2frac{pi (y-L_1)}{2L}+cosfrac{pi (y-3L_1)}{2L}-cosfrac{pi (y+L_1)}{2L}+cotfrac{pi L_1}{L}csc ^2frac{pi (y-L_1)}{2L}cscfrac{pi (y+L_1)}{2L}+csc ^2frac{pi (y+L_1)}{2L}Big)$
We can see that the energy jumps to infinity at $x=L_1$, so $F_1(L_1)$ contains the core energy and the constant energy $frac{4a^2}{L}$. Also, notice that the size of the core becomes very big, and I think it is due to the Gibbs phenomenon.
(2) I think $F_3(L_1)$ can be regarded as the interaction between the fattened core and the function $f$ on $[0,L_1)cup (L_1,L]$, so it should be thrown away.
I have posted more about it here https://figshare.com/articles/Some_examples_of_calculus_of_variations_of_discontinuous_functions_and_their_implications_pdf/7379861
answered Nov 25 '18 at 17:27
user573229user573229
5115
$endgroup$
add a comment |
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$begingroup$
My attempted answer is:
$F(L_1)=sum_{k=1}^{infty}frac{(b_kkpi)^2}{2L}=F_1(L_1)+F_2(L_1)+F_3(L_1)$
where
$F_1(L_1)=frac{1}{2L}cdotsum_{k=1}^{infty}Big(4acosfrac{kpi L_1}{L}Big)^2$
$F_2(L_1)=frac{1}{2L}cdotsum_{k=1}^{infty}Big[frac{2aL}{kpi}Big(frac{1}{L-L_1}-frac{1}{L_1}Big)sinfrac{kpi L_1}{L}Big]^2=frac{a^2}{L_1}+frac{a^2}{L-L_1}-frac{4a^2}{L}$
$F_3(L_1)=frac{1}{L}cdotsum_{k=1}^{infty}Big(4acosfrac{kpi L_1}{L}Big)cdotBig[frac{2aL}{kpi}Big(frac{1}{L-L_1}-frac{1}{L_1}Big)sinfrac{kpi L_1}{L}Big]$
To determine the ground state, we may throw away $F_1(L_1)$ and $F_3(L_1)$, because:
(1) $F_1(L_1)$ contains the fictitious core energy, and we can see this by finding the energy density.
$F(y, L_1)=int_{0}^{y}(frac{mathrm{d}f}{mathrm{d}x})^2mathrm{d}x=F_1(y, L_1)+F_2(y, L_1)+F_3(y, L_1)$,
where $F(L,L_1)=F(L_1)$, $F_1(L,L_1)=F_1(L_1)$, $F_2(L,L_1)=F_2(L_1)$, and $F_3(L,L_1)=F_3(L_1)$.
After some calculations, we have
$F_{1}(y, L_1)
=frac{2a^2y}{L^2}Big(frac{cosfrac{2Npi L_1}{L}-cosfrac{2(N+1)pi L_1}{L}}{1-cosfrac{2pi L_1}{L}}-1Big)-frac{2a^2}{pi L}cdotfrac{sinfrac{pi y}{L}}{cosfrac{pi L_1}{L}-cosfrac{pi y}{L}}+frac{a^2}{pi L}cdotfrac{cosfrac{(2N+1)pi L_1}{L}}{sinfrac{pi L_1}{L}}lnBigg (frac{1-cosfrac{pi (y-L_1)}{L}}{1-cosfrac{pi (y+L_1)}{L}}Bigg)+frac{4a^2 y}{L^2}\ \ +begin{cases}
0+frac{2a^2y}{L^2}Big(1-frac{sinfrac{(2N+1)pi L_1}{L}}{sinfrac{pi L_1}{L}}Big), & text{if } 0<y<L_1\
frac{4(N-1)a^2}{L}+frac{2a^2(L-y)}{L^2}Big(-1+frac{sinfrac{(2N+1)pi L_1}{L}}{sinfrac{pi L_1}{L}}Big), & text{if } L_1<y<L
end{cases}\=-frac{2a^2}{pi L}cdotfrac{sinfrac{pi y}{L}}{cosfrac{pi L_1}{L}-cosfrac{pi y}{L}}+frac{a^2}{pi L}cdotfrac{cosfrac{pi L_1}{L}}{sinfrac{pi L_1}{L}}lnBigg (frac{1-cosfrac{pi (y-L_1)}{L}}{1-cosfrac{pi (y+L_1)}{L}}Bigg)+frac{4a^2 y}{L^2}+begin{cases}
0, & text{if } 0<y<L_1\
frac{4(N-1)a^2}{L}, & text{if } L_1<y<L
end{cases}$
where $Ntoinfty$. And I assume $Ncdotfrac{L_1}{L}in mathbb{Z}$, because that seems to make sure the core size for different $L_1$ and finite $N$ are the same (I don't know a better way to justify it). Now we can see $F_1(L,L_1)=F_1(L_1)$ is infinite but independent of $L_1$. Its energy density profile is
$mathcal{F}_1(y, L_1)=frac{4a^2}{L^2}+frac{a^2}{2L^2}Big(csc ^2frac{pi (y-L_1)}{2L}+cosfrac{pi (y-3L_1)}{2L}-cosfrac{pi (y+L_1)}{2L}+cotfrac{pi L_1}{L}csc ^2frac{pi (y-L_1)}{2L}cscfrac{pi (y+L_1)}{2L}+csc ^2frac{pi (y+L_1)}{2L}Big)$
We can see that the energy jumps to infinity at $x=L_1$, so $F_1(L_1)$ contains the core energy and the constant energy $frac{4a^2}{L}$. Also, notice that the size of the core becomes very big, and I think it is due to the Gibbs phenomenon.
