12.2 The Wave Equation

In this section we consider initial-boundary value problems of the form

where \(a\) is a constant and \(f\) and \(g\) are given functions of \(x\).

The partial differential equation \(u_{tt}=a^2u_{xx}\) is called the wave equation. It is necessary to specify both \(f\) and \(g\) because the wave equation is a second order equation in \(t\) for each fixed \(x\).

This equation and its generalizations

to two and three space dimensions have important applications to the the propagation of electromagnetic, sonic, and water waves.

The Vibrating String

We motivate the study of the wave equation by considering its application to the vibrations of a string – such as a violin string – tightly stretched in equilibrium along the \(x\)-axis in the \(xu\)-plane and tied to the points \((0,0)\) and \((L,0)\) (Figure 12.2.1).

If the string is plucked in the vertical direction and released at time \(t=0\), it will oscillate in the \(xu\)-plane. Let \(u(x,t)\) denote the displacement of the point on the string above (or below) the abscissa \(x\) at time \(t\).

We’ll show that it’s reasonable to assume that \(u\) satisfies the wave equation under the following assumptions:

1. The mass density (mass per unit length) \(\rho\) of the string is constant throughout the string.

2. The tension \(T\) induced by tightly stretching the string along the \(x\)-axis is so great that all other forces, such as gravity and air resistance, can be neglected.

3. The tension at any point on the string acts along the tangent to the string at that point, and the magnitude of its horizontal component is always equal to \(T\), the tension in the string in equilibrium.

4. The slope of the string at every point remains sufficiently small so that we can make the approximation

   \begin{equation} \sqrt{1+u_x^2}\approx1. \tag{12.2.2}\end{equation}

Figure 12.2.2 shows a segment of the displaced string at a time \(t>0\). (Don’t think that the figure is necessarily inconsistent with Assumption 4; we exaggerated the slope for clarity.)

The vectors \({\bf T}_1\) and \({\bf T}_2\) are the forces due to tension, acting along the tangents to the segment at its endpoints. From Newton’s second law of motion, \({\bf T}_1-{\bf T}_2\) is equal to the mass times the acceleration of the center of mass of the segment. The horizontal and vertical components of \({\bf T}_1-{\bf T}_2\) are

respectively. Since

by assumption, the net horizontal force is zero, so there’s no horizontal acceleration. Since the initial horizontal velocity is zero, there’s no horizontal motion.

Applying Newton’s second law of motion in the vertical direction yields

where \(\Delta s\) is the length of the segment and \(\overline x\) is the abscissa of the center of mass; hence,

From calculus, we know that

however, because of (12.2.2), we make the approximation

so (12.2.4) becomes

Therefore

Recalling (12.2.3), we divide by \(T\) to obtain

Since \(\tan\theta_1=u_x(x,t)\) and \(\tan\theta_2=u_x(x+\Delta x,t)\), (12.2.5) is equivalent to

Letting \(\Delta x\to 0\) yields

which we rewrite as \(u_{tt}=a^2u_{xx}\), with \(a^2=T/\rho\).

The Formal Solution

As in Section 12.1, we use separation of variables to obtain a suitable definition for the formal solution of (12.2.1). We begin by looking for functions of the form \(v(x,t)=X(x)T(t)\) that are not identically zero and satisfy

for all \((x,t)\). Since

\(v_{tt}=a^2v_{xx}\) if and only if

which we rewrite as

For this to hold for all \((x,t)\), the two sides must equal the same constant; thus,

which is equivalent to

and

Since \(v(0,t)=X(0)T(t)=0\) and \(v(L,t)=X(L)T(t)=0\) and we don’t want \(T\) to be identically zero, \(X(0)=0\) and \(X(L)=0\). Therefore \(\lambda\) must be an eigenvalue of

and \(X\) must be a \(\lambda\)-eigenfunction. From Theorem 11.1.2, the eigenvalues of (12.2.7) are \(\lambda_n=n^2\pi^2/L^2\), with associated eigenfunctions

Substituting \(\lambda=n^2\pi^2/L^2\) into (12.2.6) yields

which has the general solution

where \(\alpha_n\) and \(\beta_n\) are constants. Now let

Then

so

Therefore \(v_n\) satisfies (12.2.1) with \(f(x)=\alpha_n \sin n\pi x/L\) and \(g(x)=\beta_n\cos n\pi x/L\). More generally, if \(\alpha_1\), \(\alpha_2\), …, \(\alpha_m\) and \(\beta_1\), \(\beta_2\),…, \(\beta_m\) are constants and

then \(u_m\) satisfies (12.2.1) with

This motivates the next definition.

Since there are no convergence-producing factors in (12.2.8) like the negative exponentials in \(t\) that appear in formal solutions of initial-boundary value problems for the heat equation, it isn’t obvious that (12.2.8) even converges for any values of \(x\) and \(t\), let alone that it can be differentiated term by term to show that \(u_{tt}=a^2u_{xx}\). However, the next theorem guarantees that the series converges not only for \(0\le x\le L\) and \(t\ge0\), but for \(-\infty<x<\infty\) and \(-\infty<t<\infty\).

Proof Setting \(A=n\pi x/L\) and \(B=n\pi at/L\) in the identities

and

yields

and

From (12.2.10),

Since it can be shown that a Fourier sine series can be integrated term by term between any two limits, (12.2.11) implies that

This and (12.2.12) imply (12.2.9), which completes the proof.

As we’ll see below, if \(S_g\) is differentiable and \(S_f\) is twice differentiable on \((-\infty,\infty)\), then (12.2.9) satisfies \(u_{tt}=a^2u_{xx}\) for all \((x,t)\). We need the next theorem to formulate conditions on \(f\) and \(g\) such that \(S_f\) and \(S_g\) to have these properties.

Proof Throughout this proof, \(k\) denotes an integer. Since \(f\) is differentiable on the open interval \((0,L)\), both \(p\) and \(q\) are differentiable on every open interval \(((k-1)L,kL)\). Thus, we need only to determine whether \(p\) and \(q\) are differentiable at \(x=kL\) for every \(k\).

(a) From Figure 12.2.3, \(p\) is discontinuous at \(x=2kL\) if \(h(0)\ne0\) and discontinuous at \(x=(2k-1)L\) if \(h(L)\ne0\). Therefore \(p\) is not differentiable on \((-\infty,\infty)\) unless \(h(0)=h(L)=0\). From Figure 12.2.4, if \(h(0)=h(L)=0\), then

for every \(k\); therefore, \(p\) is differentiable on \((-\infty,\infty)\).

