Boundary-Value Problems and Solvability, differential equations lesson

A second-order equation leaves two constants free, and a boundary-value problem supplies two conditions, so it seems the conditions should always pick out one solution. Yet the problem $y'' + y = \sin x$ on $[0, \pi]$ with $y(0) = y(\pi) = 0$ has no solution at all, and replacing the forcing by $\cos x$ gives infinitely many. Here $x$ is the independent variable and $y' = dy/dx$ . This lesson writes the boundary conditions as a linear system for the free constants, shows how the kernel of the homogeneous problem separates the unique case from the resonant one, and derives the condition the forcing has to meet in the resonant case.

Key terms

Solvability of boundary-value problems

Definition Let $p$ , $q$ and $f$ be continuous on a closed interval $[a, b]$ , and write

L[y] = y'' + p(x)\,y' + q(x)\,y.

A linear two-point boundary-value problem asks for a twice differentiable function $y$ on $[a, b]$ with

\begin{gathered} L[y] = f \ \text{ on } [a, b], \\[0.4em] B_1[y] = \gamma_1, \qquad B_2[y] = \gamma_2, \end{gathered}

where each $B_i[y]$ is a fixed linear combination of the four numbers $y(a)$ , $y'(a)$ , $y(b)$ and $y'(b)$ , and $\gamma_1$ and $\gamma_2$ are given numbers. Solvability asks whether a solution exists and, if so, how many there are. Such a problem always has no solution, exactly one or infinitely many. The reason is that every solution of $L[y] = f$ has the form $y = y_p + c_1y_1 + c_2y_2$ , where $y_p$ is one solution and $y_1$ , $y_2$ form a fundamental set of $L[y] = 0$ , so the two boundary conditions become two linear equations for $c_1$ and $c_2$ . The classification needs both the equation and the boundary conditions to be linear. A nonlinear problem can have, for example, exactly two solutions.

In plain words The equation leaves two constants free, and the boundary conditions are two linear equations for them. Two linear equations in two unknowns can have no solution, exactly one or infinitely many, and the boundary-value problem inherits exactly these three outcomes.

Example The equation $y'' = 0$ has the general solution $y = c_1 + c_2x$ . With $y(0) = \alpha$ and $y(1) = \beta$ the conditions read $c_1 = \alpha$ and $c_1 + c_2 = \beta$ , so there is exactly one solution, $y = \alpha + (\beta - \alpha)x$ . With $y'(0) = 0$ and $y'(1) = \gamma$ both conditions concern $c_2$ alone. They read $c_2 = 0$ and $c_2 = \gamma$ , so for $\gamma \ne 0$ there is no solution, and for $\gamma = 0$ every constant function is a solution. Now keep the equation but make one condition nonlinear, $y(0)^2 = 1$ and $y(1) = 0$ . The conditions give $c_1 = \pm 1$ and $c_1 + c_2 = 0$ , so the solutions are exactly $y = 1 - x$ and $y = x - 1$ . Exactly two solutions can never happen in a linear problem.

Linear boundary map

Definition Let $y_1$ and $y_2$ be a fundamental set of solutions of $L[y] = 0$ . The linear boundary map sends each homogeneous solution $y = c_1y_1 + c_2y_2$ to its pair of boundary values $(B_1[y], B_2[y])$ . In terms of the coefficient vector $c = (c_1, c_2)^{T}$ it is multiplication by the matrix

M = \begin{pmatrix} B_1[y_1] & B_1[y_2] \\ B_2[y_1] & B_2[y_2] \end{pmatrix}.

Because $B_1$ and $B_2$ are linear, the general solution $y = y_p + c_1y_1 + c_2y_2$ of the forced equation meets the boundary conditions exactly when

\begin{gathered} Mc = d, \\[0.4em] d = \begin{pmatrix} \gamma_1 - B_1[y_p] \\ \gamma_2 - B_2[y_p] \end{pmatrix}. \end{gathered}

A different fundamental set multiplies $M$ on the right by an invertible matrix, so the rank of $M$ , and in particular whether $M$ is invertible, does not depend on that choice.

In plain words The boundary map records what the free part of the solution does at the two ends. If its matrix is invertible, any required pair of endpoint values can be reached, and in exactly one way.

