# Toronto Math Forum

## APM346-2012 => APM346 Math => Misc Math => Topic started by: Thomas Nutz on October 05, 2012, 03:57:42 PM

Title: proof of proposition 2a) in 9th lecture notes
Post by: Thomas Nutz on October 05, 2012, 03:57:42 PM
we are given
$$U(x,t)=\int_{-\infty}^xu(x,t)dx$$
and are to prove that $U$ satisfies $U_t=kU_{xx}$.

The proof given is "one can see easily that as $x\rightarrow -\infty$ and that therefore $U$ and all its derivatives have to be zero". But the integral over any function with the upper bound approaching the lower bound goes to zero!

For instance I take the function $f(x)=x^5$, which obviously does not satisfy the heat equation for $x\neq 0$. Then isn't
$$lim_{x\rightarrow -\infty}\int_{-\infty}^{x}x^5dx=0$$, and according to this "proof" $\int_{-\infty}^{x}x^5dx$ would satisfy the heat equation?

This does not make sense to me...

Thanks!
Title: Re: proof of proposition 2a) in 9th lecture notes
Post by: Victor Ivrii on October 05, 2012, 04:46:44 PM
we are given
$$U(x,t)=\int_{-\infty}^xu(x,t)dx$$
and are to prove that $U$ satisfies $U_t=kU_{xx}$.

The proof given is "one can see easily that as $x\rightarrow -\infty$ and that therefore $U$ and all its derivatives have to be zero". But the integral over any function with the upper bound approaching the lower bound goes to zero!

Thanks!

First, a correct citation:
"as $x\to -\infty$ $U$ is fast decaying with all its derivatives".

So, this is true for $R= U_u-kU_{xx}$, right? But we know that $R$ does not depend on $x$ (fact, your 'counterexample' misses) and  therefore $R=0$