(b) From Figure 12.2.5,

so \(q\) is differentiable at \(x=2kL\) if and only if \(h'_+(0)=0\). Also,

so \(q\) is differentiable at \(x=(2k-1)L\) if and only if \(h'_-(L)=0\). Therefore \(q\) is differentiable on \((-\infty,\infty)\) if and only if \(h'_+(0)=h'_-(L)=0\), as in Figure 12.2.6. This completes the proof.

Proof We first show that \(S_g\) is differentiable and \(S_f\) is twice differentiable on \((-\infty,\infty)\). We’ll then differentiate (12.2.9) twice with respect to \(x\) and \(t\) and verify that (12.2.9) is an actual solution of (12.2.1).

Since \(f\) and \(g\) are continuous on \((0,L)\), Theorem 11.3.2 implies that \(S_f(x)=f(x)\) and \(S_g(x)=g(x)\) on \([0,L]\). Therefore \(S_f\) and \(S_g\) are the odd periodic extensions of \(f\) and \(g\). Since \(f\) and \(g\) are differentiable on \([0,L]\), (12.2.15), (12.2.16), and Theorem 12.2.3(a) imply that \(S_f\) and \(S_g\) are differentiable on \((-\infty,\infty)\).

Since \(S_f'(x)=f'(x)\) on \([0,L]\) (one-sided derivatives at the endpoints), and \(S_f'\) is even (the derivative of an odd function is even), \(S_f'\) is the even periodic extension of \(f'\). By assumption, \(f'\) is differentiable on \([0,L]\). Because of (12.2.17), Theorem 12.2.3(b) with \(h=f'\) and \(q=S_f'\) implies that \(S_f''\) exists on \((-\infty,\infty)\).

Now we can differentiate (12.2.9) twice with respect to \(x\) and \(t\):

and

Comparing (12.2.18) and (12.2.20) shows that \(u_{tt}(x,t)=a^2u_{xx}(x,t)\) for all \((x,t)\).

From (12.2.8), \(u(0,t)=u(L,t)=0\) for all \(t\). From (12.2.9), \(u(x,0)=S_f(x)\) for all \(x\), and therefore, in particular,

From (12.2.19), \(u_t(x,0)=S_g(x)\) for all \(x\), and therefore, in particular,

Therefore \(u\) is an actual solution of (12.2.1). This completes the proof.

Eqn (12.2.9) is called d’Alembert’s solution of (12.2.1). Although d’Alembert’s solution was useful for proving Theorem 12.2.4 and is very useful in a slightly different context (Exercises 63-68), (12.2.8) is preferable for computational purposes.

Solution We leave it to you to verify that \(f\) and \(g\) satisfy the assumptions of Theorem 12.2.4.

From Exercise 11.3. 39,

From Exercise 11.3. 36,

From (12.2.8),

Theorem 12.1.2 implies that \(u_{xx}\) and \(u_{tt}\) can be obtained by term by term differentiation, for all \((x,t)\), so \(u_{tt}=a^2u_{xx}\) for all \((x,t)\) (Exercise 62). Moreover, Theorem 11.3.2 implies that \(S_f(x)=f(x)\) and \(S_g(x)=g(x)\) if \(0\le x\le L\). Therefore \(u(x,0)=f(x)\) and \(u_t(x,0)=g(x)\) if \(0\le x\le L\). Hence, \(u\) is an actual solution of the initial-boundary value problem.

The Plucked String

If \(f\) and \(g\) don’t satisfy the assumptions of Theorem 12.2.4, then (12.2.8) isn’t an actual solution of (12.2.1) in fact, it can be shown that (12.2.1) doesn’t have an actual solution in this case. Nevertheless, \(u\) is defined for all \((x,t)\), and we can see from (12.2.18) and (12.2.20) that \(u_{tt}(x,t)=a^2u_{xx}(x,t)\) for all \((x,t)\) such that \(S_f''(x\pm at)\) and \(S_g'(x\pm at)\) exist. Moreover, \(u\) may still provide a useful approximation to the vibration of the string; a laboratory experiment can confirm or deny this.

We’ll now consider the initial-boundary value problem (12.2.1) with

and \(g\equiv0\). Since \(f\) isn’t differentiable at \(x=L/2\), it does’nt satisfy the assumptions of Theorem 12.2.4, so the formal solution of (12.2.1) can’t be an actual solution. Nevertheless, it’s instructive to investigate the properties of the formal solution.

The graph of \(f\) is shown in Figure 12.2.7. Intuitively, we are plucking the string by half its length at the middle. You’re right if you think this is an extraordinarily large displacement; however, we could remove this objection by multiplying the function in Figure 12.2.7 by a small constant. Since this would just multiply the formal solution by the same constant, we’ll leave \(f\) as we’ve defined it. Similar comments apply to the exercises.

From Exercise 11.3. 15, the Fourier sine series of \(f\) on \([0,L]\) is

which converges to \(f\) for all \(x\) in \([0,L]\), by Theorem 11.3.2. Therefore

This series converges absolutely for all \((x,t)\) by the comparison test, since the series

converges. Moreover, (12.2.22) satisfies the boundary conditions

and the initial condition

However, we can’t justify differentiating (12.2.22) term by term even once, and formally differentiating it twice term by term produces a series that diverges for all \((x,t)\). (Verify.). Therefore we use d’Alembert’s form

for \(u\) to study its derivatives. Figure 12.2.8 shows the graph of \(S_f\), which is the odd periodic extension of \(f\). You can see from the graph that \(S_f\) is differentiable at \(x\) (and \(S_f'(x)=\pm1\)) if and only if \(x\) isn’t an odd multiple of \(L/2\).

In Figure 12.2.9 the dashed and solid curves are the graphs of \(y=S_f(x-at)\) and \(y=S_f(x+at)\) respectively, for a fixed value of \(t\). As \(t\) increases the dashed curve moves to the right and the solid curve moves to the left. For this reason, we say that the functions \(u_1(x,t)=S_f(x+at)\) and \(u_2(x,t)=S_f(x-at)\) are traveling waves. Note that \(u_1\) satisfies the wave equation at \((x,t)\) if \(x+at\) isn’t an odd multiple of \(L/2\) and \(u_2\) satisfies the wave equation at \((x,t)\) if \(x-at\) isn’t an odd multiple of \(L/2\). Therefore (12.2.23) (or, equivalently, (12.2.22)) satisfies \(u_{tt}(x,t)=a^2u_{xx}(x,t)=0\) for all \((x,t)\) such that neither \(x-at\) nor \(x+at\) is an odd multiple of \(L/2\).

We conclude by finding an explicit formula for \(u(x,t)\) under the assumption that

To see how this formula can be used to compute \(u(x,t)\) for \(0\le x\le L\) and arbitrary \(t\), we refer you to Exercise 16.

From Figure 12.2.10,

and

if \((x,t)\) satisfies (12.2.24).

