Solve for $a^2$ in $u_{xx} = u_{tt}$ using $u_1$ and $u_2$

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Discussion Overview

The discussion revolves around the differential equation $a^2 u_{xx} = u_{tt}$, specifically exploring whether the functions $u_1$ and $u_2$ can be considered solutions. Participants examine the implications of the parameter $a^2$ in the context of these functions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents the functions $u_1\left(x, t\right)=\sin\left({\lambda x}\right)\sin\left({\lambda at}\right)$ and $u_2 \left(x, t\right)={e}^{-a^2 \lambda ^2 t}\sin\left({\lambda x}\right)$, expressing uncertainty about how to handle $a^2$.
  • Another participant questions the intent behind "finding" the differential equation, asking if the goal is to solve it or to verify the functions as solutions.
  • A participant calculates $u_{xx}$ and finds it to be $-\lambda^2\sin\left({\lambda x}\right)$, noting that a term involving $\alpha$ has disappeared.
  • Another participant corrects this by stating that the term $\sin\left( \lambda a t \right)$ is treated as a constant when differentiating with respect to $x$.
  • One participant asserts that since $u_{xx}=-\lambda^2 \sin\left({\lambda x}\right)\sin\left({a\lambda t}\right)$ and $u_{tt}=-a^2 \lambda^2 \sin\left({\lambda x}\right)\sin\left({a\lambda t}\right)$, it follows that $a^2 u_{xx}=u_{tt}$.
  • A later reply confirms this assertion as correct and prompts further exploration of the problem.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the functions as solutions and the treatment of the parameter $a^2$. While some calculations are confirmed, the overall discussion remains unresolved regarding the broader implications of these findings.

Contextual Notes

There are limitations regarding the assumptions made about the functions and the treatment of constants during differentiation. The discussion also reflects uncertainty about the broader context of the problem.

karush
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Find $a^2 u_{xx} = u_{tt} $ if $\lambda$ a real number.
If
$u_1\left(x, t\right)=\sin\left({\lambda x}\right)\sin\left({\lambda at}\right)$
And
$u_2 \left(x, t\right)={e}^{-a^2 \lambda ^2 t}\sin\left({\lambda x}\right)$

I didn't know how to deal with the $a^2$
 
Last edited:
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karush said:
Find $a^2 u_{xx} = u_{tt} $ if $\lambda$ a real number.
If
$u_1\left(x, t\right)=\sin\left({\lambda x}\right)\sin\left({\lambda at}\right)$
And
$u_2 \left(x, t\right)={e}^{-a^2 \lambda ^2 t}\sin\left({\lambda x}\right)$

I didn't know how to deal with the $a^2$

What do you mean "find" this differential equation? Are you trying to solve it? Or are you trying to show that u1 and u2 are possible solutions?
 
Yes sorry
Are $u^1$ and $u^2$ possible solutions
On the calculator I got
$${u}_{xx}=-\lambda^2\sin\left({\lambda x}\right) $$
So $\alpha$ disappeared
 
Last edited:
karush said:
Yes sorry
Are $u^1$ and $u^2$ possible solutions
On the calculator I got
$${u}_{xx}=-\lambda^2\sin\left({\lambda x}\right) $$
So $\alpha$ disappeared

It can't disappear. The entire $\displaystyle \begin{align*} \sin{ \left( \lambda \, a\, t \right) } \end{align*}$ term is a constant when differentiating partially with respect to x.
 
Since $u_{xx}=-\lambda^2 \sin\left({\lambda x}\right)\sin\left({a\lambda t}\right)$

and

$u_{tt}=-a^2 \lambda^2 \sin\left({\lambda x}\right)\sin\left({a\lambda t}\right)$

Then $a^2 u_{xx}=u_{tt } $
I hope
 
karush said:
Since $u_{xx}=-\lambda^2 \sin\left({\lambda x}\right)\sin\left({a\lambda t}\right)$

and

$u_{tt}=-a^2 \lambda^2 \sin\left({\lambda x}\right)\sin\left({a\lambda t}\right)$

Then $a^2 u_{xx}=u_{tt } $
I hope

Yes that is correct :)

Can you do the other part of the question?
 

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