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612 LINEAR PARTIAL DIFFERENTIAL EQUATIONS Given the transform f(p), the function f(t) can be found by means of the inverse Laplace transform f(t)=L –1 f(p) ,whereL –1 f(p) ≡ 1 2πi c+i∞ c–i∞ f(p)e pt dp,(14.5.1.2) where the integration path is parallel to the imaginary axis and lies to the right of all singularities of f(p), which corresponds to c > σ 0 . In order to solve nonstationary boundary value problems, the following Laplace trans- form formulas for derivatives will be required: L f (t) = p f(p)–f(0), L f (t) = p 2 f(p)–pf(0)–f (0), (14.5.1.3) where f (0)andf (0) are the initial conditions. More details on the properties of the Laplace transform and the inverse Laplace transform can be found in Section 11.2. The Laplace transforms of some functions are listed in Section T3.1. Tables of inverse Laplace transforms are listed in Section T3.2. Such tables are convenient to use in solving linear problems for partial differential equations. 14.5.1-2. Solution procedure for linear problems using the Laplace transform. Figure 14.2 shows schematically how one can utilize the Laplace transforms to solve boundary value problems for linear parabolic or hyperbolic equations with two independent variables in the case where theequation coeffi cients are independent of t (the same procedure can be applied for solving linear problems characterized by higher-order equations). Here and henceforth, the short notation w(x, p)=L{w(x, t)} will be used; the arguments of w may be omitted. It is significant that with the Laplace transform, the original problem for a partial differential equation is reduced to a simpler problem for an ordinary differential equation with parameter p; the derivatives with respect to t are replaced by appropriate algebraic expressions taking into account the initial conditions; see formulas (14.5.1.3). 14.5.1-3. Solving linear problems for parabolic equations with the Laplace transform. Consider a linear nonstationary boundary value problem for the parabolic equation ∂w ∂t = a(x) ∂ 2 w ∂x 2 + b(x) ∂w ∂x + c(x)w + Φ(x, t)(14.5.1.4) with the initial condition (14.4.1.3) and the general nonhomogeneous boundary conditions s 1 ∂ x w + k 1 w = g 1 (t)atx = x 1 , s 2 ∂ x w + k 2 w = g 2 (t)atx = x 2 . (14.5.1.5) The application of the Laplace transform results in the problem defined by the ordinary differential equation for x (p is treated as a parameter) a(x) ∂ 2 w ∂x 2 + b(x) ∂ w ∂x +[c(x)–p]w + f 0 (x)+ Φ(x, p)=0 (14.5.1.6) 14.5. INTEGRAL TRANSFORMS METHOD 613 Original problem = partial differential equation for initial conditions and boundary conditions, wwxt=(,), { Application of the Laplace transform PL.5.1. with respect to (1) t Solution of the ordinary differential equation with the boundary conditions Application of the inverse Laplace transform PL.5.1.