Function

[Anonymous No.1982926]

Hey /<span class="math">\psi[/spoiler]/,

I need any function <span class="math">r\left( x,t\right)[/spoiler] that satisfies the following:
<div class="math">
r\left( 0,t\right) = 0, \; \frac{\partial r}{\partial x}\left( L,t\right) +h\left[ r\left( L,t\right) - B\left( t\right)\right] = 0.
</div>
It shouldn't be too hard, yet I can't seem to find one :/

Thanks in advance.

[Anonymous No.1982933]

what's B(t) ? h ? L ?

[Anonymous No.1983014]

>>1982933
That doesn't matter: they are all independent of <span class="math">x[/spoiler].

[Anonymous No.1983045]

I dunno, lol, try some superposition of sine and cosine?

[Anonymous No.1983050]

I got
r(x, t) = -B(t)*(-1+exp(-h*x))

When changing all r(L,t) to r(x,t). It's too specific, you know.

[Anonymous No.1983089]

>>1983050
You're a genius, this one works. I just need one function to satisfy the given conditions as it is an intermediate step in solving a PDE with inhomogeneous boundary conditions.

Thanks dude!

[Anonymous No.1983098]

>>1983089
I just put it in Maple. If it's only an intermediate step, shouldn't you be able to do it yourself?

[Anonymous No.1983105]

so you gave us boundary conditions without specifying what happens in the middle?

Here's something:
<span class="math">r(x,t) = \begin{cases}0 & x = 0\\ B(t) & x > 0 \end{cases}[/spoiler]

[Anonymous No.1983107]

>>1983105
that should be:

<span class="math">r(x,t) = 0[/spoiler] if <span class="math">x = 0[/spoiler]
<span class="math">r(x,t) = B(t)[/spoiler] otherwise

[Anonymous No.1983188]

>>1983098
Yeah, well I hadn't got to use Maple yet and previous exercises had very simple prescriptions: I thought I was missing a very simple thing but it turns out the function isn't very trivial.
>>1983107
That function isn't properly differentiable due to the fact that it's not continuous. I don't see it satisfying the second condition anyway. Thanks for trying though.