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Partial derivatives after a transformationby MarkovMarakov
Tags: pde 
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#1
Nov2512, 04:27 AM

P: 34

Suppose I have a transformation
[tex](x'_1,x'_2)=(f(x_1,x_2), g(x_1,x_2))[/tex] and I wish to find [tex]\partial x'_1\over \partial x'_2[/tex] how do I do it? If it is difficult to find a general expression for this, what if we suppose [itex]f,g[/itex] are simply linear combinations of [itex]x_1,x_2[/itex] so something like [itex]ax_1+bx_2[/itex] where [itex]a,b[/itex] are constants? 


#2
Nov2512, 05:39 AM

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[tex]x'_1=f(x_1,x_2)[/tex] [tex]x'_2=g(x_1,x_2)[/tex] [tex]x_1=F(x'_1,x'_2)[/tex] [tex]x_2=G(x'_1,x'_2)[/tex] Then, we clearly have, for example the identity: [tex]x'_1=f(F(x'_1,x'_2),G(x'_1,x'_2))[/tex] Since the pair of marked variables (and the unmarked) are independent from each other, we have the requirement on the functional relationship: [tex]\frac{\partial{x}_{1}^{'}}{\partial{x}_{2}^{'}}=0=\frac{\partial{f}}{\p artial{x}_{1}}\frac{\partial{F}}{\partial{x}_{2}^{'}}+\frac{\partial{f} }{\partial{x}_{2}}\frac{\partial{G}}{\partial{x}_{2}^{'}}[/tex] 


#3
Nov2512, 06:44 AM

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Suppose you have, say:
x=u^2v^2 y=u+v How can you utilize those implied requirements in the previous post to derive the the correct representations of u and v in terms of x and y? Note that in this case, it is fairly trivial to find it by algrebrac means; dividing the first relationship by the second, we have x/y=uv, and thus: u=1/2(x/y+y), v=1/2(yx/y) However, using instead the four identities gained from the above requirements, we have [tex]\frac{\partial{x}}{\partial{x}}=1=2u\frac{\partial{u}}{\partial{x}}2v\frac{\partial{v}}{\partial{x}}[/tex] [tex]\frac{\partial{x}}{\partial{y}}=0=2u\frac{\partial{u}}{\partial{y}}2v\frac{\partial{v}}{\partial{y}}[/tex] [tex]\frac{\partial{y}}{\partial{y}}=1=\frac{\partial{u}}{\partial{y}}+\frac {\partial{v}}{\partial{y}}[/tex] [tex]\frac{\partial{y}}{\partial{x}}=0=\frac{\partial{u}}{\partial{x}}+\frac {\partial{v}}{\partial{x}}[/tex] from these, you should be able to derive the above relations. Note, for example, that by combining the first and fourth and the known relation y=u+v, you get: [tex]\frac{\partial{u}}{\partial{x}}=\frac{1}{2y}[/tex] meaning that we must have u(x,y)=x/2y+h(y), where h(y) is some function of y. Applying the fourth, you get v(x,y)=b(y)x/2y, for some b(y) By using known and gained information, you will be able to determine h(y) and b(y) from the second and third relationships. 


#4
Nov2512, 11:38 AM

P: 34

Partial derivatives after a transformation
Thank you, arildno.



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