How to Find the Eigenfunctions of a Differential Operator

To find eigenfunctions, turn the operator equation into a differential equation and solve it subject to the conditions that define the operator. Those conditions matter: the same differential expression can have different eigenvalues and eigenfunctions on different intervals or with different boundary conditions.

What counts as an eigenfunction?

For a linear operator $L$, an eigenfunction is a nonzero function $f$ in the domain of $L$ for which

$$Lf=\lambda f$$

for some scalar $\lambda$, called the eigenvalue. If, for example, $L$ is a second-derivative operator, this is an ordinary differential equation with $\lambda$ as a parameter.

A differential expression such as $-\frac{d^2}{dx^2}$ does not by itself specify a complete operator. You also need to know the interval, the space of functions being used, and the boundary conditions—or, more generally, the operator’s domain. These choices determine which solutions count.

A practical method

1. Write down the full operator. Record its formula, interval, function space if specified, and domain or boundary conditions.
2. Set up the eigenvalue equation. Replace $Lf$ with $\lambda f$ to get a differential equation.
3. Solve the equation in terms of $\lambda$. Keep the general solution at first; do not impose the boundary conditions prematurely.
4. Apply the domain conditions. They give equations for the integration constants. A value of $\lambda$ is an eigenvalue only if those equations permit a nonzero solution.
5. Check the result. Confirm that each resulting function is nonzero, satisfies the boundary conditions, and belongs to the operator’s domain.

For a regular Sturm–Liouville operator, the equation often has the form

$$-(p(x)y')'+q(x)y=\lambda w(x)y,$$

with specified endpoint conditions. It is still the same procedure: solve the equation and use the endpoint conditions to determine the allowed values of $\lambda$.

Worked example: the negative second derivative

Consider

$$Lf=-f''(x),\qquad 0<x<L,$$

with Dirichlet boundary conditions $f(0)=f(L)=0$. Find nonzero solutions of

$$-f''=\lambda f.$$

The boundary conditions are part of the problem. We consider the possible signs of $\lambda$.

Case 1: $\lambda=0$

The equation becomes $f''=0$, so $f(x)=A+Bx$. From $f(0)=0$, we get $A=0$. From $f(L)=0$, we get $BL=0$, hence $B=0$. There is no nonzero solution, so $0$ is not an eigenvalue.

Case 2: $\lambda<0$

Write $\lambda=-\mu^2$ with $\mu>0$. Then $f''=\mu^2 f$, whose solutions have the form

$$f(x)=A\cosh(\mu x)+B\sinh(\mu x).$$

The condition $f(0)=0$ gives $A=0$. Since $\sinh(\mu L)\ne 0$, the condition $f(L)=0$ then gives $B=0$. Again, only the zero function satisfies both conditions, so there are no negative eigenvalues.

Case 3: $\lambda>0$

Write $\lambda=k^2$ with $k>0$. The equation becomes $f''+k^2f=0$, with general solution

$$f(x)=A\cos(kx)+B\sin(kx).$$

The condition $f(0)=0$ forces $A=0$. For a nonzero solution, $B$ must be nonzero; then $f(L)=0$ requires $\sin(kL)=0$. Thus $kL=n\pi$ for a positive integer $n$, giving

$$\lambda_n=\left(\frac{n\pi}{L}\right)^2, \qquad f_n(x)=\sin\left(\frac{n\pi x}{L}\right), \qquad n=1,2,3,\ldots$$

Any nonzero constant multiple of $f_n$ is also an eigenfunction. The boundary conditions select these particular frequencies from the general family of solutions.

Common mistakes

• Ignoring the domain or boundary conditions. Solving the differential equation alone usually gives many functions that are not eigenfunctions of the specified operator.
• Counting the zero function. It solves $Lf=\lambda f$ for every $\lambda$, but eigenfunctions must be nonzero.
• Assuming every $\lambda$ is an eigenvalue. The boundary conditions typically allow only particular values.
• Using the wrong sign convention. The operators $\frac{d^2}{dx^2}$ and $-\frac{d^2}{dx^2}$ have eigenvalues with opposite signs under the same boundary conditions.
• Forgetting the function space. In an operator-theoretic setting, a candidate must lie in the operator’s domain, not merely solve the equation formally.

What changes with other conditions?

Changing the boundary conditions changes the eigenfunctions. For example, for the same operator on $(0,L)$, Neumann conditions $f'(0)=f'(L)=0$ admit the constant function as an eigenfunction with eigenvalue $0$, as well as cosine eigenfunctions for positive eigenvalues. Periodic conditions produce yet another set.

For systems of coupled differential equations, the eigenfunction is vector-valued and the same principle applies: solve the system $Lf=\lambda f$ and impose the operator’s domain conditions. If the coefficients or conditions are complicated, the eigenvalues may need to be found numerically, but the setup remains the same.