### Author Topic: HA3-P3  (Read 3394 times)

#### Victor Ivrii ##### HA3-P3
« on: September 28, 2015, 01:04:39 PM »

#### Zaihao Zhou

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« Reply #1 on: October 03, 2015, 01:45:04 PM »
Separate the operators and u:
\begin{equation} (A\partial_{t}^{2} + 2B\partial_{t}\partial_{x} + C\partial_{x}^{2})u = 0   \end{equation}
Provided \begin{equation} 4B^{2} - 4AC >= 0 \end{equation} we have roots r1, r2 s.t. \begin{equation} \partial_{t} = r_1\partial_x,  \partial_t = r_2\partial_x\end{equation}
if equal, then r1 = r2.

then we can transform the original equation to form: \begin{equation}(\partial_{t} - r_1\partial_x)(\partial_t - r_2\partial_x)u = 0\end{equation}
we can solve this equation in the way taught in class, namely substitute the partial part and u to v and w then get u. The answer is: \begin{equation} u = f(x+r_1t) + g(x+r_2t) \end{equation}
compare to the form given by the question, we have \begin{equation}r_1 = -c_1, r_2 = -c_2 \end{equation} are the roots.
a). so they satisfy \begin{equation} c_1 + c_2 = \frac{2B}{A}, c_1c_2 = \frac{C}{A}\end{equation}.
b). to have \begin{equation} c_1 < c_2\end{equation} we need \begin{equation} 4B^2 - 4AC = B^2 - AC > 0\end{equation}
« Last Edit: October 03, 2015, 02:50:04 PM by Zaihao Zhou »

#### Emily Deibert ##### Re: HA3-P3
« Reply #2 on: October 03, 2015, 05:52:31 PM »
Just wondering, is there a typo in the original problem? The equation given is: \begin{equation}
Au_{tt} + 2Bu_{tx} + Cu_{tt} \end{equation}
But I think the last term should be with respect to x: \begin{equation}
Au_{tt} + 2Bu_{tx} + Cu_{xx}
\end{equation}

#### Zaihao Zhou

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« Reply #3 on: October 03, 2015, 06:42:45 PM »
Just wondering, is there a typo in the original problem? The equation given is: \begin{equation}
Au_{tt} + 2Bu_{tx} + Cu_{tt} \end{equation}
But I think the last term should be with respect to x: \begin{equation}
Au_{tt} + 2Bu_{tx} + Cu_{xx}
\end{equation}
Yes it has to be a typo or the whole question does not make sense.

#### Yumeng Wang

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« Reply #4 on: October 03, 2015, 09:33:33 PM »
For (b), I am thinking (c1 + c2)2 - 4c1c2 = (c1 - c2)2 which is always larger or equal to zero. And plug in expression(11),  it equals to (c1 - c2)2.
Given A2 always larger than 0 (ignore case of equal 0) so B2- 4AC always larger or equal to 0. We dont need to emphasize on this condition.
Therefore, is it possible the answer is A â‰  0 ?? I am not sure of this.

#### Zaihao Zhou

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« Reply #5 on: October 03, 2015, 11:02:35 PM »
For (b), I am thinking (c1 + c2)2 - 4c1c2 = (c1 - c2)2 which is always larger or equal to zero. And plug in expression(11),  it equals to (c1 - c2)2.
Given A2 always larger than 0 (ignore case of equal 0) so B2- 4AC always larger or equal to 0. We dont need to emphasize on this condition.
Therefore, is it possible the answer is A â‰  0 ?? I am not sure of this.
We can't even write c1 and c2 if the relationship between A, B, C is unknown. The very fact that you have written both implicitly require 4B2- 4AC >= 0, that's why you concluded it always greater or equal to zero.

#### Victor Ivrii ##### Re: HA3-P3
« Reply #6 on: October 04, 2015, 04:21:46 PM »
We definitely can solve equation and find $c_1$ and $c_2$ no matter what $AC-B^2$ is but if $AC-B^2>0$ these $c_{1,2}$ are complex conjugate, solution in this form does not make sense and PDE is elliptic (rather tan hyperbolic) with completely different properties.

As $c_1=c_2 \iff AC-B^2=0$ equation sometimes is called parabolic and sometimes hyperbolic degenerate but anyway, it is not the same as $AC-B^2<0$.
« Last Edit: November 01, 2015, 07:59:40 AM by Victor Ivrii »