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cookiemonster
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I think my lack of background in proofs is showing.
From "Partial Differential Equations of Mathematical Physics and Integral Equations" by Ronald B. Guenther and John W. Lee, pp. 107 problem 4-3.10.
Prove:
For the inhomogeneous, initial, boundary value problem
[tex]\left\{\begin{array}{lll}
u_{tt} - c^2u_{xx} = F(x,t), & 0 < x < L, & t > 0, \\
u(x,0) = 0, & u_t(x,0) = 0, & 0 \leq x \leq L, \\
u(0,t) = 0, & u(L,t) = 0, & t \geq 0
\end{array} \right.[/tex]
Let F(x,t) have continuous third-order partial derivatives for [itex]0 \leq x \leq L[/itex] and [itex]t \geq 0[/itex]. Assume that [itex]F(0,t) = F(L,t) = F_{xx}(0,t) = F_{xx}(L,t) = 0 \qquad \forall t \geq 0[/itex]. Then the problem above has a unique solution given by
[tex]u(x,t) = \sum_{n=1}^\infty u_n(t) \sin{(\lambda_n x)}[/tex]
where
[tex]u_n(t) = \frac{1}{c\lambda_n}\int_0^t F_n(\tau) \sin{[c\lambda_n (t - \tau)]}\,d\tau[/tex]
and
[tex]\lambda_n = \frac{n\pi}{L}[/tex]
using the following steps. Fix T > 0 and restrict x and t to [itex]0 \leq x \leq L[/itex] and [itex] 0 \leq t \leq T[/itex].
(a) Write out the expressions for [itex]u_x,u_{xx},u_t,u_{tt}[/itex] assuming that term-by-term differentiation is permissible.
This I can do.
[tex]u_x = \sum_{n=1}^\infty u_n(t) \lambda_n \cos{(\lambda_n x)}[/tex]
[tex]u_{xx} = -\sum_{n=1}^\infty u_n(t) \lambda_n^2 \sin{(\lambda_n x)}[/tex]
[tex]u_t = \sum_{n=1}^\infty \sin{(\lambda_n x)}\int_0^t F(\tau) \cos{[c\lambda_n (t-\tau)]}\,d\tau[/tex]
[tex]u_{tt} = \sum_{n=1}^\infty \sin{(\lambda_n x)}[F(t) - c^2\lambda_n^2 u_n(t)][/tex]
(b) Use the definition of [itex]u_n[/itex] to confirm that
[tex]|\lambda_n^2 u_n(t)| \leq T \lambda_n \frac{||F_n||}{c}[/tex]
[tex]|u_n'(t)| \leq T||F_n||[/tex]
[tex]|u_n''(t)| \leq |F_n(t)| + cT\lambda_n ||F_n||[/tex]
where
[tex]||F_n|| = \max_{0 \leq t \leq T} |F_n(t)|[/tex]
Can do that, too.
(c) Assume that [itex]F(0,t) = F(L,t) = F_{xx}(0,t) = F_{xx}(L,t) = 0 \qquad \forall t \geq 0[/itex] and deduce that
[tex]|F_n(t)| \leq \frac{2}{\lambda_n^3 L}\int_0^L |F_{xxx}(x,t)\,dx[/tex]
This I haven't pulled off yet. I don't quite see how the partial derivatives of u and the derivatives of [itex]u_n[/itex] relate to [itex]F_{xxx}[/itex].
The Fourier sine series expansion of F(x,t) relates F(x,t) with [itex]F_n(t)[/itex], where [itex]F_n(t)[/itex] is the sine coefficient of the Fourier series. This puts the two on the same page, but I couldn't take it anywhere. It seems to me like bringing u(x,t) and un(t) into the proof is going out of the scope of F and Fn.
Honestly, I really can't find much use to the three inequalities in part (b).
Anyway, I'd appreciate any help. This is homework due tomorrow, but frankly I don't care if it gets done by then because I have nothing on the line in this class (I have a running bet for $50 that I'll just have to retake it in 2 or 3 years). I'm more concerned with figuring out how the proof works (for 2-3 years down the road).
