The Totally Inhomogeneous Boundary Value Problem

The utility of the Green's function extends to the inhomogeneous boundary value problem where the Dirichlet-Neumann boundary conditions are inhomogeneous.

$$Lu = -f(x) ~~~~ a<x<b$$

$$B_1(u) = d$$

$$B_2(u) = e ~~.$$

The solution is expressed in terms of the Green's function in the previous section, and it is given by

$$u(x) =\int^b_a G(x;\xi)~f(\xi)~d\xi +c_1u_1(x) +c_2u_2(x)~~,$$

where, as before, $u_1$ and $u_2$ are two independent solutions to the homogeneous differential equation, and $c_1$ and $c_2$ are adjusted so as to satisfy the given boundary condition,

$$\begin{array}{ccccccccc} d&=&B_1(u)&=&0&+& 0&+& c_2B_1(u_2)\\ e&=&B_2(u)&=&0&+& c_1B_2(u_1)&+& 0 \end{array}$$

Consequently, the solution to the problem is

$$u(x) =\int^b_a G(x;\xi)~f(\xi)~d\xi +\frac{e}{B_2(u_1)}u_1(x) + \frac{d}{B_1(u_2)}u_2(x)~~,$$

The mathematically most perspicuous aspect of this expression is the fact that it can be written as

$$u(x)=u_p(x)~+~u_h(x)~~.$$

Here $u_p(x)$ is a particular solution to the inhomogeneous differential equation,

$$Lu_p(x) = -f(x)~~,$$

while $u_h(x)$ is a complementary function which satisfies the homogeneous differential equation,

$$Lu_h(x) = 0~~.$$

The motivation for adding the appropriate solution of this equation to a particular solution is precisely the one already stated, namely to satisfy the given boundary conditions at the endpoints.

Exercise 47.1 (NON-SELFADJOINT BOUNDARY CONDITIONS)

Let $L = -{d^2\over {dx^2}}$ with boundary conditions $u(0)=0$ , $u'(0) = u(1)$ , so that the domain of $L$ is ${\cal S} = \lbrace u: Lu$ is square integrable; $u(0)=0$ , $u'(0) = u(1)\rbrace$ .

(a)

For the above differential operator FIND ${\mathcal S}^\ast$ for the adjoint with respect to

$$\langle v,u\rangle = \int^1_0 \bar v\,u\,dx\,.$$

and compare ${\mathcal S}$ with ${\mathcal S}^\ast$ .

(b)

COMPARE the eigenvalues $\lambda_n$ of

$$Lu_n = \lambda_n u_n \qquad n = 0,1,2,\dots$$

with the eigenvalues $\lambda^\ast_n$ of

$$L^\ast v_n = \lambda^\ast_n v_n \qquad n = 0,1,2,\dots$$

If the two sequences of eigenvalues are different, point out the distinction; if you find they are the same, justify that result.

(c)

EXHIBIT the corresponding eigenfunctions.

(d)

Is $\lambda =0$ an eigenvalue? Why or why not?

(e)

VERIFY that $\int^1_0 \bar v_n u_m\,dx = 0$ for $n\ne m$ .

Exercise 47.2 (TWO-COMPONENT EIGENVALUE PROBLEM)
Attack the eigenvalue problem

$$-u''(x) = \lambda u(x) \qquad 0 < x < 1$$

$$u'(1) = \lambda u(1)$$

$$u(0) =$$ 0

as follows:

Let $U = \left({u(x)\atop u_1}\right)$ be a two-component vector whose first component is a twice differentiable function $u(x)$ , and whose second component is a real number $u_1$ . Consider the corresponding vector space ${\mathcal H}$ with inner product

$$\langle U,V\rangle\equiv\int^1_0 u(x) v(x) dx + u_1 v_1$$

Let ${\mathcal S}\subset {\mathcal H}$ be the subspace

$${\cal S} = \lbrace U: U = \left({u(x)\atop u(1)}\right); \quad u(0) = 0\rbrace$$

and let

$$LU = \left({-u''(x)\atop u'(1)}\right)\,.$$

The above eigenvalue problem can now be rewritten in standard form

$$LU = \lambda U \quad {\rm with}\quad U\in{\mathcal S}\,.$$

(a)

PROVE or DISPROVE that $L$ is self adjoint, i.e. that $\langle V, LU\rangle = \langle LV, U\rangle$ .

