2

I've calculated a lot and checked the first derivatives with wolframalpha. Still I'm not sure if I have done everything correctly, could someone have a look please?

Original PDE: \begin{align*} u_t- \frac{\lambda}{2}u_{zz}=0 \end{align*} New coordinate: \begin{align*} y_i=\frac{1}{\pi} \arctan(\gamma z+c)+\frac{1}{2} \Leftrightarrow z=\frac{\tan(\frac{1}{2}(2 \pi y- \pi))-c}{\gamma} \end{align*} First derivatives: \begin{align*} y_z= \frac{\gamma}{\pi((c+\gamma z)^2+1)}\\ z_y=-\frac{2 \pi}{\gamma (\cos(2 \pi y)-1)} \end{align*} u differentiated using chaining rule, derivation of differential operators: \begin{align*} u_z=u_y y_z=u_y \frac{\gamma}{\pi((c+\gamma z)^2+1)}\\ \Rightarrow \frac{\partial}{\partial z}= \frac{\partial}{\partial y}\frac{\gamma}{\pi((c+\gamma z)^2+1)} \end{align*} \begin{align*} u_y= u_z z_y = u_z (-\frac{2 \pi}{\gamma (\cos(2 \pi y)-1)})\\ \Rightarrow \frac{\partial}{\partial y} = \frac{\partial}{\partial z} (-\frac{2 \pi}{\gamma (\cos(2 \pi y)-1)}) \end{align*} To ease notiation \begin{align*} A:= y_z= \frac{\gamma}{\pi((c+\gamma z)^2+1)}. \end{align*} 2nd derivative using product rule: \begin{align*} u_{zz} &= \frac{\partial}{\partial z} u_z = \frac{\partial}{\partial z} (u_y \frac{\gamma_i}{\pi((c+\gamma z)^2+1)})\\ &=A(\frac{\partial}{\partial y}(u_y A)) =A(u_{yy} A+u_yA_y)=A(u_{yy}A+u_y\cdot 0)=A^2 u_{yy} \end{align*} Inserting into original PDE: \begin{align*} u_t- \frac{\lambda}{2}A^2 u_{yy}=0 \Leftrightarrow u_t - \frac{\lambda}{2} (\frac{\gamma}{\pi((c+\gamma z)^2+1)})^2 u_{yy}=0 \end{align*} Now $z=\frac{\tan(\frac{1}{2}(2 \pi y- \pi))-c}{\gamma}$ should be inserted into the equation.

Thank you in advance! Nina

Nina
  • 21
  • I've at least displayed the image on the same page, but if you wrote it in $\LaTeX$ you'll find you can easily copy the code here, between $...$ or $$...$$ for displayed formulas. This isn't foolproof since MathJax isn't $\LaTeX$, but what you've written looks compatible. On a different matter, I'm curious as to why you would use this change of variables? – GPerez Jun 05 '15 at 15:01
  • Thanks for editing my Post. The computational domain of the original equation is unbounded and I want to avoid using artificial boundaries, so I want to map it to (0,1). I've tried to copy my latex into the post with dollardollar...dollardollar, but it doesn't recognize the align-environments, so if that's not somehow against the rules I would just keep it as a picture. – Nina Jun 05 '15 at 15:12
  • It's much preferable to post it as code because image links can break, and also it just looks nicer! Why don't you edit the plain code in? I might be able to make it work (but while I'm doing so I'll rollback the edit to the image version so people can still see the question). – GPerez Jun 05 '15 at 15:17
  • Didn't know it would just work like that :) Seems to be fine now, I've deleted the picture. Thanks for your help. – Nina Jun 05 '15 at 15:34
  • Okay now another question, what are the sub-indices $i$ for? Do you have more than just $(t,z)$ as variables? – GPerez Jun 05 '15 at 15:40
  • z is actually a vector, i should have put $z_i$ everywhere in the equations. So there's only t and z as variables for this equation. I happen to have other equations where I need other components of z and that is some kind of leftover. I'll remove the subindices. – Nina Jun 05 '15 at 15:55
  • You made a mistake, where you said $A_y=0$, because while your expression only depends on $z$, $z$ in turn does depend on $y$. – GPerez Jun 09 '15 at 13:54

0 Answers0