Solution to PDE U(x,t) = y^2e^{-3x} + h(x)

[nicksauce]

I have a solution for a PDE

$$U(x,t)=y^2e^{-3x} + h(x)$$

Where h(x) is any function such that h(0) = 1

What notation can I use for this clause on h(x)?

My guess is:
$$h(x)\in \{f(x)|f(0)=1\}$$

Does that make sense?

 

[arildno]

Sure, it's a fancy-pants way of putting it.

 

[nicksauce]

Sure, it's a fancy-pants way of putting it.

I like doing things the fancy-pants way :tongue:

 

[DeadWolfe]

Well if you want to be fancy pants, you might well want to mention what sort of hypotheses are needed on h, ie there probably should be some differentiability condition.

 

[nicksauce]

True; is there a symbolic way to write 'continuous'?

 

[axeae]

$$\text{yeah, if } f(x) \text{ is continuous on an interval } [a,b] \text{ then } f(x) \in C[a,b]$$

 

[HallsofIvy]

I am just a tiny bit concerned that your formula,
$$U(x,t)=y^2e^{-3x} + h(x)$$
has U(x,t) on the left but a "y" and no "t" on the right!

 

[nicksauce]

I am just a tiny bit concerned that your formula,
$$U(x,t)=y^2e^{-3x} + h(x)$$
has U(x,t) on the left but a "y" and no "t" on the right!

Lol woops, should be U(x,y)

 

[axeae]

off topic but does anyone know why my latex doesn't work? i tried doing it with a bunch of little [tex] and no \text but that didnt work, so i tried what i have now and just gave up and left it like that.

 

[nicksauce]

Try using \textnormal

$$\textnormal{This is a test}$$

 

[bomba923]

You do not need \textnormal

[tex] \text{yeah, if } f(x) \text{ is continuous on an interval } [a,b] \text{ then } f(x) \in C[a,b] [\tex]

Simply finish with "/tex", not with "\tex", between the brackets

 

[axeae]

oh wow, i guess I've been doing latex so much i forgot not everything else uses \ instead of /