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Boundary value problems

Two conditions at two ends, the eigenvalues at which a non-zero solution exists, and why the answer may be absent, unique or a whole family.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to apply two boundary conditions in turn to a general solution, decide which of the three outcomes a boundary value problem has, find the eigenvalues and eigenfunctions of $y'' + \lambda y = 0$ on an interval with zero ends, and say why every eigenvalue of that problem is positive.

2. What you already have

You can solve a constant-coefficient second-order equation from its characteristic roots, and you have applied two initial conditions at one point to fix the two constants. This lesson moves the two conditions to opposite ends of an interval, and almost everything reliable about the initial value problem stops being true.

3. Words this lesson uses

4. Two conditions, two ends

An initial value problem for a linear second-order equation is completely predictable: on any interval where the coefficients are continuous it has exactly one solution, always. Move one condition to the far end of the interval and that guarantee is gone. A boundary value problem may have no solution, exactly one, or infinitely many, and which of the three it is depends on the interval as much as on the equation.

The reason is easy to see once stated. Two initial conditions at one point read the two constants off directly. Two boundary conditions give two linear equations in $c_1$ and $c_2$, and a two-by-two linear system has one solution, none, or infinitely many depending on whether its determinant vanishes and whether it is consistent when it does.

The eigenvalue problem. The case that matters most is

$$y'' + \lambda y = 0, \qquad y(0) = y(L) = 0.$$

The left condition leaves $c\sin\mu x$ with $\lambda = \mu^{2}$, and the right one needs $\sin\mu L = 0$. So $\mu L = n\pi$, giving

$$\lambda_n = \left(\frac{n\pi}{L}\right)^{2}, \qquad y_n = \sin\frac{n\pi x}{L}.$$

Only at those values is there anything but zero. That is the whole of the phenomenon, and it is where the vibration frequencies of a string, the energy levels of a particle in a box and the modes of a heated rod all come from.

5. Why the sign of the parameter matters

It is worth checking that no other $\lambda$ can work, because the argument is short and the conclusion is used everywhere afterwards.

If $\lambda < 0$, write $\lambda = -\nu^{2}$. The solutions are built from $e^{\nu x}$ and $e^{-\nu x}$, equivalently $\sinh$ and $\cosh$. The hyperbolic sine vanishes only at zero, so the two conditions force both constants to zero and nothing survives.

If $\lambda = 0$ the equation is $y'' = 0$, whose solutions are straight lines; a line through zero at both ends is the zero line.

So every eigenvalue of this problem is positive, and the discrete list above is the complete answer. Oscillation is not one possibility among several — for these boundary conditions it is the only one that produces anything.

6. Where this goes wrong

Accepting the trivial solution. $c = 0$ satisfies every homogeneous boundary condition, and a calculation that stops there has found the answer to no question. The whole point is the values of $\lambda$ at which something else is possible.

Reporting the frequency as the eigenvalue. $\mu = n\pi/L$ is the frequency; $\lambda = \mu^{2}$ is the eigenvalue. Squaring is not optional.

Expecting a solution to exist. Unlike an initial value problem, this one may simply have none, and no solution is a complete and correct answer.

Assuming the boundary conditions are always zero. They need not be, and when they are not the problem is usually of the have-exactly-one kind rather than an eigenvalue problem at all.

7. The method, step by step, and how to check it

A boundary value problem gives conditions at two different points, and that one change is why its answer can be one solution, none, or infinitely many. The procedure handles all three.

1. Solve the equation in general. Write the general solution with both constants free, including a particular solution if the equation is forced.
2. Apply the condition at the left end. Usually one function vanishes there and one constant is fixed at once.
3. Apply the condition at the right end. This produces an equation of the form (number) $\times$ (constant) $=$ (value). Look at the number before solving.
4. Read the outcome. If the number is not zero, divide by it: exactly one solution. If it is zero and the value is zero, any constant works: infinitely many. If it is zero and the value is not, there is no solution.

