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Topics - Yifei Hu

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Chapter 2 / Textbook Chapter 2.5 Example 2
« on: February 21, 2022, 11:29:41 AM »
Hi Professor,

I think one coefficient is wrong in the solution.

In this problem the solution to the inhomogeneous equation with homogeneous initial condition should start with coefficient $\frac{1}{2c} = \frac{1}{6}$. After cancelling the 36 in the $f(x,t)$, we should arrive at $6\int_0^t \int_{x-3(t-t')}^{x+3(t-t')}\frac{1}{t^2+1}dx'dt'$ instead of $3\int_0^t \int_{x-3(t-t')}^{x+3(t-t')}\frac{1}{t^2+1}dx'dt'$.

Can you help me confirm if I missed something or the textbook has a typo?

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Chapter 2 / Text Book 2.4 Example 2.1
« on: February 19, 2022, 09:16:55 PM »
In this problem, $-1 < t < -\infty$, when applying D' Alembert's formula, why we integrate from 0 to t instead of from -1 to t?
Thanks in advance!

3
In text book 2.4, we had one line in the derivation: $\tilde{u_{\xi}}=-\frac{1}{4c^2} \int^\xi f(\xi,\eta')d\eta'=-\frac{1}{4c^2} \int_\xi^\eta\tilde f(\xi,\eta')d\eta' + \phi'(\xi)$.
Why can we replace the definite integral with the indefinite integral? Why we choose $\xi$ as lower limit and $\eta$ as upper limit?

4
Chapter 2 / S2.2 Q1
« on: February 02, 2022, 04:35:19 PM »
The problem asks for general solution of the equation $U_t+yU_x-xU_y=0; U(0,x,y)=f(x,y)$
I proceed as usual:
$$\frac{dt}{1}=\frac{dx}{y}=\frac{-dy}{x}=\frac{du}{0}$$
Integrate and this gives: $x^2+y^2=C$, $t-\int \frac{1}{\sqrt{c-x^2}}dx=D$
Hence, I conclude that $U=\phi(C,D)=\phi(x^2+y^2,t-\int \frac{1}{\sqrt{c-x^2}}dx)$. However, this involves an integral that I can not calculated by hand, can anyone give me a hint on how to do this integral?
Also, I see that in the solution we can also solve this system with a nice trigonometry form $U=f(xcos(t)-ysin(t),xsin(t)+ycos(t))$but the solution does not specify how to reach that, can anyone shed lights on how the solution is reached?
Thanks in advance.

5
Chapter 2 / Impose Initial Condition on Inhomogeneous Equations
« on: February 01, 2022, 11:34:15 AM »
Consider problem: $U_x+3U_y=xy, U(0,y)=0$, I proceed as follow:
$$\frac{dx}{1}=\frac{dy}{3}=\frac{du}{xy}$$
Integrate on first two terms:
$$x-\frac{1}{3}y=C \qquad \color{red}{(*)}$$
Integrate on $\frac{du}{xy} = \frac{dx}{1}$: Error! You must remember that in the equation of characteristics $x$ and $y$ are not independent but  connected by (*).
$$U =x^3-\frac{C}{2}x^2+D = \frac{3}{2}x^3+\frac{1}{2}yx^2+D $$
D must be constant along the integral curve hence $D=\phi(x-1/3y)$
Hence, the general solution is $U =\frac{3}{2}x^3+\frac{1}{2}yx^2+\phi(x-1/3y)$.
Impose initial condition: $U(0,y)=0$ we have $\phi(-1/3y)=0$.

Update:
Hi professor, I have fixed the integration part, but I still have question about the constant D when integrating on U. Should I include this $D=\phi(x-\frac{1}{3}y)$ in the solution to IVBP?

6
Chapter 2 / Ut+xUx=0
« on: February 01, 2022, 10:08:42 AM »
When solving this problem, I proceed as follow:
$$\frac{dt}{1}=\frac{dx}{x}=\frac{du}{0}$$
Hence, U does not depend on x and t, integrate on first part of equation:
$$t=ln(x)+C$$
I did not take exponential on both sides to get $e^t=Cx$ but I directly use $C=t-ln(x)$ and got $U=f(t-ln(x))$. Can anyone help me identify why this calculation is wrong?

7
Chapter 2 / f(x) in Method of Characteristics
« on: February 01, 2022, 09:47:20 AM »
In solving the problem $2U_t+3U_x=0, U(0,x)=f(x)$, tutorial gave answer $U=f(x-\frac{3}{2}t)$ while I have $U=f(3t-2x)$. Are both answers valid? We can use different form of C as long as it represent the same family of characteristics right?

8
Quiz 1 / Second Order PDE Classification on Quiz 1?
« on: January 23, 2022, 10:31:07 PM »
Hi professor, in homework 1, classification problem does not touch upon second order PDE classifications such as elliptic, parabolic and hyperbolic. Will these be tested on Quiz 1?

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Chapter 1 / HW1-Problem2
« on: January 20, 2022, 01:33:08 PM »
Can anyone help me check the answer? Thanks

(11) Second order, linear, homogeneous
(12) Second order, quasilinear
(13) Third order, linear, homogeneous
(14) Third order, quasi linear
(15) Forth order, linear homogeneous
(16) Forth order, linear homogeneous
(17) Forth order, linear inhomogeneous
(18) Forth order, general non-linear

Can we say that, to classify whether an equation is linear, we don't need to move the L.O.T to right hand side, but when classifying non-linear equations , we need to do so?

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Chapter 1 / HW1-Problem 1
« on: January 20, 2022, 01:17:24 PM »
Can anyone help me check the answer? Thanks

(1) linear-homogeneous
(2) Quasi-linear
(3) Semi-linear
(4)Quasi-linear
(5) Semi-linear
(6)Quasi-linear
(7)Quasi-linear
(8) Semi-linear
(9)Quasi-linear
(10) Quasi-linear

11
Chapter 2 / Transport Equation Derivation
« on: January 18, 2022, 07:34:32 PM »
Can anyone help explain where does the Ut term in the second last line come from? Thanks

12
Chapter 1 / Second Order canonical Form
« on: January 13, 2022, 02:37:35 PM »
What is the definition here (when classifying the second order PDEs) for the second order canonical form? what are the Xi and Eta here? Is the operation here defined as taking derivative? e.g: Eta^2 = second derivative of Eta?

13
Question: let w= 2-i , find w^3 + w
Are we suppose to do multiplications directly or are we suppose to use Euler's formula? Since in this case, \theta = arctan(-1/2), we can't directly come out the sin and cos of n \theta.
Are there any other alternative methods to apply to such complex numbers with a general arguments that can take advantage of Euler's formula's easy computations of power? Can we give the answer to this question as a polynomial of e^iarctan(c)?

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