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

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Homework Help Overview

The discussion revolves around a partial differential equation, specifically examining the function v(x,t) defined as v(x,t) = u(x+ct,t). Participants are tasked with demonstrating that this function satisfies the equation ut = κ uxx, while also considering the implications of the initial condition u(x,0) = φ(x).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the differentiation of the function v(x,t) and the implications of treating it as a multi-variable function. There are attempts to clarify the notation used for derivatives, particularly regarding the use of prime symbols versus partial derivatives.

Discussion Status

The conversation is ongoing, with participants providing insights into the differentiation process and questioning the correctness of initial assumptions. Some guidance has been offered regarding the notation and the relationship between variables, but no consensus has been reached on the overall approach to the problem.

Contextual Notes

There is a note that the initial condition u(x,0) = φ(x) may be disregarded for the purpose of the discussion. Participants are also navigating the complexities of multi-variable calculus in the context of the problem.

jpcjr
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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)
 
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Any help on any part will be GREATLY appreciated!
 
Welcome to PF,

jpcjr said:
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:


jpcjr said:
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
 
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)
 
What I was saying was that v_x = u_w, with underscores denoting subscripts.
 

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