# [SOLVED]change of variables in a function

#### dwsmith

##### Well-known member
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

##### Well-known member
MHB Math Helper
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

#### dwsmith

##### Well-known member
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

##### Well-known member
MHB Math Helper
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

##### Well-known member
MHB Math Helper
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
$$u_{\xi \eta} = (u_{\xi})_{\eta} = 0$$
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

#### dwsmith

##### Well-known member

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.