(2) I think $F_3(L_1)$ can be regarded as the interaction between the fattened core and the function $f$ on $[0,L_1)cup (L_1,L]$, so it should be thrown away.
I have posted more about it here https://figshare.com/articles/Some_examples_of_calculus_of_variations_of_discontinuous_functions_and_their_implications_pdf/7379861
answered Nov 25 '18 at 17:27
user573229user573229
5115
$endgroup$
add a comment |
$begingroup$
My attempted answer is:
$F(L_1)=sum_{k=1}^{infty}frac{(b_kkpi)^2}{2L}=F_1(L_1)+F_2(L_1)+F_3(L_1)$
where
$F_1(L_1)=frac{1}{2L}cdotsum_{k=1}^{infty}Big(4acosfrac{kpi L_1}{L}Big)^2$
$F_2(L_1)=frac{1}{2L}cdotsum_{k=1}^{infty}Big[frac{2aL}{kpi}Big(frac{1}{L-L_1}-frac{1}{L_1}Big)sinfrac{kpi L_1}{L}Big]^2=frac{a^2}{L_1}+frac{a^2}{L-L_1}-frac{4a^2}{L}$
$F_3(L_1)=frac{1}{L}cdotsum_{k=1}^{infty}Big(4acosfrac{kpi L_1}{L}Big)cdotBig[frac{2aL}{kpi}Big(frac{1}{L-L_1}-frac{1}{L_1}Big)sinfrac{kpi L_1}{L}Big]$
To determine the ground state, we may throw away $F_1(L_1)$ and $F_3(L_1)$, because:
(1) $F_1(L_1)$ contains the fictitious core energy, and we can see this by finding the energy density.
$F(y, L_1)=int_{0}^{y}(frac{mathrm{d}f}{mathrm{d}x})^2mathrm{d}x=F_1(y, L_1)+F_2(y, L_1)+F_3(y, L_1)$,
where $F(L,L_1)=F(L_1)$, $F_1(L,L_1)=F_1(L_1)$, $F_2(L,L_1)=F_2(L_1)$, and $F_3(L,L_1)=F_3(L_1)$.
