### Author Topic: When HA marks will be available?  (Read 5033 times)

#### Bowei Xiao

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• Karma: 2 ##### When HA marks will be available?
« on: October 03, 2012, 03:41:39 PM »
Btw..when r we suppose to see our marks for the hw?

#### Victor Ivrii ##### Re: When HA marks will be available?
« Reply #1 on: October 03, 2012, 04:11:02 PM »
Btw..when r we suppose to see our marks for the hw?

First, don't hijack topics--your post is unrelated to "Posting solutions" topic and I moved it to a separate one.

Second, the answer to your question is obvious "When your papers are graded and marks are posted". I wish to have a bit less precise but more helpful answer, but I don't; so I give you a free advice: bug both instructors #### Martin Munoz

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« Reply #2 on: October 03, 2012, 10:48:43 PM »
Many students' marks for HA1 have been posted on Portal. If yours isn't visible yet don't panic! It should be up within the next day or so.

We marked only problems 1 and 4, using the rubric:

Problem 1 (10 marks): (a) 2, (b) 2, (c) 3, (d) 3
Problem 4 (10 marks): (a) 3, (b) 3, (c) 4

Very few people got full marks for 4 (c).
« Last Edit: October 03, 2012, 10:56:34 PM by Martin MuÃ±oz »

#### Martin Munoz

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« Reply #3 on: October 03, 2012, 11:30:28 PM »
With regards to problem 4 (c)...

Let me begin with an example. Consider the function
$$f(x) = \begin{cases}x^2 \sin\left(\frac{1}{x}\right) & x \ne 0, \\ 0 & x = 0. \end{cases}$$
Although not immediately obvious, it is a simple exercise to show that this function is differentiable at all points on the real line. Indeed, when $x \ne 0$ the "problem term" $\frac{1}{x}$ is differentiable, and so we are free to apply the chain/product rule: $$f'(x) = 2 x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)$$
The naive student might conclude that the derivative does not exist at $x = 0$ because this formula no longer makes sense in that region. On the contrary, if we appeal to the definition of a derivative we see that
$$f'(0) = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{h}\right)}{h} = \lim_{h \to 0} h \sin \left(\frac{1}{h}\right) = 0$$
by the squeeze theorem.

A similar fallacy is appearing in your solutions to problem 4 (c). In short, the failure of the method of characteristics, while certainly does indicate that some interesting phenomenon is occurring, does not imply that there isn't some other method which may be able to solve the problem. In particular, the main problem with 4 (b) is not so much the vanishing of the denominator in the arcsin/arctan term, but rather the fact that it is multivalued.

Here is a quick proof sketch that there is no "global" solution to 4 (b): Consider the circle $x^2 + y^2 = 1$. Rewriting the PDE as $$(x, -y) \cdot \nabla u = x^2 + y^2$$ we see that the characteristics curves are circles. Suppose we begin our journey at the point $(1, 0) \in \mathbb{R}^2$. As we move along the characteristic curve (ie. the circle of radius one) we see from the PDE that the function $u(x,y)$ is increasing at a rate of $x^2 + y^2$ which is exactly $1$ because we are moving along the unit circle! Hence, after a full revolution we will end up with a contradiction that $u(1, 0) = u(1, 0) + 2 \pi$, where the $2 \pi$ comes from integrating the constant $1$ around the unit circle (aka the circumference).

On the other hand, if we're careful about our domain, the method of characteristics can work. For example, you could solve an IVP of 4 (b) on the upper right quadrant if I gave you the initial conditions on the positive x-axis, ie. $u(x, 0) = \phi(x)$ for $x > 0$.

To summarize, if your characteristics curves "loop back on themselves" you need to ensure that the inhomogeneous term of the equation is "compatible" with curves. The function $xy$ worked, but $x^2 + y^2$ didn't.
« Last Edit: October 03, 2012, 11:41:29 PM by Martin MuÃ±oz »

#### Levon Avanesyan

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« Reply #4 on: October 04, 2012, 02:21:41 PM »

We marked only problems 1 and 4, using the rubric:

Problem 1 (10 marks): (a) 2, (b) 2, (c) 3, (d) 3
Problem 4 (10 marks): (a) 3, (b) 3, (c) 4

Does this mean that Problem 2 and 3 are not going to be graded? And also, when and where can we pick up our Home Assignments?

#### Victor Ivrii ##### Re: When HA marks will be available?
« Reply #5 on: October 04, 2012, 04:16:23 PM »

Does this mean that Problem 2 and 3 are not going to be graded? And also, when and where can we pick up our Home Assignments?

Yes, unfortunately we do not have enough TA office hours and it was explained from the beginning that home assignments will be only partially graded.

I believe that the HA1 for day section will be brought to either Monday or Wednesday class (you wrote your  Section on the paper, right?) Night section' papers will be available during my office hours October 10 Wed http://forum.math.toronto.edu/index.php?topic=46.0