Math 5440 Problem Set 6 Solutions

Transcription

1 Math 544 Math 544 Problem Set 6 Solutions Aaron Fogelson Fall, 5 : (Logan,.6 # ) Solve the following problem using Laplace transforms. u tt c u xx g, x >, t >, u(, t), t >, u(x, ), u t (x, ), x >. The solution shows what happens to a falling cable lying on a table that is suddenly removed. Draw some time snapshots of the solution. Compute the Laplace transform in t of both sides of the PDE to obtain s U(x, s) su(x, ) u t (x, ) c U xx (x, s) g s. Using the initial conditions, this reduces to the ODE, U xx (x, s) s c U(x, s) + g c s. A particular solution to this is the function U p (x, s) g s 3. The homogeneous ODE U xx (x, s) s c U(x, s), has general solution U(x, s) a(s) exp(sx/c) + b(s) exp( sx/c). For the solution to be bounded for x large, we set a(s) for all s. Hence the general bounded solution of the nonhomogeneous ODE is U(x, s) b(s)e sx c g s 3. To determine b(s) we compute the Laplace Transform of the boundary condition and obtain U(, s). To satisfy this, we must set b(s) g/s 3, so U(x, s) g s 3 e sx/c. This is the product of two functions which is the Laplace transform of t /, and s 3 exp( sx/c) which is the Laplace transform of δ(t) δ(t x/c). By the convolution theorem, Z t (t τ) u(x, t) g [δ(τ) δ(τ x/c)] dτ, so 8 n < g t u(x, o t x c x < ct t) o : g x > ct. n t This can also be written t u(x, t) g t x x H(t c c ).

2 R t : (Logan,.6 # 3) Show that L f (τ)dτ F(s)/s. L( f ) F(s) and L() /s, so by the Convolution theorem L F(s) Z t Z t f (τ)dτ f (τ)dτ. s R t So F(s)/s L f (τ)dτ.

3 3: (Logan,.6 # 4) Show that L (H(t a) f (t a)) e as F(s) where H is the unit step function defined by H(x) for x <, and H(x), for x. L (H(t a) f (t a)) tt +a a H(t a) f (t a)e st dt H(t ) f (t )e s(t +a) dt e sa H(t ) f (t )e dt st a e sa H(t ) f (t )e dt st e sa f (t )e dt st e sa F(s). In the above, we used that H(t ) for t <. 3

4 4: (Logan,.6 # 6) Derive the solution u(x, t) H(t x/c)g(t x/c) to the problem u tt c u xx, x, t >, u(x, ) u t (x, ), x >, u(, t) g(t), t >. Compute the Laplace transform of the PDE and use the given initial conditions to obtain the ODE U xx s c U. The general solution to this is U(x, s) a(s) exp( sx/c) + b(s) exp(sx/c), and for this to be bounded for x >, we must set b(s). So U(x, t) a(s) exp( sx/c). To determine a(s), compute the Laplace transform of the boundary condtition to obtain U(, s) G(s). For our solution to match this condition, we set a(s) G(s), and so U(x, s) G(s) exp( sx/c). Since L(δ(t x/c)) exp( sx/c), the Convolution theorem tells us that That is, u(x, t) u(x, t) Z t g(t τ)δ(τ x/c)dτ. ( g(t x/c) if x/c < t if x/c > t. This can also be written u(x, t) g(t x/c)h(t x/c). 4

5 5: (Logan,.7 # 3) Verify the following properties of the Fourier transform: a) (Fu)(ξ) (F u)( ξ). b) F (e iax u)(ξ) û(ξ + a). c) F (u(x + a))(ξ) e iaξ û(ξ). a) (Fu)(ξ) u(y)eiξy dy u(x)eiξx dx. On the other hand, (F u)( ξ) (Fu)(ξ). Multiplying by verifies the assertion. R b) F (e iax R u) eiax u(x)e iξx dx u(x)ei(a+ξ)x dx û(a + ξ). c) F (u(x + a))(ξ) e iξa û(ξ). u(x + a)eiξx dx x x+a u(y)e i( ξ)y dy u(x )e iξ(x a) dx e iξa u(x )e iξx dx 5

