The equation $Lu = f$ — a linear operator applied to an unknown equals a given datum — is the central problem of linear PDE theory. When $L$ is coercive (or can be shifted to be so), the [Lax–Milgram theorem](/theorems/91) guarantees a unique solution for every $f$. But many natural problems are not coercive: the operator may have a nontrivial kernel, and existence fails for generic right-hand sides. The question is then not *whether* the equation is solvable, but *exactly when* it is solvable, and *how large* the failure of uniqueness and existence can be. In finite dimensions, the answer is given by the rank–nullity theorem: a square matrix $A$ is either invertible (trivial kernel, full range) or singular (nontrivial kernel, range of codimension equal to the dimension of the kernel), and solvability of $Ax = b$ in the singular case is controlled by orthogonality to the kernel of $A^T$. In infinite dimensions, this tidy picture breaks down for general bounded operators — the range can fail to be closed, the left and right kernels can have different dimensions, and injectivity need not imply surjectivity (the right shift on $\ell^2$ is injective but not surjective). The Fredholm Alternative is the theorem that restores the finite-dimensional picture for operators of the form $I - K$ with $K$ [compact](/page/Linear%20Operators%20on%20Banach%20Spaces). Compactness is the hypothesis that forces the operator to behave like a matrix: the kernel is finite-dimensional, the range is closed, the defect is the same on both sides, and the dichotomy between invertibility and finite-dimensional failure is exact. ## Motivation [motivation] ### The Vibrating Membrane Consider a membrane stretched over a bounded domain $U \subset \mathbb{R}^2$, fixed at the [boundary](/page/Boundary). The eigenvalue problem for the Laplacian, \begin{align*} -\Delta w &= \lambda w \quad \text{in } U, \\ w &= 0 \quad \text{on } \partial U, \end{align*} has a discrete [sequence](/page/Sequence) of eigenvalues $0 < \lambda_1 < \lambda_2 \leq \lambda_3 \leq \cdots \to \infty$ with corresponding eigenfunctions $w_k$ — the resonant modes of the membrane. Now suppose we apply an external forcing $f$ and ask whether the steady-state problem \begin{align*} -\Delta u - \lambda u &= f \quad \text{in } U, \\ u &= 0 \quad \text{on } \partial U \end{align*} has a solution. If $\lambda$ is not an eigenvalue, one expects a unique solution for every $f$ — the system is not at resonance, and the forcing produces a well-defined response. If $\lambda = \lambda_k$ for some $k$, the homogeneous equation $-\Delta u - \lambda_k u = 0$ has nontrivial solutions (the eigenfunctions $w_k$), so uniqueness fails. But existence does not fail completely: a solution still exists provided the forcing $f$ does not "feed energy into" the resonant mode, which is made precise by the condition $\int_U f \, w_k \, d\mathcal{L}^2 = 0$ for every eigenfunction $w_k$ associated to $\lambda_k$. ### What Could Go Wrong in Infinite Dimensions The vibrating membrane example might suggest that the situation is always this clean. But for a general bounded operator $T$ on an infinite-dimensional Hilbert space, several pathologies can occur that have no finite-dimensional analogue. The kernel of $T$ might be infinite-dimensional. The range of $T$ might fail to be closed, in which case the orthogonal complement of the range does not capture the solvability obstruction. The dimensions of the kernels of $T$ and $T^*$ might differ — the "defect" on the left and right sides could be different. Any of these would make a clean solvability criterion impossible. ### Compactness Restores the Finite-Dimensional Picture The hypothesis that saves the theory is compactness. If $K: H \to H$ is a [compact operator](/page/Linear%20Operators%20on%20Banach%20Spaces) and $T = I - K$, then $T$ behaves like a finite-dimensional perturbation of the identity: its kernel is finite-dimensional, its range is automatically closed, the kernel dimensions of $T$ and $T^*$ are equal, and injectivity implies surjectivity. The analogy with matrices is not a coincidence — compact operators are precisely those that can be approximated by finite-rank operators, and the Fredholm Alternative is the [limit](/page/Limit) of the rank–nullity theorem along this approximation. In the PDE context, compactness enters through the [Rellich–Kondrachov theorem](/theorems/64): the embedding $H^1_0(U) \hookrightarrow L^2(U)$ is compact for bounded $U$, which makes the resolvent operator $K = j \circ L_\gamma^{-1} \circ \iota$ a compact operator on $L^2(U)$. Rewriting the elliptic equation $Lu = f$ in the form $(I - \gamma K)u = Kf$ puts it into the framework of the abstract Fredholm Alternative. The membrane example above is the simplest instance of this reduction. [/motivation] ## Setup The Fredholm Alternative concerns equations of the form $Tu = f$ where $T = I - K$ for a compact operator $K$ on a Hilbert space. We recall the definitions of the two key ingredients; full developments are on the [Linear Operators on Banach Spaces](/page/Linear%20Operators%20on%20Banach%20Spaces) and [Hilbert Space](/page/Hilbert%20Space) pages. A bounded linear operator $K: H \to H$ on a Hilbert space is **compact** if every bounded sequence $\{u_n\} \subseteq H$ has a subsequence for which $\{Ku_{n_k}\}$ converges in $H$. Compact operators form a closed two-sided ideal $\mathcal{K}(H) \subseteq \mathcal{L}(H)$ in the operator norm. The **adjoint** $K^*: H \to H$ is the unique bounded linear operator satisfying $(Ku, v)_H = (u, K^*v)_H$ for all $u, v \in H$; if $K$ is compact, so is $K^*$. ## The Theorem [quotetheorem:72] The four conclusions of the theorem have distinct roles, and it is worth understanding each in turn. **Conclusion (1): finite-dimensionality of the kernel.