Infinite String: consider the initial/boundary value problem = c2 dx2 u(x, 0) = f(x), хER, t > 0, x E R ди (x, 0) = g(x), 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) = If(x + ct) + f(x – ct)] + [G(x + ct) – G(x 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) |x| < 1 |x| > 1 sin Tx f(x) = 0 g(x) = {
Infinite String: consider the initial/boundary value problem = c2 dx2 u(x, 0) = f(x), хER, t > 0, x E R ди (x, 0) = g(x), 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) = If(x + ct) + f(x – ct)] + [G(x + ct) – G(x 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) |x| < 1 |x| > 1 sin Tx f(x) = 0 g(x) = {
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Hi Can you please graph Question -A, B with step by step how you could you get coordinate. Please show step by step answer
![Infinite String: consider the initial/boundary value problem
Pu
x E R,
t > 0,
dx2'
u(х, 0) %3D f(x),
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
1
u(x, t) = [f(x+ ct) + f(x – ct)] + IG(x + ct) – G(x – c
|
2c
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)
x2
f(x) = {
|x| < 1
I지 > 1’
g(x) =
b)
sin TX
f(x) = 0 g(x) = {"
|x| < 1
|지 > 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8e69957f-abc4-42a5-ae54-6067c6a171e5%2F3940d490-393a-464f-b4a3-b71846abdf31%2Fzjmf49k_processed.png&w=3840&q=75)
Transcribed Image Text:Infinite String: consider the initial/boundary value problem
Pu
x E R,
t > 0,
dx2'
u(х, 0) %3D f(x),
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
1
u(x, t) = [f(x+ ct) + f(x – ct)] + IG(x + ct) – G(x – c
|
2c
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)
x2
f(x) = {
|x| < 1
I지 > 1’
g(x) =
b)
sin TX
f(x) = 0 g(x) = {"
|x| < 1
|지 > 1
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