Determine whether ## S[y] ## has a maximum or a minimum

  • #1
Math100
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Homework Statement
Let ## a, b ## and ## w ## be constants, with ## w\neq 0 ##.
a) Show that the Euler-Lagrange equation for the functional ## S[y]=\int_{1}^{2}(y'^2+w^2y^2+2y(a \sin(wx)+b \sinh(wx)))dx, y(1)=0, y(2)=1 ##, is ## y''-w^2y=a \sin(wx)+b \sinh(wx), y(1)=0, y(2)=1 ##.
b) Solve this equation to show that the stationary path is ## y=\alpha\sinh(w(x-1))+\beta\sinh(w(x-2))-\frac{a \sin(wx)}{2w^2}+\frac{bx \cosh(wx)}{2w} ##, where ## \alpha=\frac{1}{\sinh(w)}[1+\frac{a \sin(2w)}{2w^2}-\frac{b \cosh(2w)}{2}], \beta=\frac{1}{\sinh(w)}[\frac{b\cosh(w)}{2w}-\frac{a \sin(w)}{2w^2}] ##.
c) Determine whether ## S[y] ## has a maximum or a minimum on this stationary path, and whether it is global or local.
Relevant Equations
Euler-Lagrange equation:
For the functional ## S[y]=\int_{a}^{b}F(x, y, y')dx, y(a)=A, y(b)=B ##, the Euler-Lagrange equation is ## \frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0, y(a)=A, y(b)=B ##.
a) The Euler-Lagrange equation is of the form ## \frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0, y(a)=A, y(b)=B ##.
Let ## F(x, y, y')=(y'^2+w^2y^2+2y(a \sin(wx)+b \sinh(wx))) ##.
Then ## \frac{\partial F}{\partial y'}=2y' ## and ## \frac{\partial F}{\partial y}=2w^2y+2(a \sin(wx)+b \sinh(wx)) ##.
Note that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0\implies 2\frac{d^2y}{dx^2}-2w^2y-2(a \sin(wx)+b \sinh(wx))=0 ##, so we have ## \frac{d^2y}{dx^2}-w^2y=a \sin(wx)+b \sinh(wx) ##.
Therefore, the Euler-Lagrange equation is ## y''-w^2y=a \sin(wx)+b \sinh(wx), y(1)=0, y(2)=1 ##.

b) Consider the differential equation ## y''-w^2y=a \sin(wx)+b \sinh(wx), y(1)=0, y(2)=1 ##.
The characteristic equation is ## r^2-w^2=0\implies y_{h}=c_{1}e^{x}+c_{2}e^{-x} ##.
Let the particular solution be of the form ## y_{p}=A \sin(wx)+Bx \cosh(wx) ##.
Then ## y_{p}'=Aw \cos(wx)+Bwx \sinh(wx)+B \cosh(wx) ## and ## y_{p}''=-Aw^2 \sin(wx)+Bw^2x \cosh(wx)+Bw \sinh(wx)+Bw \sinh(wx) ##.
Substituting ## y_{p}'', y_{p} ## into our original differential equation produces:
## -Aw^2 \sin(wx)+Bw^2x \cosh(wx)+Bw \sinh(wx)+Bw \sinh(wx)-w^2[A \sin(wx)+Bx \cosh(wx)]=a \sin(wx)+b \sinh(wx) ##.
Now we have ## -2Aw^2 \sin(wx)=a \sin(wx) ## and ## 2Bw \sinh(wx)=b \sinh(wx) ##, so ## A=\frac{a}{-2w^2}, B=\frac{b}{2w} ##.
Thus ## y=c_{1}e^{x}+c_{2}e^{-x}-\frac{a \sin(wx)}{2w^2}+\frac{bx \cosh(wx)}{2w} ##.
By our boundary conditions, we have ## y(1)=0\implies 0=c_{1}e+c_{2}e\implies c_{1}=-c_{2} ## and ## y(2)=1\implies 1=c_{1}e^{2}+c_{2}e^{-2}-\frac{a \sin(2w)}{2w^2}+\frac{2b \cosh(2w)}{2w} ##.
Observe that ## c_{2}(-e^{2}+e^{-2})-\frac{a \sin(2w)}{2w^2}+\frac{b(e^{2w}+e^{-2w})}{2w}=1 ##.