Sorry for dragging anybody through crap they never wanted to remember.
cookiemonster
From "Partial Differential Equations of Mathematical Physics and Integral Equations" by Ronald B. Guenther and John W. Lee, pp. 107 problem 4-3.10.
Prove:
For the inhomogeneous, initial, boundary value problem
[tex]\left\{\begin{array}{lll}
u_{tt} - c^2u_{xx} = F(x,t), & 0 < x < L, & t > 0, \\
u(x,0) = 0, & u_t(x,0) = 0, & 0 \leq x \leq L, \\
u(0,t) = 0, & u(L,t) = 0, & t \geq 0
\end{array} \right.[/tex]
Let F(x,t) have continuous third-order partial derivatives for [itex]0 \leq x \leq L[/itex] and [itex]t \geq 0[/itex]. Assume that [itex]F(0,t) = F(L,t) = F_{xx}(0,t) = F_{xx}(L,t) = 0 \qquad \forall t \geq 0[/itex]. Then the problem above has a unique solution given by
[tex]u(x,t) = \sum_{n=1}^\infty u_n(t) \sin{(\lambda_n x)}[/tex]
where
[tex]u_n(t) = \frac{1}{c\lambda_n}\int_0^t F_n(\tau) \sin{[c\lambda_n (t - \tau)]}\,d\tau[/tex]
and
[tex]\lambda_n = \frac{n\pi}{L}[/tex]
using the following steps. Fix T > 0 and restrict x and t to [itex]0 \leq x \leq L[/itex] and [itex] 0 \leq t \leq T[/itex].
(a) Write out the expressions for [itex]u_x,u_{xx},u_t,u_{tt}[/itex] assuming that term-by-term differentiation is permissible.
This I can do.
[tex]u_x = \sum_{n=1}^\infty u_n(t) \lambda_n \cos{(\lambda_n x)}[/tex]
[tex]u_{xx} = -\sum_{n=1}^\infty u_n(t) \lambda_n^2 \sin{(\lambda_n x)}[/tex]
[tex]u_t = \sum_{n=1}^\infty \sin{(\lambda_n x)}\int_0^t F(\tau) \cos{[c\lambda_n (t-\tau)]}\,d\tau[/tex]
[tex]u_{tt} = \sum_{n=1}^\infty \sin{(\lambda_n x)}[F(t) - c^2\lambda_n^2 u_n(t)][/tex]
(b) Use the definition of [itex]u_n[/itex] to confirm that
[tex]|\lambda_n^2 u_n(t)| \leq T \lambda_n \frac{||F_n||}{c}[/tex]
[tex]|u_n'(t)| \leq T||F_n||[/tex]
[tex]|u_n''(t)| \leq |F_n(t)| + cT\lambda_n ||F_n||[/tex]
where
[tex]||F_n|| = \max_{0 \leq t \leq T} |F_n(t)|[/tex]
Can do that, too.
(c) Assume that [itex]F(0,t) = F(L,t) = F_{xx}(0,t) = F_{xx}(L,t) = 0 \qquad \forall t \geq 0[/itex] and deduce that
[tex]|F_n(t)| \leq \frac{2}{\lambda_n^3 L}\int_0^L |F_{xxx}(x,t)\,dx[/tex]
This I haven't pulled off yet. I don't quite see how the partial derivatives of u and the derivatives of [itex]u_n[/itex] relate to [itex]F_{xxx}[/itex].
The Fourier sine series expansion of F(x,t) relates F(x,t) with [itex]F_n(t)[/itex], where [itex]F_n(t)[/itex] is the sine coefficient of the Fourier series. This puts the two on the same page, but I couldn't take it anywhere. It seems to me like bringing u(x,t) and un(t) into the proof is going out of the scope of F and Fn.
Honestly, I really can't find much use to the three inequalities in part (b).
Anyway, I'd appreciate any help. This is homework due tomorrow, but frankly I don't care if it gets done by then because I have nothing on the line in this class (I have a running bet for $50 that I'll just have to retake it in 2 or 3 years). I'm more concerned with figuring out how the proof works (for 2-3 years down the road).
Sorry for dragging anybody through crap they never wanted to remember.
cookiemonster