For an eigenvalue problem the conditions are all zero and the equation carries a parameter $\lambda$. Step 4 is then turned around: the question is for which $\lambda$ the number in step 3 is zero, because only then can a non-zero solution exist. Treat the three signs of $\lambda$ separately: $\lambda = \mu^{2} > 0$ gives sines and cosines, $\lambda = 0$ gives straight lines, and $\lambda = -\nu^{2} < 0$ gives hyperbolic functions. Each case must be checked, because which of them produce eigenvalues depends on the conditions: slope conditions admit $\lambda = 0$, value conditions do not.

Why the three outcomes happen. Two conditions on two constants form a two-by-two linear system. Such a system has one solution when its determinant is non-zero, and none or infinitely many when the determinant is zero. An initial value problem always has a non-zero determinant (it is the Wronskian); a boundary value problem does not, and that is the whole difference.

How to check the answer. Substitute your solution into the equation, then evaluate it at both ends and confirm both conditions. For an eigenvalue, check one eigenfunction the same way, and remember that the eigenvalue is the square of the frequency: for $y = \sin\frac{n\pi x}{L}$ it is $\frac{n^{2}\pi^{2}}{L^{2}}$, not $\frac{n\pi}{L}$.

8. In the world: why a guitar string plays only certain notes

A string fixed at both ends, like a guitar or piano string, can only vibrate in shapes that are zero at the two ends. Its shape in a steady vibration satisfies $y'' + \lambda y = 0$ with $y(0) = y(L) = 0$, the boundary value problem of this lesson, and only the eigenvalues $\lambda_n = \left(\frac{n\pi}{L}\right)^{2}$ allow a shape other than a flat string. Each shape $\sin\frac{n\pi x}{L}$ vibrates at the frequency $f_n = \frac{c\sqrt{\lambda_n}}{2\pi} = \frac{nc}{2L}$, where $c$ is the speed of waves along the string.

A guitar string $0.65$ metres long with wave speed $286$ metres a second has $f_1 = \frac{286}{2 \times 0.65} = 220$ hertz, the A below middle C, and its overtones at $440$, $660$, $880$ hertz are whole multiples: the discrete list of eigenvalues is the harmonic series that gives the instrument its tone. Pressing the string against the fifth fret shortens it to about $0.487$ metres, and $f_1 = \frac{286}{0.974} \approx 294$ hertz, the D above. Tightening a string raises $c$ and every $f_n$ with it, which is how it is tuned.

The eigenvalue problem answers the question a musician cares about without solving for the string's motion in time: which notes can this string play? Any other frequency would need a shape that is not zero at both ends.

9. In the world: when a column buckles

A slender column carrying a load $P$ along its length stays straight until the load reaches a critical value, and then bows out sideways. Euler found the critical value in 1744 with this lesson's mathematics. A small sideways deflection $y$ of a column pinned at both ends obeys $EIy'' + Py = 0$ with $y(0) = y(L) = 0$, where $EI$ measures the column's stiffness in bending.

Written as $y'' + \lambda y = 0$ with $\lambda = \frac{P}{EI}$, a bent shape other than $y = 0$ exists only at the eigenvalues $\lambda_n = \frac{n^{2}\pi^{2}}{L^{2}}$. The smallest gives the critical load

$$P_1 = \frac{\pi^{2}EI}{L^{2}}.$$

A steel rod with $EI = 2000$ newton-square-metres and length $2$ metres buckles at $P_1 = \frac{9.87 \times 2000}{4} \approx 4935$ newtons, about half a tonne. Doubling the length quarters the load it can carry, which is why long slender struts are braced in the middle: a brace forces a node at $\frac{L}{2}$, ruling out the first shape and making the second, four times stronger, the one that decides.

Below the critical load the problem has only the trivial solution: the column stays straight. That is the outcome this lesson calls exactly one solution, and it is the one an engineer wants.

