Let v(x,t) = u(x+ct) and show that

  • Thread starter jpcjr
  • Start date
  • #1
17
0
1. Homework Statement

NOTE: I realize the following is a partial differential equation, but I believe the answer to my question is a straight forward multi-variate calculus question.

Let v(x,t) = u(x+ct,t) and show that v(x,t) solves the following...

ut = κ uxx ; (-∞<x<∞)

u(x,0) = [itex]\phi[/itex](x)

NOTE: You may disregard the following, if necessary for you to answer this question:

u(x,0) = [itex]\phi[/itex](x)

and simply show that v(x,t) solves the following...

ut = κ uxx

ALTERNATIVELY, you may help me by commenting on the correctness of my work below.




2. Homework Equations

See above and below...




3. The Attempt at a Solution

v(x,t) = u(x+ct,t)

v'(x,t) = c u[itex]_{x}[/itex](x+ct,t) + u[itex]_{t}[/itex](x+ct,t)

v[itex]_{x}[/itex](x,t) = c u[itex]_{x}[/itex](x+ct,t)

v[itex]_{xx}[/itex](x,t) = c[itex]^{2}[/itex] u[itex]_{xx}[/itex](x+ct,t)

v[itex]_{t}[/itex](x,t) = u[itex]_{t}[/itex](x+ct,t)
 

Answers and Replies

  • #2
17
0
Any help on any part will be GREATLY appreciated!!!
 
  • #3
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
38
Welcome to PF,

1. Homework Statement

NOTE: I realize the following is a partial differential equation, but I believe the answer to my question is a straight forward multi-variate calculus question.

Let v(x,t) = u(x+ct,t) and show that v(x,t) solves the following...

ut = κ uxx ; (-∞<x<∞)


Just before we start, are you sure it was not meant to be simply v(x,t) = u(x+ct)? Because I know that a function in which the temporal and spatial dependence are related in this way will satisfy a wave equation, which is given by utt = κuxx. This is different from the PDE that you gave. But anyway, taking what you wrote at face value:


ALTERNATIVELY, you may help me by commenting on the correctness of my work below.



Homework Equations



See above and below...

The Attempt at a Solution



v(x,t) = u(x+ct,t)

v'(x,t) = c u[itex]_{x}[/itex](x+ct,t) + u[itex]_{t}[/itex](x+ct,t)

v[itex]_{x}[/itex](x,t) = c u[itex]_{x}[/itex](x+ct,t)

v[itex]_{xx}[/itex](x,t) = c[itex]^{2}[/itex] u[itex]_{xx}[/itex](x+ct,t)

v[itex]_{t}[/itex](x,t) = u[itex]_{t}[/itex](x+ct,t)

First of all, I don't think that v' (v "prime") is meaningful here. In single-variable calculus, a prime symbol typically means "derivative with respect to the argument." But this function has two arguments. Since it is a multi-variable function, any derivative is going to be a partial derivative, and you must explicitly specify which variable you are differentiating with respect to. As for your attempt at partial derivatives, my approach would be to write:

u = u(w,t) where w = x+ct

then,

[tex]\frac{\partial u}{\partial x} = \frac{\partial u}{\partial w}\frac{\partial w}{\partial x}[/tex][tex]= \frac{\partial u}{\partial w}[/tex]

Et cetera for the other derivatives
 
  • #4
17
0
To your first point...

I am sure it is:

Let v(x,t) = u(x+ct,t)

to your second point...

That would mean the following is incorrect, right?

v.sub.x(x,t) = c u.sub.x(x+ct,t)

and should have been...

v.sub.x(x,t) = u.sub.x(x+ct,t)
 
  • #5
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
38
What I was saying was that v_x = u_w, with underscores denoting subscripts.
 

Related Threads on Let v(x,t) = u(x+ct) and show that

Replies
2
Views
1K
  • Last Post
Replies
21
Views
1K
Replies
1
Views
4K
Replies
4
Views
10K
  • Last Post
Replies
7
Views
4K
Replies
2
Views
1K
Replies
3
Views
3K
Replies
1
Views
1K
Replies
3
Views
3K
Top