MHB Eigenvalue problem of the form Sturm-Liouville

[mathmari]

Hey!

I have the following exercise and I need some help..

$"\text{The eigenvalue problem } Ly=(py')'+qy=λy, a \leq x \leq b \text{ is of the form Sturm-Liouville if it satisfies the boundary conditions } p(a)W(u(a),v^*(a))=p(b)W(u(b),v^*(b)). \text{ Show that the boundary conditions of the form: }$
$$ \begin{pmatrix}
y(b)\\
y'(b)
\end{pmatrix}=S\begin{pmatrix}
y(a)\\
y'(a)
\end{pmatrix}, S=\begin{pmatrix}
A & B\\
C & D
\end{pmatrix} $$
$\text{ lead to an eigenvalue problem of the form Sturm-Liouville if } detS=\frac{p(a)}{p(b)}."$

Could you give a hint what to do?

 

[mathmari]

My idea is the following:

Let $u, v$ two solutions of the eigenvalue problem.
These two solutions satisfy the following relation:
$u(b)=Su(a)$ and $u'(b)=Su'(a)$
$v^*(b)=Sv^*(a)$ and $ v^{*'} (b)=S v^{*'} (a)$

So $p(b)W(u(b), v^*(b))=p(b)(u(b)v^{*'}(b)-u'(b)v^*(b))=p(b)(S u(a) Sv^{*'}(a)-Su'(a) Sv^*(a))=p(b)S^2(u(a)v^{*'}(a)-u'(a) v^*(a))=p(b)S^2W(u(a), v^*(a))$.

Is it correct do far?

But how does $detS$ appear?

So that this relation is equal to $p(a)W(u(a), v^*(a))$, shouldn't it be $S^2=\frac{p(a)}{p(b)}$? How can I use the determinant of $S$?

 

[Opalg]

My idea is the following:

Let $u, v$ two solutions of the eigenvalue problem.
These two solutions satisfy the following relation:
$u(b)=Su(a)$ and $u'(b)=Su'(a)$
$v^*(b)=Sv^*(a)$ and $ v^{*'} (b)=S v^{*'} (a)$

So $p(b)W(u(b), v^*(b))=p(b)(u(b)v^{*'}(b)-u'(b)v^*(b))=p(b)(S u(a) Sv^{*'}(a)-Su'(a) Sv^*(a))=p(b)S^2(u(a)v^{*'}(a)-u'(a) v^*(a))=p(b)S^2W(u(a), v^*(a))$.

Is it correct do far?

But how does $detS$ appear?

So that this relation is equal to $p(a)W(u(a), v^*(a))$, shouldn't it be $S^2=\frac{p(a)}{p(b)}$? How can I use the determinant of $S$?

$S = \begin{pmatrix}A&B \\ C&D \end{pmatrix}$ is a matrix, so it makes no sense to talk about $Su(a)$ or $S{v^*}'(a)$. In fact, $\begin{pmatrix}u(b) \\u'(b) \end{pmatrix} = S\begin{pmatrix}u(a) \\u'(a) \end{pmatrix} = \begin{pmatrix}A&B \\ C&D \end{pmatrix}\begin{pmatrix}u(a) \\u'(a) \end{pmatrix} = \begin{pmatrix}Au(a) + Bu'(a) \\Cu(a) + Du'(a) \end{pmatrix}.$ Therefore $u(b) = Au(a) + Bu'(a)$, $u'(b) = Cu(a) + Du'(a)$, and similarly for the other solution $v$. Substitute those values into the given boundary condition, and you should end up with an equation involving $\det S = AD-BC.$

 

[mathmari]

$S = \begin{pmatrix}A&B \\ C&D \end{pmatrix}$ is a matrix, so it makes no sense to talk about $Su(a)$ or $S{v^*}'(a)$. In fact, $\begin{pmatrix}u(b) \\u'(b) \end{pmatrix} = S\begin{pmatrix}u(a) \\u'(a) \end{pmatrix} = \begin{pmatrix}A&B \\ C&D \end{pmatrix}\begin{pmatrix}u(a) \\u'(a) \end{pmatrix} = \begin{pmatrix}Au(a) + Bu'(a) \\Cu(a) + Du'(a) \end{pmatrix}.$ Therefore $u(b) = Au(a) + Bu'(a)$, $u'(b) = Cu(a) + Du'(a)$, and similarly for the other solution $v$. Substitute those values into the given boundary condition, and you should end up with an equation involving $\det S = AD-BC.$

Ok! I got it! (Yes)

I have also an other question.

