15.2 Laplace Transform

INTRODUCTION

In Chapter 4 we defined the Laplace transform of a function f(t), t ≥ 0, to be

whenever the improper integral converges. This integral transforms a function f(t) into another function F of the transform parameter s, that is, {f(t)} = F(s). The main application of the Laplace transform in Chapter 4 was to the solution of certain types of initial-value problems involving linear ordinary differential equations with constant coefficients. Recall, the Laplace transform of such an equation reduces the ODE to an algebraic equation. In this section we are going to apply the Laplace transform to linear partial differential equations. We will see that this transform reduces a PDE to an ODE.

Transform of Partial Derivatives

The boundary-value problems considered in this section will involve either the one-dimensional wave and heat equations or slight variations of these equations. These PDEs involve an unknown function of two independent variables u(x, t), where the variable t represents time t ≥ 0. We define the Laplace transform of u(x, t) with respect to t by

where x is treated as a parameter. Throughout this section we shall assume that all the operational properties of Sections 4.2, 4.3, and 4.4 apply to functions of two variables. For example, by Theorem 4.2.2, the transform of the partial derivative ∂u/∂t is

that is, (1)

Similarly, (2)

Since we are transforming with respect to t, we further suppose that it is legitimate to interchange integration and differentiation in the transform of 2u/∂x2:

that is, (3)

In view of (1) and (2) we see that the Laplace transform is suited to problems with initial conditions—namely, those problems associated with the heat equation or the wave equation. We will see in Section 15.4 that boundary-value problems involving Laplace’s equation in which one (or both) of the spatial variables is defined on an unbounded interval can often be solved using different integral transforms.

EXAMPLE 1 Laplace Transform of a PDE

Find the Laplace transform of the wave equation

SOLUTION

From (2) and (3),

becomes

or (4)

The Laplace transform with respect to t of either the wave equation or the heat equation eliminates that variable, and for the one-dimensional equations the transformed equations are then ordinary differential equations in the spatial variable x. In solving a transformed equation, we treat s as a parameter.

EXAMPLE 2 Using the Laplace Transform to Solve a BVP

Solve

subject to

SOLUTION

The partial differential equation is recognized as the wave equation with a = 1. From (4) and the given initial conditions, the transformed equation is

(5)

where U(x, s) = {u(x, t)}. Since the boundary conditions are functions of t, we must also find their Laplace transforms:

{u(0, t)} = U(0, s) = 0 and {u(1, t)} = U(1, s) = 0. (6)

The results in (6) are boundary conditions for the ordinary differential equation (5). Since (5) is defined over a finite interval, its complementary function is

Uc(x, s) = c1 cosh sx + c2 sinh sx.

The method of undetermined coefficients yields a particular solution

Hence

But the conditions U(0, s) = 0 and U(1, s) = 0 yield, in turn, c1 = 0 and c2 = 0. We conclude that

Therefore

EXAMPLE 3 Using the Laplace Transform to Solve a BVP

A very long string is initially at rest on the nonnegative x-axis. The string is secured at x = 0, and its distant right end slides down a frictionless vertical support. The string is set in motion by letting it fall under its own weight. Find the displacement u(x, t).

SOLUTION

Since the force of gravity is taken into consideration, it can be shown that the wave equation has the form

where g is the acceleration due to gravity. The boundary and initial conditions are, respectively,

The second boundary condition lim ∂u/∂x = 0 indicates that the string is horizontal at a great distance from the left end. Now from (2) and (3),

becomes

or, in view of the initial conditions,

The transforms of the boundary conditions are

With the aid of undetermined coefficients, the general solution of the transformed equation is found to be

The boundary condition lim dU/dx = 0 implies c2 = 0, and U(0, s) = 0 gives c1 = g/s3. Therefore

Now by the second translation theorem we have

or

To interpret the solution, let us suppose t > 0 is fixed. For 0 ≤ xat, the string is the shape of a parabola passing through the points (0, 0) and (at, − gt2). For x > at, the string is described by the horizontal line u = gt2. See FIGURE 15.2.1.

In an x u coordinate plane, a very long string is secured on the left end at the point (0, 0), and its distant right end slides down a frictionless vertical support indicated in the graph by a dotted vertical line. A point a t is marked out on the x axis such that a t > 0. The curve of the string is as follows: the curve begins at the point (0, 0) drops down to the point marked out and labeled (a t, negative 1 over 2 g t^2), remains steady and horizontal then moves up to form a peak then moves down to form a trough and then rises to the same level as when it was steady and horizontal which then is marked out in dotted line till it connects to the vertical support. The dotted vertical line is indicated by an arrow and labeled vertical support at infinity.

FIGURE 15.2.1 A long string falling under its own weight in Example 3

Observe that the problem in the next example could be solved by the procedure in Section 13.6. The Laplace transform provides an alternative solution.

