Infinite String: consider the initial/boundary value problem x E R, t > 0, dx2 f(x), и(х, 0) x E R ди (x, 0) = g(x), dt x E R where f and g are two given twice differentiable functions. We showed in...


Infinite String: consider the initial/boundary value problem<br>x E R,<br>t > 0,<br>dx2<br>f(x),<br>и(х, 0)<br>x E R<br>ди<br>(x, 0) = g(x),<br>dt<br>x E R<br>where f and g are two given twice differentiable functions. We showed in class that this problem is solved by<br>he d'Alembert's solution<br>u(x, t) = LS(x + ct)<br>- ct) + f(x – ct)] +<br>2c<br>IG(x + ct) – G(x – ct)],<br>where G is an antiderivative of g. Sketch the solution for t = 0, 1, 2, 3, where f(x) and g(x) are given below.<br>a)<br>[1 – x² |x| < 1<br>|x| > 1 '<br>f(x) =<br>g(x) = 0<br>b)<br>sin Tx<br>f(x) = 0 g(x) ={<br>|x| < 1<br>I지 > 1<br>

Extracted text: Infinite String: consider the initial/boundary value problem x E R, t > 0, dx2 f(x), и(х, 0) x E R ди (x, 0) = g(x), dt x E R where f and g are two given twice differentiable functions. We showed in class that this problem is solved by he d'Alembert's solution u(x, t) = LS(x + ct) - ct) + f(x – ct)] + 2c IG(x + ct) – G(x – ct)], where G is an antiderivative of g. Sketch the solution for t = 0, 1, 2, 3, where f(x) and g(x) are given below. a) [1 – x² |x| < 1="" |x|=""> 1 ' f(x) = g(x) = 0 b) sin Tx f(x) = 0 g(x) ={ |x| < 1="" i지=""> 1

Jun 04, 2022
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