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I am given the equation $$ u(x,t)_t = u(x,t)_x $$ and i have to apply coordinate transformation with $$ x=x(\xi,\theta), \quad t=\theta $$ to get an equation of the form $$ \alpha u_{\theta} +\beta x_{\theta} = \gamma u_{\xi}$$

I am having trouble applying the chain rule here. I tried: $$ \frac{\partial u(x,t)}{\partial t} = \frac{\partial u(\xi,\theta)}{\partial \xi} \frac{\partial \xi}{\partial t} + \frac{\partial u(\xi,\theta)}{\partial \theta} \frac{\partial \theta}{\partial t} = u_{\xi}\frac{\partial \xi}{\partial t} + u_{\theta}\cdot1 $$ and $$ \frac{\partial u(x,t)}{\partial x} = \frac{\partial u(\xi,\theta)}{\partial \xi} \frac{\partial \xi}{\partial x} + \frac{\partial u(\xi,\theta)}{\partial \theta} \frac{\partial \theta}{\partial x} = u_{\xi}\frac{\partial \xi}{\partial x} + u_{\theta}\cdot0 $$ Thanks for the answer.

My next problem is the Jacobian determinant because i don't know how to to interprete the coordinate transformation. It is not as usual, e.g., $(x,y) \to (u,v)$ or $(x,y) \to (r,\phi)$. The matrix is probably as follows $$ \begin{pmatrix} \frac{\partial x}{\partial x(\xi,\theta)} & \frac{\partial x}{\partial \theta} \\ \frac{\partial t}{\partial x(\xi,\theta)} & \frac{\partial t}{\partial \theta} \end{pmatrix} = \begin{pmatrix} \frac{\partial x}{\partial x(\xi,\theta)} & x_{\theta} \\ \frac{\partial t}{\partial x(\xi,\theta)} & 1 \end{pmatrix} $$ but how do i interprete $\frac{\partial x}{\partial x(\xi,\theta)}$ and $\frac{\partial t}{\partial x(\xi,\theta)}$?

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The answer is as Joriki stated.

With new transformations $$u(x,t)\Rightarrow u\big(x(\xi,\theta),t(\theta)\big)$$ The new partial derivatives are $$\frac{\partial u}{\partial \xi}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \xi}\qquad (1)$$ $$\frac{\partial u}{\partial \theta}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial t}\frac{dt}{d\theta}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial t}\qquad (2)$$ From $(1)$ it follows that $$\frac{\frac{\partial u}{\partial \xi}}{\frac{\partial x}{\partial \xi}}=\frac{\partial u}{\partial x}\qquad (3)$$ If we substitute $(3)$ into $(2)$ $$\frac{\partial u}{\partial \theta}=\frac{\frac{\partial u}{\partial \xi}}{\frac{\partial x}{\partial \xi}}\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial t}\Rightarrow \frac{\partial u}{\partial t}=\frac{\partial u}{\partial \theta}-\frac{\frac{\partial u}{\partial \xi}}{\frac{\partial x}{\partial \xi}}\frac{\partial x}{\partial \theta}\qquad (4)$$ And the original equation becomes $${\frac{\partial u}{\partial x}}={\frac{\partial u}{\partial t}}\Rightarrow \frac{\frac{\partial u}{\partial \xi}}{\frac{\partial x}{\partial \xi}}=\frac{\partial u}{\partial \theta}-\frac{\frac{\partial u}{\partial \xi}}{\frac{\partial x}{\partial \xi}}\frac{\partial x}{\partial \theta}$$ $$\frac{\partial u}{\partial \xi}=\frac{\partial u}{\partial \theta}\frac{\partial x}{\partial \xi}-\frac{\partial u}{\partial \xi}\frac{\partial x}{\partial \theta}$$

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Coordinate transformation is given by $$x=x(\xi,\theta)$$ $$t=t(\theta)=\theta$$ the the Jacobian can be defined as $$J(\xi,\theta)=\begin{pmatrix} \frac{\partial x}{\partial \xi} & \frac{\partial x}{\partial \theta} \\ \frac{\partial t}{\partial \xi} & \frac{\partial t}{\partial \theta} \end{pmatrix}=\begin{pmatrix} \frac{\partial x}{\partial \xi} & \frac{\partial x}{\partial \theta} \\ 0 & 1 \end{pmatrix}$$

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  • $\begingroup$ Thank you. Nicely written. Do you maybe have some thoughts about the Jacobian. I added something to my question. $\endgroup$ – UrošSlovenija May 7 '13 at 17:35
  • $\begingroup$ Have a look at this en.wikipedia.org/wiki/Jacobian_matrix_and_determinant#Examples $\endgroup$ – AnilB May 7 '13 at 18:00
  • $\begingroup$ Is what i did in the question above ok? Is there a way to interprete $\frac{\partial x}{\partial x(\xi,\theta)}$ and $\frac{\partial t}{\partial x(\xi,\theta)}$? $\endgroup$ – UrošSlovenija May 7 '13 at 18:07
  • $\begingroup$ Actually for which function you derive Jacobian; for coordinate transformation? If it is the case you have the wrong answer. $\endgroup$ – AnilB May 7 '13 at 18:08
  • $\begingroup$ Jacobian for coordinate transformation. Why is it wrong? $\endgroup$ – UrošSlovenija May 7 '13 at 18:13
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What you did is right, but it requires you to obtain $\partial\xi/\partial t$ and $\partial\xi/\partial x$ given $x(\xi,\theta)$ – it would probably be preferable to use the chain rule the other way around to express $u_t$ and $u_x$ in terms of $u_\xi$ and $u_\theta$; that would involve $x_\xi$ and $x_\theta$ instead $\xi_t$ and $\xi_x$; and then you can solve for $u_t$ and $u_x$.

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  • $\begingroup$ Very helpful indeed. Thanks. Do you have some thoughts about the Jacobian. I cannot get the feel for it. $\endgroup$ – UrošSlovenija May 7 '13 at 17:36

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