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June 17th, 2018, 11:47 AM   #1
Joined: Jul 2014
From: Amherst, MA

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Math Focus: Calculus, Differential Geometry, Physics, Topology
Question: Getting rid of Constants?

I was taking a physics class once, my prof was solving a PDE, and he did some trick that got rid of all the constants in the equation.

For example, if he had something like $\displaystyle \alpha y_{tt} - \beta y_{xx} = 0$, he would end up with something like $\displaystyle u_{tt}-u_{xx}=0$, then after he solved the thing, he would substitute back, and get the solution to the first equation.

Can you do this with any equation? I don't know if the method depended on some other property like linearity, I wasn't really paying attention. He did it at least under some conditions, and I can't figure out how to do it. Any ideas? Could make some work cleaner and easier to keep straight.


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June 18th, 2018, 02:52 AM   #2
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I can't say for sure what your professor did but I would take, not a new dependent variable, but two new independent variables, $\displaystyle x'= \frac{x}{\sqrt{\alpha}}$ and $\displaystyle t'= \frac{t}{\sqrt{\beta}}$. Then, by the chain rule, $\displaystyle \frac{\partial y}{\partial x}= \frac{\partial y}{\partial x'}\frac{dx'}{dx}= \frac{\frac{\partial y}{\partial x'}}{\sqrt{\alpha}}$ and then $\displaystyle \frac{\partial^2 y}{\partial x^2}= \frac{\partial}{\partial x}\left(\frac{\partial y}{\partial x}\right)= \frac{\frac{\partial}{\partial x'}}{\sqrt{\alpha}}= \frac{\frac{\partial^2 y}{\partial x'^2}}{\alpha}$.

Similarly for t.
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