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### Messages - Shentao YANG

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16
##### TT1 / Re: TT1-P2
« on: October 19, 2016, 10:54:06 PM »
My solution is:
(a)
$$u(x,t) = - {e^{ - {{{x^2}} \over 2}}}$$
As:
$$u = {1 \over 2}( - {e^{ - {{{{(x + t)}^2}} \over 2}}} - {e^{ - {{{{(x - t)}^2}} \over 2}}}) + \underbrace {{1 \over {2c}}\int_0^t {\int_{x - t + t'}^{x + t - t'} {({y^2} - 1)} } {e^{ - {{{y^2}} \over 2}}}dydt'}_{(3)}$$
The inner integral of $(3)$ yields:
$$\int_{x - t + t'}^{x + t - t'} {{y^2}} {e^{ - {{{y^2}} \over 2}}} - {e^{ - {{{y^2}} \over 2}}}dy = \underbrace {\int_{x - t + t'}^{x + t - t'} {{y^2}} {e^{ - {{{y^2}} \over 2}}}dy}_{(4)} - \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy$$
$$(4) = \int_{x - t + t'}^{x + t - t'} y {e^{ - {{{y^2}} \over 2}}}ydy = \left. { - {e^{ - {{{y^2}} \over 2}}}y} \right|_{y = x - t + t'}^{y = x + t - t'} + \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy$$
Therefore
$$\int_{x - t + t'}^{x + t - t'} {{y^2}} {e^{ - {{{y^2}} \over 2}}} - {e^{ - {{{y^2}} \over 2}}}dy = \left. { - {e^{ - {{{y^2}} \over 2}}}y} \right|_{y = x - t + t'}^{y = x + t - t'} + \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy - \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy = (x - t + t'){e^{ - {{{{(x - t + t')}^2}} \over 2}}} - (x + t - t'){e^{ - {{{{(x + t - t')}^2}} \over 2}}}$$
so $(3)$ becomes:
$${1 \over 2}\int_0^t {(x - t + t'){e^{ - {{{{(x - t + t')}^2}} \over 2}}} - (x + t - t'){e^{ - {{{{(x + t - t')}^2}} \over 2}}}} dt'\matrix{ {} & {} \cr } (c = 1)$$
$$= {1 \over 2}\int_{x - t}^x {z{e^{ - {{{z^2}} \over 2}}}} dz + {1 \over 2}\int_{x + t}^x {z{e^{ - {{{z^2}} \over 2}}}} dz$$
$$= - {e^{ - {{{x^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x - t)}^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x + t)}^2}} \over 2}}}$$
Finally,
$$u = {1 \over 2}( - {e^{ - {{{{(x + t)}^2}} \over 2}}} - {e^{ - {{{{(x - t)}^2}} \over 2}}})- {e^{ - {{{x^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x - t)}^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x + t)}^2}} \over 2}}}$$
$$\Rightarrow u(x,t) = - {e^{ - {{{x^2}} \over 2}}}$$
(b) Since u does not depend on t, so we basically get the same equation:
$$\mathop {\lim }\limits_{t \to \infty } = u(x,t) = - {e^{ - {{{x^2}} \over 2}}}$$

17
##### Chapter 2 / Re: HA #4 general
« on: October 18, 2016, 09:13:21 PM »
I guess the meaning is: even if you take the constant as $0$ (which indeed should be $\pm {c \over 2}$ for some constant $c$), $u$ function you get will still satisfy the defining equations in the problem, hence, be a solution.
And, in fact, I think you will always get the same solution to $u$ for any integration constant you put (since the constant in $\psi$ and $\phi$ will cancel each other).

18
##### Chapter 3 / Re: A tentative solution to Assignment 5, Problem3(g)
« on: October 16, 2016, 04:43:13 PM »
Your function $w(x,t)$ is not continuous, so if you solve for $u$ in terms of $w$, it wouldn't be continuous either.

Yes, on each subinterval will be continuous, but not the whole. However, when I applied this same method on Problem3(f), the solution is really continuous (hopefully I am correct, you may check it yourself), so I am not sure whether this discontinuousness is the "nature" of this particular problem or just because my method is wrong...

There is a section in the textbook (http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter3/S3.2.html#sect-3.2.2) regarding inhomogeneous boundary conditions, but this has neither been covered in lectures nor tutorials.

Yes, I am thinking of a way to do it alternatively, that's why I titled it as "a tentative solution"...