** The kernel $\ker(I - K)$ is the space of solutions to the homogeneous equation $u = Ku$ — the "eigenspace at eigenvalue $1$" of $K$. For a general bounded operator, this kernel could be infinite-dimensional (consider the identity operator, whose kernel under $I - I = 0$ is the entire space). Compactness of $K$ forces the kernel to be finite-dimensional: the closed unit ball of $\ker(I - K)$ coincides with its image under $K$ (since $u = Ku$ on the kernel), so it is precompact, which by the Riesz lemma characterisation means the space is finite-dimensional. In the PDE context, this says that the eigenspace of the elliptic operator at any given eigenvalue is finite-dimensional — there are only finitely many linearly independent resonant modes at each frequency. **Conclusion (2): closed range and the solvability condition.** The identity $\operatorname{Range}(I - K) = \ker(I - K^*)^\perp$ is the heart of the theorem from the applications point of view. It says that the equation $(I - K)u = f$ is solvable if and only if $f$ is orthogonal to every solution of the adjoint homogeneous equation $(I - K^*)v = 0$. This is a concrete, checkable criterion: to determine whether a given $f$ admits a solution, one solves the adjoint homogeneous problem, finds a basis $v_1, \ldots, v_d$ for $\ker(I - K^*)$, and checks whether $(f, v_j)_H = 0$ for each $j$. (The adjoint kernel is finite-dimensional because $K^*$ is compact whenever $K$ is, so conclusion (1) applies to $T^* = I - K^*$ as well.) In the PDE context, this becomes the condition $\int_U f \, v_j \, d\mathcal{L}^n = 0$ where $v_j$ are the eigenfunctions of the adjoint operator — exactly the orthogonality condition in the membrane example. The closedness of the range is what makes this work: for a general operator, the range need not be closed, and the orthogonal complement of the range would not capture the full solvability obstruction. **Conclusion (3): dimension equality.** The equality $\dim \ker(I - K) = \dim \ker(I - K^*)$ ties conclusions (1) and (2) together quantitatively. Conclusion (2) says the range of $T = I - K$ has codimension $\dim \ker(I - K^*)$ inside $H$ (since $\operatorname{Range}(T) = \ker(T^*)^\perp$, the codimension is $\dim \ker(T^*)$). Conclusion (3) says this codimension equals $\dim \ker(T)$. The practical consequence is: if the homogeneous equation $(I - K)u = 0$ has a $d$-dimensional solution space, then the solvability of $(I - K)u = f$ is controlled by exactly $d$ orthogonality conditions on $f$ (one for each basis element of $\ker(I - K^*)$), and when those conditions are satisfied, the solution is determined up to an arbitrary element of the $d$-dimensional kernel. The number of "degrees of freedom lost" in uniqueness is exactly the number of "conditions gained" for existence — no more, no less. When $K$ is self-adjoint ($K = K^*$), conclusion (3) is immediate because $\ker(T) = \ker(T^*)$. In the non-self-adjoint case, the equality requires the full power of compactness — the proof constructs a finite-rank perturbation that converts a hypothetical dimension mismatch into a contradiction via Step 4 (injectivity implies surjectivity). **Conclusion (4): the dichotomy.** The two alternatives are mutually exclusive and exhaustive. Either $\ker(I - K) = \{0\}$ — in which case the three preceding conclusions combine to give that $I - K$ is bijective with bounded inverse (the [Open Mapping Theorem](/theorems/631) supplies the bounded inverse) — or $\ker(I - K) \neq \{0\}$, and solvability is governed by the finitely many orthogonality conditions from conclusion (2). There is no third possibility: the equation cannot be "partially solvable" with an infinite-dimensional obstruction, nor can the kernel be nontrivial but the range still be all of $H$. In the membrane language: at a non-resonant frequency, the problem is uniquely solvable for every forcing; at a resonant frequency, the problem is solvable precisely when the forcing is orthogonal to the resonant modes. ## Examples ### All Four Conclusions in a PDE The following example shows how each of the four conclusions manifests concretely for an elliptic boundary value problem. [example: Helmholtz Equation At Resonance In One Dimension] Let $U = (0, \pi) \subset \mathbb{R}$ and consider the Dirichlet eigenvalue problem for $-u''$ on $H^1_0((0, \pi))$. The eigenvalues and normalised eigenfunctions are \begin{align*} \lambda_k &= k^2, \quad w_k(x) = \sqrt{\frac{2}{\pi}} \sin(kx), \quad k = 1, 2, 3, \ldots \end{align*} Now consider the Helmholtz equation at the first eigenvalue $\lambda_1 = 1$: \begin{align*} -u'' - u &= f \quad \text{in } (0, \pi), \\ u(0) = u(\pi) &= 0, \end{align*} for a given $f \in L^2((0, \pi))$. The operator $L = -\partial_{xx}$ is self-adjoint on $H^1_0((0, \pi))$, and the equation $Lu - u = f$ can be rewritten in Fredholm form $(I - \tilde{K})u = h$ by shifting: choose $\gamma = 2$ so that $L_\gamma = -\partial_{xx} + 2I$ is coercive. The resolvent $K = j \circ L_\gamma^{-1} \circ \iota: L^2 \to L^2$ is compact and self-adjoint (since $L$ is self-adjoint), with $\tilde{K} = (\gamma + 1)K = 3K$. The four conclusions of the Fredholm Alternative give the following. **Conclusion (1)** (finite-dimensional kernel): The kernel of $I - \tilde{K}$ corresponds to the eigenspace of $-\partial_{xx}$ at $\lambda_1 = 1$. The homogeneous equation $-u'' - u = 0$ with Dirichlet conditions has general solution $u(x) = A\sin(x) + B\cos(x)$; the boundary condition $u(0) = 0$ forces $B = 0$, and $u(\pi) = 0$ is automatically satisfied. So \begin{align*} \ker(I - \tilde{K}) &= \operatorname{span}\{\sin\}, \end{align*} which is one-dimensional, as guaranteed by the theorem. **Conclusion (2)** (solvability condition): The range of $I - \tilde{K}$ is the orthogonal complement of the adjoint kernel. Since $L$ is self-adjoint, $\tilde{K}$ is self-adjoint, so $\ker(I - \tilde{K}^*) = \ker(I - \tilde{K}) = \operatorname{span}\{\sin\}$. Therefore \begin{align*} \operatorname{Range}(I - \tilde{K}) &= \{f \in L^2((0, \pi)) : (f, \sin)_{L^2} = 0\} = \{\sin\}^\perp. \end{align*} The equation $-u'' - u = f$ is solvable if and only if \begin{align*} \int_0^\pi f(x) \sin(x) \, d\mathcal{L}^1(x) &= 0. \end{align*} For example, $f(x) = \sin(2x)$ satisfies this condition (since $\int_0^\pi \sin(2x) \sin(x) \, d\mathcal{L}^1(x) = 0$ by orthogonality of the eigenfunctions), so $-u'' - u = \sin(2x)$ is solvable. One can verify directly: substituting $u(x) = \frac{1}{3}\sin(2x)$ gives $-u'' - u = \frac{4}{3}\sin(2x) - \frac{1}{3}\sin(2x) = \sin(2x)$, confirming the solution. On the other hand, $f(x) = \sin(x)$ fails the condition ($\int_0^\pi \sin^2(x) \, d\mathcal{L}^1(x) = \pi/2 \neq 0$), so $-u'' - u = \sin(x)$ has no solution — the forcing resonates with the eigenmode. **Conclusion (3)** (dimension equality): Here $\dim \ker(I - \tilde{K}) = 1$ and $\dim \ker(I - \tilde{K}^*) = 1$, so there is exactly one solvability condition on $f$ (orthogonality to $\sin$), and when it is satisfied the solution is unique up to a one-dimensional family. Concretely, if $u_0$ is any particular solution of $-u'' - u = f$, then the general solution is \begin{align*} u &= u_0 + c \sin, \quad c \in \mathbb{R}. \end{align*} One free parameter in the solution, one condition on the data — these are matched by conclusion (3). **Conclusion (4)** (the dichotomy): Since $\ker(I - \tilde{K}) = \operatorname{span}\{\sin\} \neq \{0\}$, we are in the resonance case. If instead we considered $\lambda = 1.5$ (not an eigenvalue), the kernel would be trivial, and the theorem would give unique solvability of $-u'' - 1.5 \, u = f$ for every $f \in L^2((0, \pi))$. [/example] ### Alternative A: Unique Solvability [example: Volterra Integral Equation] Let $H = L^2([0, 1])$ and define the Volterra [integral](/page/Integral) operator \begin{align*} K: L^2([0, 1]) &\to L^2([0, 1]) \\ u &\mapsto \int_0^x u(y) \, d\mathcal{L}^1(y). \end{align*} The operator $K$ is compact (it maps bounded [sets](/page/Set) in $L^2$ to bounded equicontinuous subsets of $C([0,1])$, which are precompact in $L^2$ by the Arzelà–Ascoli theorem and the compactness of the embedding $C \hookrightarrow L^2$). We check whether $\ker(I - K) = \{0\}$. Suppose $(I - K)u = 0$, i.e. \begin{align*} u(x) &= \int_0^x u(y) \, d\mathcal{L}^1(y) \quad \text{for a.e. } x \in [0,1]. \end{align*} The right-hand side is absolutely [continuous](/page/Continuity), so $u$ is absolutely continuous with $u(0) = 0$. Differentiating both sides gives $u'(x) = u(x)$ with initial condition $u(0) = 0$. The unique solution to this ODE is $u \equiv 0$, so $\ker(I - K) = \{0\}$. By the Fredholm Alternative (conclusion 4, first case), $I - K: L^2([0,1]) \to L^2([0,1])$ is a bijection with bounded inverse. The equation \begin{align*} u(x) - \int_0^x u(y) \, d\mathcal{L}^1(y) &= f(x) \end{align*} has a unique solution $u \in L^2([0,1])$ for every $f \in L^2([0,1])$. [/example] ### Alternative B: Finite-Dimensional Obstruction [example: Symmetric Integral Operator At Resonance] Let $H = L^2([0, \pi])$ and define the symmetric integral operator \begin{align*} K: L^2([0, \pi]) &\to L^2([0, \pi]) \\ u &\mapsto \frac{2}{\pi} \int_0^\pi \sin(x) \sin(y) \, u(y) \, d\mathcal{L}^1(y). \end{align*} This is a rank-one operator: $Ku = \frac{2}{\pi} \langle u, \sin \rangle_{L^2} \sin$, where $\langle u, \sin \rangle_{L^2} = \int_0^\pi u(y) \sin(y) \, d\mathcal{L}^1(y)$. Rank-one operators are compact (their range is one-dimensional). Since $K$ is defined by a symmetric kernel, $K^* = K$. The eigenvalue equation $Ku = \lambda u$ reduces to $\frac{2}{\pi} \langle u, \sin \rangle_{L^2} \sin = \lambda u$. If $\langle u, \sin \rangle_{L^2} = 0$, then $\lambda u = 0$, so either $\lambda = 0$ or $u = 0$. If $\langle u, \sin \rangle_{L^2} \neq 0$, then $u$ must be a multiple of $\sin$, and substituting $u = \sin$ gives $\lambda = \frac{2}{\pi} \|\sin\|_{L^2}^2 = \frac{2}{\pi} \cdot \frac{\pi}{2} = 1$. So $K$ has a single nonzero eigenvalue $\lambda = 1$. The kernel of $I - K$ is therefore $\ker(I - K) = \operatorname{span}\{\sin\}$, which