EXAMPLE 4 A Solution in Terms of erf (x)

Solve the heat equation

subject to

SOLUTION

From (1) and (3) and the given initial condition,

becomes (7)

The transforms of the boundary conditions are

(8)

Since we are concerned with a finite interval on the x-axis, we choose to write the general solution of (7) as

Applying the two boundary conditions in (8) yields, respectively, . Thus

Now the inverse transform of the latter function cannot be found in most tables. However, by writing

and using the geometric series

we find

If we assume that the inverse Laplace transform can be done term by term, it follows from entry 3 of Table 15.1.1 that

(9)

The solution (9) can be rewritten in terms of the error function using erfc(x) = 1 − erf (x):

(10)

Also see Problem 8 in Exercises 15.1

FIGURE 15.2.2(a), obtained with the aid of the 3D plot function in a CAS, shows the surface over the rectangular region 0 ≤ x ≤ 1, 0 ≤ t ≤ 6 defined by the partial sum S10(x, t) of the solution (10). It is apparent from the surface and the accompanying two-dimensional graphs that at a fixed value of x (the curve of intersection of a plane slicing the surface perpendicular to the x-axis) on the interval [0, 1], the temperature u(x, t) increases rapidly to a constant value as time increases. See Figure 15.2.2(b) and 15.2.2(c). For a fixed time (the curve of intersection of a plane slicing the surface perpendicular to the t-axis), the temperature u(x, t) naturally increases from 0 to 100. See Figure 15.2.2(d) and 15.2.2(e).

A surface and corresponding curves. A. u subscript 0 = 100. An image of a 3 D plot function has a surface over the rectangular region 0 <= x <= 1, 0 <= t <= 6. The surface is graphed on an x t u (x, t) coordinate system. X ranges from 0 to 1 with intervals of 0.2. t ranges from 0 to 6 with intervals of 6. u (x, t) ranges from 0 to 100 with intervals of 25. The surface is made of the parallel ascending lines from the x axis and perpendicularly ascending lines. B. x = 0.2. The curve is graphed on a t u (0.2, t) coordinate plane. It begins at (0, 0), goes up and to the right a little and then goes straight to the right till (6, 20). C. x = 0.7. The curve is graphed on a t u (0.7, t) coordinate plane. It begins at approximately (0, 20), goes up and to the right a little and then goes straight to the right till (6, 70). D. t = 0.1. The curve is graphed on a t u (x, 0.1) coordinate plane. It begins at (0, 0), goes up and to the right up with increasing steepness up to (1, 100). E. t = 4. The line is graphed on a t u (x, 4) coordinate plane. It begins at (0, 0), goes up and to the right up to (1, 100).

FIGURE 15.2.2 Graph of solution given in (10). In (b) and (c), x is held constant. In (d) and (e), t is held constant.

15.2 Exercises Answers to selected odd-numbered problems begin on page ANS-38.

In the following problems use tables as necessary.

  1. A string is secured to the x-axis at (0, 0) and (L, 0). Find the displacement u(x, t) if the string starts from rest in the initial position A sin(πx/L).
  2. Solve the boundary-value problem
  3. The displacement of a semi-infinite elastic string is determined from

    Solve for u(x, t).

  4. Solve the boundary-value problem in Problem 3 when

    Sketch the displacement u(x, t) for t > 1.

  5. In Example 3, find the displacement u(x, t) when the left end of the string at x = 0 is given an oscillatory motion described by f(t) = A sin ωt.
  6. The displacement u(x, t) of a string that is driven by an external force is determined from

    Solve for u(x, t).

    1. A uniform elastic bar is clamped at and is initially at rest. If a constant unit force is applied to the free end at the longitudinal displacement of a cross section is determined from

      Solve for [Hint: Expand in a geometric series. See Example 4.]

    2. For a fixed value of x satisfying sketch the graph of as a function of t for
    3. Sketch the graph of as a function of t for
  7. A uniform semi-infinite elastic beam moving along the x-axis with a constant velocity −v0 is brought to a stop by hitting a wall at time t = 0. See FIGURE 15.2.3. The longitudinal displacement u(x, t) is determined from

    Solve for u(x, t).

    A uniform semi-infinite elastic beam moving toward the left along the x axis is brought to a stop by hitting a wall. The top edge of the beam is indicated by an arrow and labeled beam. The point at which the beam hits the wall is labeled x = 0. An arrow pointing to the left labeled v subscript 0 indicates the velocity and direction of the movement of the beam.

    FIGURE 15.2.3 Moving elastic beam in Problem 8

  8. Solve the boundary-value problem
  9. Solve the boundary-value problem

In Problems 11–18, use the Laplace transform to solve the heat equation uxx = ut, x > 0, t > 0 subject to the given conditions.

  1. [Hint: Use the convolution theorem.]

  2. Solve the boundary-value problem

  3. Show that a solution of the boundary-value problem

    where r is a constant, is given by

  4. Solve the boundary-value problem
  5. Solve the boundary-value problem
  6. A rod of length L is held at a constant temperature u0 at its ends x = 0 and x = L. If the rod’s initial temperature is u0 + u0 sin(/L), solve the heat equation uxx = ut, 0 < x < L, t > 0 for the temperature u(x, t).
  7. If there is a heat transfer from the lateral surface of a thin wire of length L into a medium at constant temperature um, then the heat equation takes on the form

    where h is a constant. Find the temperature u(x, t) if the initial temperature is a constant u0 throughout and the ends x = 0 and x = L are insulated.