Therefore, from (12.2.23),

if \((x,t)\) satisfies (12.2.24). Figure 12.2.11 is the graph of this function on \([0,L]\) for a fixed \(t\) in \((0,L/2a)\).

Using Technology

Although the formal solution

of (12.2.1) is defined for all \((x,t)\), we’re mainly interested in its behavior for \(0\le x\le L\) and \(t\ge0\). In fact, it’s sufficient to consider only values of \(t\) in the interval \(0\le t<2L/a\), since

for all \((x,t)\) if \(k\) is an integer. (Verify.)

You can create an animation of the motion of the string by performing the following numerical experiment.

Let \(m\) and \(k\) be positive integers. Let

thus, \(t_0\), \(t_1\), … \(t_k\) are equally spaced points in \([0,2L/a]\). For each \(j=0\), \(1\) ,\(2\), … \(k\), graph the partial sum

on \([0,L]\) as a function of \(x\). Write your program so that each graph remains displayed on the monitor for a short time, and is then deleted and replaced by the next. Repeat this procedure for various values of \(m\) and \(k\).

We suggest that you perform experiments of this kind in the exercises marked C , without other specific instructions. (These exercises were chosen arbitrarily; the experiment is worthwhile in all the exercises dealing with specific initial-boundary value problems.) In some of the exercises the formal solutions have other forms, defined in Exercises 17, 34, and 49; however, the idea of the experiment is the same.

12.2 Exercises

In Exercises 1-15 solve the initial-boundary value problem. In some of these exercises, Theorem 11.3.5(b) or Exercise \(11.3\). 35 will simplify the computation of the coefficients in the Fourier sine series.

1. \(u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0\),
   \(u(0,t)=0,\quad u(1,t)=0,\quad t>0\),
   \(u(x,0)=0,\quad u_t(x,0)= \left\{\begin{array}{cl} x,&0\le x\le{1\over2},\\1-x,&{1\over2}\le x\le1 \end{array}\right.,\quad0\le x\le1\)

   Show answer

   \(u(x,t)=\dst{4\over3\pi^3}\sum_{n=1}^\infty {(-1)^{n+1}\over(2n-1)^3}\sin3(2n-1)\pi t\sin(2n-1)\pi x\)

2. \(u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0\),
   \(u(0,t)=0,\quad u(1,t)=0,\quad t>0\),
   \(u(x,0)=x(1-x),\quad u_t(x,0)=0,\quad0\le x\le 1\)

   Show answer

   \(u(x,t)=\dst{8\over\pi^3}\sum_{n=1}^\infty{1\over(2n-1)^3} \cos3(2n-1)\pi t\sin(2n-1)\pi x\)

3. \(u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0\),
   \(u(0,t)=0,\quad u(1,t)=0,\quad t>0\),
   \(u(x,0)=x^2(1-x),\quad u_t(x,0)=0,\quad0\le x\le1\)

   Show answer

   \(u(x,t)=\dst-{4\over\pi^3}\sum_{n=1}^\infty{(1+(-1)^n2)\over n^3} \cos n\sqrt7\,\pi t\,\sin n\pi x\)

4. C \(u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0\),
   \(u(0,t)=0,\quad u(1,t)=0,\quad t>0\),
   \(u(x,0)=0,\quad u_t(x,0)=x(1-x),\quad0\le x\le 1\)

   Show answer

   \(u(x,t)=\dst{8\over3\pi^4}\sum_{n=1}^\infty{1\over(2n-1)^4} \sin3(2n-1)\pi t\sin(2n-1)\pi x\)

5. \(u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0\),
   \(u(0,t)=0,\quad u(1,t)=0,\quad t>0\),
   \(u(x,0)=0\quad u_t(x,0)=x^2(1-x),\quad0\le x\le1\)

   Show answer

   \(u(x,t)=\dst-{4\over\sqrt7\,\pi^4}\sum_{n=1}^\infty{(1+(-1)^n2)\over n^4} \sin n\sqrt7\,\pi t\,\sin n\pi x\)

6. \(u_{tt}=64u_{xx},\quad 0<x<3,\quad t>0\),
   \(u(0,t)=0,\quad u(3,t)=0,\quad t>0\),
   \(u(x,0)=x(x^2-9),\quad u_t(x,0)=0,\quad0\le x\le 3\)

   Show answer

   \(u(x,t)=\dst{324\over\pi^3}\sum_{n=1}^\infty{(-1)^n\over n^3}\cos{8n\pi t\over3}\sin{n\pi x\over3}\)

7. \(u_{tt}=4u_{xx},\quad 0<x<1,\quad t>0\),
   \(u(0,t)=0,\quad u(1,t)=0,\quad t>0\),
   \(u(x,0)=x(x^3-2x^2+1),\quad u_t(x,0)=0,\quad0\le x\le1\)

   Show answer

   \(u(x,t)=\dst{96\over\pi^5}\sum_{n=1}^\infty{1\over(2n-1)^5} \cos2(2n-1)\pi t\sin(2n-1)\pi x\)

8. C \(u_{tt}=64u_{xx},\quad 0<x<3,\quad t>0\),
   \(u(0,t)=0,\quad u(3,t)=0,\quad t>0\),
   \(u(x,0)=0,\quad u_t(x,0)=x(x^2-9),\quad0\le x\le 3\)

   Show answer

   \(u(x,t)=\dst{243\over2\pi^4}\sum_{n=1}^\infty{(-1)^n\over n^4}\sin{8n\pi t\over3}\sin{n\pi x\over3}\)

9. \(u_{tt}=4u_{xx},\quad 0<x<1,\quad t>0\),
   \(u(0,t)=0,\quad u(1,t)=0,\quad t>0\),
   \(u(x,0)=0,\quad u_t(x,0)=x(x^3-2x^2+1),\quad0\le x\le1\)

   Show answer

   \(u(x,t)=\dst{48\over\pi^6}\sum_{n=1}^\infty{1\over(2n-1)^6} \sin2(2n-1)\pi t\sin(2n-1)\pi x\).

