Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Edit: the images are from the paper by Sigurd Angenent called Parabolic Equations for Curves on Surfaces. You shouldn't need any more information to answer the question (I think..)

They define the spaces: enter image description here

And use this argument: enter image description here

I have a confusion with the spaces they use. What does $u_{a,b}$ depend on $C^{m-1, 1}$ on a and b mean? I understand if it's just $C^{m-1}$ on $a$ and $b$ means that it's continuously differentiable $m-1$ times in $a$ and $b$, but I don't know about the other. Can someone explain how they can make that statement regarding the first equation involving the partial derivatives? How does that imply the next equation and show full smoothness?


The old version of this post:

I am reading a book called "The Curve Shortening Problem" by Kai Seng Chou and Xi-Ping Zhu.

It has a result that states that if $u$ solves the PDE $$u_t = F(x,t,u,u_x,u_{xx})$$ for $F$ smooth and $u(x,0) \in C^{k+2}$ for some $k$, then also $u \in C^{k+2}$. Also, if $u_0$ depends smoothly on a parameter, then so does $u$.

I am trying to understand a proof that this implies that the solution to the PDE automatically belongs to $C^\infty$ ("instant smoothness").

Their argument is as follows.


For any solution $u$, set $$u_{a,b} = u(x+at, bt),$$ then this solves the PDE $$v_t = F_{a,b}(x, t, v, v_x, v_{xx}) = av_x + bF(x + at, bt, v, v_x, v_{xx})$$ and $F_{a,b}$ depends smoothly on $(a,b)$. According the result above, $$\frac{\partial^{j+i}u_{a,b}}{\partial a^j \partial b^i}\bigg|_{(a,b) \equiv (0,1)} = t^{j+i}\frac{\partial^{j+i}u}{\partial x^j \partial t^i}\bigg|_{(x,t)}$$ (actually they write $k$ instead of $i$ here, but I think that might be a mistake?). Anyway they conclude with that "$u$ is smooth for $t > 0$."


Can someone explain this? How does this give me control of higher derivatives since $u_{a,b}$ is defined in terms of $u$..

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.