Cauchy Problem and Wave Equations PDF

# The Cauchy Problem and Wave Equations ## Homogeneous Wave Equations - Vibrations of an Infinite String: - Let $c^2u_{xx} = 0, x \in R, t > 0$ - $u(x,0) = f(x), x \in R$ - $u_t(x,0) = g(x), x \in R$ $Solution$ - $ u(x,t) = \frac{1}{2}[f(x+ct)+f(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x...

# The Cauchy Problem and Wave Equations ## Homogeneous Wave Equations - Vibrations of an Infinite String: - Let $c^2u_{xx} = 0, x \in R, t > 0$ - $u(x,0) = f(x), x \in R$ - $u_t(x,0) = g(x), x \in R$ $Solution$ - $ u(x,t) = \frac{1}{2}[f(x+ct)+f(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(z)dz$ - $ = \frac{1}{2}[f(x+ct)+f(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(z)dz$ ### Find the solution of the initial value problem: - $ u_{tt} - c^2 u_{xx} = 0, x \in R, t > 0$ - $ u(x, 0) = 0, u_t(x,0) = 1$ $Sol$ - Using d' Alembert's Solution of the given Cauchy problem. - $u_{tt} - c^2u_{xx} = 0, x \in R, t > 0$ - $u(x,0) = f(x), x \in R$ - $u_t(x,0) = g(x), x \in R$ $u(x,t) = \frac{1}{2}[f(x+ct)+f(x-ct)] + \frac{1}{2c}\int_{x-ct}^{x+ct} g(z)dz$ Where f(x)=0 & g(x)=1 - $ u(x,t) = f(0) + \frac{ct - x}{c} \int_{x-ct}^{x + ct} 1dz$ - $ = \frac{1}{c} \int_{x-ct}^{x+ct} (ct-x)dz$ - $ = \frac{1}{c} \int_{x-ct}^{x+ct} (x+ct) - (x-ct)dz$ - $ = \frac{1}{c} \int_{x-ct}^{x+ct} 2ct dt = t$ ## Initial Boundary Value Problems ### Find the solution of the initial value problem: - $u_{tt} - c^2u_{xx} = 0, u(x,0) = sin(x), u_t(x,0) = 2x^2$ - $f(x) = sin(x)$ - $g(x) = 2x^2$ ### The Solution of given Initial-Value Problem using d' Alembert's Solution - $u(x,t) = \frac{1}{2}[f(x+ct)+f(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct}g(z)dz$ - $u(x,t) = \frac{1}{2}[sin(x+ct)+ sin(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} 2z^2 dz$ - $u(x,t) = \frac{1}{2}[sin(x+ct) +sin(x-ct)] + \frac{1}{3c}[(x+ct)^3 - (x-ct)^3]$ - $u(x,t) = \frac{1}{2} [sin(x+ct) +sin(x-ct)] + \frac{1}{3c} [6x^2ct + 6ct^3]$ - $u(x,t) = sin(x)cos(ct) + x^2t + \frac{1}{3}ct^3$ ### Find the solution of the initial value problem: - $u_{tt} - c^2u_{xx} = 0, u(x,0) = log(1+x^2), u_t(x,0) = 2x$ - $f(x) = log(1+x^2)$ - $g(x) = 2x$ ### The Solution of the given problem: - $u(x,t) = \frac{1}{2}[f(x+ct) + f(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(z)dz$ In this case, we have: - $f(x) = log(1+x^2), g(x)=2x$ $u(x,t) = \frac{1}{2}[log(1+x^2+2xct +c^2t^2)+ log(1+ x^2 -2xct +c^2t^2)] + \frac{1}{2c} \int_{x-ct}^{x+ct}2zdz $ - $u(x,t) = \frac{1}{2}[log(1+x^2+2xct+c^2t^2) +log(1+x^2-2xct+c^2t^2)]+ \frac{1}{c} \int_{x-ct}^{x +ct}(z)dz$ <start_of_image> Problems with Initial Value Problems - How would you solve for the following Cauchy problem? $u_{tt} - c^2u_{xx} = 0, x \in R, t>0, f(x)$ and $u_t(x,0) = g(x)$ for the given conditions? - *(a) c = 3, f(x) = cos(x), g(x) = sin(2x)$ - *(b) c = 7, f(x) = cos(2x), g(x) = x$ - *(c) c = 4, f(x) = cos(x), g(x) = x.e^{-x}$ **Solve each of the following problems using the d'Alembert's Solution.