  8. A rod of unit length is insulated at x = 0 and is kept at temperature zero at x = 1. If the initial temperature of the rod is a constant u0, solve kuxx = ut, 0 < x < 1, t > 0 for the temperature u(x, t). [Hint: Expand 1/(1 + ) in a geometric series. See Example 4.]
  9. An infinite porous slab of unit width is immersed in a solution of constant concentration c0. A dissolved substance in the solution diffuses into the slab. The concentration c(x, t) in the slab is determined from

    where D is a constant. Solve for c(x, t).

  10. A very long telephone transmission line is initially at a constant potential u0. If the line is grounded at x = 0 and insulated at the distant right end, then the potential u(x, t) at a point x along the line at time t is determined from

    where R, C, and G are constants known as resistance, capacitance, and conductance, respectively. Solve for u(x, t). [Hint: See Problem 7 in Exercises 15.1.]

  11. Show that a solution of the boundary-value problem

    is

  12. Starting at t = 0, a concentrated load of magnitude F0 moves with a constant velocity along a semi-infinite string. In this case, the wave equation becomes

    where is the Dirac delta function. Solve the above PDE subject to

    when

  13. Solve the boundary-value problem in Problem 29 when
  14. The temperature in a semi-infinite solid is modeled by the boundary-value problem

    where is a nonzero constant. Solve for Use the solution to determine analytically the value of

  15. In Problem 31, if there is a constant flux of heat into the solid at its left-hand boundary, then the boundary condition is

    where is a positive constant. Solve for Use the solution to determine analytically the value of

In Problems 33–36, the partial differential equation is the heat equation in spherical coordinates

Take the Laplace transform of this PDE with respect to the variable t. To solve the transformed equation, use the substitution where

  1. In Problem 9 of Exercises 14.3 you were asked to find the time-dependent temperature inside a unit sphere centered at the origin. Find the temperature outside a unit sphere if the boundary and initial conditions are
  2. Find the temperature outside a unit sphere if the boundary and initial conditions are

    where and are nonzero constants.

  3. Find the temperature inside a unit sphere if the boundary and initial conditions are

    where and are nonzero constants.

  4. Find the temperature outside a unit sphere if the boundary and initial conditions are

    where is a nonzero constant.

Mathematical Model

  1. Pulsating Sphere A sphere of radius centered at the origin is submerged in an ideal compressible fluid. Starting at the surface of the sphere is caused to pulsate with radial velocity where is a nonzero constant. The pulsation continues indefinitely. A cross section of the sphere is shown in FIGURE 15.2.4. A simplified model for the velocity potential of points in the surrounding fluid is given by the boundary-value problem

    Solve for Then determine the radial velocity of points in the fluid.

    A cross-section of a sphere of radius a is submerged in a liquid labeled fluid. Twelve equidistant two-sided arrows go radially outward such that the arrow pointing inward is inside the sphere and the arrow pointing outward is outside the sphere in the fluid region. These arrows indicate the pulsating movement of the sphere.

    FIGURE 15.2.4 Pulsating sphere in Problem 37

  2. Find the radial velocity of points in the fluid surrounding the sphere of radius r = a in Problem 37 if the boundary and initial conditions are

    In this case, the sphere is expanding for a period of time but is not pulsating.

Computer Lab Assignments

  1. Use a CAS to obtain the graph of in Problem 31 over the rectangular region defined by Assume and Indicate the two boundary conditions and initial condition on your graph. Use 2D and 3D plots of to verify the value of
  2. Use a CAS to obtain the graph of in Problem 32 over the rectangular region defined by Assume and Use 2D and 3D plots of to verify the value of
  3. Chemical Messages Humans gather most of their information on the outside world through sight and sound. But many creatures use chemical signals as their primary means of communication; for example, honeybees, when alarmed, emit a substance and fan their wings feverishly to relay the warning signal to the bees that attend to the queen. These molecular messages between members of the same species are called pheromones. The signals may be carried by moving air or water or by a diffusion process in which the random movement of gas molecules transports the chemical away from its source. FIGURE 15.2.5 shows an ant emitting an alarm chemical into the still air of a tunnel. If c(x, t) denotes the concentration of the chemical x centimeters from the source at time t, then c(x, t) satisfies

    and k is a positive constant. The emission of pheromones as a discrete pulse gives rise to a boundary condition of the form

    where δ(t) is the Dirac delta function.

    1. Solve the boundary-value problem if it is further known that c(x, 0) = 0, x > 0, and lim c(x, t) = 0, t > 0.
    2. Use a CAS to plot the graph of the solution in part (a) for x > 0 at the fixed times t = 0.1, t = 0.5, t = 1, t = 2, t = 5.
    3. For a fixed time t, show that (x, t) dx = Ak. Thus Ak represents the total amount of chemical discharged.
      A thin tunnel above an axis contains three ants. One ant on the left end of the tunnel and two ants on the right end. An arrow pointing to the right is displayed in front of the ant on the left.

      FIGURE 15.2.5 Ants in Problem 41