# PDE, Second Order Differential

• Jan 15th 2009, 09:48 AM
PDE, Second Order Differential
Sorry for the poor lay-out to follow (i'm useless at latex) but I think I'm missing a trick here, I've got the answer but I can't get to it.

The function u(x, y) satisfies the partial differential equation;

6d^2u/dx^2 + y(d^2/dxdy) - y^2 (d^2u/dy^2) + 2(du/dx) = 0.

Using the method of characteristics (told to) I got

p = x - 2ln (y) r =x + 3ln(y) (just use those letters as I don't have the greek letters on my computer that are used.)

Using the chain rule show that this equation transforms to 5(d^2u/d!d?) + du/di = 0.

I've got it down to

d^2u/dx^2 = d^2u/dp^2 + 2(d^2u/drdp) + d^2u/dr^2

d^2u/dy^2 = 9/y^2(d^2u/dr) - 12/y^2(du/drdp) + 4/y^2(d^2u/dp^2) - 3/y^2 (du/dr) + 2/y^2 (du/dp)

d^2u/dxdy = 3/y(d^2u/dr^2) + 1/y(d^2u/dpdr) - 2/y(d^2u/dp)

When substituting these values back into the original equation the terms I want to cancel out do and as its an hyperbolic I want to be left with the du/dxdy terms.

But instead of getting 5 as the co-efficient like the answer, I keep getting 25, I can't for the life of me see where I'm dropping the minuses. Spent about an hour and a half on this question so far and driving me crazy.
• Jan 15th 2009, 10:05 AM
Jester
Quote:

Sorry for the poor lay-out to follow (i'm useless at latex) but I think I'm missing a trick here, I've got the answer but I can't get to it.

The function u(x, y) satisfies the partial differential equation;

6d^2u/dx^2 + y(d^2/dxdy) - y^2 (d^2u/dy^2) + 2(du/dx) = 0.

Using the method of characteristics (told to) I got

p = x - 2ln (y) r =x + 3ln(y) (just use those letters as I don't have the greek letters on my computer that are used.)

Using the chain rule show that this equation transforms to 5(d^2u/d!d?) + du/di = 0.

I've got it down to

d^2u/dx^2 = d^2u/dp^2 + 2(d^2u/drdp) + d^2u/dr^2

d^2u/dy^2 = 9/y^2(d^2u/dr) - 12/y^2(du/drdp) + 4/y^2(d^2u/dp^2) - 3/y^2 (du/dr) + 2/y^2 (du/dp)

d^2u/dxdy = 3/y(d^2u/dr^2) + 1/y(d^2u/dpdr) - 2/y(d^2u/dp)

When substituting these values back into the original equation the terms I want to cancel out do and as its an hyperbolic I want to be left with the du/dxdy terms.

But instead of getting 5 as the co-efficient like the answer, I keep getting 25, I can't for the life of me see where I'm dropping the minuses. Spent about an hour and a half on this question so far and driving me crazy.

When you substitute your three transformed derivatives (I checked - they are correct) into the second order derivatives of your PDE, you should get

$\displaystyle 25 u_{rp} + 3 u_r - 2 u_p.$

This, with the first order derivative gives the entire PDE as

$\displaystyle 25 u_{rp} + 5 u_r = 0,$

or $\displaystyle u_{rp} + \frac{1}{5}\,u_r = 0.$

BTW - this PDE is totally solvable.
• Jan 15th 2009, 10:11 AM
I see where I went wrong now, it was all in the first order terms with what I was being cocky about and ignoring (Doh).

All I needed to do was pay more attention to the signs there and realised I had to divide it by 5!!

Quote:

BTW - this PDE is totally solvable.
Aye, thats the next part, I've solved it as it gives you the answer in the question but it was driving me crazy as I couldn't see where I was going wrong to get the pde.

Cheers.
• Jan 15th 2009, 10:28 AM
ThePerfectHacker
Quote:

You want to solve: $\displaystyle u_{rp} + \tfrac{1}{5}\,u_r = 0$.
Let $\displaystyle v = u_r$ and that gives: $\displaystyle v_p + \tfrac{1}{5}v = 0$.
Now multiply by $\displaystyle e^{p/5}$ so: $\displaystyle v_p e^{p/5} + \tfrac{1}{5}e^{p/5} v = 0 \implies \left( v e^{p/5} \right)_p = 0$