is one-dimensional. Since $K = K^*$, conclusion (3) gives $\ker(I - K^*) = \ker(I - K) = \operatorname{span}\{\sin\}$. By conclusion (2), \begin{align*} \operatorname{Range}(I - K) &= \ker(I - K^*)^\perp = \{f \in L^2([0, \pi]) : \langle f, \sin \rangle_{L^2} = 0\}. \end{align*} So the equation $(I - K)u = f$ is solvable if and only if \begin{align*} \int_0^\pi f(y) \sin(y) \, d\mathcal{L}^1(y) &= 0. \end{align*} For instance, $f(y) = \cos(y)$ satisfies this condition (since $\int_0^\pi \cos(y) \sin(y) \, d\mathcal{L}^1(y) = 0$ by the orthogonality of $\sin$ and $\cos$ on $[0, \pi]$), so $(I - K)u = \cos$ is solvable. On the other hand, $f(y) = \sin(y)$ does not satisfy the condition ($\int_0^\pi \sin^2(y) \, d\mathcal{L}^1(y) = \pi/2 \neq 0$), so $(I - K)u = \sin$ has no solution — the forcing resonates with the mode $\sin$ and no steady state exists. [/example] ### Application to Elliptic PDE [example: Helmholtz Equation On A Bounded Domain] Let $U \subset \mathbb{R}^n$ be a bounded [open set](/page/Open%20Set) with smooth boundary. The Helmholtz equation with Dirichlet boundary conditions is \begin{align*} -\Delta u - \mu \, u &= f \quad \text{in } U, \\ u &= 0 \quad \text{on } \partial U, \end{align*} where $\mu \in \mathbb{R}$ and $f \in L^2(U)$. To cast this in Fredholm form, choose $\gamma > 0$ large enough that the shifted operator $L_\gamma = -\Delta + \gamma I$ is coercive on $H^1_0(U)$ (by [Gårding's inequality](/theorems/92), any $\gamma$ exceeding the Gårding constant suffices). The equation becomes \begin{align*} (-\Delta + \gamma I)u - (\gamma + \mu) u &= f, \\ L_\gamma u &= f + (\gamma + \mu) u, \\ u &= L_\gamma^{-1} f + (\gamma + \mu) L_\gamma^{-1} u = L_\gamma^{-1} f + (\gamma + \mu) K u, \end{align*} where $K := j \circ L_\gamma^{-1} \circ \iota: L^2(U) \to L^2(U)$ is the resolvent operator (compact by the [Rellich–Kondrachov theorem](/theorems/64), as described on the [Second-Order Elliptic Equations](/page/Second-Order%20Elliptic%20Equations) page). Setting $\tilde{K} := (\gamma + \mu) K$, which is compact, the equation takes the standard Fredholm form \begin{align*} (I - \tilde{K})u &= L_\gamma^{-1} f. \end{align*} The Fredholm Alternative now gives: either $\mu$ is not an eigenvalue of $-\Delta$ (i.e. $\ker(I - \tilde{K}) = \{0\}$), in which case the Helmholtz equation has a unique solution for every $f \in L^2(U)$; or $\mu = \lambda_k$ for some eigenvalue of $-\Delta$, in which case a solution exists if and only if $f$ is $L^2$-orthogonal to every eigenfunction $w$ of $-\Delta$ with eigenvalue $\lambda_k$. Since $-\Delta$ is self-adjoint on $H^1_0(U)$, the adjoint condition reduces to $\int_U f \, w \, d\mathcal{L}^n = 0$ for each such eigenfunction — exactly the orthogonality condition from the membrane example in the motivation. [/example] ## Problems ### Guided Problems The first two problems walk through each conclusion of the Fredholm Alternative for a concrete boundary value problem. Problem 1 treats the self-adjoint case, where the adjoint kernel coincides with the kernel. Problem 2 treats a non-[self-adjoint operator](/page/Self-Adjoint%20Operators), where the solvability condition involves different functions from those spanning the kernel — a distinction that is invisible in the self-adjoint setting and therefore easy to miss. [problem] **Problem 1 (Self-adjoint ODE at resonance).** Consider the Dirichlet boundary value problem \begin{align*} -u'' - 4u &= f \quad \text{in } (0, \pi), \\ u(0) = u(\pi) &= 0, \end{align*} where $f \in L^2((0, \pi))$. The eigenvalues of $-\partial_{xx}$ on $H^1_0((0, \pi))$ are $\lambda_k = k^2$ for $k = 1, 2, 3, \ldots$, so $\lambda = 4 = \lambda_2$ is the second eigenvalue. (a) Verify that the operator $L = -\partial_{xx}$ is self-adjoint on $H^1_0((0, \pi))$ by showing that the associated bilinear form $B[u, v] = \int_0^\pi u' v' \, d\mathcal{L}^1$ is symmetric. (b) Find $\ker(T)$ where $T = L - 4I$ by solving $-u'' - 4u = 0$ with Dirichlet conditions. State which conclusion of the Fredholm Alternative guarantees this kernel is finite-dimensional. (c) Identify $\ker(T^*)$ using self-adjointness. State the solvability condition for $-u'' - 4u = f$ and name which conclusions you are using. (d) Let $f(x) = \sin(x) + 3\sin(2x)$. Determine whether the equation is solvable. (e) Let $f(x) = 5\sin(3x)$. Verify solvability, find the general weak solution, and verify it satisfies the equation. [/problem] [solution] **Step 1 (Part a: self-adjointness).** The weak formulation of $Lu = -u''$ arises from [integration by parts](/theorems/210). For $u, v \in H^1_0((0, \pi))$, integration by parts gives \begin{align*} \int_0^\pi (-u'') v \, d\mathcal{L}^1 &= \int_0^\pi u' v' \, d\mathcal{L}^1 - [u' v]_0^\pi = \int_0^\pi u' v' \, d\mathcal{L}^1, \end{align*} where the boundary term vanishes because $v(0) = v(\pi) = 0$. The associated bilinear form is therefore \begin{align*} B[u, v] &= \int_0^\pi u'(x) \, v'(x) \, d\mathcal{L}^1(x), \end{align*} which is symmetric: $B[u, v] = B[v, u]$ for all $u, v \in H^1_0((0, \pi))$. Hence $L$ is self-adjoint. **Step 2 (Part b: kernel computation).