Example For $y'' + y = f$ on $[0, \pi]$ with $y(0) = 0$ and $y(\pi) = 0$ , take $y_1 = \sin x$ and $y_2 = \cos x$ . The homogeneous solution $A\sin x + B\cos x$ has the value $B$ at $0$ and the value $-B$ at $\pi$ , so

M = \begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix}.

Both rows constrain $B$ only, the rank is $1$ , and $M$ is singular. On the shorter interval $[0, \pi/2]$ with $y(0) = 0$ and $y(\pi/2) = 0$ , the two values are $B$ and $A$ , so

M = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},

which is invertible, with determinant $-1$ .

Kernel of the homogeneous problem

Definition The homogeneous problem keeps the operator and the boundary operators and sets the forcing and the data to zero,

\begin{gathered} L[y] = 0, \\[0.4em] B_1[y] = 0, \qquad B_2[y] = 0. \end{gathered}

Its solutions form a vector space, called the kernel of the homogeneous problem. It always contains the zero function. A homogeneous solution $c_1y_1 + c_2y_2$ lies in it exactly when $Mc = 0$ , so the kernel has dimension $2 - \operatorname{rank} M$ , which is $0$ , $1$ or $2$ . If the forced problem has a solution $y^{*}$ , then its solutions are exactly the functions $y^{*} + u$ with $u$ in the kernel, because the difference of two solutions of the forced problem solves the homogeneous one.

In plain words The kernel collects the shapes that satisfy the equation without forcing and are invisible to the boundary conditions. Any of them can be added to a solution without changing the forcing or the endpoint data, so the kernel measures exactly how uniqueness fails.

Example For $y'' + y = 0$ with $y(0) = y(\pi) = 0$ , the matrix above forces $B = 0$ and leaves $A$ free, so the kernel consists of the functions $A\sin x$ and has dimension $1$ . On $[0, \pi/2]$ the invertible matrix forces $A = B = 0$ , and the kernel is $\{0\}$ . With the periodic conditions $y(0) = y(2\pi)$ and $y'(0) = y'(2\pi)$ on $[0, 2\pi]$ , every function $A\sin x + B\cos x$ satisfies both conditions, the matrix is the zero matrix and the kernel has dimension $2$ .

Resonant forcing

Definition A linear boundary-value problem is resonant when its homogeneous problem has a nonzero solution, that is, when $M$ is singular. Forcing applied to a resonant problem is called resonant forcing. A resonant problem never has exactly one solution. It has none when the vector $d$ lies outside the column space of $M$ , and infinitely many when $d$ lies inside it, since every kernel function can then be added to a solution. Which of the two happens depends on the forcing and the boundary data, not on the operator alone.

In plain words A nonzero kernel function is a free mode that the boundary conditions cannot detect, much like the natural frequency of an undamped oscillator. Forcing can drive that mode in a way the boundaries cannot absorb, and then no solution exists. Resonance alone tells you only that a unique solution is impossible.

Example Take $y'' + y = \sin x$ on $[0, \pi]$ with $y(0) = y(\pi) = 0$ , a problem whose kernel contains $\sin x$ . A particular solution is $y_p = -\tfrac{1}{2}x\cos x$ , because

\begin{aligned} y_p' &= -\tfrac{1}{2}\cos x + \tfrac{1}{2}x\sin x, \\[0.4em] y_p'' &= \sin x + \tfrac{1}{2}x\cos x, \end{aligned}

and so $y_p'' + y_p = \sin x$ . The general solution of the equation is $y = -\tfrac{1}{2}x\cos x + A\sin x + B\cos x$ . The left condition gives $y(0) = B = 0$ . At the right end

y(\pi) = \frac{\pi}{2} - B = \frac{\pi}{2},

and no choice of $A$ changes this, because $\sin\pi = 0$ . So the boundary-value problem has no solution, although the differential equation alone has a two-parameter family of them. The growing factor $x$ in $y_p$ is the same one that appears when an undamped oscillator is driven at its natural frequency.