(2) { Problem for transform = ordinary differential equation for boundary conditionswwxp=(,), ~~ ~~ Finding the transform wwxt=(,) Obtaining solution to the original problem: wwxt=(,) Figure 14.2. Solution procedure for linear boundary value problems using the Laplace transform. with the boundary conditions s 1 ∂ x w + k 1 w = g 1 (p)atx = x 1 , s 2 ∂ x w + k 2 w = g 2 (p)atx = x 2 . (14.5.1.7) Notation employed: Φ(x, p)=L Φ(x, t) and g n (p)=L g n (t) (n = 1, 2). On solving problem (14.5.1.6)–(14.5.1.7), one should apply to the resulting solution w = w(x, p)the inverse Laplace transform (14.5.1.2) to obtain the solution, w = w(x, t), of the original problem. Example 1. Consider the first boundary value problem for the heat equation: ∂w ∂t = ∂ 2 w ∂x 2 (x > 0, t > 0), w = 0 at t = 0 (initial condition), w = w 0 at x = 0 (boundary condition), w → 0 at x →∞ (boundary condition). We apply the Laplace transform with respect to t. Let us multiply the equation, the initial condition, and the boundary conditions by e –pt and then integrate with respect to t fromzerotoinfinity. Taking into account the relations L{∂ t w} = pw – w| t=0 = pw (used are first property (14.5.1.3) and the initial condition), L{w 0 } = w 0 L{1} = w 0 /p (used are property 1 of Subsection T3.1.1 and the relation L{1} = 1/p, see Property 1 of Subsection T3.1.2) we arrive at the following problem for a second-order linear ordinary differential equation with parameter p: w xx – pw = 0, w = w 0 /p at x = 0 (boundary condition), w → 0 at x →∞ (boundary condition). 614 LINEAR PARTIAL DIFFERENTIAL EQUATIONS Integrating the equation yields the general solution w = A 1 (p)e –x √ p + A 2 (p)e x √ p . Using the boundary conditions, we determine the constants, A 1 (p)=w 0 /p and A 2 (p)=0. Thus, we have w = w 0 p e –x √ p . Let us apply the inverse Laplace transform to both sides of this relation. We refer to the table in Subsec- tion T3.2.5, row 20 (where x must be replaced by t and then a by x 2 ), to find the inverse transform of the right-hand side. Finally, we obtain the solution of the original problem in the form w = w 0 erfc x 2 √ t . 14.5.1-4. Solving linear problems for hyperbolic equations by the Laplace transform. Consider a linear nonstationary boundary value problem defined by the hyperbolic equation ∂ 2 w ∂t 2 + ϕ(x) ∂w ∂t = a(x) ∂ 2 w ∂x 2 + b(x) ∂w ∂x + c(x)w + Φ(x, t)(14.5.1.8) with the initial conditions (14.4.1.3), (14.4.1.4) and the general boundary conditions (14.5.1.5). The application of the Laplace transform results in the problem defined by the ordinary differential equation for x (p is treated as a parameter) a(x) ∂ 2 w ∂x 2 +b(x) ∂ w ∂x +[c(x)–p 2 –pϕ(x)]w+f 0 (x)[p+ϕ(x)]+f 1 (x)+ Φ(x, p)=0 (14.5.1.9) with the boundary conditions (14.5.1.7). On solving this problem, one should apply the inverse Laplace transform to the resulting solution, w = w(x, p). 14.5.2. Fourier Transform and Its Application in Mathematical Physics 14.5.2-1. Fourier transform and its properties. The Fourier transform is defi ned as follows: f(u)=F f(x) ,whereF f(x) ≡ 1 √ 2π ∞ –∞ f(x)e –iux dx, i 2 =–1.(14.5.2.1) This relation is meaningful for any function f (x) absolutely integrable on the interval (–∞, ∞). Given f(u), the function f(x) can be found by means of the inverse Fourier transform f(x)=F –1 f(u) ,whereF –1 f(u) ≡ 1 √ 2π ∞ –∞ f(u)e iux du,(14.5.2.2) where the integral is understood in the sense of the Cauchy principal value. The main properties of the correspondence between functions and their Fourier trans- forms are gathered in Table 11.4. 