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### Topics - Emily Deibert

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16
##### APM346--Lectures / Should we study section 2.8?
« on: October 17, 2015, 03:49:46 PM »
Hi Professor,

Please move this topic if it should be posted elsewhere. I am just wondering if section 2.8 will be covered on term test 1, since we didn't cover it in lecture notes or home assignments.

Thank you!

17
##### HA4 / HA4-P1
« on: October 09, 2015, 05:07:44 PM »
Problem 1: http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.P.html#problem-2.6.1

(a) Here we have a Dirichlet boundary condition. In the domain ${t>0, x \geqslant ct}$, the solution is given by the D'Alembert formula:

u(x,t) = \frac{1}{2}[g(x+ct) + g(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} h(y)dy

So plugging in the initial conditions as specified in the problem, we arrive at the solution:

u(x,t) = \phi(x+ct)

In the domain ${0<x<ct}$, the solution is given by:

u(x,t) = \frac{1}{2}g(x+ct) + \frac{1}{2c} \int_{0}^{x+ct} h(y)dy + p\bigg(t - \frac{x}{c}\bigg) - \frac{1}{2}g(ct-x) - \frac{1}{2c} \int_0^{ct-x} h(y)dy

So plugging in the initial and boundary conditions as specified in the problem, we arrive at the solution:

u(x,t) = \phi (x+ct) + \chi\bigg(t-\frac{x}{c}\bigg) - \phi(ct-x)

So the solution to (a) is

u(x,t) =
\begin{cases}
\phi(x+ct) & \text{for } \{t>0, x \geqslant ct\}\\
\phi (x+ct) + \chi\bigg(t-\frac{x}{c}\bigg) - \phi(ct-x) & \text{for } \{0<x<ct\}
\end{cases}

(b) In this case we have a Neumann boundary condition. The solution for the domain ${t>0, x \geqslant ct}$ is the same as in part (a), namely:

u(x,t) = \phi(x+ct)

For the domain ${0<x<ct}$, the solution is given by:

u(x,t) = \frac{1}{2}g(x+ct) + \frac{1}{2c} \int_0^{x+ct} h(y)dy - c \int_0^{t-\frac{x}{c}} q(t')dt' + \frac{1}{2}g(ct-x) + \frac{1}{2c} \int_0^{ct-x} h(y)dy

Plugging in our initial and boundary conditions as specified in the problem, we arrive at the solution:

u(x,t) = \phi (x+ct) - \phi (0) - cX\bigg(t-\frac{x}{c}\bigg) + cX(0) + \phi(ct-x)

where

\int\chi(t')dt' = X(t')

So the solution to (b) is

u(x,t) =
\begin{cases}
\phi(x+ct) & \text{for } \{t>0, x \geqslant ct\}\\
\phi (x+ct) - \phi (0) - cX\bigg(t-\frac{x}{c}\bigg) + cX(0) + \phi(ct-x) & \text{for } \{0<x<ct\}
\end{cases}

18
##### Quiz 2 / Solution to Quiz 2
« on: October 08, 2015, 10:57:27 PM »
Here is my solution. The process is the same as HA3-Problem2 (http://forum.math.toronto.edu/index.php?topic=627.0) so I will not post that. Please let me know if you notice any errors or typos.

Professor, please excuse my ugly graph---I do not know how to make one as nice as you did in Problem 2! (http://forum.math.toronto.edu/index.php?topic=627.0) I also don't know how to put the image in the post as you did, so I will attach it here.

19
##### Textbook errors / Broken link in web bonus problem week 4 #6
« on: October 08, 2015, 03:37:58 PM »

20
##### Textbook errors / Typo in HA3 P3?
« on: October 03, 2015, 05:54:34 PM »
I asked this in the board for HA3 P3 as well but I am cross-posting here.

Quote
Just wondering, is there a typo in the original problem? The equation given is:
Au_{tt} + 2Bu_{tx} + Cu_{tt}
But I think the last term should be with respect to x:
Au_{tt} + 2Bu_{tx} + Cu_{xx}

Indeed, corrected

21
##### Quiz 1 / Quiz 1 - P3
« on: October 02, 2015, 12:35:31 AM »
This problem was: \begin{cases}
u_x + 3u_y = xy \\
u|_{x=0} = 0
\end{cases}

And my solution is:
\frac{dx}{1} = \frac{dy}{3} = \frac{du}{xy}

3dx = dy

3x - y = C

du=xydx

du = x(3x-C)dx

du = (3x^2-Cx)dx

u = x^3 - \frac{C}{2}x^2 + \phi(3x-y)

u = x^3 - \frac{3x-y}{2}x^2 + \phi(3x-y)

With initial condition, we have:

u|_{x=0}=\phi(-y)=0

So: $$\phi = 0$$

So the final solution will be:
u = x^3 - \frac{3x-y}{2}x^2

22
##### Textbook errors / Error in HA 2 Problem 1?
« on: October 01, 2015, 01:22:09 PM »
Hi Professor,

HA2 Problem 1 a) has as the fifth problem,
u_x + x^3u_x = 0

Did you perhaps mean
u_t + x^3u_x = 0

to make an equation involving partial derivatives with respect to t AND x, like the other problems?