From here, I cannot see why does the stationary path is ## y=\alpha\sinh(w(x-1))+\beta\sinh(w(x-2))-\frac{a \sin(wx)}{2w^2}+\frac{bx\cosh(wx)}{2w} ##, even with the given forms of both ## \alpha ## and ## \beta ##. I know that the hyperbolic functions of ## \sinh(x), \cosh(x) ## are ## \sinh(x)=\frac{e^{x}-e^{-x}}{2} ## and ## \cosh(x)=\frac{e^{x}+e^{-x}}{2} ## but I still don't see how this gives our desired answer.

c) Does anyone know how to find the second variation? If the second variation is positive, then the given functional ## S[y] ## has a minimum on the stationary path and if the second variation is negative, then the given functional ## S[y] ## has a maximum on the stationary path.
 
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  • #2
On bounded domains, [itex]\cosh wx[/itex] and [itex]\sinh wx[/itex] are linearly independent solutions of [itex]y'' - w^2 y = 0[/itex] which are easier to work with than [itex]e^{\pm wx}[/itex]. You should know the basic identities [tex]\begin{split}
\sinh (u \pm v) &\equiv \sinh u \cosh v \pm \sinh v \cosh u \\
\cosh (u \pm v) &\equiv \cosh u \cosh v \mp \sinh u \sinh v \end{split}
[/tex] The point here is that [itex]\sinh (w(x- 1))[/itex] and [itex]\sinh (w(x-2))[/itex] are linearly independent solutions of [itex]y'' - w^2 y = 0[/itex] which make it easy to apply boundary conditions on [itex]y(1)[/itex] and [itex]y(2)[/itex], since if [tex]
y(x) = \alpha \sinh (w(x - 1)) + \beta \sinh (w(x - 2)) + y_p(x)[/tex] then [tex]\begin{split}
y(1) &= - \beta \sinh w + y_p(1) \\
y(2) &= \alpha \sinh w + y_p(2). \end{split}[/tex]
 
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  • #3
Math100 said:
c) Does anyone know how to find the second variation? If the second variation is positive, then the given functional ## S[y] ## has a minimum on the stationary path and if the second variation is negative, then the given functional ## S[y] ## has a maximum on the stationary path.

You have jumped straight to the E-L formula. The text does not want you to do that; that is why it is asking questions which cannot be answered that way, but which are easily answered if you derive the E-L equation from the Gateaux derivative for the specific case, and in the course of so doing calculate [itex]S[y + \epsilon h] - S[y].[/itex]

Try that now. Write [itex]S[y] = \int_1^2 y'^2 + w^2y^2 + 2yf(x)\,dx.[/itex] Then calculate [itex]S[y + \epsilon h] - S[y][/itex], remembering to integrate [itex]h'y' = (y'h)' - y''h[/itex] by parts. You know that the boundary term disappears because [itex]h[/itex] must satisfy [itex]h(1) = h(2) = 0[/itex], and that on the stationarry path [itex]-y'' + w^2 y + f(x) = 0[/itex]. What's left? Is it necessarily positive or negative?
 