10. \(u_{tt}=5u_{xx},\quad 0<x<\pi,\quad t>0\),
    \(u(0,t)=0,\quad u(\pi,t)=0,\quad t>0\),
    \(u(x,0)=x\sin x,\quad u_t(x,0)=0,\quad0\le x\le \pi\)

    Show answer

    \(u(x,t)=\dst{\pi\over2}\cos\sqrt{5}\,t\sin x-{16\over\pi} \sum_{n=1}^\infty{n\over(4n^2-1)^2}\cos2n\sqrt5\,t\sin 2nx\)

11. \(u_{tt}=u_{xx},\quad 0<x<1,\quad t>0\),
    \(u(0,t)=0,\quad u(1,t)=0,\quad t>0\),
    \(u(x,0)=x(3x^4-5x^3+2),\quad u_t(x,0)=0,\quad0\le x\le1\)

    Show answer

    \(u(x,t)=\dst-{240\over\pi^5}\sum_{n=1}^\infty{1+(-1)^n2\over n^5} \cos n\pi t\sin n\pi x\)

12. C \(u_{tt}=5u_{xx},\quad 0<x<\pi,\quad t>0\),
    \(u(0,t)=0,\quad u(\pi,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=x\sin x,\quad0\le x\le \pi\)

    Show answer

    \(u(x,t)=\dst{\pi\over2\sqrt5}\sin\sqrt{5}\,t\sin x-{8\over\pi\sqrt5} \sum_{n=1}^\infty{1\over(4n^2-1)^2}\sin2n\sqrt5\,t\sin 2nx\)

13. \(u_{tt}=u_{xx},\quad 0<x<1,\quad t>0\),
    \(u(0,t)=0,\quad u(1,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=x(3x^4-5x^3+2),\quad0\le x\le1\)

    Show answer

    \(u(x,t)=\dst-{240\over\pi^6}\sum_{n=1}^\infty{1+(-1)^n2\over n^6} \sin n\pi t\sin n\pi x\)

14. \(u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0\),
    \(u(0,t)=0,\quad u(1,t)=0,\quad t>0\),
    \(u(x,0)=x(3x^4-10x^2+7),\quad u_t(x,0)=0,\quad0\le x\le1\)

    Show answer

    \(u(x,t)=\dst-{720\over\pi^5}\sum_{n=1}^\infty{(-1)^n\over n^5} \cos3n\pi t\sin n\pi x\)

15. C \(u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0\),
    \(u(0,t)=0,\quad u(1,t)=0,\quad t>0\),
    \(u(x,0)=0\quad u_t(x,0)=x(3x^4-10x^2+7),\quad0\le x\le1\)

    Show answer

    \(u(x,t)=\dst-{240\over\pi^6}\sum_{n=1}^\infty{(-1)^n\over n^6} \sin3n\pi t\sin n\pi x\)

16. We saw that the displacement of the plucked string is, on the one hand,

    \[ u(x,t)=\dst{4L\over\pi^2}\sum_{n=1}^\infty{(-1)^{n+1}\over(2n-1)^2} \cos{(2n-1)\pi at\over L}\sin{(2n-1)\pi x\over L},\; 0\le x\le L,\; t\ge 0, \tag*{\rm(A)} \]

    and, on the other hand,

    \[ u(x,\tau)= \left\{\begin{array}{cl} x,&0\le x\le{L\over2}-a\tau,\\[6pt] {L\over2}-a\tau,&{L\over2}-a\tau\le x\le{L\over2}+a\tau,\\[6pt] L-x,&{L\over2}-a\tau\le x\le L. \end{array}\right. \tag*{\rm(B)} \]

    if \(0\le \tau\le L/2a\). The first objective of this exercise is to show that (B) can be used to compute \(u(x,t)\) for \(0\le x\le L\) and all \(t>0\).

    1. Show that if \(t>0\), there’s a nonnegative integer \(m\) such that either

       \[ \mbox{\bf(i)}\quad t={mL\over a}+\tau\mbox{\quad or \quad} \mbox{\bf(ii)}\quad t={(m+1)L\over a}-\tau, \]

       where \(0\le \tau\le L/2a\).

    2. Use (A) to show that \(u(x,t)=(-1)^mu(x,\tau)\) if (i) holds, while \(u(x,t)=(-1)^{m+1}u(x,\tau)\) if (ii) holds.

    3. L   Perform the following experiment for specific values of \(L\) and \(a\) and various values of \(m\) and \(k\): Let

       \[ t_j={Lj\over 2ka},\quad j=0,1,\dots k; \]

       thus, \(t_0\), \(t_1\), …, \(t_k\) are equally spaced points in \([0,L/2a]\). For each \(j=0\), \(1\) , \(2\),…, \(k\), graph the \(m\)th partial sum of (A) and \(u(x,t_j)\) computed from (B) on the same axis. Create an animation, as described in the remarks on using technology at the end of the section.

17. If a string vibrates with the end at \(x=0\) free to move in a frictionless vertical track and the end at \(x=L\) fixed, then the initial-boundary value problem for its displacement takes the form

    \[ \begin{array}{c} u_{tt}=a^2u_{xx},\quad 0<x<L,\quad t>0,\\ u_x(0,t)=0,\quad u(L,t)=0,\quad t>0,\\ u(x,0)=f(x),\quad u_t(x,0)=g(x),\quad 0\le x\le L. \end{array} \tag*{\rm(A)} \]

    Justify defining the formal solution of (A) to be

    \[ u(x,t)=\sum_{n=1}^\infty \left(\alpha_n\cos{(2n-1)\pi a t\over2L}+{2L\beta_n\over(2n-1)\pi a}\sin{(2n-1)\pi at\over2L}\right) \cos{(2n-1)\pi x\over2L}, \]

    where

    \[ C_{M\!f}(x)=\sum_{n=1}^\infty\alpha_n\cos{(2n-1)\pi x\over2L} \mbox{\quad and \quad} C_{M\!g}(x)=\sum_{n=1}^\infty\beta_n\cos{(2n-1)\pi x\over2L} \]

    are the mixed Fourier cosine series of \(f\) and \(g\) on \([0,L]\); that is,

    \[ \alpha_n={2\over L}\int_0^Lf(x)\cos{(2n-1)\pi x\over2L}\,dx \mbox{\quad and \quad} \beta_n={2\over L}\int_0^Lg(x)\cos{(2n-1)\pi x\over2L}\,dx. \]

18. \(u_{tt}=9u_{xx},\quad 0<x<2,\quad t>0\),
    \(u_x(0,t)=0,\quad u(2,t)=0,\quad t>0\),
    \(u(x,0)=4-x^2,\quad u_t(x,0)=0,\quad0\le x\le2\)

    Show answer

    \(u(x,t)=-\dst {128\over\pi^3}\sum_{n=1}^\infty{(-1)^n\over (2n-1)^3}\cos{3(2n-1)\pi t\over 4}\,\cos{(2n-1)\pi x\over4}\)

19. \(u_{tt}=4u_{xx},\quad 0<x<1,\quad t>0\),
    \(u_x(0,t)=0,\quad u(1,t)=0,\quad t>0\),
    \(u(x,0)=x^2(1-x),\quad u_t(x,0)=0,\quad0\le x\le 1\)

    Show answer

    \(u(x,t)=-\dst{64\over\pi^3}\sum_{n=1}^\infty{1\over(2n-1)^3}\left[ (-1)^n+{3\over(2n-1)\pi}\right]\cos(2n-1)\pi t\,\cos{(2n-1)\pi x\over2}\)