10. In the world: the lowest energy of a particle in a box

Quantum mechanics describes a particle confined to a region of length $L$ by a wave function $\psi$ that must vanish at the walls. Inside, the Schrödinger equation reduces to $\psi'' + \lambda\psi = 0$ with $\psi(0) = \psi(L) = 0$, where $\lambda$ is proportional to the particle's energy. It is the guitar string's problem exactly, and it has the same answer: only the eigenvalues $\lambda_n = \frac{n^{2}\pi^{2}}{L^{2}}$ allow a particle to exist at all.

So the energy of a confined particle is quantised: it can take the values $E_n = n^{2}E_1$ and nothing between, and it can never be zero, because $\lambda = 0$ gives only the trivial solution. Halving the box quadruples every energy. This is why electrons in very small semiconductor crystals, quantum dots a few nanometres across, emit light of a colour set by the crystal's size: smaller dots have larger energy gaps and glow bluer. The trivial solution of this lesson is a particle that is not there, and the eigenvalues are the only energies nature allows it.

11. A boundary value problem is not an initial value problem with the conditions moved

The two look like the same question asked in a different order, and they are not. The initial value theorem is a genuine existence and uniqueness theorem: continuity of the coefficients is enough, and the answer exists and is one. Nothing of the sort holds at two ends, and the reason is structural rather than technical — two conditions at one point evaluate the solution and its derivative where both constants are directly readable, while two conditions at two points are a linear system whose determinant can vanish. The habit to build is to stop expecting an answer and start asking which of the three outcomes this is, because no solution is as legitimate a result here as a formula.

12. The same equation, three different answers

1. Solve $y'' + y = 0$ and apply $y(0) = 0$, which all three problems share.

   $y = c_1\cos x + c_2\sin x, \qquad y(0) = c_1 \cdot 1 + c_2 \cdot 0 = 0 \;\Rightarrow\; c_1 = 0$

   One step settles the left end everywhere, leaving $y = c_2\sin x$.

2. For the first problem, with $y(\pi) = 0$, substitute $x = \pi$.

   $c_2\sin\pi = c_2 \cdot 0 = 0 \quad \text{holds for every } c_2$

   Infinitely many solutions: the condition says nothing about $c_2$.

3. For the second problem, with $y(\pi) = 3$, substitute $x = \pi$.

   $c_2 \cdot 0 = 3 \quad \text{has no solution}$

   The left side is zero whatever $c_2$ is, so the problem has none.

4. For the third problem, with $y(\pi/2) = 3$, substitute $x = \pi/2$.

   $c_2\sin\dfrac{\pi}{2} = c_2 \cdot 1 = 3 \;\Rightarrow\; c_2 = 3, \quad y = 3\sin x$

   A unique solution: the interval is part of the problem.

5. Name what decided the three outcomes.

   $\sin L = 0: \ \text{none or infinitely many}; \qquad \sin L \ne 0: \ \text{exactly one}$

   Everything hangs on the coefficient $\sin L$ that multiplies $c_2$ at the right end: when it is zero, $c_2$ cannot be solved for.

13. Eigenvalues on an interval of length two

1. Solve $y'' + \lambda y = 0$ with $\lambda = \mu^{2} > 0$ and apply $y(0) = 0$.

   $y = c_1\cos\mu x + c_2\sin\mu x, \quad y(0) = c_1 = 0 \;\Rightarrow\; y = c_2\sin\mu x$

   The cosine goes first, because it is $1$ at the left end.

2. Apply the right-hand condition $y(2) = 0$, keeping $c_2 \ne 0$.

   $c_2\sin 2\mu = 0 \quad\Rightarrow\quad \sin 2\mu = 0$

   With $c_2 = 0$ the solution would be zero, which is the answer to no question. So the sine must vanish instead.

3. Solve the sine equation for the frequency $\mu$.

   $2\mu = n\pi \quad\Rightarrow\quad \mu = \dfrac{n\pi}{2}, \qquad n = 1, 2, 3, \dots$

   The sine is zero exactly at whole multiples of $\pi$. $n = 0$ gives $\mu = 0$, which is not in this case.

4. Square to get the eigenvalues, with their eigenfunctions.

   $\lambda_n = \left(\dfrac{n\pi}{2}\right)^{2} = \dfrac{n^{2}\pi^{2}}{4}, \qquad y_n = \sin\dfrac{n\pi x}{2}$

   Doubling the interval quarters the lowest eigenvalue, which is why a longer string sounds lower.