  • #4
pasmith said:
On bounded domains, [itex]\cosh wx[/itex] and [itex]\sinh wx[/itex] are linearly independent solutions of [itex]y'' - w^2 y = 0[/itex] which are easier to work with than [itex]e^{\pm wx}[/itex]. You should know the basic identities [tex]\begin{split}
\sinh (u \pm v) &\equiv \sinh u \cosh v \pm \sinh v \cosh u \\
\cosh (u \pm v) &\equiv \cosh u \cosh v \mp \sinh u \sinh v \end{split}
[/tex] The point here is that [itex]\sinh (w(x- 1))[/itex] and [itex]\sinh (w(x-2))[/itex] are linearly independent solutions of [itex]y'' - w^2 y = 0[/itex] which make it easy to apply boundary conditions on [itex]y(1)[/itex] and [itex]y(2)[/itex], since if [tex]
y(x) = \alpha \sinh (w(x - 1)) + \beta \sinh (w(x - 2)) + y_p(x)[/tex] then [tex]\begin{split}
y(1) &= - \beta \sinh w + y_p(1) \\
y(2) &= \alpha \sinh w + y_p(2). \end{split}[/tex]
I didn't know about this before. It's been several years since I took the course "Ordinary Differential Equations" and since then, I never worked with hyperbolic functions so these are all new to me. But thank you for telling me all these identities.
 
  • #5
pasmith said:
You have jumped straight to the E-L formula. The text does not want you to do that; that is why it is asking questions which cannot be answered that way, but which are easily answered if you derive the E-L equation from the Gateaux derivative for the specific case, and in the course of so doing calculate [itex]S[y + \epsilon h] - S[y].[/itex]

Try that now. Write [itex]S[y] = \int_1^2 y'^2 + w^2y^2 + 2yf(x)\,dx.[/itex] Then calculate [itex]S[y + \epsilon h] - S[y][/itex], remembering to integrate [itex]h'y' = (y'h)' - y''h[/itex] by parts. You know that the boundary term disappears because [itex]h[/itex] must satisfy [itex]h(1) = h(2) = 0[/itex], and that on the stationarry path [itex]-y'' + w^2 y + f(x) = 0[/itex]. What's left? Is it necessarily positive or negative?
I got the desired answer for part b). However, for part c), I found out that ## S[y+\epsilon h]-S[y]=\int_{1}^{2}(2\epsilon y'h'+\epsilon^2h'^2-2\epsilon hw^2y-\epsilon^2w^2h^2+2\epsilon h''y+2\epsilon hy''+2\epsilon^2hh'')dx ## and I factored out the ##\epsilon##, to get ## \epsilon\int_{1}^{2}(2y'h'+\epsilon h'^2-2w^2yh-\epsilon w^2h^2+2yh''+2y''h+2\epsilon hh'')dx ##. From here, how to integrate each term inside this integrand? I know that I have to use the method of integration by parts, but I don't know what to get after I integrate. For example, the first term I got is ## \int_{1}^{2}2y'h'dx=[y'h]_{1}^{2}-\int_{1}^{2}2y''hdx=y'h(2)-y'h(1)-[2y''h(2)-2y''h(1)]=0 ## since ## h(1)=h(2)=0 ## but is this correct?
 
  • #6
If [itex]y_c[/itex] is the stationary path, then any other path [itex]y[/itex] can be expressed as [itex]y = y_c + (y -y_c) = y_c + h[/itex] where [itex]h(1) = h(2) = 0[/itex]. Then [tex]\begin{split}
S[y] &= S[y_c + h] \\
&= \int_1^2 (y_c' + h')^2 + w^2(y_c + h)^2 + 2(y_c + h)f(x)\,dx \\
&= S[y_c] + \int_1^2 2y_c'h' + h'^2 + 2w^2 y_ch + w^2h^2 + 2h f(x)\,dx\\
&= S[y_c] + \underbrace{[2hy_c']_1^2}_{\mbox{$= 0$}} + \int_1^2 -2y_c''h +h'^2 + 2w^2 y_ch + w^2h^2 + 2h f(x)\,dx \\
&= S[y_c] + \int_1^2 2h (-y_c'' + w^2 y_c + f(x)) + h'^2 + w^2h^2\,dx.\end{split}
[/tex] We already know that [itex]-y''_c + w^2 y_c + f(x) = 0[/itex], so we are left with [tex]
S[y] = S[y_c] + \int_1^2 h'^2 + w^2 h^2\,dx.[/tex] What can we conclude about the sign of the integand for arbitrary [itex]h[/itex]? What follows from this?
 