** We solve the homogeneous ODE $u'' + 4u = 0$. The characteristic equation is $r^2 + 4 = 0$, which has roots $r = \pm 2i$. The general solution is \begin{align*} u(x) &= A\cos(2x) + B\sin(2x). \end{align*} Imposing the boundary condition $u(0) = 0$ gives $A = 0$. The second condition $u(\pi) = B\sin(2\pi) = 0$ is satisfied for every $B \in \mathbb{R}$. Therefore \begin{align*} \ker(T) &= \operatorname{span}\{\sin(2x)\}, \end{align*} which is one-dimensional. **Conclusion (1)** of the Fredholm Alternative guarantees that $\ker(T)$ is finite-dimensional — the computation confirms this and identifies the kernel explicitly. **Step 3 (Part c: solvability condition).** Since $L$ is self-adjoint, $T^* = (L - 4I)^* = L^* - 4I = L - 4I = T$. Therefore \begin{align*} \ker(T^*) &= \ker(T) = \operatorname{span}\{\sin(2x)\}. \end{align*} By **conclusion (3)**, $\dim \ker(T) = 1 = \dim \ker(T^*)$, so there is exactly one solvability condition. By **conclusion (2)**, the equation $Tu = f$ is solvable if and only if $f \in \ker(T^*)^\perp$, which gives \begin{align*} \int_0^\pi f(x) \sin(2x) \, d\mathcal{L}^1(x) &= 0. \end{align*} **Step 4 (Part d: solvability check for $f = \sin(x) + 3\sin(2x)$).** We compute \begin{align*} \int_0^\pi \bigl(\sin(x) + 3\sin(2x)\bigr) \sin(2x) \, d\mathcal{L}^1(x) &= \int_0^\pi \sin(x)\sin(2x) \, d\mathcal{L}^1(x) + 3\int_0^\pi \sin^2(2x) \, d\mathcal{L}^1(x). \end{align*} The first integral vanishes by orthogonality of the Dirichlet eigenfunctions: $\int_0^\pi \sin(jx)\sin(kx) \, d\mathcal{L}^1 = \frac{\pi}{2}\delta_{jk}$ for $j, k \geq 1$, and here $j = 1 \neq 2 = k$. The second integral equals $3 \cdot \frac{\pi}{2}$. Therefore \begin{align*} \int_0^\pi f(x)\sin(2x) \, d\mathcal{L}^1(x) &= 0 + \frac{3\pi}{2} = \frac{3\pi}{2} \neq 0. \end{align*} The solvability condition fails. The component $3\sin(2x)$ in $f$ is a nonzero multiple of the resonant eigenfunction $\sin(2x)$, so the forcing resonates with the mode and no solution exists. We are in the resonance case of **conclusion (4)**, but with $f \notin \ker(T^*)^\perp$. **Step 5 (Part e: solving for $f = 5\sin(3x)$).** First, verify solvability. By orthogonality of Dirichlet eigenfunctions ($k = 3 \neq 2$): \begin{align*} \int_0^\pi 5\sin(3x)\sin(2x) \, d\mathcal{L}^1(x) &= 5 \cdot 0 = 0. \quad \checkmark \end{align*} Next, find a particular solution by the ansatz $u_p(x) = C\sin(3x)$. Substituting into the equation: \begin{align*} -u_p'' - 4u_p &= 9C\sin(3x) - 4C\sin(3x) = 5C\sin(3x). \end{align*} Setting $5C = 5$ gives $C = 1$. The particular solution is $u_p(x) = \sin(3x)$. By **conclusion (4)** (resonance case), the general solution is $u_p$ plus an arbitrary element of $\ker(T)$: \begin{align*} u(x) &= \sin(3x) + c\sin(2x), \quad c \in \mathbb{R}. \end{align*} One free parameter (from the one-dimensional kernel), matching the one solvability condition on $f$ — this is conclusion (3) in action. **Verification.** For any $c \in \mathbb{R}$: \begin{align*} -u'' - 4u &= -\bigl(-9\sin(3x) - 4c\sin(2x)\bigr) - 4\bigl(\sin(3x) + c\sin(2x)\bigr) \\ &= 9\sin(3x) + 4c\sin(2x) - 4\sin(3x) - 4c\sin(2x) \\ &= 5\sin(3x). \quad \checkmark \end{align*} [/solution] [problem] **Problem 2 (Non-self-adjoint ODE at resonance).** Consider the advection–diffusion problem with Dirichlet conditions: \begin{align*} -u'' + 2u' - 2u &= f \quad \text{in } (0, \pi), \\ u(0) = u(\pi) &= 0, \end{align*} where $f \in L^2((0, \pi))$. Define $L = -\partial_{xx} + 2\partial_x$, so the equation reads $(L - 2I)u = f$. (a) Find the eigenvalues and eigenfunctions of $L$ on $H^1_0((0, \pi))$ by solving $-u'' + 2u' = \lambda u$ with Dirichlet conditions. (b) Compute the adjoint operator $L^*$ by integration by parts. Find the eigenvalues and eigenfunctions of $L^*$ on $H^1_0((0, \pi))$. (c) Identify $\ker(L - 2I)$ and $\ker(L^* - 2I)$. Verify that conclusion (3) holds, and explain why the solvability condition involves the adjoint eigenfunctions rather than the original eigenfunctions. (d) State the solvability condition for $-u'' + 2u' - 2u = f$. (e) For $f(x) = e^x \sin(2x)$, verify solvability and find the general solution. (f) For $f(x) = e^x \sin(x)$, show the equation is not solvable. [/problem] [solution] **Step 1 (Part a: eigenvalues and eigenfunctions of $L$).** The eigenvalue equation $-u'' + 2u' = \lambda u$ rearranges to the constant-coefficient ODE \begin{align*} u'' - 2u' + \lambda u &= 0. \end{align*} The characteristic equation is $r^2 - 2r + \lambda = 0$, giving $r = 1 \pm \sqrt{1 - \lambda}$. For $\lambda > 1$, write $\omega = \sqrt{\lambda - 1}$ so that $r = 1 \pm i\omega$. The general solution is \begin{align*} u(x) &= e^x\bigl(A\cos(\omega x) + B\sin(\omega x)\bigr). \end{align*} Imposing $u(0) = 0$ gives $A = 0$. The condition $u(\pi) = Be^\pi \sin(\omega \pi) = 0$ requires $\sin(\omega \pi) = 0$ (since $Be^\pi \neq 0$ for nontrivial solutions), i.e. $\omega = k$ for $k \in \mathbb{N}$. Therefore the eigenvalues and eigenfunctions are \begin{align*} \lambda_k &= 1 + k^2, \quad u_k(x) = e^x \sin(kx), \quad k = 1, 2, 3, \ldots \end{align*} In particular, $\lambda_1 = 2$ and $u_1(x) = e^x \sin(x)$. **Step 2 (Part b: adjoint operator and its eigenfunctions).