Compatibility condition

Definition A compatibility condition is a condition on the forcing and the boundary data that is necessary and sufficient for a resonant problem to have a solution. In terms of the boundary map it says that $d$ lies in the column space of $M$ , which amounts to one scalar equation when the kernel has dimension $1$ and two when it has dimension $2$ . For the problem on $[0, \pi]$

\begin{gathered} y'' + y = f(x), \\[0.4em] y(0) = 0, \qquad y(\pi) = 0, \end{gathered}

with $f$ continuous on $[0, \pi]$ , the compatibility condition is

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

The condition is necessary. Multiplying the equation by $\sin x$ and integrating by parts twice gives

\begin{aligned} &\int_0^\pi f(x)\sin x\,dx \\[0.4em] &\quad = \int_0^\pi (y'' + y)\sin x\,dx \\[0.4em] &\quad = \Big[\, y'\sin x - y\cos x \,\Big]_0^\pi, \end{aligned}

and the bracket is zero because $\sin 0 = \sin\pi = 0$ and $y(0) = y(\pi) = 0$ . The condition is also sufficient. By variation of parameters, the solution of the equation with $y(0) = y'(0) = 0$ is

y_0(x) = \int_0^x \sin(x - t)\,f(t)\,dt

and every solution with $y(0) = 0$ has the form $y_0 + A\sin x$ . Since $\sin(\pi - t) = \sin t$ , the value $y_0(\pi)$ equals the integral of $f\sin x$ , so when that integral vanishes every function $y_0 + A\sin x$ solves the problem. The test says that $f$ is orthogonal to the kernel function $\sin x$ . For self-adjoint problems with zero boundary data the same pattern holds in general, with orthogonality to every kernel function as the test, and this is the content of the Fredholm alternative later in the course.

In plain words Before solving anything, measure how much of the forcing points along the kernel function. A nonzero amount drives the free mode in a way the boundaries cannot absorb, and a zero amount lets solutions through.

Example For $f = \sin x$ the test integral is

\int_0^\pi \sin^2 x\,dx = \frac{\pi}{2} \ne 0,

which confirms that there is no solution. For $f = \cos x$ it is $\int_0^\pi \sin x\cos x\,dx = 0$ , so solutions exist. One of them is $u = \tfrac{1}{2}x\sin x$ , since

\begin{aligned} u' &= \tfrac{1}{2}\sin x + \tfrac{1}{2}x\cos x, \\[0.4em] u'' &= \cos x - \tfrac{1}{2}x\sin x, \end{aligned}

so $u'' + u = \cos x$ , and $u$ vanishes at $0$ and at $\pi$ . Adding the kernel gives all solutions,

y = \tfrac{1}{2}x\sin x + A\sin x,

one for each real $A$ . Compatibility restores existence, but it does not restore uniqueness. Nonzero boundary data enter the condition too. With $y(0) = \alpha$ and $y(\pi) = \beta$ the bracket above equals $\alpha + \beta$ , and the condition becomes

\int_0^\pi f(x)\sin x\,dx = \alpha + \beta.

Unique solvability

Definition A linear boundary-value problem is uniquely solvable when it has exactly one solution for every forcing $f$ continuous on $[a, b]$ and all boundary values $\gamma_1$ and $\gamma_2$ . The following statements are equivalent.

The problem is uniquely solvable.
The boundary matrix $M$ is invertible, that is, $\det M \ne 0$ .
The homogeneous problem has only the zero solution.

Indeed, an invertible $M$ gives exactly one $c$ for every $d$ , unique solvability applied to zero forcing and zero data leaves only the zero function, and a zero kernel means $Mc = 0$ only for $c = 0$ . So exactly one of two cases occurs. Either the kernel is $\{0\}$ and every forced problem has exactly one solution, or the kernel is nonzero and every forced problem has none or infinitely many, with both outcomes occurring for suitable data.

In plain words Test the problem with zero forcing and zero endpoint data. If only the zero function survives, every version of the problem has exactly one answer.

Example On $[0, \pi/2]$ the problem $y'' + y = 1$ with $y(0) = y(\pi/2) = 0$ has the invertible boundary matrix found above, so it has exactly one solution. The constant $1$ is a particular solution, and $y = 1 + A\sin x + B\cos x$ gives $1 + B = 0$ at $0$ and $1 + A = 0$ at $\pi/2$ , so

y = 1 - \sin x - \cos x.