** - $u_{xx} +2u_{xy} - 3u_{yy} = 0$ - $u(x,0) = sin(x), u_y(x,0)=x$ - $u_{tt} - c^2u_{xx}=0, u(x,0) = f(x), u_t(x,0) = g(x), u(0,t) = 0$ - What is the solution of the Initial-Boundary Value Problem obtained using the d' Alembert's Solution? - $u_{tt} - c^2u_{xx} = 0$ - $ u(x,0) = f(x), u_t(x,0) = g(x), u(0,t) = 0$ ### Determine the solution of each of the following initial value problem- - $u_{tt} - 4u_{xx} = 0, u(x,0) = 2x^4, u_t(x,0) = 0, u(0,t) = 0$ **Solution:** $u(x,t) = \frac{1}{2}[f(x+ct) + f(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(z)dz$, $for x < ct$ Where $c = 2, f(x) = 2x^4, g(x) = 0$ ### Determine the solution of the Initial-boundary value problem - $u_{tt} - 9u_{xx} = 0, u(x,0) = 0 u_t(x,0) = x^3, u_x(0,t) = 0 $ **Solution:** - $u(x,t) = \frac{1}{2}\Big[f(x+3t) + f(x -3t)\Big] + \frac{1}{6} \int_{x-3t}^{x+3t} g(z) dz$, $for x \ge 3t$ Where $c = 3, f(x) = 0, g(x) = x^3$ ### Determine the solution of the Initial-boundary value problem: - $u_{tt} = 4 u_{xx}$ - $u(x,0) = sin(x), u_t(x,0) = 0, u(0,t) = 0$ **Solution:** - $u(x,t) = \frac{1}{2} \Big[f(x+ct) + f(x-ct)\Big] + \frac{1}{2c}\int_{x-ct}^{x+ct} g(z)dz$, $for x \ge ct$ Where c = 2, f(x) = sin(x), g(x)=0 ### Determine the solution of the Initial-boundary value problem: - $ u_{tt} = c^2u_{xx} $ - $u(x,0) = f(x), u_t(x,0) = g(x)$ - $ u(0,t) = 0, t>0$ **Solution:** - $u(x,t) = \frac{1}{2} \Big[f(x+ct) + f(x-ct) \Big] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(z)dz$ - $u(x,t) = \frac{1}{2} \Big[f(x + ct) - f(ct - x) \Big] + \frac{1}{2c} \int_{ct - x}^{x+ct} g(z) dz$ - $u(x,t) = \frac{1}{2} \Big[f(x +ct) - f(ct-x) \Big] + \frac{1}{2c} \int_{ct - x}^{x+ct} g(z) dz$, $for x \le ct$ ### Determine the solution of the Initial-boundary value problem: - $u_{tt} = c^2u_{xx} $ - $u(x,0) = f(x), u_t(x,0) = g(x), u(0,t) = p(t) $ **Solution:** - $u(x,t) = \frac{1}{2}[f(x+ct) + f(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(z) dz + \frac{1}{2c}\int_{x-ct}^{x+ct} p(t-(z/c))dz $, $for x \le ct$ ### Determine the solution to the wave equation governed by the following conditions: - $u_{tt} = c^2u_{xx}, x > 0, t > 0$ - $ u(x,0) = 0, u_t(x,0) = 0, t > 0$ - $u(0,t) = u(t), x>0$ **Solution:** - $u(x,t) = u(t - x/c)H(t-x/c), x>0$ **Where H is the Heaviside unit step function.** ### Explain the Initial Boundary Value Problem with the given conditions - $u_{tt} = c^2u_{xx}, 0 < x \le L$ - $u(x,0) = f(x), 0 \le x \le L$ - $u_t(x,0) = g(x), 0 \le x \le L$ - $u(0,t) = 0, t \ge 0$ **Solution** - The solution of the wave equation are $u(x,t) = \frac{1}{2}[f(x+ct) + f( x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(z) dz$ - we have $f(x) = u(x,0) = \Phi(x) + \Psi(x), 0 \le x \le L$ - $g(x) = u_t(x,0) = c \Phi'(x) - c \Psi'(x), 0 \le x \le L$ - $u(0,t) = 0$ - $u(x,t) = \frac{1}{2}[f(x+ct) + f(x - ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(z) dz$ ### Find the solution of the Initial boundary value problem: - $u_{tt} = u_{xx}$ - $u(x,0) = sin( \frac{\pi x}{L}), u_t(x,0) = 0$, $0 \le x \le L$ - $u(0,t) = 0, u(L,t) = 0, t \ge 0$ **Solution** - $u(x,t) = \frac{1}{2}\Big[f(x+ct) + f(x - ct) \Big] + \frac{1}{2c} \int_{x-ct}^{x +ct} g(z)dz$ Where $c=1, f(x) = sin(\frac{\pi x}{L}), g(x) = 0$ ### Find the solution of the initial boundary value problem - $u_{tt} = c^2 u_{xx}, 0 < x < \infty , t > 0$. - $u(x,0) = f(x), 0 < x < \infty $ - $u_t(x,0) = 0, 0 < x < \infty$ - $u(0,t) = p(t), t > 0$ **Solution** - $u(x,t) = \frac{1}{2} [f(x+ct) + f(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(z)dz$ - $u(x,t) = \frac{1}{2} [f(x+ct) + f(ct-x)] + \frac{1}{2c} \int_{ct-x}^{x+ct} g(z)dz + \frac{1}{2c} \int_{ct-x}^{x+ct} p(t-(z/c)) dz$ - $u(x,t) = \frac{1}{2}[f(x+ct) + f(ct - x)] + \frac{1}{2c} \int_{ct-x}^{x+ct} g(z) dz + \frac{1}{2c} \int_{ct-x}^{x+ct}p(t -(z/c))dz$ - $u(x,t) = \frac{1}{2}[f(x+ct) + f(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(z)dz$ ### Find the solution of the initial boundary value problem - $u_{tt} = u_{xx}, 0 < x < \infty , t > 0$. - $u(x,0) = 0, 0 < x < \infty$ - $u_t(x,0) = 0, 0 < x < \infty$ - $u(0,t) = p(t), t > 0$ **Solution** - $u(x,t) = p(t-x)H(t-x), x > 0$ **Where H is the Heaviside unit step function** ### Find the solution of the initial boundary value problem: - $u_{tt} = 4u_{xx}, x > 0, t > 0$ - $u(x,0) = sin(x), x > 0$ - $u_t (x,0) = 0, x > 0$ - $u(0,t) = 0, t > 0$ **Solution** - $u(x,t) = \frac{1}{2}\Big[f(x+ct) + f(x-ct) \Big] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(z) dz$, $for x > ct$ ### Determine the solution of the initial boundary value problem: - $u_{tt} = c^2 u_{xx}, 0 < x < \infty, t> 0$ - $u(x,0) = f(x), 0 < x < \infty$ - $u_t(x,0) = 0, 0 < x < \infty$ **Solution** - $u(x,t) = \frac{1}{2}[f(x+ct) + f(x-ct)]$, $for x >= ct$ ## Equations with Non-homogeneous Boundary Conditions - $u_{tt} = c^2u_{xx}, x > 0, t > 0$ - $u(x,0) = f(x), x > 0$ - $u_t (x,0) = g(x), x > 0$ - $u(0,t) = p(t), t > 0$ **Solution** $u(x,t) = \frac{1}{2}[f(x+ct)+f(x-ct)] +\frac{1}{2c} \int_{x-ct}^{x+ct} g(z)dz + \frac{1}{2c} \int_{x-ct}^{x+ct} p(t-(z/c)) dz $, $for x \le ct$ ## Explain why the solution to the wave equation is odd. If $f$ and $g$ are extended as odd functions, show that $u(x,t)$ is given by the solution. - $u(x,t) = \frac{1}{2}[f(x+ct)+f(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(z) dz$, $for x < ct$ When the solution is given by - $u(x,t) = \frac{1}{2}[f(x+ct) + f(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(z) dz$, $for x>ct$ and the solution: - $u(x,t) = \frac{1}{2} [f(x+ct) + f(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(z) dz$, $for x < ct$