How should I find the nontrivial stationary paths?

[Math100]

Homework Statement

Consider the functional $S[y]=\alpha y(1)^2+\int_{0}^{1}\beta y'^2dx, y(0)=0$, with a natural boundary condition at $x=1$ and subject to the constraint $C[y]=\gamma y(1)^2+\int_{0}^{1}w(x)y^2dx=1$, where $\alpha, \beta$ and $\gamma$ are nonzero constants.
a) Show that the stationary paths of this system satisfy the Euler-Lagrange equation $\beta\frac{d^2y}{dx^2}+\lambda w(x)y=0, y(0)=0, (\alpha-\gamma\lambda)y(1)+\beta y'(1)=0$, where $\lambda$ is a Lagrange multiplier.
b) Let $w(x)=1$ and $\alpha=\beta=\gamma=1$. Find the nontrivial stationary paths, stating clearly the eigenfunctions $y$ (normalized so that $C[y]=1$) and the values of the associated Lagrange multiplier.

Relevant Equations

None.

a) Proof:
Let $\lambda$ be the Lagrange multiplier.
Then the auxiliary functional is $\overline{S}[y]=\alpha y(1)^2+\int_{0}^{1}\beta y'^2dx-\lambda (\gamma y(1)^2+\int_{0}^{1}w(x)y^2dx-1)$.
This gives $\overline{S}[y+\epsilon h]=\alpha (y(1)+\epsilon h(1))^2+\int_{0}^{1}\beta (y'+\epsilon h')^2dx-\lambda (\gamma(y(1)+\epsilon h(1))^2+\int_{0}^{1}w(x)(y+\epsilon h)^2dx-1)$, where $y+\epsilon h$ is an admissible perturbation, so that $h(0)=0$.
Note that the Gateaux differential $\triangle\overline{S}[y, h]$ is given by $\frac{d}{d\epsilon}\overline{S}[y+\epsilon h]\vert_{\epsilon=0}$.
Thus $\frac{d}{d\epsilon}\overline{S}[y+\epsilon h]\vert_{\epsilon=0}=2\alpha y(1)h(1)+2\int_{0}^{1}\beta y'h'dx-2\lambda (\gamma y(1)h(1)+\int_{0}^{1}wyhdx)$.

From here, how should I show that the stationary paths of this system satisfy the given Euler-Lagrange equation?

b) Let $w(x)=1$ and $\alpha=\beta=\gamma=1$.
Consider the Euler-Lagrange equation $\beta\frac{d^2y}{dx^2}+\lambda w(x)y=0, y(0)=0, (\alpha-\gamma\lambda)y(1)+\beta y'(1)=0$, where $\lambda$ is a Lagrange multiplier.
Then we have $\frac{d^2y}{dx^2}+\lambda y=0, y(0)=0, (1-\lambda)y(1)+y'(1)=0$, where $\lambda$ is a Lagrange multiplier.
This gives $y=c_{1}sin(\sqrt{\lambda}x)+c_{2}cos(\sqrt{\lambda}x)$.

From here, how should I find the nontrivial stationary paths?

 

[pasmith]

Homework Statement: Consider the functional $S[y]=\alpha y(1)^2+\int_{0}^{1}\beta y'^2dx, y(0)=0$, with a natural boundary condition at $x=1$ and subject to the constraint $C[y]=\gamma y(1)^2+\int_{0}^{1}w(x)y^2dx=1$, where $\alpha, \beta$ and $\gamma$ are nonzero constants.
a) Show that the stationary paths of this system satisfy the Euler-Lagrange equation $\beta\frac{d^2y}{dx^2}+\lambda w(x)y=0, y(0)=0, (\alpha-\gamma\lambda)y(1)+\beta y'(1)=0$, where $\lambda$ is a Lagrange multiplier.
b) Let $w(x)=1$ and $\alpha=\beta=\gamma=1$. Find the nontrivial stationary paths, stating clearly the eigenfunctions $y$ (normalized so that $C[y]=1$) and the values of the associated Lagrange multiplier.
Relevant Equations: None.