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  • #7
pasmith said:
If [itex]y_c[/itex] is the stationary path, then any other path [itex]y[/itex] can be expressed as [itex]y = y_c + (y -y_c) = y_c + h[/itex] where [itex]h(1) = h(2) = 0[/itex]. Then [tex]\begin{split}
S[y] &= S[y_c + h] \\
&= \int_1^2 (y_c' + h')^2 + w^2(y_c + h)^2 + 2(y_c + h)f(x)\,dx \\
&= S[y_c] + \int_1^2 2y_c'h' + h'^2 + 2w^2 y_ch + w^2h^2 + 2h f(x)\,dx\\
&= S[y_c] + \underbrace{[2hy_c']_1^2}_{\mbox{$= 0$}} + \int_1^2 -2y_c''h +h'^2 + 2w^2 y_ch + w^2h^2 + 2h f(x)\,dx \\
&= S[y_c] + \int_1^2 2h (-y_c'' + w^2 y_c + f(x)) + h'^2 + w^2h^2\,dx.\end{split}
[/tex] We already know that [itex]-y''_c + w^2 y_c + f(x) = 0[/itex], so we are left with [tex]
S[y] = S[y_c] + \int_1^2 h'^2 + w^2 h^2\,dx.[/tex] What can we conclude about the sign of the integand for arbitrary [itex]h[/itex]? What follows from this?
I got ## \int_{1}^{2}(h'^2+w^2h^2)dx=(h(2))^2-(h(1))^2+w^2(\frac{(h(2))^3}{3}-\frac{(h(1))^3}{3})=0 ## but is this correct?
 
  • #8
Math100 said:
I got ## \int_{1}^{2}(h'^2+w^2h^2)dx=(h(2))^2-(h(1))^2+w^2(\frac{(h(2))^3}{3}-\frac{(h(1))^3}{3})=0 ## but is this correct?
No this is not correct.
@pasmith asked about the sign of the integrand (what's inside the integral), i.e. ##h'^{2}+w^2 h^2##. What is the sign of that expression, and what can you conclude about its integral?
 
  • #9
renormalize said:
No this is not correct.
@pasmith asked about the sign of the integrand (what's inside the integral), i.e. ##h'^{2}+w^2 h^2##. What is the sign of that expression, and what can you conclude about its integral?
The sign of the integrand is obviously positive, but what can we conclude about this integral then?
 
  • #10
Math100 said:
The sign of the integrand is obviously positive, but what can we conclude about this integral then?
The integral can be defined as the limit of a finite Riemann sum. If you sum a sequence of all positive numbers, what sign will that sum have? What can you conclude about the sign of the limiting integral?
 
  • #11
renormalize said:
The integral can be defined as the limit of a finite Riemann sum. If you sum a sequence of all positive numbers, what sign will that sum have? What can you conclude about the sign of the limiting integral?
That sum will have a positive sign. And the sign of the limiting integral is also positive.
 
  • #12
Math100 said:
That sum will have a positive sign. And the sign of the limiting integral is also positive.
OK, so that means:$$\int_{1}^{2}\left(h'^{2}+w^{2}h^{2}\right)\,dx\geq0$$(where I've used a ##\geq## sign to include the edge case ##h\equiv0##). You can therefore rewrite the last equation in post #6 by @pasmith to read:$$S[y]-S[y_{c}]=\int_{1}^{2}\left(h'^{2}+w^{2}h^{2}\right)\,dx\geq0$$or:$$S[y]\geq S[y_{c}]$$To answer part c) of your OP, what do you finally conclude about ##S[y_{c}]## in comparison to all other possible values of the action ##S##?
 
  • #13
renormalize said:
OK, so that means:$$\int_{1}^{2}\left(h'^{2}+w^{2}h^{2}\right)\,dx\geq0$$(where I've used a ##\geq## sign to include the edge case ##h\equiv0##). You can therefore rewrite the last equation in post #6 by @pasmith to read:$$S[y]-S[y_{c}]=\int_{1}^{2}\left(h'^{2}+w^{2}h^{2}\right)\,dx\geq0$$or:$$S[y]\geq S[y_{c}]$$To answer part c) of your OP, what do you finally conclude about ##S[y_{c}]## in comparison to all other possible values of the action ##S##?
That the stationary path ## y ## is a local maximum.
 
