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

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Discussion Overview

The discussion revolves around a proposed solution to a partial differential equation (PDE) represented as U(x,t) = y^2e^{-3x} + h(x), where h(x) is a function constrained by h(0) = 1. Participants explore notation for expressing conditions on h(x) and raise concerns about the formulation of the solution.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Meta-discussion

Main Points Raised

  • One participant suggests a notation h(x) ∈ {f(x) | f(0) = 1} to express the condition on h(x).
  • Another participant agrees with the notation but humorously refers to it as "fancy-pants."
  • A different participant raises the need for additional hypotheses on h(x), specifically mentioning differentiability conditions.
  • There is a query about how to symbolically represent continuity for functions.
  • One participant expresses concern about the presence of "y" in the equation while "t" is missing, leading to a correction of U(x,t) to U(x,y).
  • Another participant discusses issues with LaTeX formatting, seeking help on why their LaTeX does not work as intended.
  • Participants provide suggestions for LaTeX formatting, including the use of \textnormal and correcting the use of brackets.

Areas of Agreement / Disagreement

Participants generally agree on the need for proper notation and conditions for h(x), but there are differing opinions on the specifics of these conditions and the LaTeX formatting issues remain unresolved.

Contextual Notes

Participants have not fully resolved the requirements for h(x) beyond the initial condition, and there are ongoing discussions about the appropriate notation and formatting conventions.

nicksauce
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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?
 
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Sure, it's a fancy-pants way of putting it.
 
arildno said:
Sure, it's a fancy-pants way of putting it.

I like doing things the fancy-pants way :-p
 
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.
 
True; is there a symbolic way to write 'continuous'?
 
\text{yeah, if } f(x) \text{ is continuous on an interval } [a,b] \text{ then } f(x) \in C[a,b]
 
Last edited:
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!
 
HallsofIvy said:
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)
 
off topic but does anyone know why my latex doesn't work? i tried doing it with a bunch of little and no \text but that didnt work, so i tried what i have now and just gave up and left it like that.
 
  • #10
Try using \textnormal

\textnormal{This is a test}
 
  • #11
You do not need \textnormal
axeae said:
\text{yeah, if } f(x) \text{ is continuous on an interval } [a,b] \text{ then } f(x) \in C[a,b] [\tex]
<br /> Simply finish with &quot;/tex&quot;, not with &quot;\tex&quot;, between the brackets
 
  • #12
oh wow, i guess I've been doing latex so much i forgot not everything else uses \ instead of /
 

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