SECTION 8.5 433 §8.5 Application of Laplace Transforms to Partial Differential Equations In Sections 8.2 and 8.3, we illustrated the effective use of Laplace transforms in solv- ing ordinary differential equations. The transform replaces a differential equation in y(t) with an algebraic equation in its transformy ˜(s). It is then a matter of finding the inverse transform ofy ˜(s) either by partial fractions and tables (Section 8.1) or by residues (Section 8.4). Laplace transforms also provide a potent technique for solving partial differential equations. When the transform is applied to the variable t in a partial differential equation for a function y(x, t), the result is an ordinary differential equation for the transformy ˜(x, s). The ordinary differential equation is solved fory ˜(x, s) and the function is inverted to yield y(x, t). We illustrate this procedure with five physical examples. The first two examples are on unbounded spatial intervals; inverse transforms are found in tables. The last three examples are on bounded spatial intervals; inverse transforms are calculated with residues. Problems on Unbounded Intervals Example 8.18 A very long cylindrical rod is placed along the positive x-axis with one end at x = 0 (Figure 8.22). The rod is so long that any effects due to its right end may be neglected. Its sides are covered with perfect insulation so that no heat can enter or escape therethrough. At time t = 0, the temperature of the rod is 0◦C throughout. Suddenly the left end of the rod has its temperature raised to U0, and maintained at this temperature thereafter. The initial, boundary-value problem describing temperature U(x, t) at points in the rod is ∂U ∂2U = k ,x>0,t>0, (8.19a) ∂t ∂x2 U(0,t)=U0,t>0, (8.19b) U(x, 0) = 0,x>0, (8.19c) where k is a constant called the thermal diffusivity of the material in the rod. Use Laplace transforms on variable t to find U(x, t). Insulation x =0 Insulation x Figure 8.22 Solution When we apply the Laplace transform to the partial differential equa- tion, and use property 8.10a, ∂2U sU˜(x, s) − U(x, 0) = k L . ∂x2 Since the integration with respect to t in the Laplace transform and the differenti- ation with respect to x are independent, we interchange the order of operations on the right, ∂2U˜ sU˜(x, s)=k , ∂x2 where we have also used initial condition 8.19c. Because only derivatives with respect to x remain, we replace the partial derivative with an ordinary derivative, 434 SECTION 8.5 d2U˜ sU˜ = k ,x>0. (8.20a) dx2 When we take Laplace transforms of boundary condition 8.19b, we obtain U U˜(0,s)= 0 , (8.20b) s a boundary condition to accompany differential equation 8.20a. A general solution of 8.20a is √ √ U˜(x, s)=Ae s/kx + Be− s/kx. Because U(x, t) must remain bounded as x becomes infinite, so also must U˜(x, s). We must therefore set A = 0, in which case 8.20b requires B = U0/s. Thus, U √ U˜(x, s)= 0 e− s/kx. (8.21) s The inverse transform of this function can be found in tables, √ − s/kx −1 (e ) x U(x, t)=U0 L = U0 erfc √ , (8.22a) s 2 kt where erfc (x) is the complementary error function x ∞ 2 2 2 2 erfc (x)=1− erf (x)=1− √ Z e−u du = √ Z e−u du. (8.22b) π 0 π x Notice that for any x>0 and any t>0, h(,x t) temperature U(x, t) is positive. This t indicates that the abrupt change in 3 >t2 >t1 ◦ temperature at the end x = 0 from 0 C t =t3 to U0 is felt instantaneously at every t =t2 point in the rod. We have shown a plot t=t1 of U(x, t) for various fixed values of t in Figure 8.23.