What Is the Solvability Condition for a Differential Operator?

A differential equation written as $Lu=f$ may have no solution, one solution, or many. The relevant solvability condition depends not only on the differential expression $L$, but also on its domain and boundary conditions.

For a broad and useful class of boundary-value problems, the Fredholm alternative gives the condition: the forcing term must be orthogonal to every solution of the homogeneous adjoint equation. The same principle also explains when a solution is unique.

The Fredholm solvability condition

Treat $L$ as an operator between suitable function spaces, with its domain specifying the boundary conditions. Its adjoint $L^*$ is defined using the inner product and that domain. For example, in an $L^2$ space the inner product is often

$$\langle u,v\rangle=\int_a^b \overline{u(x)}v(x)\,dx.$$

If $L$ is a Fredholm operator, then its range is closed and the Fredholm alternative says

$$Lu=f \text{ is solvable} \quad\Longleftrightarrow\quad \langle z,f\rangle=0 \text{ for every } z\in\ker L^*.$$

Here, $\ker L^*$ is the set of solutions of the adjoint homogeneous problem, including the adjoint boundary conditions. In a real-valued problem, the orthogonality condition is commonly written as an integral of a product. In a complex-valued problem, the inner product includes complex conjugation according to the chosen convention.

If a solution exists, it is unique if $\ker L=\{0\}$. Otherwise, adding any element of $\ker L$ to a solution produces another solution. The condition involving $\ker L^*$ is about existence; the kernel of $L$ is about uniqueness.

Why orthogonality is necessary

Suppose $Ly=f$, and let $z$ satisfy $L^*z=0$. The definition of the adjoint gives

$$\langle z,f\rangle=\langle z,Ly\rangle=\langle L^*z,y\rangle=0,$$

provided the boundary conditions make the adjoint identity valid. Thus every solution must satisfy the compatibility condition.

For a Fredholm operator, this condition is also sufficient. In general, however, orthogonality to $\ker L^*$ only characterizes the closure of the range: if $L$ does not have closed range, the condition alone may not guarantee that $f$ is actually in the range. This is why the Fredholm hypothesis matters.

Worked example: a resonant boundary-value problem

Consider

$$y''+y=f(x),\qquad 0<x<\pi,\qquad y(0)=y(\pi)=0.$$

The operator is $Ly=y''+y$. Integration by parts shows that it is self-adjoint under these Dirichlet boundary conditions: the boundary terms vanish when both functions are zero at the endpoints. Its adjoint homogeneous equation is therefore

$$z''+z=0,\qquad z(0)=z(\pi)=0.$$

The general homogeneous solution is $z(x)=A\cos x+B\sin x$. The boundary conditions force $A=0$ but allow any $B$, so the adjoint kernel is spanned by $\sin x$. The solvability condition is consequently

$$\int_0^\pi f(x)\sin x\,dx=0.$$

For $f(x)=1$, the condition fails:

$$\int_0^\pi \sin x\,dx=2\ne 0.$$

Therefore the boundary-value problem has no solution for this forcing term.

For $f(x)=\cos x$, the condition holds because

$$\int_0^\pi \cos x\sin x\,dx=0.$$

Indeed, $y(x)=\tfrac{x}{2}\sin x$ is a solution: it vanishes at both endpoints and satisfies $y''+y=\cos x$. Since $\sin x$ is also in the kernel of $L$, every solution is $\tfrac{x}{2}\sin x+C\sin x$ for a constant $C$.

How to apply the condition

For a particular differential operator, use this sequence:

1. Specify the function space, operator domain, and boundary conditions.
2. Compute the formal adjoint by integration by parts.
3. Determine the adjoint boundary conditions that eliminate the boundary terms.
4. Solve the adjoint homogeneous problem to find $\ker L^*$.
5. Check that the forcing is orthogonal to every element of that kernel.

The formal adjoint differential expression alone is not enough. The adjoint’s domain—and therefore its boundary conditions—can change the solvability test.

Common mistakes

• Using the kernel of $L$ instead of $L^*$. The compatibility condition involves the adjoint kernel. These kernels agree for a self-adjoint operator, but not in general.
• Ignoring boundary conditions. They determine the operator domain and affect the adjoint problem. Integration by parts may leave nonzero boundary terms if the conditions are not handled correctly.
• Assuming the condition guarantees a solution in every setting. Orthogonality is sufficient under the Fredholm alternative, but for operators without closed range it may only show that $f$ is in the closure of the range.
• Expecting uniqueness whenever a solution exists. A nontrivial kernel of $L$ means solutions can differ by homogeneous solutions.