20. \(u_{tt}=9u_{xx},\quad 0<x<2,\quad t>0\),
    \(u_x(0,t)=0,\quad u(2,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=4-x^2,\quad0\le x\le2\)

    Show answer

    \(u(x,t)=-\dst {512\over3\pi^4}\sum_{n=1}^\infty{(-1)^n\over (2n-1)^4}\sin{3(2n-1)\pi t\over 4}\,\cos{(2n-1)\pi x\over4}\)

21. \(u_{tt}=4u_{xx},\quad 0<x<1,\quad t>0\),
    \(u_x(0,t)=0,\quad u(1,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=x^2(1-x),\quad0\le x\le 1\)

    Show answer

    \(u(x,t)=-\dst{64\over\pi^4}\sum_{n=1}^\infty{1\over(2n-1)^4}\left[ (-1)^n+{3\over(2n-1)\pi}\right]\sin(2n-1)\pi t\,\cos{(2n-1)\pi x\over2}\)

22. C \(u_{tt}=5u_{xx},\quad 0<x<1,\quad t>0\),
    \(u_x(0,t)=0,\quad u(1,t)=0,\quad t>0\),
    \(u(x,0)=2x^3+3x^2-5,\quad u_t(x,0)=0,\quad0\le x\le1\)

    Show answer

    \(u(x,t)=\dst{96\over\pi^3}\sum_{n=1}^\infty{1\over(2n-1)^3}\left[ (-1)^n3+{4\over(2n-1)\pi}\right]\cos{(2n-1)\sqrt5\,\pi t\over2}\cos{(2n-1)\pi x\over2}\)

23. \(u_{tt}=3u_{xx},\quad 0<x<\pi,\quad t>0\),
    \(u_x(0,t)=0,\quad u(\pi,t)=0,\quad t>0\),
    \(u(x,0)=\pi^3-x^3,\quad u_t(x,0)=0,\quad0\le x\le\pi\)

    Show answer

    \(u(x,t)=-\dst96\sum_{n=1}^\infty{1\over(2n-1)^3}\left[ (-1)^n+{2\over(2n-1)\pi}\right]\cos{(2n-1)\sqrt3\, t\over2}\cos{(2n-1) x\over2}\)

24. \(u_{tt}=5u_{xx},\quad 0<x<1,\quad t>0\),
    \(u_x(0,t)=0,\quad u(1,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=2x^3+3x^2-5,\quad0\le x\le1\)

    Show answer

    \(u(x,t)=\dst{192\over\pi^4\sqrt5}\sum_{n=1}^\infty{1\over(2n-1)^4}\left[ (-1)^n3+{4\over(2n-1)\pi}\right]\sin{(2n-1)\sqrt5\,\pi t\over2}\cos{(2n-1)\pi x\over2}\)

25. C \(u_{tt}=3u_{xx},\quad 0<x<\pi,\quad t>0\),
    \(u_x(0,t)=0,\quad u(\pi,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=\pi^3-x^3,\quad0\le x\le\pi\)

    Show answer

    \(u(x,t)=-\dst{192\over\sqrt3}\sum_{n=1}^\infty{1\over(2n-1)^4}\left[ (-1)^n+{2\over(2n-1)\pi}\right]\sin{(2n-1)\sqrt3 t\over2}\sin{(2n-1) x\over2}\)

26. \(u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0\),
    \(u_x(0,t)=0,\quad u(1,t)=0,\quad t>0\),
    \(u(x,0)=x^4-2x^3+1,\quad u_t(x,0)=0,\quad0\le x\le1\)

    Show answer

    \(u(x,t)=-\dst{384\over\pi^4}\sum_{n=1}^\infty{1\over(2n-1)^4}\left[ 1+{(-1)^n4\over(2n-1)\pi}\right]\cos{3(2n-1)\pi t\over2}\cos{(2n-1)\pi x\over2}\)

27. \(u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0\),
    \(u_x(0,t)=0,\quad u(1,t)=0,\quad t>0\),
    \(u(x,0)=4x^3+3x^2-7,\quad u_t(x,0)=0,\quad0\le x\le1\)

    Show answer

    \(u(x,t)=\dst{96\over\pi^3}\sum_{n=1}^\infty{1\over(2n-1)^3}\left[ (-1)^n5+{8\over(2n-1)\pi}\right]\cos{(2n-1)\sqrt7\,\pi t\over2}\cos{(2n-1)\pi x\over2}\)

28. \(u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0\),
    \(u_x(0,t)=0,\quad u(1,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=x^4-2x^3+1,\quad0\le x\le1\)

    Show answer

    \(u(x,t)=-\dst{768\over3\pi^5}\sum_{n=1}^\infty{1\over(2n-1)^5}\left[ 1+{(-1)^n4\over(2n-1)\pi}\right]\sin{3(2n-1)\pi t\over2}\cos{(2n-1)\pi x\over2}\)

29. C \(u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0\),
    \(u_x(0,t)=0,\quad u(1,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=4x^3+3x^2-7,\quad0\le x\le1\)

    Show answer

    \(u(x,t)=\dst{192\over\pi^4\sqrt7}\sum_{n=1}^\infty{1\over(2n-1)^4}\left[ (-1)^n5+{8\over(2n-1)\pi}\right]\sin{(2n-1)\sqrt7\,\pi t\over2}\cos{(2n-1)\pi x\over2}\)

30. \(u_{tt}=u_{xx},\quad 0<x<1,\quad t>0\),
    \(u_x(0,t)=0,\quad u(1,t)=0,\quad t>0\),
    \(u(x,0)=x^4-4x^3+6x^2-3,\quad u_t(x,0)=0,\quad0\le x\le1\)

    Show answer

    \(u(x,t)=-\dst{768\over\pi^4}\sum_{n=1}^\infty{1\over(2n-1)^4}\left[ 1+{(-1)^n2\over(2n-1)\pi}\right]\cos{(2n-1)\pi t\over2}\cos{(2n-1)\pi x\over2}\)

31. \(u_{tt}=u_{xx},\quad 0<x<1,\quad t>0\),
    \(u_x(0,t)=0,\quad u(1,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=x^4-4x^3+6x^2-3,\quad0\le x\le1\)

    Show answer

    \(u(x,t)=-\dst{1536\over\pi^5}\sum_{n=1}^\infty{1\over(2n-1)^5}\left[ 1+{(-1)^n2\over(2n-1)\pi}\right]\sin{(2n-1)\pi t\over2}\cos{(2n-1)\pi x\over2}\)

32. Adapt the proof of Theorem 12.2.2 to find d’Alembert’s solution of the initial-boundary value problem in Exercise 17.