$v^*$ is the complex conjugate of $v$.
At the beggining of the solution do I have to suppose that $v$ is a solution or $v^*$?

 

[mathmari]

When $v$ is a solution of the eigenvalue problem, then is the complex conjugate $v^*$ also a solution? (Wondering)

 

[I like Serena]

When $v$ is a solution of the eigenvalue problem, then is the complex conjugate $v^*$ also a solution? (Wondering)

That depends.
Let's see...

If $v$ is a solution, then we have:
\begin{array}{}
(pv')' + qv &=& \lambda v \\

((pv')' + qv)^* &=& (\lambda v)^* \\

(p^* {v^*}')' + q^* v^* &=& \lambda^* v^* \\
\end{array}

In other words, $v^*$ is the solution of:
$$(p^*y')' + q^*y = \lambda^* y$$

So we can only expect $v^*$ to be a solution of the original ODE if $p=p^*$ and $q=q^*$ (both real-valued).
The associated eigenvalue would then be the conjugate of the original eigenvalue.

 

[mathmari]

That depends.
Let's see...

If $v$ is a solution, then we have:
\begin{array}{}
(pv)' + qv &=& \lambda v \\

((pv')' + qv)^* &=& (\lambda v)^* \\

(p^* {v^*}')' + q^* v^* &=& \lambda^* v^* \\
\end{array}

In other words, $v^*$ is the solution of:
$$(p^*y')' + q^*y = \lambda^* y$$

So we can only expect $v^*$ to be a solution of the original ODE if $p=p^*$ and $q=q^*$ (both real-valued).
The associated eigenvalue would then be the conjugate of the original eigenvalue.

Aha! Ok! So to use the sentence with the Wronskian, I have to suppose that $u$ and $v^*$ are solutions, right?

 

[I like Serena]

Aha! Ok! So to use the sentence with the Wronskian, I have to suppose that $u$ and $v^*$ are solutions, right?

I don't know, since I haven't found useful references linking the Wronskian to Sturm-Liouville (yet).
My expectation is that $v$ should be the solution.
But what does your course material say exactly?

 

[mathmari]

I don't know, since I haven't found useful references linking the Wronskian to Sturm-Liouville (yet).
My expectation is that $v$ should be the solution.
But what does your course material say exactly?

$L$ is the differential operator.

$$Lu(x)=P_o(x)\frac{d^2u}{dx^2}+P_1(x)\frac{du}{dx}+P_2(x)u$$
We suppose that $P_j(x) \in C^2(\mathbb{R})$

We define the dot product $(v(x),u(x))=\int_a^b{v^*(x) u(x)}dx$, where $v^*(x)$ is the complex conjugate.

$(v,Lu)=\int_a^b{v^*(P_o(x)\frac{d^2u}{dx^2}+P_1(x)\frac{du}{dx}+P_2(x)u)}dx= \cdots =[P_o(v^*u'-v^{*'}u)+(P_1-P_0')v^*u]_a^b+\int_a^b{[\frac{d}{dx^2}(P_0v^*)-\frac{d}{dx}(P_1v^*)+P_2v^*]u}dx$

We define the conjugate operator
$$\widetilde{L}u=\frac{d^2}{dx^2}(P_0u)-\frac{d}{dx}(P_1u)+P_2u$$

$$(v,Lu)=[P_0(v^*u'-v^{*'}u)+(P_1-P_0')v^*u]_a^b+(\widetilde{L}v,u)$$

Now we take the special case where:
$a) P_1=P_0'$
$b) \text{Boundary Conditions } [P_0(x)(v^*u'-v^{*'}u)]_a^b=0$

$$(v,Lu)=(\widetilde{L}v,u)$$

Self-adjoint is called an operator when $L=\widetilde{L}$

$\widetilde{L}u=\frac{d^2}{dx^2}(P_0u)-\frac{d}{dx}(P_1u)+P_2u=P_0u''+(2P_0'-P_1)u'+(P_0''-P_1'+P_2)u$

When $(a)$ stands, then:
$\widetilde{L}u=P_0u''+P_1u'+P_2u=Lu$

When $(a)$ and $(b)$ stand, then the operator is indeed self-adjoint.