After some calculations, we have
$F_{1}(y, L_1)
=frac{2a^2y}{L^2}Big(frac{cosfrac{2Npi L_1}{L}-cosfrac{2(N+1)pi L_1}{L}}{1-cosfrac{2pi L_1}{L}}-1Big)-frac{2a^2}{pi L}cdotfrac{sinfrac{pi y}{L}}{cosfrac{pi L_1}{L}-cosfrac{pi y}{L}}+frac{a^2}{pi L}cdotfrac{cosfrac{(2N+1)pi L_1}{L}}{sinfrac{pi L_1}{L}}lnBigg (frac{1-cosfrac{pi (y-L_1)}{L}}{1-cosfrac{pi (y+L_1)}{L}}Bigg)+frac{4a^2 y}{L^2}\ \ +begin{cases}
0+frac{2a^2y}{L^2}Big(1-frac{sinfrac{(2N+1)pi L_1}{L}}{sinfrac{pi L_1}{L}}Big), & text{if } 0<y<L_1\
frac{4(N-1)a^2}{L}+frac{2a^2(L-y)}{L^2}Big(-1+frac{sinfrac{(2N+1)pi L_1}{L}}{sinfrac{pi L_1}{L}}Big), & text{if } L_1<y<L
end{cases}\=-frac{2a^2}{pi L}cdotfrac{sinfrac{pi y}{L}}{cosfrac{pi L_1}{L}-cosfrac{pi y}{L}}+frac{a^2}{pi L}cdotfrac{cosfrac{pi L_1}{L}}{sinfrac{pi L_1}{L}}lnBigg (frac{1-cosfrac{pi (y-L_1)}{L}}{1-cosfrac{pi (y+L_1)}{L}}Bigg)+frac{4a^2 y}{L^2}+begin{cases}
0, & text{if } 0<y<L_1\
frac{4(N-1)a^2}{L}, & text{if } L_1<y<L
end{cases}$
where $Ntoinfty$. And I assume $Ncdotfrac{L_1}{L}in mathbb{Z}$, because that seems to make sure the core size for different $L_1$ and finite $N$ are the same (I don't know a better way to justify it). Now we can see $F_1(L,L_1)=F_1(L_1)$ is infinite but independent of $L_1$. Its energy density profile is
$mathcal{F}_1(y, L_1)=frac{4a^2}{L^2}+frac{a^2}{2L^2}Big(csc ^2frac{pi (y-L_1)}{2L}+cosfrac{pi (y-3L_1)}{2L}-cosfrac{pi (y+L_1)}{2L}+cotfrac{pi L_1}{L}csc ^2frac{pi (y-L_1)}{2L}cscfrac{pi (y+L_1)}{2L}+csc ^2frac{pi (y+L_1)}{2L}Big)$
We can see that the energy jumps to infinity at $x=L_1$, so $F_1(L_1)$ contains the core energy and the constant energy $frac{4a^2}{L}$. Also, notice that the size of the core becomes very big, and I think it is due to the Gibbs phenomenon.
(2) I think $F_3(L_1)$ can be regarded as the interaction between the fattened core and the function $f$ on $[0,L_1)cup (L_1,L]$, so it should be thrown away.
I have posted more about it here https://figshare.com/articles/Some_examples_of_calculus_of_variations_of_discontinuous_functions_and_their_implications_pdf/7379861
answered Nov 25 '18 at 17:27
user573229user573229
5115
$endgroup$
add a comment |
$begingroup$
My attempted answer is:
$F(L_1)=sum_{k=1}^{infty}frac{(b_kkpi)^2}{2L}=F_1(L_1)+F_2(L_1)+F_3(L_1)$
where
$F_1(L_1)=frac{1}{2L}cdotsum_{k=1}^{infty}Big(4acosfrac{kpi L_1}{L}Big)^2$
$F_2(L_1)=frac{1}{2L}cdotsum_{k=1}^{infty}Big[frac{2aL}{kpi}Big(frac{1}{L-L_1}-frac{1}{L_1}Big)sinfrac{kpi L_1}{L}Big]^2=frac{a^2}{L_1}+frac{a^2}{L-L_1}-frac{4a^2}{L}$
$F_3(L_1)=frac{1}{L}cdotsum_{k=1}^{infty}Big(4acosfrac{kpi L_1}{L}Big)cdotBig[frac{2aL}{kpi}Big(frac{1}{L-L_1}-frac{1}{L_1}Big)sinfrac{kpi L_1}{L}Big]$
To determine the ground state, we may throw away $F_1(L_1)$ and $F_3(L_1)$, because:
(1) $F_1(L_1)$ contains the fictitious core energy, and we can see this by finding the energy density.
$F(y, L_1)=int_{0}^{y}(frac{mathrm{d}f}{mathrm{d}x})^2mathrm{d}x=F_1(y, L_1)+F_2(y, L_1)+F_3(y, L_1)$,
where $F(L,L_1)=F(L_1)$, $F_1(L,L_1)=F_1(L_1)$, $F_2(L,L_1)=F_2(L_1)$, and $F_3(L,L_1)=F_3(L_1)$.