a) Proof:
Let $\lambda$ be the Lagrange multiplier.
Then the auxiliary functional is $\overline{S}[y]=\alpha y(1)^2+\int_{0}^{1}\beta y'^2dx-\lambda (\gamma y(1)^2+\int_{0}^{1}w(x)y^2dx-1)$.
This gives $\overline{S}[y+\epsilon h]=\alpha (y(1)+\epsilon h(1))^2+\int_{0}^{1}\beta (y'+\epsilon h')^2dx-\lambda (\gamma(y(1)+\epsilon h(1))^2+\int_{0}^{1}w(x)(y+\epsilon h)^2dx-1)$, where $y+\epsilon h$ is an admissible perturbation, so that $h(0)=0$.
Note that the Gateaux differential $\triangle\overline{S}[y, h]$ is given by $\frac{d}{d\epsilon}\overline{S}[y+\epsilon h]\vert_{\epsilon=0}$.
Thus $\frac{d}{d\epsilon}\overline{S}[y+\epsilon h]\vert_{\epsilon=0}=2\alpha y(1)h(1)+2\int_{0}^{1}\beta y'h'dx-2\lambda (\gamma y(1)h(1)+\int_{0}^{1}wyhdx)$.

From here, how should I show that the stationary paths of this system satisfy the given Euler-Lagrange equation?

Assuming your work is correct, you have $$(\alpha - \gamma\lambda)y(1)h(1) + \int_0^1 \beta y'h' - \lambda w y h \,dx = 0.$$ What is the next step in all of these problems? Integrate $y'h'$ by parts.

b) Let $w(x)=1$ and $\alpha=\beta=\gamma=1$.
Consider the Euler-Lagrange equation $\beta\frac{d^2y}{dx^2}+\lambda w(x)y=0, y(0)=0, (\alpha-\gamma\lambda)y(1)+\beta y'(1)=0$, where $\lambda$ is a Lagrange multiplier.
Then we have $\frac{d^2y}{dx^2}+\lambda y=0, y(0)=0, (1-\lambda)y(1)+y'(1)=0$, where $\lambda$ is a Lagrange multiplier.
This gives $y=c_{1}sin(\sqrt{\lambda}x)+c_{2}cos(\sqrt{\lambda}x)$.

This assumes $\lambda \neq 0$. What happens if $\lambda = 0$? Do you get a non-zero solution for $y$?

For $\lambda = k^2 > 0$, you know from the condition $y(0) = 0$ that $c_2 = 0$. That leaves you with the condition at $y(1)$, which takes the form $$c_1f(k) = 0.$$ We know that $c_1 \neq 0$, so that requires $f(k) = 0$. The condition $C[c_1\sin kx] = 1$ then gives you $c_1$ in terms of $k$.

 

[Math100]

Assuming your work is correct, you have $$(\alpha - \gamma\lambda)y(1)h(1) + \int_0^1 \beta y'h' - \lambda w y h \,dx = 0.$$ What is the next step in all of these problems? Integrate $y'h'$ by parts.



This assumes $\lambda \neq 0$. What happens if $\lambda = 0$? Do you get a non-zero solution for $y$?

For $\lambda = k^2 > 0$, you know from the condition $y(0) = 0$ that $c_2 = 0$. That leaves you with the condition at $y(1)$, which takes the form $$c_1f(k) = 0.$$ We know that $c_1 \neq 0$, so that requires $f(k) = 0$. The condition $C[c_1\sin kx] = 1$ then gives you $c_1$ in terms of $k$.

So for part b), I've got $y=Asin(\sqrt{\lambda}x)+Bcos(\sqrt{\lambda}x)$, where $A, B$ are constants. The condition $y(0)=0$ gives $B=0$ and the boundary condition at $x=1$ gives $y(1)=0\implies 0=Asin(\sqrt{\lambda})\implies sin(\sqrt{\lambda})=0$ since $A\neq 0$. This means $\sqrt{\lambda}=n\pi\implies y=Asin(n\pi x)$.
Thus, the constraint gives $1=\int_{0}^{1}[Asin(n\pi x)]^2dx\implies 1=A^2\int_{0}^{1}sin^2(n\pi x)dx\implies A=\sqrt{2}$.
Hence, $y=\sqrt{2}sin(\sqrt{\lambda x})$.
Is this the correct stationary path?

 

[pasmith]

Try again. The condition at $x = 1$ is $$(1 - \lambda)y(1) + y'(1) = 0.$$

 

[Math100]

Try again. The condition at $x = 1$ is $$(1 - \lambda)y(1) + y'(1) = 0.$$

I still don't get this one. How does $(1-\lambda)y(1)+y'(1)=0$ determine our another constant $A$?

 

[pasmith]

You are dealing with an eigenvalue problem. The condition at 1 tells you that either $A = 0$, which is the trivial solution, or else $\lambda$ must satisfy a certain condition. Then the constraint $C[y] = 1$ determines $A$.