  • #14
Math100 said:
That the stationary path ## y ## is a local maximum.
The problem is asking about the the stationary action (i.e., the value of ##S[y_c]##), not anything about the stationary path ##y_c## itself: "c) Determine whether ##S[y]## has a maximum or a minimum on this stationary path, and whether it is global or local." Is it's value greater or lesser than all other action values ##S[y]##? And is this inequality true only for paths sufficiently near ##y_c## (i.e., locally) or for all paths that deviate from the stationary path, no matter how far (i.e., globally)?
 
  • #15
renormalize said:
The problem is asking about the the stationary action (i.e., the value of ##S[y_c]##), not anything about the stationary path ##y_c## itself: "c) Determine whether ##S[y]## has a maximum or a minimum on this stationary path, and whether it is global or local." Is it's value greater or lesser than all other action values ##S[y]##? And is this inequality true only for paths sufficiently near ##y_c## (i.e., locally) or for all paths that deviate from the stationary path, no matter how far (i.e., globally)?
Its value is less than all other action values ##S[y]##. But how do we know if this inequality is true only for paths sufficiently near ##y_c## or for all paths that deviate from the stationary path, no matter how far?
 
  • #16
Math100 said:
Its value is less than all other action values ##S[y]##. But how do we know if this inequality is true only for paths sufficiently near ##y_c## or for all paths that deviate from the stationary path, no matter how far?

In this case, the expression [tex]
S[y_c + \epsilon h] = S[y_c] + \epsilon^2 \int_1^2 w^2h^2 + h'^2\,dx[/tex] is exact; there are no terms of order [itex]\epsilon^3[/itex] or higher which we neglected in deriving it. What can you conclude?
 
  • #17
pasmith said:
In this case, the expression [tex]
S[y_c + \epsilon h] = S[y_c] + \epsilon^2 \int_1^2 w^2h^2 + h'^2\,dx[/tex] is exact; there are no terms of order [itex]\epsilon^3[/itex] or higher which we neglected in deriving it. What can you conclude?
So ## S[y_{c}+\epsilon h]-S[y_{c}]=\epsilon^2\int_{1}^{2}(w^2h^2+h'^2)dx ##
## S[y_{c}+\epsilon h]-S[y_{c}]=\epsilon^2\int_{1}^{2}(w^2h^2+h'^2)dx\geq 0 ##
## S[y_{c}+\epsilon h]\geq S[y_{c}] ##
How is this different from the one we obtained above? I still don't know what we can conclude about this result.
 
  • #18
Math100 said:
So ## S[y_{c}+\epsilon h]-S[y_{c}]=\epsilon^2\int_{1}^{2}(w^2h^2+h'^2)dx ##
## S[y_{c}+\epsilon h]-S[y_{c}]=\epsilon^2\int_{1}^{2}(w^2h^2+h'^2)dx\geq 0 ##
## S[y_{c}+\epsilon h]\geq S[y_{c}] ##
How is this different from the one we obtained above? I still don't know what we can conclude about this result.
To derive this inequality, were you forced to put any limitation at all on ##\epsilon## (like ##\epsilon \ll 1##)? If not, then ##\epsilon## is completely arbitrary, which says that the inequality remain true, even when the difference between ##y## and ##y_c## is arbitrarily large. What do you conclude?
 
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  • #19
renormalize said:
To derive this inequality, were you forced to put any limitation at all on ##\epsilon## (like ##\epsilon \ll 1##)? If not, then ##\epsilon## is completely arbitrary, which says that the inequality remain true, even when the difference between ##y## and ##y_c## is arbitrarily large. What do you conclude?
That the value of ## S[y_{c}] ## has a global minimum.
 
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