** We compute $L^*$ by integration by parts. For $u, v \in H^1_0((0, \pi))$: \begin{align*} \int_0^\pi (Lu)\, v \, d\mathcal{L}^1 &= \int_0^\pi (-u'' + 2u')\, v \, d\mathcal{L}^1. \end{align*} For the first term, integration by parts gives $\int_0^\pi (-u'')v \, d\mathcal{L}^1 = \int_0^\pi u'v' \, d\mathcal{L}^1$ (the boundary term $[-u'v]_0^\pi$ vanishes because $v \in H^1_0$). For the second term, integration by parts gives \begin{align*} \int_0^\pi 2u' v \, d\mathcal{L}^1 &= [2uv]_0^\pi - \int_0^\pi 2u v' \, d\mathcal{L}^1 = -\int_0^\pi 2u v' \, d\mathcal{L}^1, \end{align*} where the boundary term vanishes because $u, v \in H^1_0$. Combining: \begin{align*} \int_0^\pi (Lu)\, v \, d\mathcal{L}^1 &= \int_0^\pi u'v' \, d\mathcal{L}^1 - \int_0^\pi 2uv' \, d\mathcal{L}^1. \end{align*} Integrating the first term by parts once more: $\int_0^\pi u'v' \, d\mathcal{L}^1 = -\int_0^\pi u\, v'' \, d\mathcal{L}^1$ (boundary term vanishes since $u \in H^1_0$). Therefore \begin{align*} \int_0^\pi (Lu)\, v \, d\mathcal{L}^1 &= \int_0^\pi u\bigl(-v'' - 2v'\bigr) \, d\mathcal{L}^1, \end{align*} which identifies $L^* v = -v'' - 2v'$. The drift term $+2\partial_x$ in $L$ becomes $-2\partial_x$ in $L^*$ — the advection reverses direction under the adjoint. The eigenvalue equation $L^* v = \mu v$ gives $v'' + 2v' + \mu v = 0$, with characteristic equation $r^2 + 2r + \mu = 0$ and roots $r = -1 \pm \sqrt{1 - \mu}$. For $\mu > 1$, writing $\omega = \sqrt{\mu - 1}$: \begin{align*} v(x) &= e^{-x}\bigl(A\cos(\omega x) + B\sin(\omega x)\bigr). \end{align*} Imposing $v(0) = 0$ gives $A = 0$, and $v(\pi) = 0$ requires $\omega = k$. The adjoint eigenvalues and eigenfunctions are \begin{align*} \mu_k &= 1 + k^2, \quad v_k(x) = e^{-x}\sin(kx), \quad k = 1, 2, 3, \ldots \end{align*} The eigenvalues $\mu_k = \lambda_k$ coincide (as they must — the spectra of $L$ and $L^*$ are identical), but the eigenfunctions differ: $u_k(x) = e^x \sin(kx)$ versus $v_k(x) = e^{-x}\sin(kx)$. The exponential factors point in opposite directions, reflecting the reversal of the drift. **Step 3 (Part c: kernels and conclusion (3)).** At the first eigenvalue $\lambda_1 = 2$: \begin{align*} \ker(L - 2I) &= \operatorname{span}\{e^x \sin(x)\}, \\ \ker(L^* - 2I) &= \operatorname{span}\{e^{-x}\sin(x)\}. \end{align*} Both are one-dimensional, confirming **conclusion (3)**: $\dim \ker(T) = 1 = \dim \ker(T^*)$. The solvability condition involves the adjoint eigenfunctions, not the original eigenfunctions, because **conclusion (2)** says $\operatorname{Range}(T) = \ker(T^*)^\perp$ — it is orthogonality to the *adjoint* kernel that controls solvability. In the self-adjoint case $\ker(T) = \ker(T^*)$, so this distinction is invisible. Here, the distinction is concrete: solvability requires $\int_0^\pi f \cdot e^{-x}\sin(x) \, d\mathcal{L}^1 = 0$, not $\int_0^\pi f \cdot e^x \sin(x) \, d\mathcal{L}^1 = 0$. **Step 4 (Part d: solvability condition).** By conclusion (2), the equation $(L - 2I)u = f$ is solvable if and only if \begin{align*} \int_0^\pi f(x) \, e^{-x}\sin(x) \, d\mathcal{L}^1(x) &= 0. \end{align*} **Step 5 (Part e: solving for $f(x) = e^x \sin(2x)$).** First, verify solvability. The key observation is that the eigenfunctions $\{u_k\}$ and $\{v_k\}$ satisfy the **biorthogonality relation** \begin{align*} \int_0^\pi u_j(x) \, v_k(x) \, d\mathcal{L}^1(x) &= \int_0^\pi e^x \sin(jx) \cdot e^{-x}\sin(kx) \, d\mathcal{L}^1(x) = \int_0^\pi \sin(jx)\sin(kx) \, d\mathcal{L}^1(x) = \frac{\pi}{2}\delta_{jk}. \end{align*} The exponential factors cancel, reducing to the standard $L^2$ orthogonality of sines. Since $f = u_2 = e^x\sin(2x)$, the solvability condition becomes \begin{align*} \int_0^\pi e^x \sin(2x) \cdot e^{-x}\sin(x) \, d\mathcal{L}^1(x) &= \int_0^\pi \sin(2x)\sin(x) \, d\mathcal{L}^1(x) = 0, \end{align*} by orthogonality ($j = 2 \neq 1 = k$). The condition is satisfied. To find the solution, observe that $f = u_2$ is the second eigenfunction of $L$, with $Lu_2 = \lambda_2 u_2 = 5 u_2$. A particular solution is $u_p = \alpha \, u_2$ where \begin{align*} (L - 2I)(\alpha \, u_2) &= \alpha(\lambda_2 - 2) u_2 = 3\alpha \, u_2. \end{align*} Setting $3\alpha = 1$ gives $\alpha = \frac{1}{3}$, so $u_p(x) = \frac{1}{3} e^x \sin(2x)$. The general solution is \begin{align*} u(x) &= \frac{1}{3}e^x \sin(2x) + c \, e^x \sin(x), \quad c \in \mathbb{R}. \end{align*} **Verification.** Since $Lu_k = \lambda_k u_k$: \begin{align*} (L - 2I)u &= \frac{1}{3}(\lambda_2 - 2) u_2 + c(\lambda_1 - 2)u_1 = \frac{1}{3} \cdot 3 \cdot e^x\sin(2x) + c \cdot 0 \cdot e^x\sin(x) = e^x\sin(2x). \quad \checkmark \end{align*} **Step 6 (Part f: non-solvability for $f(x) = e^x \sin(x)$).** We compute \begin{align*} \int_0^\pi e^x \sin(x) \cdot e^{-x}\sin(x) \, d\mathcal{L}^1(x) &= \int_0^\pi \sin^2(x) \, d\mathcal{L}^1(x) = \frac{\pi}{2} \neq 0. \end{align*} The solvability condition fails. The forcing $f = u_1 = e^x \sin(x)$ is exactly the eigenfunction of $L$ at eigenvalue $\lambda_1 = 2$, so we are trying to solve $(L - 2I)u = u_1$, which would require $u_1 \perp v_1$ in $L^2$. But $\int_0^\pi u_1 v_1 = \frac{\pi}{2} \neq 0$, so no solution exists — the forcing resonates with the adjoint mode. [/solution] ### Further Problems [problem] **Problem 3 (Neumann Laplacian).