    Show answer

    \(u(x,t)=\dst{1\over2}[C_{M\!f}(x+at)+C_{M\!f}(x-at)]+ {1\over2a}\int_{x-at}^{x+at}C_{M\!g}(\tau)\,d\tau\)

33. Use the result of Exercise 32 to show that the formal solution of the initial-boundary value problem in Exercise 17 is an actual solution if \(g\) is differentiable and \(f\) is twice differentiable on \([0,L]\) and

    \[ g'_+(0)=g(L)=f'_+(0)=f(L)=f''_-(L)=0. \]

    Hint: See Exercise \(11.3\). 57, and apply Theorem 12.2.3 with \(L\) replaced by \(2L\).

34. Justify defining the formal solution of the initial-boundary value problem

    \[ \begin{array}{c} u_{tt}=a^2u_{xx},\quad 0<x<L,\quad t>0,\\ u(0,t)=0,\quad u_x(L,t)=0,\quad t>0,\\ u(x,0)=f(x),\quad u_t(x,0)=g(x),\quad 0\le x\le L \end{array} \]

    to be

    \[ u(x,t)=\sum_{n=1}^\infty \left(\alpha_n\cos{(2n-1)\pi a t\over2L}+{2L\beta_n\over(2n-1)\pi a}\sin{(2n-1)\pi at\over2L}\right) \sin{(2n-1)\pi x\over2L}, \]

    where

    \[ S_{M\!f}(x)=\sum_{n=1}^\infty\alpha_n\sin{(2n-1)\pi x\over2L} \mbox{\quad and \quad} S_{M\!g}(x)=\sum_{n=1}^\infty\beta_n\sin{(2n-1)\pi x\over2L} \]

    are the mixed Fourier sine series of \(f\) and \(g\) on \([0,L]\); that is,

    \[ \alpha_n={2\over L}\int_0^Lf(x)\sin{(2n-1)\pi x\over2L}\,dx \mbox{\quad and \quad} \beta_n={2\over L}\int_0^Lg(x)\sin{(2n-1)\pi x\over2L}\,dx. \]

35. \(u_{tt}=64u_{xx},\quad 0<x<\pi,\quad t>0\),
    \(u(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0\),
    \(u(x,0)=x(2\pi-x),\quad u_t(x,0)=0,\quad0\le x\le \pi\)

    Show answer

    \(u(x,t)=\dst {32\over\pi}\sum_{n=1}^\infty{1\over (2n-1)^3}\cos4(2n-1)t\,\sin{(2n-1)x\over2}\)

36. \(u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0\),
    \(u(0,t)=0,\quad u_x(1,t)=0,\quad t>0\),
    \(u(x,0)=x^2(3-2x),\quad u_t(x,0)=0,\quad0\le x\le 1\)

    Show answer

    \(u(x,t) =\dst-{96\over\pi^3}\sum_{n=1}^\infty {1\over(2n-1)^3}\left[1+(-1)^n{4\over(2n-1)\pi}\right] \cos{3(2n-1)\pi t\over2}\sin{(2n-1)\pi x\over2}\)

37. \(u_{tt}=64u_{xx},\quad 0<x<\pi,\quad t>0\),
    \(u(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=x(2\pi-x),\quad0\le x\le \pi\)

    Show answer

    \(u(x,t)=\dst {8\over\pi}\sum_{n=1}^\infty{1\over (2n-1)^4}\sin4(2n-1)t\,\sin{(2n-1)x\over2}\)

38. C \(u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0\),
    \(u(0,t)=0,\quad u_x(1,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=x^2(3-2x),\quad0\le x\le 1\)

    Show answer

    \(u(x,t) =\dst-{64\over\pi^4}\sum_{n=1}^\infty {1\over(2n-1)^4}\left[1+(-1)^n{4\over(2n-1)\pi}\right] \sin{3(2n-1)\pi t\over2}\sin{(2n-1)\pi x\over2}\)

39. \(u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0\),
    \(u(0,t)=0,\quad u_x(1,t)=0,\quad t>0\),
    \(u(x,0)=(x-1)^3+1,\quad u_t(x,0)=0,\quad0\le x\le 1\)

    Show answer

    \(u(x,t) =\dst{96\over\pi^3}\sum_{n=1}^\infty {1\over(2n-1)^3}\left[1+(-1)^n{2\over(2n-1)\pi}\right] \cos{3(2n-1)\pi t\over2} \sin{(2n-1)\pi x\over2}\)

40. \(u_{tt}=3u_{xx},\quad 0<x<\pi,\quad t>0\),
    \(u(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0\),
    \(u(x,0)=x(x^2-3\pi^2),\quad u_t(x,0)=0,\quad0\le x\le\pi\)

    Show answer

    \(u(x,t)=\dst{192\over\pi}\sum_{n=1}^\infty{(-1)^n\over(2n-1)^4} \cos{(2n-1)\sqrt3\,t\over2}\sin{(2n-1) x\over2}\)

41. \(u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0\),
    \(u(0,t)=0,\quad u_x(1,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=(x-1)^3+1,\quad0\le x\le 1\)

    Show answer

    \(u(x,t) =\dst{64\over\pi^4}\sum_{n=1}^\infty {1\over(2n-1)^4}\left[1+(-1)^n{2\over(2n-1)\pi}\right] \sin{3(2n-1)\pi t\over2} \sin{(2n-1)\pi x\over2}\)

42. \(u_{tt}=3u_{xx},\quad 0<x<\pi,\quad t>0\),
    \(u(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=x(x^2-3\pi^2),\quad0\le x\le\pi\)

    Show answer

    \(u(x,t)=\dst{384\over\sqrt3\,\pi}\sum_{n=1}^\infty{(-1)^n\over(2n-1)^5} \sin{(2n-1)\sqrt3\,t\over2}\sin{(2n-1) x\over2}\)

43. \(u_{tt}=5u_{xx},\quad 0<x<1,\quad t>0\),
    \(u(0,t)=0,\quad u_x(1,t)=0,\quad t>0\),
    \(u(x,0)=x^3(3x-4),\quad u_t(x,0)=0,\quad0\le x\le1\)

    Show answer

    \(u(x,t)=\dst{1536\over\pi^4}\sum_{n=1}^\infty{1\over(2n-1)^4} \left[(-1)^n+{3\over(2n-1)\pi}\right] \cos{(2n-1)\sqrt5\,\pi t\over2}\sin{(2n-1)\pi x\over2}\)

44. C \(u_{tt}=16u_{xx},\quad 0<x<1,\quad t>0\),
    \(u(0,t)=0,\quad u_x(1,t)=0,\quad t>0\),
    \(u(x,0)=x(x^3-2x^2+2),\quad u_t(x,0)=0,\quad0\le x\le1\)

