Messages

16

Quiz-3 / Re: Q3-T5102

« on: February 11, 2018, 01:35:32 AM »

Given:

$$ u_{tt} - c^2u_{xx} = 0 $$ with $ t > 0, x> 0 $
$$u|_{t=0} = g(x) ,  u_t|_{t=0} = cg'(x) $$ with $ x > 0 $
$$  (u_x + \alpha u_t)  |_{x=0} = 0 $$ for $ t> 0 $

Generally:
$$ u(x,t) = \phi(x+ct) + \psi(x-ct) $$
The initial conditions lead to:
$$ \phi(x) = g(x) - \frac{g(0)}{2} $$
$$ \psi(x) = \frac{g(0)}{2} $$

It follows for the nicer region ($x > ct $) we have:

$$u(x,t) = \phi(x+ct) + \psi(x-ct) = g(x+ct) -\frac{g(0)}{2} + \frac{g(0)}{2} = g(x+ct) $$

While for the case of ($0<x<ct$) means that we have to investigate the behaviour of \psi($\gamma $) with $\gamma = x - ct < 0 $

Applying the boundary condition of  $  (u_x + \alpha u_t)  |_{x=0} = 0 $

Resulting in :

$$ 0 = \phi'(ct) +\psi'(-ct) + \alpha (c \phi'(ct) - c\psi'(-ct)) $$

Rearranging, and applying the substitution of $ x =  -ct $ this gives us:

$$ \psi '  (x)  = \phi ' (-x) \frac{ (1 + \alpha)}{\alpha - 1 } $$

This is where I usually have trouble with these sorts of problems, where I don't know where/if I am allowed to integrate both sides/ use the original expressions for $\phi$ and $\psi (x) $ above. Although I do know that I should try to manipulate it into the form of an ODE.

Tentatively, I'd start by taking $ \phi'(x) - \psi'(x) = g'(x) $  rearranging to:
$$ \phi'(x) = g'(x) + \psi'(x) $$

the issue is that substituting this expression into the above results in results in an expression of $\psi'(x)$ and $\psi'(-x)$ , so clearly this tentative step is incorrect, but I am unsure of what to do next.

b. Will be discussed later

17

Web Bonus Problems / Re: Web Bonus Problem –– Week 1

« on: January 07, 2018, 07:56:11 PM »

Would one approach to prove necessity would be to simply plug the results and show that the remaining terms demand that $f_{yy} = g_{xx}$


$ u_{xx} = f $ (1) and to keep things general, let's then write $u_{x} = \int f dx + \phi(y) $ and $u = \int\int f dxdx + x\phi(y) + \psi(y) $ (I). Then $  u_{y} = \int\int f_y dxdx + x\phi_{y}(y) + \psi_y(y) $ and $u_{yy} = \int\int f_{yy}dxdx + x\phi_{yy}(y) + \psi_{yy}(y) = g $
Then $u_{yy} = g $ , $u_{y} = \int g dy + \alpha(x) $ , $ u = \int \int g dydy  + y\alpha(x) + \beta(x) $ , $ u_x = \int\int g_x dydy + y\alpha_x (x) + \beta_x (x) $ , $ u_{xx} = \int\int g_{xx}dydy + y\alpha_{xx} (x) + \beta_{xx}(x) = f $

From there: $ f_{yy} =  u_{xxyy} = \partial{y}\partial{y} \int\int g_{xx}dydy +\partial{y}\partial{y} (y\alpha_{xx} (x)) + \partial{y}\partial{y}\beta_{xx}(x)  = g_{xx} + 0 $
and $ g_{xx} = u_{yyxx} = \partial{x}\partial{x} \int\int f_{yy}dxdx + \partial{x}\partial{x} (x\phi(y)_{yy} + \psi(y)_{yy}) = f_{yy} + 0  $

Or as Jingxuan said

$u_{xxyy}=f_{yy}=u_{yyxx}=g_{xx}$.

And we can conclude that if $f_{yy} \neq g_{xx} $ then we wouldn't be able to get this result?

18

Web Bonus Problems / Re: Web Bonus Problem –– Week 1

« on: January 05, 2018, 05:01:52 PM »

I've given the problem a try, and I'm still trying to figure out how to reconcile my results.

What I did:

Given $ u_{xx} = 0 $ (1) and $ u _{yy} = 0 $ (2), then let's work on (1) first:

$ u_{xx} = v_{x} = 0 $ which implies that $ v = \phi (y) $, then $ u_{x} = \phi (y) $ and $ u(x,y) = x\phi(y) + \psi(y) $

but then there is (2)

$ u_{yy} = w_{y} = 0 , w = \alpha(x) $ then $ u_{y} = \alpha(x), u(x,y) = y \alpha(x) + \beta(x) $

So comparing our two solutions for u:
(I) : $ u (x,y)  = x\phi(y) + \psi(y) $
(II) : $ u(x,y) = y\alpha(x) + \beta(x) $

Could we simply conclude that in this family of solutions, $\phi(y) = ay$ and similarly $\alpha(x) = bx$. Furthermore, $\psi(y) = \beta(x)$ , and since we know those terms go to 0 when the first partial is taken (see above), can we just make them an arbitrary constant?

(Note: a,b are some arbitrary constant)

Am I on the right track, or am I making too many assumptions?