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Here's the question:

Let $f(z)$ be a function with a continuous second derivative, we define:
$u(x,t)=f(x+ct)$
It is known that:
$\frac{\partial^2u}{\partial t^2}-12\frac{\partial^2 u}{\partial x \partial t} +36\frac{\partial ^2 u}{\partial x^2}=0$
Find the value of $c$.

I have some ideas but I might be doing something illegal, here's what I'm thinking of:
$u_x(x,t)=f(x+ct)*1$
$u_t(x,t)=f(x+ct)*c$
Is this true?

Another thing I thought of doing is $\frac{\partial^2u}{\partial t^2}-12\frac{\partial^2 u}{\partial x \partial y} +36\frac{\partial ^2 u}{\partial x^2}=(\frac{\partial u}{\partial t}-6 \frac{\partial u}{\partial x})^2$
I would appreciate an approval, the most thing bugging me is the $(a+b)^2=a^2+2ab+b^2$, in my equation the $2ab$ would be $\frac{\partial u}{\partial t}\frac{\partial u}{\partial x}$, which is opposite, but I'm not sure if this product of derivatives is equal to the main $2ab$ in the question which is $\frac{\partial u^2}{\partial x \partial t}$

Any feedback and corrections are really appreciated.

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1 Answer 1

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just using chain rule gives you

\begin{align} \frac{\partial u(x,t)}{\partial x} &= f'(x+ct) \\ \frac{\partial u(x,t)}{\partial t} &= cf'(x+ct) \end{align}

so for your equation it is

\begin{align} 0&= \frac{\partial^2u}{\partial t^2}-12\frac{\partial^2 u}{\partial x \partial t} +36\frac{\partial ^2 u}{\partial x^2} \\ &= (c^2-12c+36)f''(x+ct) \end{align}

I think you can take it from here ;)

EDIT: your factorizing approach is very nice too, but you have it slightly wrong. The correct equation is: $$\frac{\partial^2u}{\partial t^2}-12\frac{\partial^2 u}{\partial x \partial y} +36\frac{\partial ^2 u}{\partial x^2} =\left(\frac{\partial^2}{\partial t^2}-12\frac{\partial^2 }{\partial x \partial y} +36\frac{\partial ^2 }{\partial x^2}\right)u =\left(\frac{\partial }{\partial t}-6 \frac{\partial }{\partial x}\right)^2u$$ The thing you are factorizing is the differential operator itself, and not the derivatives of a specific $u$.

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  • $\begingroup$ Thanks alot! I have a question about the first approach, when you reached $(c-6)^2 f''(x+ct)=0$, how do I explain that $c$ must be $6$? or in another words how do I prove that $f''(x+ct)\ne0$? $\endgroup$
    – Pwaol
    May 19, 2021 at 16:07
  • $\begingroup$ you cant. If $f''$ is zero everywhere, there is no way to determine $c$ because all expressions we wrote are completey zero. But it is sufficient if you have a single point where $f''$ is non-zero. $\endgroup$
    – Simon
    May 19, 2021 at 16:13
  • $\begingroup$ I see, then would you say the question is missing that info? I'm still finding it hard to really (do tricks with differential operator) so I'm not sure if the factorization gives a solution for that. (can I say that $\frac{\partial u}{\partial t} = 6\frac{\partial u}{\partial x}$ from the factorization?) $\endgroup$
    – Pwaol
    May 19, 2021 at 16:27
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    $\begingroup$ (1) yeah, the question is missing that info (the only $f$ with $f''=0$ everywhere are linear functions of the form $f(x)=mx+b$. Any other function works). (2) Well, using the factorization of linear operators is a little tricky if you want to be mathematically rigorous. for a completey arbitrary operator $T$, the fact that $T^2u=0$ does not imply $Tu=0$. You should think about it like a linear-algebra problem with a (infinitely-big) matrix $T$. So you need to investigate the kernel of $T$ more closely. But for the purposes of this question I think my first approach is the way to go. $\endgroup$
    – Simon
    May 19, 2021 at 16:37

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