MHB 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, $$\frac{\partial u}{\partial x}= \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial x}+ \frac{\partial u}{\partial\tau}\frac{\partial \tau}{\partial x}$$ and $$\frac{\partial u}{\partial t}= \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial t}+ \frac{\partial u}{\partial\tau}\frac{\partial \tau}{\partial t}$$.

Here, we are given [math]\xi= x- Vt[/math] and [math]\tau= t[/math] so [math]\frac{\partial \xi}{\partial x}= 1[/math] and [math]\frac{\partial \tau}{\partial x}= 0[/math]. [math]\frac{\partial u}{\partial x}= \frac{\partial u}{\partial \xi}[/math].

Similarly [math]\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}[/math]. Since [math]\frac{\partial \xi}{\partial t}= -V[/math] and [math]\frac{\partial \tau}{\partial t}= 1[/math], [math]\frac{\partial u}{\partial t}= -V\frac{\partial u}{\partial \xi}+ \frac{\partial u}{\partial\tau}[/math].

Repeat that to get the second derivatives.