Manipulating Differential Equations with the Chain Rule

[abhay1]

can anyone please help me ?

View attachment 8084

 

[topsquark]

can anyone please help me ?

It would help us help you if you could tell us what you are able to do on this question.

-Dan

 

[HOI]

You "change variables" in a differential equation using the "chain rule".

That is, \(\displaystyle \frac{\partial u}{\partial x}= \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial x}+ \frac{\partial u}{\partial\tau}\frac{\partial \tau}{\partial x}\) and \(\displaystyle \frac{\partial u}{\partial t}= \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial t}+ \frac{\partial u}{\partial\tau}\frac{\partial \tau}{\partial t}\).

Here, we are given \(\displaystyle \xi= x- Vt\) and \(\displaystyle \tau= t\) so \(\displaystyle \frac{\partial \xi}{\partial x}= 1\) and \(\displaystyle \frac{\partial \tau}{\partial x}= 0\). \(\displaystyle \frac{\partial u}{\partial x}= \frac{\partial u}{\partial \xi}\).

Similarly \(\displaystyle \frac{\partial u}{\partial t}= \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial t}+ \frac{\partial u}{\partial \tau}\frac{\partial \tau}{\partial t}\). Since \(\displaystyle \frac{\partial \xi}{\partial t}= -V\) and \(\displaystyle \frac{\partial \tau}{\partial t}= 1\), \(\displaystyle \frac{\partial u}{\partial t}= -V\frac{\partial u}{\partial \xi}+ \frac{\partial u}{\partial\tau}\).

Repeat that to get the second derivatives.