(b)

PROVE or DISPROVE that $L$ is positive-definite, i.e. that $\langle U, LU\rangle > 0$ for $U\ne\vec 0$ . (Reminder: ``positive-definiteness'' applies to all vectors, not only to eigenvectors of $L$ .

(c)

FIND the (transcendental) equation for the eigenvalues of $L$ .

(d)

Denoting these eigenvalues by $\lambda_1, \lambda_2, \lambda_3,\cdots$ , EXHIBIT the orthonormalized eigenvectors $U_n$ , $n = 1,2,3,\cdots$ , associated with these eigenvalues.

Exercise 47.3 (ASYMPTOTIC EIGENVALUE SPECTRUM)
The eigenvalue equation for the Exercise on the previous page (``Non-selfadjoint Boundary Value Conditions'') is

$$\sin\lambda^{1/2}=\lambda^{1/2}$$

Prove or disprove that an asymptotic formula for the roots is

$$\lambda^{1/2}\sim (2m+{1\over{2}})\pi-{2\log(4m+1)\pi\over{(4m+1)\pi}}\pm i\log(4m+1)\pi$$

(You might put $\lambda^{1/2}=\alpha+i\beta$ so that

$$\sin\alpha\cosh\beta = \alpha\qquad (1)$$

$$\cos\alpha\sinh\beta = \beta\qquad (2)$$

For large $\alpha$ , Eq. (1) implies $\beta$ is large. If $\beta$ is large then $\beta/ \sinh\beta$ approaches zero so that $\alpha=(n+{1\over{2}})\pi+\epsilon_n$ where $\epsilon_n\to 0$ . From Eq. (1), since $\cos\epsilon_n\sim 1$ , one has $\cosh\beta=(2m+ {1\over{2}})\pi$ , where $n=2m$ )

Exercise 47.4 (REPRESENTATION VIA EIGENFUNCTIONS)
Consider the eigenvalue problem

$$Lu=\lambda u\quad L= \alpha {d^2\over {dx^2}} +\beta {d\over {dx}} +\gamma$$

$$B_1(u) = 0$$

$$B_2(u) = 0$$

and its adjoint

$$L^\ast v = \bar \lambda v$$

$$B_1^\ast (v) = 0$$

$$B_2^\ast (v) = 0$$

with respect to the inner product $\langle v, u\rangle = \int^b_a \bar v(x) u(x) dx$ . One can show, and you may safely assume, that the eigenvalue spectra of these two problems are complex conjugates of each other.

(a)

Prove that the solution $u(x;\lambda )$ for the problem

$$Lu - \lambda u = -f(x)$$

$$B_1 (u) = 0\,; ~~B_2 (u) = 0$$

is given by

$$u(x;\lambda) = \sum\limits_n {\langle v_n, f\rangle\over {\lambda - \lambda_n}} u_n (x)$$

Here $u_n$ and $v_n$ are the eigenfunctions of $L$ and $L^\ast$ and they have been normalized by the condition

$$\langle v_n,\, u_m\rangle = \delta_{nm}\,.$$

(b)

Show that the Green's function is

$$G_\lambda (x\vert\xi) = \sum\limits_n {u_n (x) \bar v_n (\xi ) \over {\lambda_n - \lambda}}$$

Exercise 47.5 (NORMAL MODE PROFILES VIA COMPLEX INTEGRATION)
Obtain the $o.n.$ set of eigenfunctions for the Sturm-Liouville problem

$$Lu \equiv - {d^2 u\over {dx^2}} = \omega^2 u$$

$$u(a) = u(b) = 0$$

by applying the complex integration technique to the Green's function $G_\omega(x;\xi)$

$$(L - \omega ^2)G \equiv - {d^2 G_\omega \over {dx^2}} - \omega^2 G_\omega = \delta (x - \xi) \qquad a < x\,, ~~\xi < b$$

$$\begin{array}{c} G_\omega(a;\xi) = 0 \\ G_\omega(b;\xi) = 0 \end{array} \qquad a < \xi < b$$

Lecture 32