14.6. SOLUTION OF THE CAUCHY PROBLEM VIA THE FUNDAMENTAL SOLUTION 615 14.5.2-2. Solving linear problems of mathematical physics by the Fourier transform. The Fourier transform is usually employed to solve boundary value problems for linear partial differential equations whose coefficients are independent of the space variable x, –∞ < x < ∞. The scheme for solving linear boundary value problems with the help of the Fourier transform is similar to that used in solving problems with help of the Laplace transform. With the Fourier transform, the derivatives with respect to x in the equation are replaced by appropriate algebraic expressions; see Property 4 or 5 in Table 11.4. In the case of two independent variables, the problem for a partial differential equation is reduced to a simpler problem for an ordinary differential equation with parameter u. On solving the latter problem, one determines the transform. After that, by applying the inverse Fourier transform, one obtains the solution of the original boundary value problem. Example 2. Consider the following Cauchy problem for the heat equation: ∂w ∂t = ∂ 2 w ∂x 2 (–∞ < x < ∞), w = f(x)att = 0 (initial condition). We apply the Fourier transform with respect to the space variable x. Setting w = F{w(x, t)} and taking into account the relation F{∂ xx w} =–u 2 w (see Property 4 in Table 11.4), we arrive at the following problem for a linear first-order ordinary differential equation in t with parameter u: w t + u 2 w = 0, w = f(u)att = 0, where f(u)isdefined by (14.5.2.1). On solving this problem for the transform w,wefind w = f(u)e –u 2 t . Let us apply the inversion formula to both sides of this equation. After some calculations, we obtain the solution of the original problem in the form w = 1 √ 2π ∞ –∞ f(u)e –u 2 t e iux du = 1 2π ∞ –∞ ∞ –∞ f(ξ)e –iuξ dξ e –u 2 t+iux du = 1 2π ∞ –∞ f(ξ) dξ ∞ –∞ e –u 2 t+iu(x–ξ) du = 1 √ 2πt ∞ –∞ f(ξ)exp – (x – ξ) 2 4t dξ. At the last stage we used the relation ∞ –∞ exp –a 2 u 2 + bu du = √ π |a| exp b 2 4a 2 . 14.6. Representation of the Solution of the Cauchy Problem via the Fundamental Solution 14.6.1. Cauchy Problem for Parabolic Equations 14.6.1-1. General formula for the solution of the Cauchy problem. Let x = {x 1 , , x n } and y = {y 1 , , y n },wherex R n and y R n . Consider a nonhomogeneous linear equation of the parabolic type with an arbitrary right-hand side, ∂w ∂t – L x,t [w]=Φ(x, t), (14.6.1.1) where the second-order linear differential operator L x,t is defined by relation (14.2.1.2). 616 LINEAR PARTIAL DIFFERENTIAL EQUATIONS The solution of the Cauchy problem for equation (14.6.1.1) with an arbitrary initial condition, w = f (x)att = 0, can be represented as the sum of two integrals, w(x, t)= t 0 R n Φ(y, τ) (x, y, t, τ) dy dτ + R n f(y) (x, y, t, 0) dy, dy = dy 1 dy n . Here, = (x, y, t, τ) is the fundamental solution of the Cauchy problem that satisfies, for t > τ ≥ 0, the homogeneous linear equation ∂ ∂t – L x,t [ ]=0 (14.6.1.2) with the nonhomogeneous initial condition of special form = δ(x – y)att = τ.(14.6.1.3) The quantities τ and y appear in problem (14.6.1.2)–(14.6.1.3) as free parameters, and δ(x)=δ(x 1 ) δ(x n )isthen-dimensional Dirac delta function. Remark 1. If the coefficients of the differential operator L x,t in (14.6.1.2) are independent of time t,then the fundamental solution of the Cauchy problem depends on only three arguments, (x, y, t,τ )= (x, y, t –τ). Remark 2. If the differential operator L x,t has constant coefficients, then the fundamental solution of the Cauchy problem depends on only two arguments, (x, y, t,τ )= (x – y, t – τ ). 