1 Wave Equation on Finite Interval
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1 1 Wave Equation on Finite Interval 1.1 Wave Equation Dirichlet Boundary Conditions u tt (x, t) = c u xx (x, t), < x < l, t > (1.1) u(, t) =, u(l, t) = u(x, ) = f(x) u t (x, ) = g(x) First we present the solution to this problem and then provide a detailed derivation. u(x, t) = (a n cos (µ n ct) + b n sin (µ n ct)) sin (µ n x) µ n = nπ l, λ n = µ n, ϕ n (x) = l sin(µ nx) f(x) ϕ n dx, b n = 1 cµ n g(x) ϕ n dx. Look for simple solutions in the form u(x, t) = ϕ(x)ψ(t). Substituting into (1.1) and dividing both sides by ϕ(x)ψ(t) gives c ψ(t) = ϕ (x) ϕ(x) Since the left side is independent of x and the right side is independent of t, it follows that the expression must be a constant: c ψ(t) = ϕ (x) ϕ(x) = λ. (Here ψ means the derivative of ψ with respect to t and ϕ means means the derivative of ϕ with respect to x.) We seek to find all possible constants λ and the corresponding nonzero functions ϕ and T. We obtain Furthermore, the boundary conditions give ϕ λϕ =, ψ c λψ =. ϕ()ψ(t) =, ϕ(l)ψ(t) = for all t. 1
2 Since ψ(t) is not identically zero we obtain the desired eigenvalue problem ϕ (x) λϕ(x) =, ϕ() =, ϕ(l) =. (1.) We have solved this problem many times and we have λ = µ so that ϕ(x) = a cos(µx) + b sin(µx). Applying the boundary conditions we have = ϕ() = a a = = ϕ(l) = b sin(µl). From this we conclude sin(µl) = which implies µ n = nπ l and therefore ( nπ λ n = µ n = l ), ϕn (x) = l sin(µ nx), n = 1,,.. (1.3) The solution of ψ c λ n ψ = is then ψ(t) = a cos (µ n ct) + b sin (µ n ct) (1.4) where a and b are arbitrary constants. Next we look for u as an infinite sum u(x, t) = (a n cos (µ n ct) + b n sin (µ n ct)) sin (µ n x) (1.5) The only problem remaining is to somehow pick the constants a n and b n so that the initial conditions u(x, ) = f(x) and u t (x, ) = g(x) are satisfied. Setting t = in (1.5), we seek to obtain {a n } satisfying f(x) = u(x, ) = a n ϕ n (x). This gives a Sine expansion for the function f(x) on the interval (, l). f(x)ϕ n (x) dx. (1.6) Next we differentiate (formally) (1.5) with respect to t to obtain u t (x, t) = (cµ n ) ( a n sin (µ n ct) + b n cos (µ n ct)) ϕ n (x).
3 Setting t = in this expression, we seek to obtain {b n } satisfying g(x) = u t (x, ) = b n (cµ n ) ϕ n (x). This is almost a Sine expansion of the function g(x) on the interval (, l). Namely we obtain ( ) 1 l b n = g(x)ϕ n (x) dx. cµ n l Our after simplifying b n = 1 µ n c g(x)ϕ n (x) dx. (1.7) 1. Wave Equation Neumann Boundary Conditions u tt (x, t) = c u xx (x, t), < x < l, t > (1.8) u x (, t) =, u x (l, t) = u(x, ) = f(x), u t (x, ) = g(x) First we present the solution to this problem and then provide a detailed derivation. u(x, t) = (a + b t)ϕ + (a n cos (cµ n t) + b n sin (cµ n t)) ϕ n (x) λ =, ϕ = 1 l, µ n = nπ l, λ n = µ n, ϕ n (x) = l cos(µ n x) b n = 1 µ n c f(x)ϕ n (x) dx, n =, 1,, a = g(x)ϕ n (x) dx, n =, 1, b = f(x)ϕ (x) dx g(x)ϕ (x) dx. Look for simple solutions in the form u(x, t) = ϕ(x)ψ(t). Substituting into (1.8) and dividing both sides by ϕ(x)ψ(t) gives c ψ(t) = ϕ (x) ϕ(x) Since the left side is independent of x and the right side is independent of t, it follows that the expression must be a constant: c ψ(t) = ϕ (x) ϕ(x) = λ. 3
4 (Here ψ means the derivative of ψ with respect to t and ϕ means means the derivative of ϕ with respect to x.) We seek to find all possible constants λ and the corresponding nonzero functions ϕ and T. We obtain Furthermore, the boundary conditions give ϕ λϕ =, ψ c λψ =. ϕ()ψ(t) =, ϕ(l)ψ(t) = for all t. Since ψ(t) is not identically zero we obtain the desired eigenvalue problem ϕ (x) λϕ(x) =, ϕ() =, ϕ(l) =. (1.9) We have solved this problem a few times. What we found was that λ = is an eigevalue with normalized eigenfunction ϕ = 1/ l and for λ = µ we have Applying the boundary conditions we have ϕ(x) = a cos(µx) + b sin(µx). = ϕ () = b a = = ϕ (l) = a sin(µl). From this we conclude sin(µl) = which implies and therefore ( nπ λ n = µ n = l µ n = nπ l ), ϕn (x) = l cos(µ nx), n = 1,,. (1.1) The solution of ψ c λ n ψ = is then ψ(t) = a cos (µ n t) + b sin (µ n t) (1.11) where a and b are arbitrary constants. For λ = we have ψ = which implies ψ(t) = a + b t. Next we look for u as an infinite sum u(x, t) = (a + b t)ϕ + (a n cos (µ n t) + b n sin (µ n t)) ϕ n (x) (1.1) The problem remaining is to find the constants a n and b n so that the initial conditions u(x, ) = f(x) and u t (x, ) = g(x) are satisfied. 4
5 Setting t = in (1.15), we seek to obtain {a n } satisfying f(x) = u(x, ) = a n ϕ n (x). This is basically a Cosine expansion of the function f(x) on the interval (, l). n= f(x)ϕ n (x) dx. (1.13) Next we differentiate (formally) (1.15) with respect to t to obtain u t (x, t) = (µ n c) ( a n sin (µ n ct) + b n cos (µ n ct)) ϕ n (x). Setting t = in this expression, we seek to obtain {b n } satisfying g(x) = u t (x, ) = We have b n (µ n c) ϕ n (x). n= b n = (µ n c) 1 g(x)ϕ n (x) dx for n 1 and b = g(x)ϕ (x) dx. Example 1.1. Consider the following example u tt (x, t) = c u xx (x, t), < x < l, t > (1.14) u x (, t) =, u x (l, t) = u(x, ) = f(x) = x u t (x, ) = g(x) = (π/ x) Here we have Dirichlet BCs and l = π so we have λ =, ϕ (x) = 1/ π, µ n = n, λ n = n, ϕ n (x) = /π cos(nx). The solution is u(x, t) = (a + b t)ϕ + (a n cos (µ n t) + b n sin (µ n t)) ϕ n (x) (1.15) where a = b = f(x)ϕ (x) dx = 1 π x dx = π5/ 3 g(x)ϕ n (x) dx = 1 (π/ x) dx = π 5
6 b n = f(x)ϕ n (x) dx = π g(x)ϕ n (x) dx = π u(x, t) = π 3 + ( 4( 1) n cos(nt) n x cos(nx) dx = π( 1) n n (π/ x) cos(nx) dx = (1 ( 1)n ) πn + (1 ) ( 1)n ) sin(nt) cos(nx). πn 6
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