After some calculations, we have
$F_{1}(y, L_1)
=frac{2a^2y}{L^2}Big(frac{cosfrac{2Npi L_1}{L}-cosfrac{2(N+1)pi L_1}{L}}{1-cosfrac{2pi L_1}{L}}-1Big)-frac{2a^2}{pi L}cdotfrac{sinfrac{pi y}{L}}{cosfrac{pi L_1}{L}-cosfrac{pi y}{L}}+frac{a^2}{pi L}cdotfrac{cosfrac{(2N+1)pi L_1}{L}}{sinfrac{pi L_1}{L}}lnBigg (frac{1-cosfrac{pi (y-L_1)}{L}}{1-cosfrac{pi (y+L_1)}{L}}Bigg)+frac{4a^2 y}{L^2}\ \ +begin{cases}
0+frac{2a^2y}{L^2}Big(1-frac{sinfrac{(2N+1)pi L_1}{L}}{sinfrac{pi L_1}{L}}Big), & text{if } 0<y<L_1\
frac{4(N-1)a^2}{L}+frac{2a^2(L-y)}{L^2}Big(-1+frac{sinfrac{(2N+1)pi L_1}{L}}{sinfrac{pi L_1}{L}}Big), & text{if } L_1<y<L
end{cases}\=-frac{2a^2}{pi L}cdotfrac{sinfrac{pi y}{L}}{cosfrac{pi L_1}{L}-cosfrac{pi y}{L}}+frac{a^2}{pi L}cdotfrac{cosfrac{pi L_1}{L}}{sinfrac{pi L_1}{L}}lnBigg (frac{1-cosfrac{pi (y-L_1)}{L}}{1-cosfrac{pi (y+L_1)}{L}}Bigg)+frac{4a^2 y}{L^2}+begin{cases}
0, & text{if } 0<y<L_1\
frac{4(N-1)a^2}{L}, & text{if } L_1<y<L
end{cases}$
where $Ntoinfty$. And I assume $Ncdotfrac{L_1}{L}in mathbb{Z}$, because that seems to make sure the core size for different $L_1$ and finite $N$ are the same (I don't know a better way to justify it). Now we can see $F_1(L,L_1)=F_1(L_1)$ is infinite but independent of $L_1$. Its energy density profile is
$mathcal{F}_1(y, L_1)=frac{4a^2}{L^2}+frac{a^2}{2L^2}Big(csc ^2frac{pi (y-L_1)}{2L}+cosfrac{pi (y-3L_1)}{2L}-cosfrac{pi (y+L_1)}{2L}+cotfrac{pi L_1}{L}csc ^2frac{pi (y-L_1)}{2L}cscfrac{pi (y+L_1)}{2L}+csc ^2frac{pi (y+L_1)}{2L}Big)$
We can see that the energy jumps to infinity at $x=L_1$, so $F_1(L_1)$ contains the core energy and the constant energy $frac{4a^2}{L}$. Also, notice that the size of the core becomes very big, and I think it is due to the Gibbs phenomenon.
(2) I think $F_3(L_1)$ can be regarded as the interaction between the fattened core and the function $f$ on $[0,L_1)cup (L_1,L]$, so it should be thrown away.
I have posted more about it here https://figshare.com/articles/Some_examples_of_calculus_of_variations_of_discontinuous_functions_and_their_implications_pdf/7379861
answered Nov 25 '18 at 17:27
user573229user573229
5115
$endgroup$
My attempted answer is:
$F(L_1)=sum_{k=1}^{infty}frac{(b_kkpi)^2}{2L}=F_1(L_1)+F_2(L_1)+F_3(L_1)$
where
$F_1(L_1)=frac{1}{2L}cdotsum_{k=1}^{infty}Big(4acosfrac{kpi L_1}{L}Big)^2$
$F_2(L_1)=frac{1}{2L}cdotsum_{k=1}^{infty}Big[frac{2aL}{kpi}Big(frac{1}{L-L_1}-frac{1}{L_1}Big)sinfrac{kpi L_1}{L}Big]^2=frac{a^2}{L_1}+frac{a^2}{L-L_1}-frac{4a^2}{L}$
$F_3(L_1)=frac{1}{L}cdotsum_{k=1}^{infty}Big(4acosfrac{kpi L_1}{L}Big)cdotBig[frac{2aL}{kpi}Big(frac{1}{L-L_1}-frac{1}{L_1}Big)sinfrac{kpi L_1}{L}Big]$
To determine the ground state, we may throw away $F_1(L_1)$ and $F_3(L_1)$, because:
(1) $F_1(L_1)$ contains the fictitious core energy, and we can see this by finding the energy density.
$F(y, L_1)=int_{0}^{y}(frac{mathrm{d}f}{mathrm{d}x})^2mathrm{d}x=F_1(y, L_1)+F_2(y, L_1)+F_3(y, L_1)$,
where $F(L,L_1)=F(L_1)$, $F_1(L,L_1)=F_1(L_1)$, $F_2(L,L_1)=F_2(L_1)$, and $F_3(L,L_1)=F_3(L_1)$.