23
##### HA1 / Problem 5
« on: September 29, 2015, 08:40:23 PM »
5.
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter1/S1.P.html#problem-1.P.5
a)
u = y\int\Bigg(\int f(x)dx\Bigg)dx + \int\Bigg(\int g(x)dx\Bigg)dx + xh(y) + j(y)

b)
u = \int \int f(x,y)dydz + \int g(x,z)dx + h(y,z)

c)
u = sin(x)sin(y) + y\int \Bigg(\int f(x)dx \Bigg)dx + \int \Bigg(\int g(x)dx\Bigg)dx + xh(y) + j(y)

d)
u = -cos(x)cos(y)cos(z) + \int \int f(x,y)dydx + \int g(x,z)dx + h(y,z)

e)
u = -yzcos(x) - xzcos(y) - xycos(z) + \int \int f(x,y)dydx + \int g(x,z)dx + h(y,z)

24
##### HA1 / Problem 2
« on: September 29, 2015, 08:39:43 PM »
2.
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter1/S1.P.html#problem-1.P.2
a) 2nd order linear homogeneous
b) 2nd order nonlinear (quasilinear) homogeneous
c) 3rd order linear homogeneous
d) 3rd order nonlinear (quasilinear) homogeneous
e) 4th order linear homogeneous
f) 4th order linear homogeneous
g) 4th order linear inhomogeneous
h) 4th order linear homogeneous 4th order semilinear

25
##### HA1 / Problem 1
« on: September 29, 2015, 08:39:18 PM »
1.
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter1/S1.P.html#problem-1.P.1
a) Linear homogeneous
b) Nonlinear (quasilinear) homogeneous
c) Linear homogeneous
d) Nonlinear (quasilinear) inhomogeneous
e) Nonlinear homogeneous (though I am not sure, as this one seems like it could also be called inhomogeneous semilinear?) I think I have decided that this one is semilinear. Indeed
f) Nonlinear inhomogeneous
g) Nonlinear inhomogeneous
h) Nonlinear homogeneous Linear homogeneous
i) Nonlinear inhomogeneous
j) Nonlinear homogeneous

26
##### HA1 / Home Assignment 1 Solutions
« on: September 26, 2015, 08:29:51 PM »
EDIT: For clarity, I have created separate topics for each of the below problems.

Hello,

Here are my solutions to Home Assignment 1. Please comment with any mistakes or typos you notice!

1.
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter1/S1.P.html#problem-1.P.1
a) Linear homogeneous
b) Nonlinear (quasilinear) homogeneous
c) Linear homogeneous
d) Nonlinear (quasilinear) inhomogeneous
e) Nonlinear homogeneous (though I am not sure, as this one seems like it could also be called inhomogeneous semilinear?)
f) Nonlinear inhomogeneous
g) Nonlinear inhomogeneous
h) Nonlinear inhomogeneous
i) Nonlinear inhomogeneous
j) Nonlinear homogeneous

2.
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter1/S1.P.html#problem-1.P.2
a) 2nd order linear homogeneous
b) 2nd order nonlinear (quasilinear) homogeneous
c) 3rd order linear homogeneous
d) 3rd order nonlinear (quasilinear) homogeneous
e) 4th order linear homogeneous
f) 4th order linear homogeneous
g) 4th order linear inhomogeneous
h) 4th order linear homogeneous

3.
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter1/S1.P.html#problem-1.P.3
a)
u = \int f(x)dx + g(y)

b)
u = (Ce^{2y})x + f(y)

Incorrect. V.I.
c)
u = e^{x+y} + \int f(x)dx + g(y)

d) I am not very confident in this answer, so hopefully someone is able to explain this question to me!
u = \frac{e^xe^y}{(1-2y)}

4.
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter1/S1.P.html#problem-1.P.4
a)
u = e^{x + g(y)}

b)
u = e^{e^2x + g(y)}

u_{xy} = u_{y}u_{x} \Rightarrow
\frac{u_{xy}}{u_{x}} = u_{y} \Rightarrow
ln(u_{x}) = u_{y} \Rightarrow

5.
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter1/S1.P.html#problem-1.P.5
a)
u = y\int\Bigg(\int f(x)dx\Bigg)dx + \int\Bigg(\int g(x)dx\Bigg)dx + xh(y) + j(y)

b)
u = \int \int f(x,y)dydz + \int g(x,z)dx + h(y,z)

c)
u = sin(x)sin(y) + y\int \Bigg(\int f(x)dx \Bigg)dx + \int \Bigg(\int g(x)dx\Bigg)dx + xh(y) + j(y)

d)
u = -cos(x)cos(y)cos(z) + \int \int f(x,y)dydx + \int g(x,z)dx + h(y,z)

e)
u = -yzcos(x) - xzcos(y) - xycos(z) + \int \int f(x,y)dydx + \int g(x,z)dx + h(y,z)

Also, I am not very good with Latex so if anyone has any suggestions on how to improve any of my formatting please let me know. Thank you!

27
##### APM346--Lectures / Clarification about Textbook Section 2.1.2
« on: September 23, 2015, 11:38:10 PM »
Hi Professor,

In section 2.1.2 of the online textbook ("Constant Coefficients") you have written in the example:

"Consider initial value condition u|t = 0 = f(x). It allows us define Ï•: Ï•(x) = f(x)"

I'm wondering: do you mean that f(x) = 0? Or should the statement read:

Consider initial value condition u|t=0 = f(x)

implying that f(x) is the value of u where t = 0?

Thank you for the help!

28
##### MAT244 Math--Lectures / Should we know derivatives of hyperbolic functions?
« on: October 06, 2014, 11:15:20 PM »
Hi,

I've noticed in quite a few of the homework problems that we need to work with hyperbolic functions. I have never discussed hyperbolic functions in any math courses at U of T before (I've taken MAT135, MAT136, MAT235, and MAT223 prior to this course). Should I memorize the derivatives and properties of these functions for the test?

Thank you.

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