6 6: (Logan,.7 # 6) Solve the following initial value problem for the inhomogeneous heat equation: u t βu xx + F(x, t), < x <, t > ; u(x, ), < x <. Compute the Fourier transform of the PDE to obtain û t βξ û + ˆF(ξ, t). Using an integrating factor, this can be written as Integrating with respect to t, we find that ûe βξ t ˆF(ξ, t)e βξt. t û(ξ, t)e βξ t û(ξ, ) Z t ˆF(ξ, s)e βξs ds, so that û(ξ, t) e βξtû(ξ, ) + Z t ˆF(ξ, s)e βξ (t s) ds. The first term on the right side is zero because of the initial condition and we see that u(x, t) F Z t ˆF(ξ, s)e βξ (t s) ds Z t ˆF is the Fourier transform of F, and recalling that F a /(4β(t s)), we see that! F p e x /4β(t s) 4πβ(t s) F ˆF(ξ, s)e βξ (t s) ds. e ax q πa e ξ /4a, and letting e βξ (t s). Using the convolution theorem, ˆF(ξ, s)e βξ (t s) is the Fourier transform of So F(x y, s) p e y /4β(t s) dy. 4πβ(t s) Z t u(x, t) Z t Z t F(x y, s) p e y /4β(t s) dyds 4πβ(t s) F(x y, t s) p e y /4βs dyds 4πβs F(x y, t s)g(y, s)dyds. 6

7 7: (Logan,.7 # ) Solve the Cuachy problem for the following convection-diffusion equation using Fourier transforms: u t Du xx cu x, < x <, t >, u(x, ) φ(x), < x <. Compute the Fourier transform of the PDE and initial conditions to obtain û t Dξ û + icξû, û(ξ, ) ˆφ(ξ). Integrating this ODE in t and using the initial condition we find that So, using the convolution theorem, û(ξ, t) ˆφ(ξ)e Dξ t+icξt. u(x, t) φ(x y)w(y, t)dy, where F (w(y, t))(ξ) exp( Dξ t icξt), that is w is the function whose Fourier Transform is the expression on the right side of this equation. By the Fourier Integral Theorem (Inverse Fourier Transform), w(y, t) e Dξ t+icξt e iξx dξ e Dξt e iξ(x ct) dξ F e Dξ t (x ct, t). But we know that so that Therefore, F p e x /4Dt e Dξt, 4πDt F e Dξ t (x ct, t) u(x, t) p 4πDt e (x ct) /4Dt. φ(x y) p e (y ct) /4Dt dy. 4πDt 7

8 8: (Logan,.7 #4) Use Fourier transforms to derive D Alembert s solution to the Cauchy problem for the wave equation u tt c u xx, < x <, t >, u(x, ) f (x), < x <, u t (x, ), < x <. Compute the Fourier transform of the PDE and initial conditions to obtain The ODE has general solution The t derivative of this is û tt c ξ û û(ξ, ) ˆf (ξ) û t (ξ, ). û(ξ, t) a(ξ)e icξt + b(ξ)e icξt. û t (ξ, t) icξa(ξ)e icξt icξb(ξ)e icξt. Using these two formulas at t along with the transformed initial data, we see that a(ξ) and b(ξ) must satisfy ˆf (ξ) a(ξ) + b(ξ) icξa(ξ) icξb(ξ). The solution of these two equations is a(ξ) b(ξ) ˆf (ξ)/, so that Therefore, u(x, t) û(ξ, t) ˆf (ξ, t)e icξt + ˆf (ξ, t)e icξt. û(ξ, t)e iξx dξ ˆf (ξ, t)e icξt + ˆf (ξ, t)e icξt e iξx dξ ˆf (ξ, t)e iξ(x ct) dξ + ˆf (ξ, t)e iξ(x+ct) dξ ( f (x ct) + f (x + ct)). 8