Question 1. Consider the second order partial differential equation (1) for an unknown real-valued function u = u(t, x), where t represents time, x represents a point in space, and c> 0 is a constant. 1. For any twice differentiable functions F = satisfies (1). G(x), show that u(t, x) = F(x + ct)+ G(x – ct) F(x) and G = Partial differential equations such as (1) the unknown function and its time derivative are provided at each point in space. Along these lines, suppose we are given that u(0, x) = g(x) and (0, x) = h(x), for some given functions g and h. are often solved as initial value problems, where the initial description of 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) – cGʻ (x). 3. By integrating the last equation for h(x), show that for any constant a ER, h(s) ds = cF(x) – cG(x) – cF(a) + cG(a), and from here solve a linear system to show that F(x) g(x) + h(s) ds + F(a) – G(a) ), and G(») = (ole) - 1 1 h(s) ds – F(a) + G(a) 4. Lastly, given that u(t, x) = F(x + ct) + G(x - value problem for (1): ct), arrive at an explicit formula for the solution to the initial 1 rx+ct u(t, x) = [g(x + ct) + g(x – ct)] + h(s)ds. 2c x-ct
Question 1. Consider the second order partial differential equation (1) for an unknown real-valued function u = u(t, x), where t represents time, x represents a point in space, and c> 0 is a constant. 1. For any twice differentiable functions F = satisfies (1). G(x), show that u(t, x) = F(x + ct)+ G(x – ct) F(x) and G = Partial differential equations such as (1) the unknown function and its time derivative are provided at each point in space. Along these lines, suppose we are given that u(0, x) = g(x) and (0, x) = h(x), for some given functions g and h. are often solved as initial value problems, where the initial description of 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) – cGʻ (x). 3. By integrating the last equation for h(x), show that for any constant a ER, h(s) ds = cF(x) – cG(x) – cF(a) + cG(a), and from here solve a linear system to show that F(x) g(x) + h(s) ds + F(a) – G(a) ), and G(») = (ole) - 1 1 h(s) ds – F(a) + G(a) 4. Lastly, given that u(t, x) = F(x + ct) + G(x - value problem for (1): ct), arrive at an explicit formula for the solution to the initial 1 rx+ct u(t, x) = [g(x + ct) + g(x – ct)] + h(s)ds. 2c x-ct
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![* 2c /
Question 1. Consider the second order partial differential equation
c2.
(1)
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.
G(x), show that u(t, x) = F(x + ct) + G(x – ct)
1. For any twice differentiable functions F
satisfies (1).
F(x) and G
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, x) = g(x) and u (0, x) = h(x), for some given functions g and h.
at
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 E R,
| h(s) ds = cF(x) – cG(x) – cF(a) + cG(a),
and from here solve a linear system to show that
F(2) =(st=) +
1
h(2) ds + F(a) – G(a)),
and
Gl2) – (ste) -
= ; (9(=) – ds – F(a) + G(a)
1
h(s)
G(x)
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
1
px+ct
u(t, x)
= lg(x + ct) + g(x – ct)] +
2
h(s)ds.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8ce1d3e8-79ce-4525-99fa-3e56bb347219%2F670e1c8c-e112-464f-9057-54d47aa29c49%2F7d2d2vn_processed.png&w=3840&q=75)
Transcribed Image Text:* 2c /
Question 1. Consider the second order partial differential equation
c2.
(1)
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.
G(x), show that u(t, x) = F(x + ct) + G(x – ct)
1. For any twice differentiable functions F
satisfies (1).
F(x) and G
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, x) = g(x) and u (0, x) = h(x), for some given functions g and h.
at
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 E R,
| h(s) ds = cF(x) – cG(x) – cF(a) + cG(a),
and from here solve a linear system to show that
F(2) =(st=) +
1
h(2) ds + F(a) – G(a)),
and
Gl2) – (ste) -
= ; (9(=) – ds – F(a) + G(a)
1
h(s)
G(x)
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
1
px+ct
u(t, x)
= lg(x + ct) + g(x – ct)] +
2
h(s)ds.
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