$=\frac{d}{dx}(P_0\frac{du}{dx})+P_2u=Lu$

$P_0=p, P_2=q$
$\frac{d}{dx}(p\frac{du}{dx})+qu=Lu=\widetilde{L}u$

The eigenvalue problem $Lu=λu$, where the operator $L$ is given by the relation:
$\frac{d}{dx}(p\frac{du}{dx})+qu=Lu=\widetilde{L}u$ and the boundary conditions satisfy $[p(uv^{*'}-u'v^*)]_a^b=0 (*)$, where $u(x), v(x)$ are solutions of the eigenvalue problem, is called Sturm-Liouville problem.

The relation $(*)$ can be written as $P_0(x)W(u(x), v^*(x)) |_a^b=0$

 

[I like Serena]

The eigenvalue problem $Lu=λu$, where the operator $L$ is given by the relation:
$\frac{d}{dx}(p\frac{du}{dx})+qu=Lu=\widetilde{L}u$ and the boundary conditions satisfy $[p(uv^{*'}-u'v^*)]_a^b=0 (*)$, where $u(x), v(x)$ are solutions of the eigenvalue problem, is called Sturm-Liouville problem.

The relation $(*)$ can be written as $P_0(x)W(u(x), v^*(x)) |_a^b=0$

That's a whole lot to digest. (Whew)

Either way, I'm reading: "$u(x), v(x)$ are solutions of the eigenvalue problem".
That clearly means the $v$ is the solution of the ODE and $v^*$ is merely its conjugate, which shows up when taking an inner product.

 

[mathmari]

That's a whole lot to digest. (Whew)

Either way, I'm reading: "$u(x), v(x)$ are solutions of the eigenvalue problem".
That clearly means the $v$ is the solution of the ODE and $v^*$ is merely its conjugate, which shows up when taking an inner product.

Ok! But at the exercise I have to use that $v^*$ satisfies the given boundary conditions, how can I use this when I suppose that $v$ is a solution? (Wondering)

 

[mathmari]

Supposing that $v$ is a solution of the eigenvalue problem, can I just use that $v^{*}$ satisfies the boundary conditions or do I have to write also something from the theory according to the dot product? (Wondering)

 

[mathmari]

Let's suppose that $v$ is a solution.
So, $$\begin{pmatrix}
v(b)\\
v'(b)
\end{pmatrix}=\begin{pmatrix}
A & B\\
C & D
\end{pmatrix} \begin{pmatrix}
v(a)\\
v'(a)
\end{pmatrix} \Rightarrow v(b)=A v(a)+B v'(a) \text{ and } v'(b)=C v(a)+D v'(a)$$

The complex conjugate of $v(b)=A v(a)+B v'(a)$ is $v^*(b)=A v^*(a)+B v^{*'}(a)$ and of $ v'(b)=C v(a)+D v'(a)$ it's $v^{*'}(b)=C v^*(a)+D v^{*'}(a)$, isn't it? (Wondering) (If we suppose that $A,B,C,D$ are real numbers)

 

[I like Serena]

Let's suppose that $v$ is a solution.
So, $$\begin{pmatrix}
v(b)\\
v'(b)
\end{pmatrix}=\begin{pmatrix}
A & B\\
C & D
\end{pmatrix} \begin{pmatrix}
v(a)\\
v'(a)
\end{pmatrix} \Rightarrow v(b)=A v(a)+B v'(a) \text{ and } v'(b)=C v(a)+D v'(a)$$

The complex conjugate of $v(b)=A v(a)+B v'(a)$ is $v^*(b)=A v^*(a)+B v^{*'}(a)$ and of $ v'(b)=C v(a)+D v'(a)$ it's $v^{*'}(b)=C v^*(a)+D v^{*'}(a)$, isn't it? (Wondering) (If we suppose that $A,B,C,D$ are real numbers)

Yes.
But indeed only if we suppose that $A,B,C,D$ are real numbers.
Can it be that there is some more context to your problem that specifies this? (Wondering)
Because it seems to me this is necessary information to solve the problem.