# PDE for Exam

• Sep 13th 2010, 05:39 AM
woody198403
PDE for Exam
I'm doing revision for an exam in 2 days and I don't quite understand one of the final steps in the problem (worked example)

$\displaystyle x u_x + yu_y = ku$ subject to $\displaystyle u(x,1) = F(x)$

Solution

Parametrization of IC's ->
$\displaystyle x(s) = s, y(s) = 1, u(s) = F(s)$

I don't quite understand why this is done. I can see how its used to solve this question....is it just a standard method to solve PDE's?

$\displaystyle \frac{dx}{dt} = x, \ \frac{dy}{dt} = y$

Then $\displaystyle x = c_1(s)e^t, \ y = c_2(s) e^t$

Apply the initial data to get $\displaystyle x = se^t, \ y = e^t$

$\displaystyle t = \ln \ y$ I'm not sure why this is....(Worried)

and $\displaystyle s = \frac{x}{e^t} = \frac{x}{y}$

PDE becomes $\displaystyle \frac{du}{dt} = ku$

$\displaystyle \frac{du}{u} = k \ dt$

and I don't understand any of the remaining steps:

$\displaystyle \ln u = kt + c$ (I'm not sure where $\displaystyle \ln u$ comes from and the RHS is just integrating with respect to t, right?

$\displaystyle u = D(s) e^{kt}$ Im not sure what $\displaystyle D(s)$ is or where the $\displaystyle e^{kt}$ term comes from

$\displaystyle u(s,t = 0) = D(s) = F(s)$. I don't understand this because I don't know what $\displaystyle D(s)$ is.

$\displaystyle u = F(s) e^{kt} = F \Big( \frac{x}{y}\Big) (e^t)^k = F(\frac{x}{y})y^k$ I actually think I understand this line.

Can someone please explain this to me? I'm worried.
• Sep 13th 2010, 08:42 AM
MattMan
Quote:

Originally Posted by woody198403
I'm doing revision for an exam in 2 days and I don't quite understand one of the final steps in the problem (worked example)

$\displaystyle x u_x + yu_y = ku$ subject to $\displaystyle u(x,1) = F(x)$

Solution

Parametrization of IC's ->
$\displaystyle x(s) = s, y(s) = 1, u(s) = F(s)$

I don't quite understand why this is done. I can see how its used to solve this question....is it just a standard method to solve PDE's? This is standard to get an ODE.

$\displaystyle \frac{dx}{dt} = x, \ \frac{dy}{dt} = y$

Then $\displaystyle x = c_1(s)e^t, \ y = c_2(s) e^t$

Apply the initial data to get $\displaystyle x = se^t, \ y = e^t$

$\displaystyle t = \ln \ y$ I'm not sure why this is....(Worried) You took the logarithm of both sides (you should be worried in a PDE class if this is not clear).

and $\displaystyle s = \frac{x}{e^t} = \frac{x}{y}$

PDE becomes $\displaystyle \frac{du}{dt} = ku$

$\displaystyle \frac{du}{u} = k \ dt$

and I don't understand any of the remaining steps:

$\displaystyle \ln u = kt + c$ (I'm not sure where $\displaystyle \ln u$ comes from and the RHS is just integrating with respect to t, right? You took the integral of both sides (first method you learn to solve a DE in Calculus 1). The left side is clearly with respect to u and the right side is with respect to t.

$\displaystyle u = D(s) e^{kt}$ Im not sure what $\displaystyle D(s)$ is or where the $\displaystyle e^{kt}$ term comes from You exponentiated both sides and the D(s) comes from the fact that your constant C is a function of s (which might just be a real number).

$\displaystyle u(s,t = 0) = D(s) = F(s)$. I don't understand this because I don't know what $\displaystyle D(s)$ is. That's why you plugged in t=0, it gives you that D(s)=F(s), where F(s) is given in the problem.

$\displaystyle u = F(s) e^{kt} = F \Big( \frac{x}{y}\Big) (e^t)^k = F(\frac{x}{y})y^k$ I actually think I understand this line.

Can someone please explain this to me? I'm worried.

To be blunt, you shouldn't expect to be lacking this severely in algebra and calculus, and to still end up understanding the material very well.
• Sep 13th 2010, 09:57 AM
wonderboy1953
"I'm doing revision for an exam in 2 days..."

Even if I knew the answer I couldn't help due to the policy of this website.
• Sep 13th 2010, 10:12 AM
Ackbeet
I could be wrong, but I think help is fine with this problem. I think the OP'er meant to say, "...review for an exam in 2 days..."
• Sep 13th 2010, 11:49 AM
MattMan
He said "worked example", so I just assumed...