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- Thread starter dwsmith
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- Aug 30, 2012

- 1,123

I'm not prepared to discuss this fully, but this is not a simple matter of a change of variables. The function u(x, y) has to solve the Partial Differential Equation:Given

$$

u(\xi,\eta) = F(\xi) + G(\eta)

$$

and

$$

\xi = x+ct\qquad\qquad \eta = x -ct.

$$

How do we get to $u(x,t) = F(x+ct) + G(x-ct)$?

I see that we can make the sub

$$

u(x+ct,x-ct) = F(x+ct) + G(x-ct)

$$

but how do I then get simply $u(x,t)$?

[tex]u_{tt} - c^2u_{xx} = 0[/tex]

This is the 1D wave equation with wave propagation speed c. The change of variable of the characteristic lines leads to the replacement

[tex]x + ct = \xi[/tex] and [tex]x - ct = \eta[/tex]

which changes the PDE to

[tex]u_{\xi \eta} = 0[/tex]

More explanation than this should probably be given by someone else until I bone up on the solution.

-Dan

- Thread starter
- #3

I already solved that. I wondering how I get back to $u(x,t)$.I'm not prepared to discuss this fully, but this is not a simple matter of a change of variables. The function u(x, y) has to solve the Partial Differential Equation:

[tex]u_{tt} - c^2u_{xx} = 0[/tex]

This is the 1D wave equation with wave propagation speed c. The change of variable of the characteristic lines leads to the replacement

[tex]x + ct = \xi[/tex] and [tex]x - ct = \eta[/tex]

which changes the PDE to

[tex]u_{\xi \eta} = 0[/tex]

More explanation than this should probably be given by someone else until I bone up on the solution.

-Dan

Here is where I am at

$$

\frac{\partial^2 u}{\partial\xi\partial\eta} = 0.

$$

Let's take the standard form of the wave equation $u_{tt} = c^2u_{xx}$.

Then by the chain rule

$$

\frac{\partial}{\partial x} = \frac{\partial\xi}{\partial x}\frac{\partial}{\partial\xi} + \frac{\partial\eta}{\partial x}\frac{\partial}{\partial\eta} = \frac{\partial}{\partial\xi} + \frac{\partial}{\partial\eta}

$$

and

$$

\frac{\partial}{\partial t} = \frac{\partial\xi}{\partial t}\frac{\partial}{\partial\xi} + \frac{\partial\eta}{\partial t}\frac{\partial}{\partial\eta} = c\frac{\partial}{\partial\xi} - c\frac{\partial}{\partial\eta}.

$$

Therefore, the second derivatives are

\begin{alignat*}{3}

\frac{\partial^2 u}{\partial x^2} & = & \frac{\partial^2 u}{\partial\xi^2} + 2\frac{\partial^2 u}{\partial\xi\partial\eta} + \frac{\partial^2 u}{\partial\eta^2}\\

\frac{\partial^2 u}{\partial t^2} & = & c^2\frac{\partial^2 u}{\partial\xi^2} - 2c^2\frac{\partial^2 u}{\partial\xi\partial\eta} + c^2\frac{\partial^2 u}{\partial\eta^2}

\end{alignat*}

Making the appropriate substitution, we have

\begin{alignat*}{3}

c^2\frac{\partial^2 u}{\partial\xi^2} - 2c^2\frac{\partial^2 u}{\partial\xi\partial\eta} + c^2\frac{\partial^2 u}{\partial\eta^2} & = & c^2\frac{\partial^2 u}{\partial\xi^2} + 2c^2\frac{\partial^2 u}{\partial\xi\partial\eta} + c^2\frac{\partial^2 u}{\partial\eta^2}\\

4c^2\frac{\partial^2 u}{\partial\xi\partial\eta} & = & 0\\

\frac{\partial^2 u}{\partial\xi\partial\eta} & = & 0

\end{alignat*}

Show that by integrating twice, once with respect to $\eta$ and once with respect to $\xi$, that the general form for $u$ is

\begin{alignat*}{3}

u(\xi,\eta) & = & F(\xi) + G(\eta)\\

u(x,t) & = & F(x + ct) + G(x - ct)

\end{alignat*}

where the forms $F$ and $G$ are arbitrary. This result lies at the heart of the method of characteristics for linear wave motion.

\begin{alignat*}{3}

\iint u_{\xi\eta} d\eta d\xi & = & \iint 0 d\eta d\xi\\

\int u_{\xi} & = & \int f(\xi)\\

u(\xi,\eta) & = & F(\xi) + G(\eta)

\end{alignat*}

- Feb 5, 2012

- 1,621

Hi dwsmith,Given

$$

u(\xi,\eta) = F(\xi) + G(\eta)

$$

and

$$

\xi = x+ct\qquad\qquad \eta = x -ct.

$$

How do we get to $u(x,t) = F(x+ct) + G(x-ct)$?

I see that we can make the sub

$$

u(x+ct,x-ct) = F(x+ct) + G(x-ct)

$$

but how do I then get simply $u(x,t)$?