14.6.1-2. Fundamental solution allowing incomplete separation of variables. Consider the special case where the differential operator L x,t in equation (14.6.1.1) can be represented as the sum L x,t [w]=L 1,t [w]+···+ L n,t [w], (14.6.1.4) where each term depends on a single space coordinate and time, L k,t [w] ≡ a k (x k , t) ∂ 2 w ∂x 2 k + b k (x k , t) ∂w ∂x k + c k (x k , t)w, k = 1, , n. Equations of this form are often encountered in applications. The fundamental solution of the Cauchy problem for the n-dimensional equation (14.6.1.1) with operator (14.6.1.4) can be represented in the product form (x, y, t, τ)= n k=1 k (x k , y k , t, τ), (14.6.1.5) where k = k (x k , y k , t, τ) are the fundamental solutions satisfying the one-dimensional equations ∂ k ∂t – L k,t [ k ]=0 (k = 1, , n) with the initial conditions k = δ(x k – y k )att = τ . In this case, the fundamental solution of the Cauchy problem (14.6.1.5) admits incom- plete separation of variables; the fundamental solution is separated in the space variables x 1 , , x n but not in time t. 14.6. SOLUTION OF THE CAUCHY PROBLEM VIA THE FUNDAMENTAL SOLUTION 617 Example 1. Consider the two-dimensional heat equation ∂w ∂t = ∂ 2 w ∂x 2 1 + ∂ 2 w ∂x 2 2 . The fundamental solutions of the corresponding one-dimensional heat equations are expressed as Equations Fundamental solutions ∂w ∂t = ∂ 2 w ∂x 2 1 =⇒ 1 (x 1 , y 1 , t, τ)= 1 2 √ π(t – τ ) exp – (x 1 – y 1 ) 2 4(t – τ ) , ∂w ∂t = ∂ 2 w ∂x 2 2 =⇒ 2 (x 2 , y 2 , t, τ)= 1 2 √ π(t – τ ) exp – (x 2 – y 2 ) 2 4(t – τ ) . Multiplying 1 and 2 together gives the fundamental solution of the two-dimensional heat equation: (x 1 , x 2 , y 1 , y 2 , t, τ)= 1 4π(t – τ ) exp – (x 1 – y 1 ) 2 +(x 2 – y 2 ) 2 4(t – τ ) . Example 2. The fundamental solution of the equation ∂w ∂t = n k=1 a k (t) ∂ 2 w ∂x 2 k , 0 < a k (t)<∞, is given by formula (14.6.1.5) with k (x k , y k , t, τ)= 1 2 √ πT k exp – (x k – y k ) 2 4T k , T k = t τ a k (η) dη. In the derivation of this formula it was taken into account that the corresponding one-dimensional equations could be reduced to the ordinary constant-coefficient heat equation by passing from x k , t to the new variables x k , T k . 14.6.2. Cauchy Problem for Hyperbolic Equations Consider a nonhomogeneous linear equation of the hyperbolic type with an arbitrary right- hand side, ∂ 2 w ∂t 2 + ϕ(x, t) ∂w ∂t – L x,t [w]=Φ(x, t), (14.6.2.1) where the second-order linear differential operator L x,t is defined by relation (14.2.1.2) with x R n . The solution of the Cauchy problem for equation (14.6.2.1) with general initial condi- tions, w = f 0 (x)att = 0, ∂ t w = f 1 (x)att = 0, can be represented as the sum w(x, t)= t 0 R n Φ(y, τ) (x, y, t, τ) dy dτ – R n f 0 (y) ∂ ∂τ (x, y, t, τ) τ=0 dy + R n f 1 (y)+f 0 (y)ϕ(y, 0) (x, y, t, 0) dy, dy = dy 1 dy n . Here, = (x, y, t, τ) is the fundamental solution of the Cauchy problem that satisfies, for t > τ ≥ 0, the homogeneous linear equation ∂ 2 ∂t 2 + ϕ(x, t) ∂ ∂t – L x,t [ ]=0 (14.6.2.2) with the semihomogeneous initial conditions of special form = 0 at t = τ , ∂ t = δ(x – y)att = τ. (14.6.2.3) The quantities τ and y appear in problem (14.6.2.2)–(14.6.2.3) as free parameters (y R n ). 