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### Topics - Ioana Nedelcu

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##### 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

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##### 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?

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##### 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)

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##### 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

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