\(u(x+ct,x-ct)\) means that your function \(u\) depends on \(x+ct\) and \(x-ct\). But since \(c\) is a constant value that is same as saying that your function is dependent upon \(x\) and \(t\). So writing \(u(x,t)\) instead of \(u(x+ct,x-ct)\) is the same except the fact that \(u(x+ct,x-ct)\) gives you the additional detail that \(x+ct\) and \(x-ct\) appears explicitly in the expression of \(u\). What I am saying is when you write, \(u(\xi,\eta)=u(x+ct,x-ct)\) you know that the function \(u\) is something like,

\[u(x+ct,x-ct)=(x-ct)^2+\sqrt{x+ct}\]

Clearly the above function is a function of \(x\) and \(t\). So you can write, \(u(x,t)\) but then you are missing the detail that \(x+ct\) and \(x-ct\) could be separated and the function could be written as \(u(\xi,\eta)\). Hope my explanation helps.

Kind Regards,

Sudharaka.

- Aug 30, 2012

- 1,123

If you already knew about this then why did you post it in Pre-Calc???I already solved that. I wondering how I get back to $u(x,t)$.

Here is where I am at

$$

\frac{\partial^2 u}{\partial\xi\partial\eta} = 0.

$$

Let's take the standard form of the wave equation $u_{tt} = c^2u_{xx}$.

Then by the chain rule

$$

\frac{\partial}{\partial x} = \frac{\partial\xi}{\partial x}\frac{\partial}{\partial\xi} + \frac{\partial\eta}{\partial x}\frac{\partial}{\partial\eta} = \frac{\partial}{\partial\xi} + \frac{\partial}{\partial\eta}

$$

and

$$

\frac{\partial}{\partial t} = \frac{\partial\xi}{\partial t}\frac{\partial}{\partial\xi} + \frac{\partial\eta}{\partial t}\frac{\partial}{\partial\eta} = c\frac{\partial}{\partial\xi} - c\frac{\partial}{\partial\eta}.

$$

Therefore, the second derivatives are

\begin{alignat*}{3}

\frac{\partial^2 u}{\partial x^2} & = & \frac{\partial^2 u}{\partial\xi^2} + 2\frac{\partial^2 u}{\partial\xi\partial\eta} + \frac{\partial^2 u}{\partial\eta^2}\\

\frac{\partial^2 u}{\partial t^2} & = & c^2\frac{\partial^2 u}{\partial\xi^2} - 2c^2\frac{\partial^2 u}{\partial\xi\partial\eta} + c^2\frac{\partial^2 u}{\partial\eta^2}

\end{alignat*}

Making the appropriate substitution, we have

\begin{alignat*}{3}

c^2\frac{\partial^2 u}{\partial\xi^2} - 2c^2\frac{\partial^2 u}{\partial\xi\partial\eta} + c^2\frac{\partial^2 u}{\partial\eta^2} & = & c^2\frac{\partial^2 u}{\partial\xi^2} + 2c^2\frac{\partial^2 u}{\partial\xi\partial\eta} + c^2\frac{\partial^2 u}{\partial\eta^2}\\

4c^2\frac{\partial^2 u}{\partial\xi\partial\eta} & = & 0\\

\frac{\partial^2 u}{\partial\xi\partial\eta} & = & 0

\end{alignat*}

Show that by integrating twice, once with respect to $\eta$ and once with respect to $\xi$, that the general form for $u$ is

\begin{alignat*}{3}

u(\xi,\eta) & = & F(\xi) + G(\eta)\\

u(x,t) & = & F(x + ct) + G(x - ct)

\end{alignat*}

where the forms $F$ and $G$ are arbitrary. This result lies at the heart of the method of characteristics for linear wave motion.

\begin{alignat*}{3}

\iint u_{\xi\eta} d\eta d\xi & = & \iint 0 d\eta d\xi\\

\int u_{\xi} & = & \int f(\xi)\\

u(\xi,\eta) & = & F(\xi) + G(\eta)

\end{alignat*}

Okay, I'm kind of reading this right out of my book, so I feel like I'm cheating. But we know that

[tex]u_{\xi \eta} = (u_{\xi})_{\eta} = 0[/tex]

which implies that [tex]u_{\xi}[/tex] is independent of [tex]\eta[/tex] so

[tex]u_{\xi} = f'(\xi)[/tex]

Thus

[tex]u(\xi, \eta) = \int f'(\xi)~d \xi + G(\eta) = F(\xi) + G(\eta)[/tex]

then put back the original x, t.

I should note two things...first that u is assumed to be convex in either pair of variables. (I really don't know what that means.) And second, my source is "Partial Differential Equations", 4 ed. by Fritz John, section 4, pg 40.

-Dan

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- #6

I put it in pre-calc since I was asking about change of variables not solving the problem (I had that under control). That is also why I didn't supply the details.If you already knew about this then why did you post it in Pre-Calc???

Okay, I'm kind of reading this right out of my book, so I feel like I'm cheating. But we know that

[tex]u_{\xi \eta} = (u_{\xi})_{\eta} = 0[/tex]

which implies that [tex]u_{\xi}[/tex] is independent of [tex]\eta[/tex] so

[tex]u_{\xi} = f'(\xi)[/tex]

Thus

[tex]u(\xi, \eta) = \int f'(\xi)~d \xi + G(\eta) = F(\xi) + G(\eta)[/tex]

then put back the original x, t.

I should note two things...first that u is assumed to be convex in either pair of variables. (I really don't know what that means.) And second, my source is "Partial Differential Equations", 4 ed. by Fritz John, section 4, pg 40.

-Dan