5. Check one eigenfunction in the equation and at both ends.

   $y_1 = \sin\dfrac{\pi x}{2}: \quad y_1'' = -\dfrac{\pi^{2}}{4}\sin\dfrac{\pi x}{2} = -\lambda_1y_1, \qquad y_1(0) = 0, \quad y_1(2) = \sin\pi = 0$

   The equation and both conditions hold, so $\lambda_1 = \frac{\pi^{2}}{4}$ is a genuine eigenvalue.

14. Eigenvalues with conditions on the slope

1. Find the eigenvalues of $y'' + \lambda y = 0$ with $y'(0) = 0$ and $y'(\pi) = 0$. Start with $\lambda = \mu^{2} > 0$ and differentiate the general solution.

   $y = c_1\cos\mu x + c_2\sin\mu x, \qquad y' = -\mu c_1\sin\mu x + \mu c_2\cos\mu x$

   The conditions are on $y'$, so the derivative is needed before anything is substituted.

2. Apply the left-hand condition $y'(0) = 0$.

   $-\mu c_1 \cdot 0 + \mu c_2 \cdot 1 = \mu c_2 = 0 \quad\Rightarrow\quad c_2 = 0$

   $\mu \ne 0$ in this case, so it can be divided out. Now the cosine survives, not the sine.

3. Apply the right-hand condition $y'(\pi) = 0$, keeping $c_1 \ne 0$.

   $-\mu c_1\sin\mu\pi = 0 \quad\Rightarrow\quad \sin\mu\pi = 0 \quad\Rightarrow\quad \mu = n, \ n = 1, 2, \dots$

   Dividing by $-\mu c_1$, which is not zero, leaves the sine.

4. Write the positive eigenvalues and eigenfunctions.

   $\lambda_n = n^{2}, \qquad y_n = \cos nx$

   Squaring $\mu = n$. Each cosine is flat at both ends, as the conditions demand.

5. Now try $\lambda = 0$, where the equation is $y'' = 0$.

   $y = a + bx, \quad y' = b; \qquad y'(0) = b = 0 \quad\Rightarrow\quad y = a$

   A constant has zero slope everywhere, so both conditions hold for any $a$.

6. Record $\lambda = 0$ as an eigenvalue.

   $\lambda_0 = 0, \qquad y_0 = 1$

   A non-zero solution exists, which is all the word eigenvalue asks. With conditions on $y$ instead, $\lambda = 0$ gave only the zero line.

7. Rule out the negative case, $\lambda = -\nu^{2} < 0$.

   $y = c_1\cosh\nu x + c_2\sinh\nu x; \quad y'(0) = \nu c_2 = 0; \quad y'(\pi) = \nu c_1\sinh\nu\pi = 0 \;\Rightarrow\; c_1 = 0$

   $\sinh$ is zero only at zero, so both constants vanish. The complete list is $\lambda = 0, 1, 4, 9, \dots$

15. Your turn: find the eigenvalues of $y'' + \lambda y = 0$ with $y'(0) = 0$ and $y(\pi) = 0$

1. Write the solution for $\lambda = \mu^{2}$ and its derivative.

   $y = c_1\cos\mu x + c_2\sin\mu x, \qquad y' = -c_1\mu\sin\mu x + c_2\mu\cos\mu x$

   This time the condition is on the derivative.

2. Your turn: work this step out. Its working is at the end of the packet.

   Apply $y'(0) = 0$.

3. Your turn: work this step out. Its working is at the end of the packet.

   Apply $y(\pi) = 0$ with $c_1 \ne 0$ and solve for $\mu$.

16. Guided practice

Sort these four boundary value problems by how many solutions each one has.

17. Guided practice

Complete the worked solution: the first two eigenvalues of $y'' + \lambda y = 0$ on $0 < x < \pi/3$ with $y = 0$ at both ends.