    Show answer

    \(u(x,t)=\dst{384\over\pi^4}\sum_{n=1}^\infty{1\over(2n-1)^4} \left[(-1)^n+{4\over(2n-1)\pi}\right]\cos(2n-1)\pi t\sin{(2n-1)\pi x\over2}\)

45. \(u_{tt}=5u_{xx},\quad 0<x<1,\quad t>0\),
    \(u(0,t)=0,\quad u_x(1,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=x^3(3x-4),\quad0\le x\le1\)

    Show answer

    \(u(x,t)=\dst{3072\over\sqrt5\,\pi^5}\sum_{n=1}^\infty{1\over(2n-1)^5} \left[(-1)^n+{3\over(2n-1)\pi}\right] \sin{(2n-1)\sqrt5\,\pi t\over2}\sin{(2n-1)\pi x\over2}\)

46. C \(u_{tt}=16u_{xx},\quad 0<x<1,\quad t>0\),
    \(u(0,t)=0,\quad u_x(1,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=x(x^3-2x^2+2),\quad0\le x\le1\)

    Show answer

    \(u(x,t)=\dst{384\over\pi^5}\sum_{n=1}^\infty{1\over(2n-1)^5} \left[(-1)^n+{4\over(2n-1)\pi}\right]\sin(2n-1)\pi t\sin{(2n-1)\pi x\over2}\)

47. Adapt the proof of Theorem 12.2.2 to find d’Alembert’s solution of the initial-boundary value problem in Exercise 34.

    Show answer

    \(\dst u(x,t)={1\over2}[S_{M\!f}(x+at)+S_{M\!f}(x-at)]+ {1\over2a}\int_{x-at}^{x+at}S_{M\!g}(\tau)\,d\tau\)

48. Use the result of Exercise 47 to show that the formal solution of the initial-boundary value problem in Exercise 34 is an actual solution if \(g\) is differentiable and \(f\) is twice differentiable on \([0,L]\) and

    \[ f(0)=f'_-(L)=g(0)=g_-'(L)=f''_+(0)=0. \]

    Hint: See Exercise \(11.3\). 58 and apply Theorem 12.2.3 with \(L\) replaced by \(2L\).

49. Justify defining the formal solution of the initial-boundary value problem

    \[ \begin{array}{c} u_{tt}=a^2u_{xx},\quad 0<x<L,\quad t>0,\\ u_x(0,t)=0,\quad u_x(L,t)=0,\quad t>0,\\ u(x,0)=f(x),\quad u_t(x,0)=g(x),\quad 0\le x\le L. \end{array} \]

    to be

    \[ u(x,t)=\alpha_0+\beta_0t+\sum_{n=1}^\infty \left(\alpha_n\cos{n\pi at\over L}+{L\beta_n\over n\pi a}\sin{n\pi at\over L}\right) \cos{n\pi x\over L}, \]

    where

    \[ C_f(x)=\alpha_0+\sum_{n=1}^\infty\alpha_n\cos{n\pi x\over L} \mbox{\quad and \quad} C_g(x)=\beta_0+\sum_{n=1}^\infty\beta_n\cos{n\pi x\over L} \]

    are the Fourier cosine series of \(f\) and \(g\) on \([0,L]\); that is,

    \[ \alpha_0={1\over L}\int_0^Lf(x)\,dx,\quad \beta_0={1\over L}\int_0^Lg(x)\,dx, \]
    \[ \alpha_n={2\over L}\int_0^Lf(x)\cos{n\pi x\over L}\,dx, \mbox{\quad and \quad} \beta_n={2\over L}\int_0^Lg(x)\cos{n\pi x\over L}\,dx,\quad n=1,2,3,\dots. \]

50. \(u_{tt}=5u_{xx},\quad 0<x<2,\quad t>0\),
    \(u_x(0,t)=0,\quad u_x(2,t)=0,\quad t>0\),
    \(u(x,0)=2x^2(3-x),\quad u_t(x,0)=0,\quad0\le x\le 2\)

    Show answer

    \(u(x,t)=\dst4-{768\over\pi^4}\sum_{n=1}^\infty{1\over (2n-1)^4}\cos{\sqrt5(2n-1)\pi t\over2}\,\cos{(2n-1)\pi x\over 2}\)

51. \(u_{tt}=5u_{xx},\quad 0<x<2,\quad t>0\),
    \(u_x(0,t)=0,\quad u_x(2,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=2x^2(3-x),\quad0\le x\le 2\)

    Show answer

    \(u(x,t)=\dst4t-{1536\over\sqrt5\,\pi^5}\sum_{n=1}^\infty{1\over (2n-1)^5}\sin{\sqrt5(2n-1)\pi t\over2}\,\cos{(2n-1)\pi x\over 2}\)

52. \(u_{tt}=4u_{xx},\quad 0<x<\pi,\quad t>0\),
    \(u_x(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0\),
    \(u(x,0)=x^3(3x-4\pi),\quad u_t(x,0)=0,\quad0\le x\le \pi\)

    Show answer

    \(u(x,t)=\dst-{2\pi^4\over5}-48\sum_{n=1}^\infty{1+(-1)^n2\over n^4}\cos2nt\cos nx\)

53. \(u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0\),
    \(u_x(0,t)=0,\quad u_x(1,t)=0,\quad t>0\),
    \(u(x,0)=3x^2(x^2-2),\quad u_t(x,0)=0,\quad0\le x\le 1\)

    Show answer

    \(u(x,t)=\dst-{7\over5}-{144\over\pi^4}\sum_{n=1}^\infty{(-1)^n\over n^4}\cos n\sqrt7\,\pi t\cos n\pi x\)

54. C \(u_{tt}=4u_{xx},\quad 0<x<\pi,\quad t>0\),
    \(u_x(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=x^3(3x-4\pi),\quad0\le x\le \pi\)

    Show answer

    \(u(x,t)=\dst-{2\pi^4t\over5}-24\sum_{n=1}^\infty{1+(-1)^n2\over n^5}\sin2n t\cos nx\)

55. \(u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0\),
    \(u_x(0,t)=0,\quad u_x(1,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=3x^2(x^2-2),\quad0\le x\le 1\)

    Show answer

    \(u(x,t)=\dst-{7t\over5}-{144\over\pi^5\sqrt7}\sum_{n=1}^\infty{(-1)^n\over n^5}\sin n\sqrt7\,\pi t\cos n\pi x\)

56. \(u_{tt}=16u_{xx},\quad 0<x<\pi,\quad t>0\),
    \(u_x(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0\),
    \(u(x,0)=x^2(x-\pi)^2,\quad u_t(x,0)=0,\quad0\le x\le \pi\)

    Show answer

    \(u(x,t)=\dst{{\pi^4\over30}-3\sum_{n=1}^\infty{1\over n^4}\cos8nt\cos2nx}\)

