# Eigenfunction Method

The Eigenfunction method is similiar to the concept of Eigenvektor in linear Algebra, where the linear Operator $\mathbf{A}$ acts on the eigenvektor $\mathbf{x}^i$ which results in a vector that points in the same direction as $\mathbf{x}^i$ but its length is changed by the factor $\lambda_i$. The eigenvalue problem reads \begin{align} \mathbf{A} \cdot \mathbf{x}^i= \lambda_i \cdot \mathbf{x}^i \end{align} In the Eigenfunction Method we consider the linear Operator $\mathcal{L}$ that acts on a function $y(x)$ \begin{align} \mathcal{L}y_i(x)= \lambda_i y_i(x) \tag{1} \end{align} $\lambda$ is called the Eigenvalue corresponding to the Eigenfunction $y(x)$. A simple example is the harmonic oscillator with the linear operator $\mathcal{L}=-\frac{d^2}{dx^2}$ \begin{align} \mathcal{L}y_i(x)=-\frac{d^2}{dx^2}y_i(x)= \omega_i^2 y_i(x) \end{align}

## Superposition of Eigenfunctions: Green's Functions

If we want to solve the inhomogenious problem \begin{align} \mathcal{L}y(x)= f(x) \tag{2} \end{align} by means of Green's function. We can use the Eigenfunctions $y_n(x)$ of the Eigenfunction Equation (1), and write $y(x)$ in terms of the eigenfunctions \begin{align} y(x)=\sum_{n=0}^\infty c_n y_n(x) \end{align} for some $c_n$ we don't know yet. Then the function (2) becomes \begin{align} f(x) = \mathcal{L}y(x) =\mathcal{L} \left( \sum_{n=0}^\infty c_n y_n(x) \right)=\sum_{n=0}^\infty c_n \lambda_n y_n(x) \end{align} Now once again orthoghonality comes into play. We muliply by $y_j$ and integrate \begin{align} \int_a^b y_j(z) f(z) \, dz= \sum_{n=0}^\infty \int_a^b c_n \lambda_n y_j(z) y_n(z) \, dz \end{align} were we used $z$ as integration variable for reasons we will see later. By orthoghonality all integrals on the RHS are zero except for $n=j$, thus we obtain the coefficients \begin{align} c_n=\frac{1}{\lambda_n} \frac{\int_a^b y_n(z) f(z) \, dz}{\int_a^b y_n(z) y_n(z) \, dz} \end{align} Now that we have found the coefficients we can write for the solution $y(x)$ \begin{align} y(x)=\sum_{n=0}^\infty c_n y_n(x)=\sum_{n=0}^\infty \frac{1}{\lambda_n} \frac{\int_a^b y_n(z) f(z) \, dz}{\int_a^b y_n(z) y_n(z) \, dz} y_n(x) \end{align} If the Eigenfunctions are already normalised and $\int_a^b y_n(z) y_n(z) \, dz=1$ the solution for $y(x)$ becomes \begin{align} y(x)=\int_a^b \left[ \sum_{n=0}^\infty \left( \frac{1}{\lambda_n} y_n(x) y_n(z) \right) \right] f(z) \, dz \end{align} the term in brackets is called the Green's function $G(x,z)$. Using this notation we have \begin{align} y(x)=\int_a^b G(x,z) f(z) \, dz \end{align} were

\begin{align} G(x,z) = \sum_{n=0}^\infty \frac{1}{\lambda_n} y_n(x) y_n(z) \end{align} $G(x,z)$ is entirely determined by the boundary conditions and the eigenfunctions $y_n$ and hence by $\mathcal{L}$ itself.

Example

Find an appropriate Green's function for the equation \begin{align} y'' + \frac{1}{4} y =f(x) \end{align} and boundary conditions $y(0)=y(\pi)=0$. Solve this ODE for $f(x)=sin2x$.

The equation \begin{align} y'' + \frac{1}{4} y =f(x) \end{align} is simply a harmonic ODE with the familiar solution \begin{align} y(x)=A \, sin[ \sqrt{1/4-\lambda} \, x] + B \, cos [\sqrt{1/4-\lambda}\, x] \end{align} If we plug in the boundary conditions $y(0)=y(\pi)=0$, we get $B=0$ and \begin{align} y(x)=A \, sin( n \, x) \hspace{2cm} \text{ with } n=\sqrt{1/4-\lambda}= \pm 1, \pm 2, \dots \end{align} Normalisation of $y_n(x)$ gives \begin{align} \int_0^\pi A_n^2 sin^2 nx \, dx=1 \Rightarrow A_n=\left( \frac{2}{\pi} \right)^{1/2} \end{align} ultimately \begin{align} G(x,z) = \sum_{n=0}^\infty \frac{1}{\lambda_n} y_n(x) y_n(z)=\frac{2}{\pi} \sum_{n=0}^{\infty} \frac{sin (nx) \, sin (nz)}{\frac{1}{4} - n^2} \end{align} now we can use Green's function to solve for the particular or complementary solution $y_p(x)$ \begin{align} y_p(x) &= \int_a^b G(x,z) f(z) \, dz\\ y_p(x) &= \frac{2}{\pi} \int_0^{\pi} \sum_{n=0}^{\infty} \frac{sin (nx) \, sin (nz)}{\frac{1}{4} - n^2} sin(2z) \, dz\\ &= \frac{2}{\pi} \sum_{n=0}^{\infty} \frac{sin (nx) }{\frac{1}{4} - n^2} \,\int_0^{\pi} sin (nz)sin(2z) \, dz \end{align} for $n \not =2$ all integrals vanish and we only have \begin{align} \int_0^{\pi}sin^2(2z) \, dz=\pi /2 \end{align} and therefore \begin{align} y_p(x) &= \frac{2}{\pi} \sum_{n=0}^{\infty} \frac{sin (nx) }{\frac{1}{4} - n^2} \frac{\pi}{2}=-\frac{4}{15} sin(2x) \end{align}

## Green's function is the response to an impuls input

If we consider the solution by means of Green's function with boudary conditions in the range $a \leq x \leq b$ we found that \begin{align} y(x)=\int_a^b G(x,z) f(z) \, dz \end{align} Applying the operator $\mathcal{L}$ we have \begin{align} \mathcal{L}y(x)=\int_a^b [\mathcal{L}G(x,z)] f(z) \, dz=f(x) \end{align} If we now use the dirac delta function to write $f(x)$ as \begin{align} f(x)=\int_a^b \delta(x-z) f(z) dz \end{align} this means that

Green's function is the response of a system to a unit impulse at $x=z$ \begin{align} \mathcal{L}G(x,z)= \delta(x-z) \end{align}