1. For $\lambda = \mu^{2} > 0$, write the general solution.

   $y = c_1\cos\mu x + c_2\sin\mu x$

   The characteristic equation $r^{2} + \mu^{2} = 0$ has roots $\pm\mu i$.

2. Substitute $x = 0$ into the solution and use $y(0) = 0$.

   $c_1\cos 0 + c_2\sin 0 = c_1 \cdot 1 + c_2 \cdot 0 = 0 \quad\Rightarrow\quad c_1 =$ _a

   $\cos 0 = 1$ and $\sin 0 = 0$, so only $c_1$ survives.

3. Substitute $x = \pi/3$, use $y(\pi/3) = 0$ with $c_2 \ne 0$, and multiply both sides by $\frac{3}{\pi}$.

   $c_2\sin\dfrac{\mu\pi}{3} = 0 \quad\Rightarrow\quad \dfrac{\mu\pi}{3} = n\pi \quad\Rightarrow\quad \mu = 3n$

   A non-zero $c_2$ needs the sine to vanish, which happens at whole multiples of $\pi$.

4. Square $\mu$ with $n = 1$ for the lowest eigenvalue.

   $\lambda_1 = \left(3 \cdot 1\right)^{2} =$ _l

   The eigenvalue is the square of the frequency, not the frequency.

5. Square $\mu$ with $n = 2$ for the next one.

   $\lambda_2 = \left(3 \cdot 2\right)^{2} =$ _t

   Two half-waves fit: doubling the frequency quadruples the eigenvalue.

18. Guided practice

How many solutions does $y'' + y = 0$ with $y(0) = 0, \ y(\pi/2) = 4$ have?

• No solution
• Exactly one solution
• Infinitely many solutions

19. Practice

For $y'' + \lambda y = 0$ on $0 < x < \pi/2$ with $y = 0$ at both ends, fill in the first four values of $\lambda$ that allow a solution other than zero.

20. Practice

What is the smallest value of $\lambda$ for which $y'' + \lambda y = 0$ on $0 < x < \pi/4$, with $y = 0$ at both ends, has a solution other than zero?

Answer:

21. Practice

Solve $y'' + 9y = 9$ on $0 < x < \frac{\pi}{6}$ with $y(0) = 6$ and $y\left(\frac{\pi}{6}\right) = 8$. Write $y$ as a formula in $x$ (type cos(...) and sin(...)).

Answer:

22. Somewhere new

Put the steps of finding the eigenvalues of $y'' + \lambda y = 0$ on $0 < x < \pi/4$, with $y = 0$ at both ends, into the order they must be done.

Number the steps in order (write the number in the box):

• Write the general solution in terms of $\mu$
• Apply the condition at the left end
• Apply the condition at the right end
• Reject the solution that is identically zero
• Solve the resulting condition on $\mu$

23. Somewhere new

A string $0.7$ metres long is fixed at both ends, and waves travel along it at $280$ metres a second. Its shape in a steady vibration solves $y'' + \lambda y = 0$ with $y(0) = y(0.7) = 0$, and a shape with eigenvalue $\lambda$ vibrates at $\dfrac{280\sqrt{\lambda}}{2\pi}$ hertz. What is the lowest frequency, in hertz, the string can play?

Answer:

24. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

25. Test question

Solve $y'' + 4y = 16$ on $0 < x < \frac{\pi}{4}$ with $y(0) = 3$ and $y\left(\frac{\pi}{4}\right) = 7$. Write $y$ as a formula in $x$ (type cos(...) and sin(...)).

Answer:

26. What you can do now

You can decide whether a boundary value problem has no solution, one, or infinitely many, and you can produce the eigenvalues for a string fixed at both ends. Say in your own words why the zero solution has to be refused before the eigenvalues appear.

Working for the steps left to you

15. Your turn: find the eigenvalues of $y'' + \lambda y = 0$ with $y'(0) = 0$ and $y(\pi) = 0$, step 2

15. Your turn: find the eigenvalues of $y'' + \lambda y = 0$ with $y'(0) = 0$ and $y(\pi) = 0$, step 3