By the way, can any one tell me will this kind of inhomogeneous boundary conditions problems be tested? (since it is not covered in both lectures and tutorials)

19
##### Chapter 3 / Re: A tentative solution to Assignment 5, Problem3(g)
« on: October 16, 2016, 12:15:00 PM »
By the way, I guess there are typos in problem (4):
The $g(t)$ in equation (10) should be $g(x)$ and the r.h.s. of equation (11) should be $h(t)$.

20
##### Chapter 3 / A tentative solution to Assignment 5, Problem3(g)
« on: October 16, 2016, 12:12:08 PM »
The problem is here: http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter3/S3.2.P.html
Here is my tentative solution, hope everyone for correction:
consider:
$$w(x,t) = \left\{ {\matrix{ {u(x,t) - 1} & {0 < x < \infty } & {t < 1} \cr {u(x,t)} & {0 < x < \infty } & {t > 1} \cr } } \right.$$
Then we have:
$${w_t} = {u_t}, {w_{xx}} = {u_{xx}}$$
Therefore:

$$\matrix{ {{w_t} = k{w_{xx}}} & {t > 0} & {x > 0} \cr {w\left| {_{t = 0} = u(x,0) - 1 = - 1} \right.} & {} & {x > 0} \cr {w\left| {_{x = 0} = \matrix{ {\left\{ {\matrix{ {u(0,t) - 1 = 1 - 1 = 0} & {t < 1} \cr {u(0,t) = 0} & {t > 1} \cr } } \right\}} & { = 0} \cr } } \right.} & {t > 0} & {} \cr }$$
which is the Dirichlet boundary condition.
Solving this problem wrt $w(x,t)$, I get (The calculation may be wrong):
$$w = 1 - erf({x \over {\sqrt {4kt} }})$$
Therefore:
$$u = \left\{ {\matrix{ {2 - erf({x \over {\sqrt {4kt} }})} & {x > 0} & {0 < t < 1} \cr {1 - erf({x \over {\sqrt {4kt} }})} & {x > 0} & {t > 1} \cr } } \right.$$
However, in this case $u(x,t)$ is not continuous at $t = 1$.
Hope everyone for correction / verification.

21
##### Chapter 3 / Re: Are there typos?
« on: October 10, 2016, 06:41:32 PM »
Quote
Quote
By the way, I cannot understand why there is a $k$ in this equation, can any one explain it to me?
Factor $k$ comes from the same factor in $ku_{xx}$ term in the heat equation.
I guess I know this $k$ is the same factor in $ku_{xx}$ term in the heat equation, I am curious where / when this $k$ is introduced in the derivation of the final formula (I cannot see any hint from the textbook...)

By the way, I guess the two unlabeled equations above http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter3/S3.2.html#mjx-eqn-eq-3.2.14 should all have zero in the position of $y$. I guess you have already plug in the value $y = 0$ inside the integral.

22
##### Chapter 3 / Are there typos?
« on: October 09, 2016, 01:29:01 PM »
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter3/S3.2.html#mjx-eqn-eq-3.2.15
I guess the second term should be ${G_N}(...)$ instead of ${G_N}_y(...)$. As far as I can understand, we have already cancelled out the ${G_N}_y(...)$ term under the context of Neumann Boundary condition.
By the way, I cannot understand why there is a $k$ in this equation, can any one explain it to me?

http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter3/S3.2.html#mjx-eqn-eq-3.2.24
I guess this one should be
${(4\pi kt)^{{{ - n} \over 2}}}$ instead of ${(2\sqrt {\pi kt} )^{{{ - n} \over 2}}}$

23
##### Q1 / Re: Q1-P1
« on: September 29, 2016, 09:43:40 PM »
Below is my solution:
$$u_t+xu_x-u= 0\text{ : linear homogeneous}$$
$$u_x^2+u_y^2-1= 0\text{ : nonlinear}$$

24
##### Chapter 2 / Re: Solving the Hopf equation
« on: September 23, 2016, 10:10:38 PM »
Here is my thinking:

Since u is a constant along the integral curve, then in solving
$$\frac{dt}{1}=\frac{dx}{u}.$$
we can treat the u as a constant, since the above equation already ensures / implies that u is on the integral line.
Therefore, we basically have
$$dx = u dt\implies\ x - ut = c .$$(probably you can think of u as some constant like 1, 2, 3...)
And noted that $$u = f(c)$$
for some function f of one variable.
Therefore, by combining the above two equation, we have: $$u = f (x-ut).$$
Which is the answer that Prof. Ivrii gave us in the class.
Waiting for Prof. Ivrii's correction...

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