• x Figure 8.23 Example 8.19 A very long taut string is supported from below so that it lies motionless on the positive x-axis. At time t = 0, the support is removed and gravity is permitted to act on the string. If the end x = 0 is fixed at the origin, the initial, boundary-value problem describing displacements y(x, t) of points in the string is ∂2y ∂2y = c2 − g, x > 0,t>0, (8.23a) ∂t2 ∂x2 y(0,t)=0,t>0, (8.23b) y(x, 0) = 0,x>0, (8.23c) yt(x, 0) = 0,x>0, (8.23d) where g =9.81 and c>0 is a constant depending on the material and tension of the string. Initial condition 8.23d expresses the fact that the initial velocity of the string is zero. Use Laplace transforms to solve this problem. SECTION 8.5 435 Solution When we apply Laplace transforms to the partial differential equation, and use property 8.10b, ∂2y g s2y˜(x, s) − sy(x, 0) − y (x, 0) = c2L − . t ∂x2 s We now use initial conditions 8.23c,d, and interchange operations on the right, d2y˜ g s2y˜ = c2 − , dx2 s or, d2y˜ s2 g − y˜ = ,x>0. (8.24a) dx2 c2 c2s This ordinary differential equation is subject to the transform of 8.23b, y˜(0,s)=0. (8.24b) A general solution of differential equation 8.24a is g y˜(x, s)=Aesx/c + Be−sx/c − . s3 For this function to remain bounded as x →∞, we must set A = 0, in which case condition 8.24b implies that B = g/s3. Thus, g g y˜(x, s)=− + e−sx/c. (8.25) s3 s3 Property 8.4b gives gt2 g x 2 x y(x, t)=− + t − h t − (8.26) 2 2 c c where h(t − x/c) is the Heaviside unit step function. What this says is that a point x in the string falls freely under gravity for 0 <t<x/c, after which it falls with constant velocity −gx/c [since for t > x/c, y(x, t)=(g/2)(−2xt/c + x2/c2)]. A picture of the string at any given time t0 is shown in Figure 8.24. It is parabolic for 0 <x<ct0 and horizontal for x>ct0.Ast0 increases, the parabolic portion lengthens and the horizontal section drops.• y ct0 x gt 2 - 0 2 2 gx gt0 y = (x-2ct0) y =- 2c2 2 Figure 8.24 Problems on Bounded Intervals The next three examples are on bounded intervals; we use residues to find inverse transforms. 436 SECTION 8.5 Example 8.20 A cylindrical rod of length L has its ends at x = 0 and x = L on the x-axis. Its sides are insulated so that no heat can enter or escape therethrough. At time t = 0, the left end (x = 0) has temperature 0◦C, and the right end (x = L) has temperature L◦C, and the temperature rises linearly between the ends. Suddenly at time t = 0, the temperature of the right end is reduced to 0◦C, and both ends are held at temperature zero thereafter. The initial, boundary-value problem describing temperature U(x, t) at points in the rod is ∂U ∂2U = k , 0 <x<L, t>0, (8.27a) ∂t ∂x2 U(0,t)=0,t>0, (8.27b) U(L, t)=0,t>0, (8.27c) U(x, 0) = x, 0 <x<L, (8.27d) k again being the thermal diffusivity of the material in the rod. Use Laplace trans- forms to find U(x, t). Solution When we take Laplace transforms of 8.27a, and use property 8.10a, d2U˜ sU˜(x, s) − x = k , dx2 or, d2U˜ s x − U˜ = − , 0 <x<L. (8.28a) dx2 k k This ordinary differential equation is subject to the transforms of 8.27b,c, U˜(0,s)=0, (8.28b) U˜(L, s)=0. (8.28c) A general solution of differential equation 8.28a is r s r s x U˜(x, s)=A cosh x + B sinh x + . k k s For Example 8.18 on an unbounded interval, we used exponentials in the solution of 8.20a. On bounded intervals, hyperbolic functions are preferable. Boundary conditions 8.28b,c require r s r s L 0=A, 0=A cosh L + B sinh L + . k k s From these, 1 L sinh ps/kx! U˜(x, s)= x − . (8.29) s sinh ps/kL Although ps/k denotes the principal square root function, U˜(x, s) in 8.29 is a solution of problem 8.28 for any branch of the root function. It remains now to find the inverse transform of U˜(x, s). We do this by cal- culating residues of estU˜(x, s) at its singularities. To discover the nature of the SECTION 8.5 437 singularityats = 0, we expand U˜(x, s) in a Laurent series around s = 0. Provided we use the same branch for the root function in numerator and denominator of 8.29, we may write that 1 ( L[ps/kx +(ps/kx)3/3! + ···] ) U˜(x, s)= x − s ps/kL +(ps/kL)3/3! + ··· 1 x + sx3/(6k)+··· = x − s 1+sL2/(6k)+··· 1 sx(L2 − x2) = + ··· s 6k x(L2 − x2) = + terms in s, s2, ···. 6k Since this is the Laurent series of U˜(x, s) valid in some annulus around s = 0, it follows that U˜(x, s) has a removable singularity at s =0.