618 LINEAR PARTIAL DIFFERENTIAL EQUATIONS Remark 1. If the coefficients of the differential operator L x,t in (14.6.2.2) are independent of time t,then the fundamental solution of the Cauchy problem depends on only three arguments, (x, y, t,τ )= (x, y, t –τ). Here, ∂ ∂τ (x, y, t, τ) τ=0 =– ∂ ∂t (x, y, t). Remark 2. If the differential operator L x,t has constant coefficients, then the fundamental solution of the Cauchy problem depends on only two arguments, (x, y, t,τ )= (x – y, t – τ ). Example 1. For the one-, two-, and three-dimensional wave equations, the fundamental solutions have the forms Equations Fundamental solutions ∂ 2 w ∂t 2 – ∂ 2 w ∂x 2 = 0 =⇒ (x, t, y, τ)= 1 2 ϑ(t – τ – |x – y|), ϑ(z)= 1 if z ≥ 0, 0 if z < 0; ∂ 2 w ∂t 2 – ∂ 2 w ∂x 2 1 – ∂ 2 w ∂x 2 2 = 0 =⇒ (x 1 , x 2 , t, y 1 , y 2 , τ)= ϑ(t – τ – ρ) 2π (t – τ ) 2 – ρ 2 ; ∂ 2 w ∂t 2 – ∂ 2 w ∂x 2 1 – ∂ 2 w ∂x 2 2 – ∂ 2 w ∂x 2 3 = 0 =⇒ (x 1 , x 2 , x 3 , t, y 1 , y 2 , y 3 , τ)= 1 2π δ (t – τ ) 2 – r 2 , where ϑ(z) is the Heaviside unit step function (ϑ = 0 for z < 0 and ϑ = 1 for z ≥ 0), ρ = (x 1 –y 1 ) 2 +(x 2 –y 2 ) 2 , r = (x 1 –y 1 ) 2 +(x 2 –y 2 ) 2 +(x 3 –y 3 ) 2 ,andδ(z) is the Dirac delta function. Example 2. The one-dimensional Klein-Gordon equation ∂ 2 w ∂t 2 = ∂ 2 w ∂x 2 – bw has the fundamental solutions (x, t, y, τ)= 1 2 ϑ t – τ – |x – y| J 0 k (t – τ) 2 –(x – y) 2 for b = k 2 > 0, (x, t, y, τ)= 1 2 ϑ t – τ – |x – y| I 0 k (t – τ ) 2 –(x – y) 2 for b =–k 2 < 0, where ϑ(z) is the Heaviside unit step function, J 0 (z) is the Bessel function, I 0 (z) is the modified Bessel function, and k > 0. Example 3. The two-dimensional Klein-Gordon equation ∂ 2 w ∂t 2 = ∂ 2 w ∂x 2 1 + ∂ 2 w ∂x 2 2 – bw has the fundamental solutions (x 1 , x 2 , t, y 1 , y 2 , τ)=ϑ(t – τ – ρ) cos k (t – τ) 2 – ρ 2 2π (t – τ ) 2 – ρ 2 for b = k 2 > 0, (x 1 , x 2 , t, y 1 , y 2 , τ)=ϑ(t – τ – ρ) cosh k (t – τ ) 2 – ρ 2 2π (t – τ ) 2 – ρ 2 for b =–k 2 < 0, where ϑ(z) is the Heaviside unit step function and ρ = (x 1 – y 1 ) 2 +(x 2 – y 2 ) 2 . 14.7. Boundary Value Problems for Parabolic Equations with One Space Variable. Green’s Function 14.7.1. Representation of Solutions via the Green’s Function 14.7.1-1. Statement of the problem (t ≥ 0, x 1 ≤ x ≤ x 2 ). In general, a nonhomogeneous linear differential equation of the parabolic type with variable coefficients in one dimension can be written as ∂w ∂t – L x,t [w]=Φ(x, t), (14.7.1.1) . f (0)andf (0) are the initial conditions. More details on the properties of the Laplace transform and the inverse Laplace transform can be found in Section 11.2. The Laplace transforms of some. axis and lies to the right of all singularities of f(p), which corresponds to c > σ 0 . In order to solve nonstationary boundary value problems, the following Laplace trans- form formulas for. conditions Application of the inverse Laplace transform PL.5.1.(2) { Problem for transform = ordinary differential equation for boundary conditionswwxp=(,), ~~ ~~ Finding the transform wwxt=(,) Obtaining
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