2266

**Home Assignment 1 / HA1-pdf**

« **on:**September 22, 2012, 01:40:44 PM »

Here is HA1.pdf - Home assignment 1 printed to pdf (updated Mon 24 Sep 2012 05:03:22 EDT)

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

2266

Here is HA1.pdf - Home assignment 1 printed to pdf (updated Mon 24 Sep 2012 05:03:22 EDT)

2267

Hi, for number 2, say we get $u = f(g(x,y))$ for our general solution when $(x,y) \ne (0,0)$. Then are we just finding a $u(0,0)$ that is equal to the limit of $u = f(g(x,y))$ as $(x,y)$ approaching $(0,0)$ in order to make u continuous at $(0,0)$?

Well, you need to be sure that this limit exists, right?

2268

I feel like I understand what is being said here, but I am confused by the second set of lecture notes, specifically equation (7) which seems to be doing the same thing as equation (4) in the reply to this post except with a + sign instead of a equal. Am I misunderstanding and these are actually different scenarios, or is one a misprint?

Miranda, thanks! There was a misprint (mistype) in Lecture 2 (equation (7)), should be =.

Now it has been corrected.

2269

In this settings you need to select a solution having certain properties

It may happen that

Your task is to determine what happens here.

It may happen that

- Is what one expects
- Unusually broad
- Unusually narrow

Your task is to determine what happens here.

2270

Thank you for your hint but I still didn't get the point..

For example if the general solution has the form $f(x/y)$, how can I make it continuous at (0,0)? Thanks!!

You are almost done (but check the general solutions!) Think - and don't post solutions!!!

2271

Heythere,

I am confused by the terms characteristic lines and integral lines. The book introduces characteristic lines as the curves along which a function is constant. Now in the notes integral lines are curves to which the vector field is tangential, i.e. in the case of the gradient vector field the lines along which the function changes most (in abs. value).

So I thought these two should be orthogonal in the case of $au_t+bu_x=0$.

So are integral lines the same as characteristic lines?

If we consider 1-st order PDEs in the form

\begin{equation}

a_0\partial_t u + a_1\partial_x u + a_2\partial_y u=0

\label{eq-1}

\end{equation}

then characteristics of the equation (\ref{eq-1}) are integral lines of the vector field $(a_0,a_1,a_2)$ i.e. curves

\begin{equation}

\frac{dt}{a_0}=\frac{dx}{a_1}=\frac{dy}{a_2}.

\label{eq-2}

\end{equation}

There could be just two variables $(t,x)$ or more ... and coefficients are not necessary constant. Yes, for (\ref{eq-1}) characteristics are lines along which $u$ is constant.

But we preserve the same definition of characteristics as integral lines for more general equation f.e.

\begin{equation}

a_0\partial_t u + a_1\partial_x u + a_2\partial_y u =f (t,x,y,u)

\label{eq-3}

\end{equation}

and here $u$ is no more constant along characteristics but solves ODE

\begin{equation}

\frac{dt}{a_0}=\frac{dx}{a_1}=\frac{dy}{a_2}=\frac{du}{f}.

\label{eq-4}

\end{equation}

Further, notion of characteristics generalizes to higher order equations. Definition

2272

2273

First, I need to apologize: this problem contained a misprint which I just corrected. The source of errors is simple: everything was done in the extreme rush and nobody checked it.

Now hint: it may happen in some problems that the solution (having certain properties) does not exist or must have a very special form. Then the your presented solution should state this.

I am not claiming that this is the case with this problem but your statement is wrong anyway: there is at least one continuous solution $u=0$ identically.

Now hint: it may happen in some problems that the solution (having certain properties) does not exist or must have a very special form. Then the your presented solution should state this.

I am not claiming that this is the case with this problem but your statement is wrong anyway: there is at least one continuous solution $u=0$ identically.

2274

This is just a test to see whether I can copy and paste from LyX:

u(x, t) = \varphi(x-ct)+\phi(x+ct)-\intop_{x-ct}^{x+ct}g(y)dy

Yes, you can but need to switch on math (inline or display respectively)

Code: [Select]

`$u(x, t) = \varphi(x-ct)+\phi(x+ct)-\intop_{x-ct}^{x+ct}g(y)dy$`

$u(x, t) = \varphi(x-ct)+\phi(x+ct)-\intop_{x-ct}^{x+ct}g(y)dy$Code: [Select]

`$$u(x, t) = \varphi(x-ct)+\phi(x+ct)-\intop_{x-ct}^{x+ct}g(y)dy$$`

$$u(x, t) = \varphi(x-ct)+\phi(x+ct)-\intop_{x-ct}^{x+ct}g(y)dy$$Actually I never saw \intop (just \int) and double dollars are deprecated (see my code), but your example works

2275

To both: for one of them solution does not exist.

There was an error in the left-hand expression, now it has been corrected

There was an error in the left-hand expression, now it has been corrected

2276

Testing how MathJax was hooked up

\begin{align*}

u(x,t)= &\underbracket{\frac{1}{2}\bigl[ g(x+ct)+g(x-ct)\bigr]+\frac{1}{2c}\int_{x-ct}^{x+ct} h(y)\,dy}_{=u_2}+\\[3pt]

&\underbracket{\frac{1}{2c}

\iint_{\Delta (x,t)} f(x',t' )\,dx\,d t' }_{=u_1}.

\label{eq-4}

\end{align*}

\begin{align*}

u(x,t)= &\underbracket{\frac{1}{2}\bigl[ g(x+ct)+g(x-ct)\bigr]+\frac{1}{2c}\int_{x-ct}^{x+ct} h(y)\,dy}_{=u_2}+\\[3pt]

&\underbracket{\frac{1}{2c}

\iint_{\Delta (x,t)} f(x',t' )\,dx\,d t' }_{=u_1}.

\label{eq-4}

\end{align*}