Its second derivative is $\sin x + \cos x$ , so $y'' + y = 1$ , and the end values are $1 - 0 - 1 = 0$ and $1 - 1 - 0 = 0$ . In a resonant problem an extra condition can single out one member of a family, but it changes the problem rather than making the original one uniquely solvable. For the family $\tfrac{1}{2}x\sin x + A\sin x$ above, imposing the extra requirement $\int_0^\pi y\sin x\,dx = 0$ and using $\int_0^\pi x\sin^2 x\,dx = \pi^2/4$ gives

\begin{gathered} \frac{\pi^2}{8} + \frac{\pi}{2}A = 0, \\[0.4em] A = -\frac{\pi}{4}. \end{gathered}

Check yourself

Does $y'' + y = x$ on $[0, \pi]$ with $y(0) = y(\pi) = 0$ have a solution? Use the compatibility condition, then confirm the answer by solving directly.

Show the answer

This is the resonant problem of this page, so the test is the integral of $x\sin x$ . Integration by parts gives

\begin{aligned} \int_0^\pi x\sin x\,dx &= \big[-x\cos x\big]_0^\pi \\[0.4em] &\quad + \int_0^\pi \cos x\,dx \\[0.4em] &= \pi + 0 = \pi, \end{aligned}

which is not zero, so there is no solution. Directly, the function $y_p = x$ is a particular solution, so every solution of the equation is $y = x + A\sin x + B\cos x$ . The left condition gives $B = 0$ , and then $y(\pi) = \pi - B = \pi \ne 0$ for every $A$ .

A classmate says that $y'' + y = \sin 2x$ with $y(0) = y(\pi) = 0$ has no solution, because the problem is resonant. Is that right? Find all solutions.

Show the answer

No. Resonance rules out exactly one solution, but it allows infinitely many. The compatibility test gives

\begin{aligned} &\int_0^\pi \sin 2x\sin x\,dx \\[0.4em] &\quad = \int_0^\pi 2\sin^2 x\cos x\,dx \\[0.4em] &\quad = \Big[\tfrac{2}{3}\sin^3 x\Big]_0^\pi = 0, \end{aligned}

so solutions exist. Trying $y_p = k\sin 2x$ gives $y_p'' + y_p = -3k\sin 2x$ , so $k = -\tfrac{1}{3}$ , and this $y_p$ vanishes at $0$ and at $\pi$ . Adding the kernel gives every solution,

y = -\tfrac{1}{3}\sin 2x + A\sin x,

for any real $A$ .

For which $b > 0$ does the problem $y'' + y = f$ with $y(0) = 0$ and $y(b) = 0$ have exactly one solution for every continuous $f$ on $[0, b]$ ? For $b = 2\pi$ , state the compatibility condition.

Show the answer

With the basis $\sin x$ , $\cos x$ , the homogeneous solution $A\sin x + B\cos x$ has the value $B$ at $0$ and the value $A\sin b + B\cos b$ at $b$ , so

\begin{gathered} M = \begin{pmatrix} 0 & 1 \\ \sin b & \cos b \end{pmatrix}, \\[0.4em] \det M = -\sin b. \end{gathered}

The problem is uniquely solvable exactly when $\sin b \ne 0$ , that is, for every $b > 0$ that is not an integer multiple of $\pi$ . For $b = 2\pi$ the kernel consists of the functions $A\sin x$ . Every solution with $y(0) = 0$ is $y_0 + A\sin x$ with $y_0$ as on this page, and since $\sin(2\pi - t) = -\sin t$ ,

y(2\pi) = -\int_0^{2\pi} f(t)\sin t\,dt.

So the compatibility condition is $\int_0^{2\pi} f(x)\sin x\,dx = 0$ , and when it holds the solutions are $y_0 + A\sin x$ for every real $A$ .

Boundary-Value Problems and Solvability

Key terms

Solvability of boundary-value problems

Linear boundary map

Kernel of the homogeneous problem

Resonant forcing

Compatibility condition

Unique solvability

Check yourself

How this lesson connects

Builds on

Used later in