Math 412-501Theory of Partial Differential EquationsLecture 4: D’Alembert’s solution (continued).Wave equation∂2u∂t2= c2∂2u∂x2, −∞ < x < ∞, −∞ < t < ∞Change of independent variables:w = x + ct, z = x − ct.Jacobian: ∂w∂x∂w∂t∂z∂x∂z∂t!=1 c1 −cHow does the equation look in new coordinates?∂∂t=∂w∂t∂∂w+∂z∂t∂∂z= c∂∂w− c∂∂z∂∂x=∂w∂x∂∂w+∂z∂x∂∂z=∂∂w+∂∂z∂2u∂t2= c2∂∂w−∂∂z∂∂w−∂∂zu= c2∂2u∂w2− 2∂2u∂w ∂z+∂2u∂z2.∂2u∂x2=∂2u∂w2+ 2∂2u∂w ∂z+∂2u∂z2.∂2u∂t2− c2∂2u∂x2= −4c2∂2u∂w ∂z.Wave equation in new coordinates:∂2u∂w ∂z= 0.∂2u∂w ∂z= 0, −∞ < w, z < ∞General solution:u(w , z) = B(z) + C (w)where B, C : R → R are arbitrary (smooth)functions.General solution of the 1D wave equation:u(x, t) = B(x − ct) + C (x + ct)(d’Alembert’s solution)u(x, t) = B(x − ct)t1= 0, t2= 1, t3= 2u(x, t) = C (x + ct)t1= 0, t2= 1, t3= 2Initial value problem∂2u∂t2= c2∂2u∂x2, −∞ < x, t < ∞,u(x, 0) = f (x),∂u∂t(x, 0) = g (x), −∞ < x < ∞.General solution: u(x, t) = B(x − ct) + C (x + ct).Functions B and C are determined by the initialconditions:f (x) = B(x) + C (x), g (x) = −cB′(x) + cC′(x).B + C = f , c(−B + C )′= g.B + C = f , c(−B + C )′= g.B + C = f , −B + C = G , where G′= g/c(G is determined up to adding a constant).It follows that B =12(f − G ), C =12(f + G ).u(x, t) =12f (x − ct) + f (x + ct)+ G (x + ct) − G (x − ct)(d’Alembert’s formula)In this formula, G may be an arbitraryanti-derivative of g/c.The solution is unique, but functions B and C are not!f = χ[−h,h]g = 0f = 0g = χ[−h,h]G′= g/cF = −Gu(x, t) =12f (x − ct) + f (x + ct)+ G (x + ct) − G (x − ct).Since G′= g/c, we haveG (x + ct) − G (x − ct) =1cZx +ctx −ctg(ξ) dξ.u(x, t) =f (x − ct) + f (x + ct)2+12cZx +ctx −ctg(ξ) dξ(d’Alembert’s formula)Example∂2u∂t2= c2∂2u∂x2, −∞ < x, t < ∞,u(x, 0) = cos 2x,∂u∂t(x, 0) = sin x, −∞ < x < ∞.According to the (2nd) d’Alembert’s formula, theunique solution isu(x, t) =12f (x − ct) + f (x + ct)+ G (x + ct) − G (x − ct),where f (x) = cos 2x, x ∈ R, and G is an arbitraryfunction such that G′(x) =sin xcfor all x ∈ R.We can take G (x) = −cos xc. Thenu(x, t) =12cos 2(x − ct) + cos 2(x + ct)+12c− cos(x + ct) + cos(x − ct).After simplifying,u(x, t) = cos 2ct · cos 2x +1csin ct · sin x.Semi-infinite stringInitial-boundary value problem∂2u∂t2= c2∂2u∂x2, x ≥ 0;u(x, 0) = f (x),∂u∂t(x, 0) = g(x), x ≥ 0;u(0, t) = 0 (fixed end).General solution: u(x, t) = B(x − ct) + C (x + ct).Initial conditions imply:f (x) = B(x) + C (x), g (x) = −cB′(x) + cC′(x),x ≥ 0.B + C = f , c(−B + C )′= g.B + C = f , −B + C = G , where G′= g/c(G is determined up to adding a constant).It follows that B =12(f − G ), C =12(f + G ).However this yields B(x) and C (x) only for x ≥ 0.Boundary condition implies:B(−ct) + C (ct) = 0 for all t ∈ R.That is, B(−x) = −C (x) and C (−x) = −B(x).This yields B(x) and C (x) for x < 0.Another approachInitial value problem:∂2u∂t2= c2∂2u∂x2, −∞ < x, t < ∞,u(x, 0) = f (x),∂u∂t(x, 0) = g(x), −∞ < x < ∞.Lemma Suppose that the functions f and g areodd, that is, f (−x) = −f (x) and g (−x) = −g (x)for all x.Then the solution satisfies the fixed-end boundarycondition at the origin: u(0, t) = 0 for all t.Proof: By the (3rd) d’Alembert’s formula,u(x, t) =f (x − ct) + f (x + ct)2+12cZx +ctx −ctg(ξ) dξ.Henceu(0, t) =f (−ct) + f (ct)2+12cZct−ctg(ξ) dξ.Since f is odd, we have f (−ct) + f (ct) = 0.Since g is odd, we haveZ0−ctg(ξ) dξ = −Zct0g(ξ) dξ=⇒Zct−ctg(ξ) dξ = 0Initial-boundary value problem:∂2u∂t2= c2∂2u∂x2, x ≥ 0;u(x, 0) = f (x),∂u∂t(x, 0) = g(x), x ≥ 0;u(0, t) = 0 (fixed end).The problem can be solved as follows:• extend f and g to the whole line so that theyare odd;• solve the initial value problem in the whole plane;• restrict the solution to the half-plane x ≥ 0.Initial-boundary value problem:∂2u∂t2= c2∂2u∂x2, x ≥ 0;u(x, 0) = f (x),∂u∂t(x, 0) = g(x), x ≥ 0;∂u∂x(0, t) = 0 (free end).The problem can be solved as follows:• extend f and g to the whole line so that they areeven: f (−x) = f (x) and g (−x) = g (x) for all x;• solve the initial value problem in the whole plane;• restrict the solution to the half-plane x ≥ 0 (theboundary condition should h