# Change of Variable in a second-order PDE

• Mar 14th 2011, 12:28 PM
matt.qmar
Change of Variable in a second-order PDE
Hello!

I am working on a question involving a P.D.E. in a adv. calc. course.

Consider the general homogenous second-order partial differential equation
$a\frac{\delta^2u}{\delta x^2} + 2b\frac{\delta^2 u}{\delta x\delta y} + c\frac{\delta^2 u}{\delta y^2} = 0$

with constant coefficients (a,b,c).

If $ac - b^2 = 0$, show that the substitution $s = bx - ay, \ t = y$ reduces the above equation to $\frac{\delta^2 u}{\delta t^2} = 0$

So we must consider the change of variable from $u(x,y)$ to $u(s,t)$, which is $u(bx-ay, y)$ and take partial derivatives? I can't seem to derive the desired result, however, and I have double-checked my messy algebra...

Any help appreciated. Thanks!
• Mar 14th 2011, 01:33 PM
Ackbeet
I recommend solving for x = x(s,t) and y = y(s,t). Compute their partial derivatives with respect to s and t. Then simply compute

$u_{t}=u_{x}x_{t}+u_{y}y_{t},$ and the subsequent $u_{tt}.$ Using both facts that are given to you (that is, the original pde, and the condition on a,b,c), you can show the desired result.

What do you get?
• Mar 14th 2011, 02:38 PM
matt.qmar
I see! Let me try to fill in the blanks... you wrote $x = x(s,t) = \frac{s + at}{b}$ and $y = y(s,t) = t$

Then by the chain rule,
$u_t = u_xx_t + u_yy_t$

but we can see that $x_t = \frac{a}{b}$ and $y_t = 1$, so
$u_t = u_x\frac{a}{b} + u_y$

then another chain rule yields

$u_{tt} = (u_{xx}\frac{a}{b})x_t + u_{yy}y_t$

If that is correct, then...
$u_{tt} = (u_{xx}\frac{a}{b})\frac{a}{b} + u_{yy}$
or $u_{tt} = (u_{xx}\frac{a^2}{b^2}) + u_{yy}$
am I close enough here? I don't see any algebraic manipulation I could do with the condition and PDE to derive the result...

I appreciate your post, thank you!
• Mar 14th 2011, 02:47 PM
Ackbeet
When you compute your second derivative, you need to do this:

$u_{tt}=\dfrac{\partial}{\partial x}\left(u_{t}\right)\,\dfrac{\partial x}{\partial t}+\dfrac{\partial}{\partial y}\left(u_{t}\right)\,\dfrac{\partial y}{\partial t}=\dfrac{\partial}{\partial x}\left(\dfrac{a}{b}\,u_{x}+u_{y}\right)\,\dfrac{\ partial x}{\partial t}+\dfrac{\partial}{\partial y}\left(\dfrac{a}{b}\,u_{x}+u_{y}\right)\,\dfrac{\ partial y}{\partial t}\dots$

Once you've brought that correction through, why not solve the pde for $a u_{xx}$ and plug into your expression for $u_{tt}?$ Also, don't forget to use your $b^{2}=ac$ relation.
• Mar 14th 2011, 03:11 PM
matt.qmar
OK! Two chain rules "compounds" those partials.... so
$u_{tt} = (\frac{a}{b}u_{xx} + u_{yx})(\frac{a}{b}) + (\frac{a}{b}u_{xy} + u_{yy})$

Then,
$u_{tt} = \frac{a^2}{b^2}u_{xx} + 2\frac{a}{b}u_{xy} + u_{yy}$

But $b^2 = ac$

$u_{tt} = \frac{a}{c}u_{xx} + 2\frac{a}{b}u_{xy} + u_{yy}$

Also, $a = \frac{b^2}{c}$

$u_{tt} = \frac{a}{c}u_{xx} + 2\frac{b}{c}u_{xy} + u_{yy}$

Multiply through by c?

$cu_{tt} = au_{xx} + 2bu_{xy} + cu_{yy}$

RHS = 0 from assumption

$cu_{tt} = 0$

$c \neq 0$ so

$u_{tt} = 0$

Woo-hoo! Thank you so much!!! I was really fudging the chain rule. I think everything I have done here is OK?
• Mar 14th 2011, 03:33 PM
Ackbeet
Yep, that looks fine. My route to the end was a bit different, but definitely mathematically equivalent. Good job.