5. Let L>1/2. Consider the diffusion equation Uų = Kurr, -L 0, u(-L,t) = u(L, t) = 0, u(x, 0) = 4 3 By using the maximum principle to show 1 (a) Ju(x, t)| < for -L

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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I need help on 5

1. Use the coordinate method to solve \( 2u_x - u_y + (x + 2y)u = 0 \) with \( u(x, 0) = e^{-2x^2/5} \).

2. Show that the steady state of the diffusion equation

   \[
   \begin{cases} 
   u_t = Ku_{xx}, & 0 < x < L, \, t > 0, \\
   u_x(0, t) = u_x(L, t) = 0, \\
   u(x, 0) = \phi(x)
   \end{cases}
   \]

   is given by

   \[
   u_s(x) = \frac{1}{L} \int_0^L \phi(x) \, dx.
   \]

3. Classify equation \( u_{xx} + u_{xt} - 20u_{tt} = 0 \), and solve the equation with initial conditions: \( u(x, 0) = \phi(x) \) and \( u_t(x, 0) = \psi(x) \).

4. Find the general solution of \( w_{tt} = w_{xx} + \sin(x + t) \) by using the coordinate method.

5. Let \( L > 1/2 \). Consider the diffusion equation

   \[
   \begin{cases} 
   u_t = Ku_{xx}, & -L < x < L, \, t > 0, \\
   u(-L, t) = u(L, t) = 0, \\
   u(x, 0) = \frac{x}{4} - \frac{x^3}{3}
   \end{cases}
   \]

   By using the maximum principle, show:
   
   (a) \( |u(x,t)| \le \frac{1}{12} \) for \( -L \le x \le L \) and \( t \ge 0 \);
   
   (b) \( u(-x,t) = -u(x,t) \) for \( -L \le x \le L \) and \( t \ge 0 \).
Transcribed Image Text:1. Use the coordinate method to solve \( 2u_x - u_y + (x + 2y)u = 0 \) with \( u(x, 0) = e^{-2x^2/5} \). 2. Show that the steady state of the diffusion equation \[ \begin{cases} u_t = Ku_{xx}, & 0 < x < L, \, t > 0, \\ u_x(0, t) = u_x(L, t) = 0, \\ u(x, 0) = \phi(x) \end{cases} \] is given by \[ u_s(x) = \frac{1}{L} \int_0^L \phi(x) \, dx. \] 3. Classify equation \( u_{xx} + u_{xt} - 20u_{tt} = 0 \), and solve the equation with initial conditions: \( u(x, 0) = \phi(x) \) and \( u_t(x, 0) = \psi(x) \). 4. Find the general solution of \( w_{tt} = w_{xx} + \sin(x + t) \) by using the coordinate method. 5. Let \( L > 1/2 \). Consider the diffusion equation \[ \begin{cases} u_t = Ku_{xx}, & -L < x < L, \, t > 0, \\ u(-L, t) = u(L, t) = 0, \\ u(x, 0) = \frac{x}{4} - \frac{x^3}{3} \end{cases} \] By using the maximum principle, show: (a) \( |u(x,t)| \le \frac{1}{12} \) for \( -L \le x \le L \) and \( t \ge 0 \); (b) \( u(-x,t) = -u(x,t) \) for \( -L \le x \le L \) and \( t \ge 0 \).
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