uestion 1. Consider the second order partial differential equation = c? dx² (1) r an unknown real-valued function u = u(t, x), where t represents time, x represents a point in space, and c> 0 is constant. 1. For any twice differentiable functions F = F(x) and G = G(x), show that u(t, x) = F(x + ct) + G(x – ct) satisfies (1). artial differential equations such as (1) are often solved as initial value problems, where the initial description of e unknown function and its time derivative are provided at each point in space. Along these lines, suppose we are ven that u(0, x) = g(x) and (0, x) = h(x), for some given functions g and h. 2. Assume that u(t, x) = F(x + ct) + G(x - ct) for some functions F and G, as described in problem 1.1. If u = u(t, x) solves the initial value problem described above, show that g(x) F(x) + G(x) and h(x) = cF'(x) – cG'(x). %3D 3. By integrating the last equation for h(x), show that for any constant a E R, h(s) ds = cF(x) – CG(x) – cF(a)+ cG(a), and from here solve a linear system to show that F(x) = ; (9(2) +[ h(0) ds + F(a) – G(a)), - and (9(=) -[ ) . G(x) h(s) ds – F(a) + G(a) 4. Lastly, given that u(t, x) = F(x + ct) + G(x – ct), arrive at an explicit formula for the solution to the initial value problem for (1): 1 cr+ct u(t, x) = , [9(x + ct) + g(x – ct)] + h(s)ds. 2c
uestion 1. Consider the second order partial differential equation = c? dx² (1) r an unknown real-valued function u = u(t, x), where t represents time, x represents a point in space, and c> 0 is constant. 1. For any twice differentiable functions F = F(x) and G = G(x), show that u(t, x) = F(x + ct) + G(x – ct) satisfies (1). artial differential equations such as (1) are often solved as initial value problems, where the initial description of e unknown function and its time derivative are provided at each point in space. Along these lines, suppose we are ven that u(0, x) = g(x) and (0, x) = h(x), for some given functions g and h. 2. Assume that u(t, x) = F(x + ct) + G(x - ct) for some functions F and G, as described in problem 1.1. If u = u(t, x) solves the initial value problem described above, show that g(x) F(x) + G(x) and h(x) = cF'(x) – cG'(x). %3D 3. By integrating the last equation for h(x), show that for any constant a E R, h(s) ds = cF(x) – CG(x) – cF(a)+ cG(a), and from here solve a linear system to show that F(x) = ; (9(2) +[ h(0) ds + F(a) – G(a)), - and (9(=) -[ ) . G(x) h(s) ds – F(a) + G(a) 4. Lastly, given that u(t, x) = F(x + ct) + G(x – ct), arrive at an explicit formula for the solution to the initial value problem for (1): 1 cr+ct u(t, x) = , [9(x + ct) + g(x – ct)] + h(s)ds. 2c
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Question 1. Consider the second order partial differential equation
и
(1)
dx² '
for an unknown real-valued function u =
u(t, x), wheret represents time, x represents a point in space,
and c > 0 is
a constant.
1. For any twice differentiable functions F
satisfies (1).
F(x) and G = G(x), show that u(t, x) = F(x + ct) + G(x – ct)
%3D
Partial differential equations such as (1) are often solved as initial value problems, where the initial description of
the unknown function and its time derivative are provided at each point in space. Along these lines, suppose we are
given that u(0, æ) = g(x) and " (0, x) = h(x), for some given functions
du
and h.
2. Assume that u(t, x)
u = u(t, x) solves the initial value problem described above, show that
F(x + ct) + G(x – ct) for some functions F and G, as described in problem 1.1. If
g(x) = F(x)+ G(x)
and
h(x).
cF'(x) – cG' (x).
3. By integrating the last equation for h(x), show that for any constant a € R,
| h(s) ds = cF(æ) – cG(x) – cF(a)+ cG(a),
and from here solve a linear system to show that
1
F(x) = ; (g(x)
(o(e) + Mo) ds + F(a) – G(a)
and
G(e) = } (s(e) }
:| h(s) ds – F(a) + G(a)
2
4. Lastly, given that u(t, x) = F(x + ct) + G(x – ct), arrive at an explicit formula for the solution to the initial
value problem for (1):
%3D
1
1
px+ct
u(t, x) = ; [g(x + ct) + g(x – ct)] +
h(s)ds.
2c
x-ct](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2ef31c9f-1e44-437f-a71e-09574dc73129%2F021c6f4d-7467-44af-a6c9-e1400b1a7067%2Fu233utj_processed.png&w=3840&q=75)
Transcribed Image Text:Question 1. Consider the second order partial differential equation
и
(1)
dx² '
for an unknown real-valued function u =
u(t, x), wheret represents time, x represents a point in space,
and c > 0 is
a constant.
1. For any twice differentiable functions F
satisfies (1).
F(x) and G = G(x), show that u(t, x) = F(x + ct) + G(x – ct)
%3D
Partial differential equations such as (1) are often solved as initial value problems, where the initial description of
the unknown function and its time derivative are provided at each point in space. Along these lines, suppose we are
given that u(0, æ) = g(x) and " (0, x) = h(x), for some given functions
du
and h.
2. Assume that u(t, x)
u = u(t, x) solves the initial value problem described above, show that
F(x + ct) + G(x – ct) for some functions F and G, as described in problem 1.1. If
g(x) = F(x)+ G(x)
and
h(x).
cF'(x) – cG' (x).
3. By integrating the last equation for h(x), show that for any constant a € R,
| h(s) ds = cF(æ) – cG(x) – cF(a)+ cG(a),
and from here solve a linear system to show that
1
F(x) = ; (g(x)
(o(e) + Mo) ds + F(a) – G(a)
and
G(e) = } (s(e) }
:| h(s) ds – F(a) + G(a)
2
4. Lastly, given that u(t, x) = F(x + ct) + G(x – ct), arrive at an explicit formula for the solution to the initial
value problem for (1):
%3D
1
1
px+ct
u(t, x) = ; [g(x + ct) + g(x – ct)] +
h(s)ds.
2c
x-ct
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