** Consider the Neumann boundary value problem on $(0, 1)$: \begin{align*} -u'' &= f \quad \text{in } (0, 1), \\ u'(0) = u'(1) &= 0, \end{align*} where $f \in L^2((0, 1))$. (a) Write the weak formulation, explain why the bilinear form $B[u, v] = \int_0^1 u'v' \, d\mathcal{L}^1$ is not coercive on the function space $H^1((0, 1))$, and reduce the problem to Fredholm form $(I - K)u = h$ by shifting. (b) Find $\ker(I - K)$ and state the solvability condition. (c) Let $f(x) = \cos(\pi x)$. Verify the solvability condition, find the general solution, and verify it directly. (d) Let $f(x) = 1$. Show the equation is not solvable and give a physical interpretation. [/problem] [solution] **Step 1 (Part a: weak formulation and Fredholm reduction).** The function space for the Neumann problem is $H^1((0, 1))$ (no boundary condition imposed on $u$ itself — only on $u'$, which enters via integration by parts). The weak formulation asks: find $u \in H^1((0, 1))$ such that \begin{align*} \int_0^1 u'v' \, d\mathcal{L}^1 &= \int_0^1 f v \, d\mathcal{L}^1 \quad \text{for all } v \in H^1((0, 1)). \end{align*} (This arises from multiplying $-u'' = f$ by $v$, integrating by parts, and noting that the boundary terms $[u'v]_0^1$ vanish when $u'(0) = u'(1) = 0$.) The bilinear form $B[u, v] = \int_0^1 u'v' \, d\mathcal{L}^1$ is **not coercive** on $H^1((0, 1))$: for constant [functions](/page/Function) $u(x) \equiv c$ with $c \neq 0$, \begin{align*} B[c, c] &= \int_0^1 0 \cdot 0 \, d\mathcal{L}^1 = 0, \quad \text{but} \quad \|c\|_{H^1}^2 = \int_0^1 (c^2 + 0) \, d\mathcal{L}^1 = c^2 > 0. \end{align*} No constant $\beta > 0$ satisfies $B[u, u] \geq \beta \|u\|_{H^1}^2$ for all $u$. To restore coercivity, shift by $\gamma = 1$. Define \begin{align*} B_1[u, v] &:= \int_0^1 (u'v' + uv) \, d\mathcal{L}^1 = (u, v)_{H^1}, \end{align*} which is the $H^1$ inner product — coercive with constant $\beta = 1$. By the [Lax–Milgram theorem](/theorems/91), the shifted operator $L_1 := -\partial_{xx} + I$ (with Neumann conditions) is an isomorphism from $H^1((0, 1))$ to its dual. The resolvent \begin{align*} K := j \circ L_1^{-1} \circ \iota : L^2((0, 1)) \to L^2((0, 1)) \end{align*} (inclusion into the dual, inversion, then compact Sobolev embedding back into $L^2$) is a compact self-adjoint operator. The original equation $-u'' = f$ rewrites as $(-\partial_{xx} + I)u = f + u$, i.e. $u = K(f + u) = Kf + Ku$. Rearranging: \begin{align*} (I - K)u &= Kf. \end{align*} This is in standard Fredholm form. **Step 2 (Part b: kernel and solvability).** The condition $u \in \ker(I - K)$ means $u = Ku$, i.e. $(-\partial_{xx} + I)u = u$, which simplifies to $-u'' = 0$. The general solution of $u'' = 0$ is $u(x) = ax + b$. Imposing Neumann conditions: $u'(0) = a = 0$ and $u'(1) = a = 0$. Therefore $a = 0$ and \begin{align*} \ker(I - K) &= \{u(x) = b : b \in \mathbb{R}\} = \operatorname{span}\{1\}. \end{align*} Since $K$ is self-adjoint ($L$ is self-adjoint with Neumann conditions), $\ker(I - K^*) = \ker(I - K) = \operatorname{span}\{1\}$. By conclusion (2), the equation $-u'' = f$ is solvable if and only if \begin{align*} \int_0^1 f(x) \cdot 1 \, d\mathcal{L}^1(x) &= 0, \end{align*} i.e. $f$ has zero mean. Conclusion (3) confirms $\dim \ker = 1 = \dim \ker^*$: one solvability condition, and the solution (when it exists) is unique up to one additive constant. **Step 3 (Part c: solving for $f(x) = \cos(\pi x)$).** Verify solvability: \begin{align*} \int_0^1 \cos(\pi x) \, d\mathcal{L}^1(x) &= \left[\frac{\sin(\pi x)}{\pi}\right]_0^1 = \frac{\sin(\pi) - \sin(0)}{\pi} = 0. \quad \checkmark \end{align*} Solve by ansatz. Try $u_p(x) = C\cos(\pi x)$. Then \begin{align*} -u_p''(x) &= C\pi^2 \cos(\pi x). \end{align*} Setting $C\pi^2 = 1$ gives $C = \frac{1}{\pi^2}$, so $u_p(x) = \frac{1}{\pi^2}\cos(\pi x)$. Verify the Neumann conditions: $u_p'(x) = -\frac{1}{\pi}\sin(\pi x)$, so $u_p'(0) = 0$ and $u_p'(1) = -\frac{1}{\pi}\sin(\pi) = 0$. $\checkmark$ The general solution is \begin{align*} u(x) &= \frac{1}{\pi^2}\cos(\pi x) + c, \quad c \in \mathbb{R}. \end{align*} **Verification.** $-u'' = -\bigl(-\cos(\pi x)\bigr) = \cos(\pi x) = f(x)$, and the additive constant contributes $0$ to $-u''$. $\checkmark$ **Step 4 (Part d: non-solvability for $f(x) = 1$).** Compute \begin{align*} \int_0^1 1 \, d\mathcal{L}^1(x) &= 1 \neq 0. \end{align*} The solvability condition fails. Physically, the Neumann condition $u'(0) = u'(1) = 0$ means the boundary is insulated — no flux enters or leaves the interval. If $-u'' = f > 0$ represents a steady-state [heat equation](/page/Heat%20Equation) with a strictly positive source $f$, heat is being generated everywhere but cannot escape through the insulated endpoints. No equilibrium temperature [distribution](/page/Distribution) exists. The solvability condition $\int_0^1 f = 0$ expresses the physical requirement that the total heat generated equals the total heat absorbed — which is impossible when $f$ is a positive constant. [/solution] [problem] **Problem 4 (Periodic boundary conditions with two-dimensional kernel).** Consider the periodic eigenvalue problem \begin{align*} -u'' - u &= f \quad \text{in } (0, 2\pi), \\ u(0) &= u(2\pi), \\ u'(0) &= u'(2\pi), \end{align*} where $f \in L^2((0, 2\pi))$. (a) Find $\ker(T)$ where $T = -\partial_{xx} - I$ with periodic boundary conditions. Show it is two-dimensional and identify a basis. (b) The operator $-\partial_{xx}$ with periodic boundary conditions is self-adjoint. State the solvability conditions and describe the solution set when they are satisfied. (c) Let $f(x) = \cos(3x) + 2\sin(5x)$. Verify both solvability conditions, find the general solution, and verify it satisfies the equation. (d) Let $f(x) = 1 + \cos(x)$. Determine which solvability condition(s) fail and explain why no solution exists. [/problem] [solution] **Step 1 (Part a: two-dimensional kernel).