 

[Math100]

You are dealing with an eigenvalue problem. The condition at 1 tells you that either $A = 0$, which is the trivial solution, or else $\lambda$ must satisfy a certain condition. Then the constraint $C[y] = 1$ determines $A$.

How can $\lambda$ satisfy a certian condition? And how to find the constant $A$?

 

[Math100]

You are dealing with an eigenvalue problem. The condition at 1 tells you that either $A = 0$, which is the trivial solution, or else $\lambda$ must satisfy a certain condition. Then the constraint $C[y] = 1$ determines $A$.

Since the constant $B=0$, we have $y=Asin(\sqrt{\lambda}x)$ and using the boundary condition at $x=1$ gives $(1-\lambda)y(1)+y'(1)=0\implies (1-\lambda)Asin(\sqrt{\lambda})+A\sqrt{\lambda}cos(\sqrt{\lambda})=0$. But then this means $sin(\sqrt{\lambda})=0\implies \sqrt{\lambda}=n\pi$ for $n\neq 0$ and $cos(\sqrt{\lambda})=0\implies \sqrt{\lambda}=(n+\frac{1}{2})\pi$ for some $n\in\mathbb{Z}$. What's wrong in here?

 

[renormalize]

Since the constant $B=0$, we have $y=Asin(\sqrt{\lambda}x)$ and using the boundary condition at $x=1$ gives $(1-\lambda)y(1)+y'(1)=0\implies (1-\lambda)Asin(\sqrt{\lambda})+A\sqrt{\lambda}cos(\sqrt{\lambda})=0$. But then this means $sin(\sqrt{\lambda})=0\implies \sqrt{\lambda}=n\pi$ for $n\neq 0$ and $cos(\sqrt{\lambda})=0\implies \sqrt{\lambda}=(n+\frac{1}{2})\pi$ for some $n\in\mathbb{Z}$. What's wrong in here?

What's wrong is your assumption that the $\sin$ and $\cos$ terms must vanish individually. Following the suggestion of @pasmith, set $\lambda=k^2$ and write your boundary (eigenvalue) condition as $\left(1-k^{2}\right)\sin k+k\cos k=0$. Beyond the the trivial solution $k=0$, a plot of the function on the left-side suggests that the condition has an infinity of roots, the first few of which are (using Mathematica): $k_1=1.20779,k_2=3.44824,k_3=6.44095,k_4=9.53048,k_5=12.6458$. The squares of these are the first five allowed values of the Lagrange multiplier $\lambda$ in your variational problem part b). All that remains to do now is to plug your eigensolutions $y_\lambda (x)$ into $C[y_\lambda]=1$ and calculate the normalization factors $A_\lambda$.

 

[Math100]

What's wrong is your assumption that the $\sin$ and $\cos$ terms must vanish individually. Following the suggestion of @pasmith, set $\lambda=k^2$ and write your boundary (eigenvalue) condition as $\left(1-k^{2}\right)\sin k+k\cos k=0$. Beyond the the trivial solution $k=0$, a plot of the function on the left-side suggests that the condition has an infinity of roots, the first few of which are (using Mathematica): $k_1=1.20779,k_2=3.44824,k_3=6.44095,k_4=9.53048,k_5=12.6458$. The squares of these are the first five allowed values of the Lagrange multiplier $\lambda$ in your variational problem part b). All that remains to do now is to plug your eigensolutions $y_\lambda (x)$ into $C[y_\lambda]=1$ and calculate the normalization factors $A_\lambda$.

I don't understand. If the condition has an infinity of roots, then how are we supposed to plug those $k$ values into the eigensolutions $y_\lambda (x)$ into $C[y_\lambda]=1$ in order to find the constant $A$?

 

[renormalize]

I don't understand. If the condition has an infinity of roots, then how are we supposed to plug those $k$ values into the eigensolutions $y_\lambda (x)$ into $C[y_\lambda]=1$ in order to find the constant $A$?

It's exactly analogous to what you would do for the simple boundary/eigenvalue condition $\sin k_n=0$. Of course, for that case you know that the infinity of roots are explicitly given by $k_n=n\pi$, where $n$ is any natural number, and you use those values to express the normalization $A_n$ as a function of $n\pi$. But what if you didn't know that explicit solution? You'd simply replace $n\pi$ in $A_n$ by $k_n$, along with the statement than the allowed eigenvalues of $k_n$ are the roots of $\sin k_n=0$. (And perhaps display one or more of the eigenvalues that you find numerically.) Just do the same thing for your variational problem.