57. C \(u_{tt}=u_{xx},\quad 0<x<1,\quad t>0\),
    \(u_x(0,t)=0,\quad u_x(1,t)=0,\quad t>0\),
    \(u(x,0)=x^2(3x^2-8x+6),\quad u_t(x,0)=0,\quad0\le x\le 1\)

    Show answer

    \(u(x,t)=\dst{3\over5}-{48\over\pi^4}\sum_{n=1}^\infty{2+(-1)^n\over n^4}\cos n\pi t\cos n\pi x\)

58. \(u_{tt}=16u_{xx},\quad 0<x<\pi,\quad t>0\),
    \(u_x(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=x^2(x-\pi)^2,\quad0\le x\le \pi\)

    Show answer

    \(u(x,t)=\dst{{\pi^4t\over30}-{3\over8}\sum_{n=1}^\infty{1\over n^5}\sin8nt\cos2nx}\)

59. C \(u_{tt}=u_{xx},\quad 0<x<1,\quad t>0\),
    \(u_x(0,t)=0,\quad u_x(1,t)=0,\quad t>0\),
    \(u(x,0)=0,\quad u_t(x,0)=x^2(3x^2-8x+6),\quad0\le x\le 1\)

    Show answer

    \(u(x,t)=\dst{3t\over5}-{48\over\pi^5}\sum_{n=1}^\infty{2+(-1)^n\over n^5}\sin n\pi t\cos n\pi x\)

60. Adapt the proof of Theorem 12.2.2 to find d’Alembert’s solution of the initial-boundary value problem in Exercise 49.

    Show answer

    \(u(x,t)={1\over2}[C_f(x+at)+C_f(x-at)]+ {1\over2a}\int_{x-at}^{x+at}C_g(\tau)\,d\tau\)

61. Use the result of Exercise 60 to show that the formal solution of the initial-boundary value problem in Exercise 49 is an actual solution if \(g\) is differentiable and \(f\) is twice differentiable on \([0,L]\) and

    \[ f'_+(0)=f'_-(L)=g'_+(0)=g_-'(L)=0. \]

62. Suppose \(\lambda\) and \(\mu\) are constants and either \(p_n(x)= \cos n\lambda x\) or \(p_n(x)=\sin n\lambda x\), while either \(q_n(t)=\cos n\mu t\) or \(q_n(t)=\sin n\mu t\) for \(n=1\), \(2\), \(3\), …. Let

    \[ u(x,t)=\sum_{n=1}^\infty k_np_n(x)q_n(t), \tag*{\rm(A)} \]

    where \(\{k_n\}_{n=1}^\infty\) are constants.

    1. Show that if \(\sum_{n=1}^\infty |k_n|\) converges then \(u(x,t)\) converges for all \((x,t)\).

    2. Use Theorem 12.1.2 to show that if \(\sum_{n=1}^\infty n|k_n|\) converges then (A) can be differentiated term by term with respect to \(x\) and \(t\) for all \((x,t)\); that is,

       \[ u_x(x,t)= \sum_{n=1}^\infty k_np_n'(x)q_n(t) \]

       and

       \[ u_t(x,t)= \sum_{n=1}^\infty k_np_n(x)q_n'(t). \]

    3. Suppose \(\sum_{n=1}^\infty n^2|k_n|\) converges. Show that

       \[ u_{xx}(x,y)= \sum_{n=1}^\infty k_np_n''(x)q_n(t) \]

       and

       \[ u_{tt}(x,y)= \sum_{n=1}^\infty k_np_n(x)q_n''(t) \]

    4. Suppose \(\sum_{n=1}^\infty n^2|\alpha_n|\) and \(\sum_{n=1}^\infty n|\beta_n|\) both converge. Show that the formal solution

       \[ u(x,t)=\sum_{n=1}^\infty\left(\alpha_n\cos{n\pi at\over L}+{\beta_nL\over n\pi a}\sin{n\pi at\over L}\right) \sin{n\pi x\over L} \]

       of Equation 12.2.1 satisfies \(u_{tt}=a^2u_{xx}\) for all \((x,t)\).

       This conclusion also applies to the formal solutions defined in Exercises 17, 34, and 49.

63. Suppose \(g\) is differentiable and \(f\) is twice differentiable on \((-\infty,\infty)\), and let

    \[ u_0(x,t)={f(x+at)+f(x-at)\over2}\mbox{\quad and \quad} u_1(x,t)={1\over2a}\int_{x-at}^{x+at}g(u)\,du. \]

    1. Show that

       \[ {\partial^2 u_0\over\partial t^2}=a^2{\partial^2u_0\over\partial x^2},\quad-\infty<x<\infty,\quad t>0, \]

       and

       \[ u_0(x,0)=f(x),\quad {\partial u_0\over\partial t}(x,0)=0,\quad -\infty<x<\infty. \]

    2. Show that

       \[ {\partial^2 u_1\over\partial t^2}=a^2{\partial^2u_1\over\partial x^2},\quad-\infty<x<\infty,\quad t>0, \]

       and

       \[ u_1(x,0)=0,\quad {\partial u_1\over\partial t}(x,0)=g(x),\quad -\infty<x<\infty. \]

    3. Solve

       \[ u_{tt}=a^2u_{xx},\quad-\infty<t<\infty,\quad t>0, \]
       \[ u(x,0)=f(x),\quad u_t(x,0)=g(x),\quad-\infty<x<\infty. \]
    Show answer

    (c) \(u(x,t)=\dst{f(x+at)+f(x-at)\over2}+ {1\over2a}\int_{x-at}^{x+at}g(u)\,du\)

64. \(u(x,0)=x\),  \(u_t(x,0)=4ax\),  \(-\infty<x<\infty\)

    Show answer

    \(u(x,t)=x(1+4at)\)

65. \(u(x,0)=x^2\),  \(u_t(x,0)=1\),  \(-\infty<x<\infty\)

    Show answer

    \(u(x,t)=x^2+a^2t^2+t\)

66. \(u(x,0)=\sin x\),  \(u_t(x,0)=a\cos x\),  \(-\infty<x<\infty\)

    Show answer

    \(u(x,t)=\sin(x+at)\)

67. \(u(x,0)=x^3\),  \(u_t(x,0)=6x^2\),  \(-\infty<x<\infty\)

    Show answer

    \(u(x,t)=x^3+6tx^2+3a^2t^2x+2a^2t^3\)

68. \(u(x,0)=x\sin x\),  \(u_t(x,0)=\sin x\),  \(-\infty<x<\infty\)

    Show answer

    \(u(x,t)=\dst x\sin x\cos at+at\cos x\sin at+{\sin x\sin at\over a}\)