** Solve $u'' + u = 0$. The general solution is \begin{align*} u(x) &= A\cos(x) + B\sin(x). \end{align*} Verify the periodic boundary conditions. For $u(0) = u(2\pi)$: \begin{align*} A = A\cos(2\pi) + B\sin(2\pi) = A \cdot 1 + B \cdot 0 = A. \quad \checkmark \end{align*} For $u'(0) = u'(2\pi)$: $u'(x) = -A\sin(x) + B\cos(x)$, so \begin{align*} B = -A\sin(2\pi) + B\cos(2\pi) = 0 + B = B. \quad \checkmark \end{align*} Both conditions are satisfied for all $A, B \in \mathbb{R}$. Therefore \begin{align*} \ker(T) &= \operatorname{span}\{\cos(x),\, \sin(x)\}, \end{align*} which is two-dimensional. **Conclusion (1)** guarantees finite-dimensionality; the explicit computation shows the dimension is $2$. This is the essential difference from the Dirichlet problems in Problems 1 and 2: the periodic eigenvalue $\lambda = 1$ has **multiplicity two** (both $\cos(x)$ and $\sin(x)$ are eigenfunctions), whereas Dirichlet eigenvalues are always simple. **Step 2 (Part b: solvability conditions and solution set).** Since $-\partial_{xx}$ with periodic conditions is self-adjoint, $\ker(T^*) = \ker(T) = \operatorname{span}\{\cos(x), \sin(x)\}$. Conclusion (3) gives $\dim \ker(T) = 2 = \dim \ker(T^*)$. By conclusion (2), the equation $Tu = f$ is solvable if and only if $f \perp \ker(T^*)$, which gives **two** independent conditions: \begin{align*} \int_0^{2\pi} f(x)\cos(x) \, d\mathcal{L}^1(x) &= 0, \\ \int_0^{2\pi} f(x)\sin(x) \, d\mathcal{L}^1(x) &= 0. \end{align*} In Fourier-analytic terms, these conditions require that the first Fourier cosine coefficient and the first Fourier sine coefficient of $f$ both vanish — equivalently, the first complex Fourier coefficient $\hat{f}(1) = 0$. When the conditions are satisfied, conclusion (4) (resonance case) gives the general solution as $u = u_p + A\cos(x) + B\sin(x)$ for arbitrary $A, B \in \mathbb{R}$. Two solvability conditions on $f$, two free parameters in $u$ — matched by conclusion (3). **Step 3 (Part c: solving for $f(x) = \cos(3x) + 2\sin(5x)$).** Verify the first condition, using orthogonality of the trigonometric system on $[0, 2\pi]$ (i.e. $\int_0^{2\pi} \cos(jx)\cos(kx) \, d\mathcal{L}^1 = \pi \delta_{jk}$ for $j, k \geq 1$, and $\int_0^{2\pi} \sin(jx)\cos(kx) \, d\mathcal{L}^1 = 0$ for all $j, k$): \begin{align*} \int_0^{2\pi} \bigl(\cos(3x) + 2\sin(5x)\bigr)\cos(x) \, d\mathcal{L}^1(x) &= \int_0^{2\pi}\cos(3x)\cos(x) \, d\mathcal{L}^1 + 2\int_0^{2\pi}\sin(5x)\cos(x) \, d\mathcal{L}^1 \\ &= \pi\delta_{3,1} + 2 \cdot 0 = 0. \quad \checkmark \end{align*} Verify the second condition: \begin{align*} \int_0^{2\pi} \bigl(\cos(3x) + 2\sin(5x)\bigr)\sin(x) \, d\mathcal{L}^1(x) &= \int_0^{2\pi}\cos(3x)\sin(x) \, d\mathcal{L}^1 + 2\int_0^{2\pi}\sin(5x)\sin(x) \, d\mathcal{L}^1 \\ &= 0 + 2\pi\delta_{5,1} = 0. \quad \checkmark \end{align*} Both conditions pass. Find a particular solution by the ansatz $u_p(x) = C_1 \cos(3x) + C_2 \sin(5x)$: \begin{align*} -u_p'' - u_p &= 9C_1\cos(3x) + 25C_2\sin(5x) - C_1\cos(3x) - C_2\sin(5x) \\ &= 8C_1\cos(3x) + 24C_2\sin(5x). \end{align*} Matching coefficients with $f = \cos(3x) + 2\sin(5x)$: $8C_1 = 1$ and $24C_2 = 2$, giving $C_1 = \frac{1}{8}$ and $C_2 = \frac{1}{12}$. The general solution is \begin{align*} u(x) &= \frac{1}{8}\cos(3x) + \frac{1}{12}\sin(5x) + A\cos(x) + B\sin(x), \quad A, B \in \mathbb{R}. \end{align*} **Verification.** The kernel elements contribute $-(\cos(x))'' - \cos(x) = \cos(x) - \cos(x) = 0$, and similarly for $\sin(x)$. For the particular solution: \begin{align*} -u_p'' - u_p &= \frac{9}{8}\cos(3x) + \frac{25}{12}\sin(5x) - \frac{1}{8}\cos(3x) - \frac{1}{12}\sin(5x) \\ &= \frac{8}{8}\cos(3x) + \frac{24}{12}\sin(5x) = \cos(3x) + 2\sin(5x). \quad \checkmark \end{align*} **Step 4 (Part d: non-solvability for $f(x) = 1 + \cos(x)$).** Compute the first condition: \begin{align*} \int_0^{2\pi} (1 + \cos(x))\cos(x) \, d\mathcal{L}^1(x) &= \int_0^{2\pi}\cos(x) \, d\mathcal{L}^1 + \int_0^{2\pi}\cos^2(x) \, d\mathcal{L}^1 = 0 + \pi = \pi \neq 0. \end{align*} The first solvability condition fails: $f$ has a nonzero projection $\cos(x)$ onto the kernel element $\cos(x)$. This component resonates with the eigenmode and prevents the existence of a solution. For completeness, the second condition does pass: \begin{align*} \int_0^{2\pi}(1 + \cos(x))\sin(x) \, d\mathcal{L}^1(x) &= \int_0^{2\pi}\sin(x) \, d\mathcal{L}^1 + \int_0^{2\pi}\cos(x)\sin(x) \, d\mathcal{L}^1 = 0 + 0 = 0. \end{align*} But both conditions must hold simultaneously. The failure of even one is enough to preclude solvability. The constant term $1$ in $f$ is not the obstruction (it is orthogonal to both $\cos(x)$ and $\sin(x)$); the obstruction is entirely the $\cos(x)$ component. [/solution] ## References 1. L. C. Evans, *Partial Differential Equations*, 2nd ed. (2010). 2. H. Brezis, *Functional Analysis, [Sobolev Spaces](/page/Sobolev%20Space) and Partial Differential Equations* (2011). 3. F. Riesz and B. Sz.-Nagy, *Functional Analysis* (1955). 4. K. Yosida, *Functional Analysis*, 6th ed. (1980).