Change of variables in a function

[Dustinsfl]

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 $u(x,t)$?

 

[topsquark]

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 $u(x,t)$?

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:
$$u_{tt} - c^2u_{xx} = 0$$

This is the 1D wave equation with wave propagation speed c. The change of variable of the characteristic lines leads to the replacement
$$x + ct = \xi$$ and $$x - ct = \eta$$

which changes the PDE to
$$u_{\xi \eta} = 0$$

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

-Dan

 

[Dustinsfl]

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:
$$u_{tt} - c^2u_{xx} = 0$$

This is the 1D wave equation with wave propagation speed c. The change of variable of the characteristic lines leads to the replacement
$$x + ct = \xi$$ and $$x - ct = \eta$$

which changes the PDE to
$$u_{\xi \eta} = 0$$

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

-Dan

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*}

 

[Sudharaka]

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 $u(x,t)$?

Hi dwsmith, :)

\(u(x+ct,x-ct)\) 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.

 

[topsquark]

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*}

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
$$u_{\xi \eta} = (u_{\xi})_{\eta} = 0$$
which implies that $$u_{\xi}$$ is independent of $$\eta$$ so
$$u_{\xi} = f'(\xi)$$

Thus
$$u(\xi, \eta) = \int f'(\xi)~d \xi + G(\eta) = F(\xi) + G(\eta)$$

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

 

[Dustinsfl]

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
$$u_{\xi \eta} = (u_{\xi})_{\eta} = 0$$
which implies that $$u_{\xi}$$ is independent of $$\eta$$ so
$$u_{\xi} = f'(\xi)$$

Thus
$$u(\xi, \eta) = \int f'(\xi)~d \xi + G(\eta) = F(\xi) + G(\eta)$$

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

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.