After some calculations, we have
$F_{1}(y, L_1)
=frac{2a^2y}{L^2}Big(frac{cosfrac{2Npi L_1}{L}-cosfrac{2(N+1)pi L_1}{L}}{1-cosfrac{2pi L_1}{L}}-1Big)-frac{2a^2}{pi L}cdotfrac{sinfrac{pi y}{L}}{cosfrac{pi L_1}{L}-cosfrac{pi y}{L}}+frac{a^2}{pi L}cdotfrac{cosfrac{(2N+1)pi L_1}{L}}{sinfrac{pi L_1}{L}}lnBigg (frac{1-cosfrac{pi (y-L_1)}{L}}{1-cosfrac{pi (y+L_1)}{L}}Bigg)+frac{4a^2 y}{L^2}\ \ +begin{cases}
0+frac{2a^2y}{L^2}Big(1-frac{sinfrac{(2N+1)pi L_1}{L}}{sinfrac{pi L_1}{L}}Big), & text{if } 0<y<L_1\
frac{4(N-1)a^2}{L}+frac{2a^2(L-y)}{L^2}Big(-1+frac{sinfrac{(2N+1)pi L_1}{L}}{sinfrac{pi L_1}{L}}Big), & text{if } L_1<y<L
end{cases}\=-frac{2a^2}{pi L}cdotfrac{sinfrac{pi y}{L}}{cosfrac{pi L_1}{L}-cosfrac{pi y}{L}}+frac{a^2}{pi L}cdotfrac{cosfrac{pi L_1}{L}}{sinfrac{pi L_1}{L}}lnBigg (frac{1-cosfrac{pi (y-L_1)}{L}}{1-cosfrac{pi (y+L_1)}{L}}Bigg)+frac{4a^2 y}{L^2}+begin{cases}
0, & text{if } 0<y<L_1\
frac{4(N-1)a^2}{L}, & text{if } L_1<y<L
end{cases}$
where $Ntoinfty$. And I assume $Ncdotfrac{L_1}{L}in mathbb{Z}$, because that seems to make sure the core size for different $L_1$ and finite $N$ are the same (I don't know a better way to justify it). Now we can see $F_1(L,L_1)=F_1(L_1)$ is infinite but independent of $L_1$. Its energy density profile is
$mathcal{F}_1(y, L_1)=frac{4a^2}{L^2}+frac{a^2}{2L^2}Big(csc ^2frac{pi (y-L_1)}{2L}+cosfrac{pi (y-3L_1)}{2L}-cosfrac{pi (y+L_1)}{2L}+cotfrac{pi L_1}{L}csc ^2frac{pi (y-L_1)}{2L}cscfrac{pi (y+L_1)}{2L}+csc ^2frac{pi (y+L_1)}{2L}Big)$
We can see that the energy jumps to infinity at $x=L_1$, so $F_1(L_1)$ contains the core energy and the constant energy $frac{4a^2}{L}$. Also, notice that the size of the core becomes very big, and I think it is due to the Gibbs phenomenon.
(2) I think $F_3(L_1)$ can be regarded as the interaction between the fattened core and the function $f$ on $[0,L_1)cup (L_1,L]$, so it should be thrown away.
I have posted more about it here https://figshare.com/articles/Some_examples_of_calculus_of_variations_of_discontinuous_functions_and_their_implications_pdf/7379861
answered Nov 25 '18 at 17:27
user573229user573229
5115
edited Nov 26 '18 at 19:55
answered Nov 25 '18 at 17:27
user573229user573229
5115
answered Nov 25 '18 at 17:27
user573229user573229
5115
answered Nov 25 '18 at 17:27
user573229user573229
5115
5115
add a comment |
add a comment |
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Required, but never shown
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It seems to me that you should get the right result if you consider the mollified problem with $f(0)=0,f(L_1-delta)=-a,f(L_1+delta)=a,f(L)=0$ and then ultimately send $delta to 0$. Euler-Lagrange cannot be derived rigorously on the whole interval in this situation, but it can be in the mollified problem (on each of the three subdomains, of course). The only weird thing here is that you are basically looking at the $L^2$ norm of a non-$L^2$ distribution, which is a somewhat weird thing to do in the first place.
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– Ian
Nov 25 '18 at 19:49
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Do you assume $f$ is continuous on $(L_1-delta, L_1+delta)$? I'm not quite sure of how to do it in this way. But I think this problem can be done in separate regions. Then I'm wondering if there is a way to "glue" them. I agree that it is weird.
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– user573229
Nov 25 '18 at 19:55
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Yes, the point is to loosely take into account the jump discontinuity and how it should affect the $L^2$ norm of the derivative, even though the derivative in the real problem isn't $L^2$.
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– Ian
Nov 25 '18 at 21:01