Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.


Topics - Ioana Nedelcu

Pages: [1]
1
APM346––Home Assignments / Boundary condition
« on: February 05, 2018, 02:05:49 AM »
For section 2.6, question 1, how do we use the boundary condition at x = 0?

Solving using the initial conditions with d'Alembert's formula, the solution is $ u = \phi(x + ct) $, for x>ct and 0<x<ct since it has a positive argument always

2
APM346––Home Assignments / Section 2.3, Q 5
« on: January 31, 2018, 12:17:06 AM »
Since we found the solution for $ v(r,t) $ and $ u = v/r $, then the general solution will be

 $$ u = \frac{1}{2r}\Phi (r+ct)+\Phi (r-ct)+\frac{1}{2rc}\int_{r-ct}^{r+ct}\Psi (s) \,ds $$

Clearly, it is not continuous at r = 0 so do we only consider as r approaches 0?

3
APM346--Lectures / Uniqueness/ continuity of solutions
« on: January 24, 2018, 12:29:49 AM »
I'm still trying to understand how the initial and boundary conditions affect the existence and types of solutions.

For example, in one of the tutorials the solution to $$ xu_{t} + u_{x} = 0 $$ $$ u(x, 0) = f(x) $$ is not uniquely determined. What exactly determines this?

(It's also possible I understood the question completely wrong but I'm generally confused about uniqueness)

4
APM346--Lectures / Overdetermined systems
« on: January 10, 2018, 09:34:55 PM »
I'm having some trouble solving the overdetermined systems in the homework assignment.

$$\begin{align} & u_{xx}=2xy \label{A}\\ &u_{y}=x^2 \label{B} \end{align}$$
Solving the last equation gives $ u = x^2y + f(x) $

Then using the first equation: $$ u_{xx} = 2xy = 2y + f_{xx}(x) $$ Then solving for $ f(x) $ gives $ f(x) = x^3y/3 -x^2y $ so the overall solution is $$ u = x^3y/3 $$

However, this clearly doesn't satisfy the partial derivative equation with respect to $y$.

Just wondering what I'm doing wrong? I haven't had much experience with systems, even in ODE


Pages: [1]