Partial differential equation problem

[eljose]

If we call $$U_{xx}= \partial _{x} \partial _{x} U$$ the second partial differential derivative so we have for the Laplace operator:

$$\nabla ^{2} U = U_{xx}+U_{yy}+U_{zz}$$ then let,s suppose we have the differential equation:

$$aU_{xx}+bU_{yy}+cU_{zz}+dU_{xy}+eU_{xz}+fU_{yz}=0$$

then my question is if we can use a linear transform to choose another coordinate system so the equation read: $$\nabla^{2} U=0$$

thanks.

 

[Tantoblin]

I think not.

Think of quadratic forms. In three dimensions, can you use a coordinate transform to bring an arbitrary quadratic form into the form $$\xi_1^2 + \xi_2^2 + \xi_3^2$$? What quantities are invariant under linear transforms?

 

[eljose]

-But if the form is quadratic?...take into account that if the function is "smooth" then $$U_{xy}=U_{yx}$$ so the differential equation remains unchanged under changing x by y or z and so on...

 

[Tantoblin]

Well, what I wanted to say was (in two dimensions, say), that it doesn't make sense to try to change the PDE $$U_{xx} - U_{yy} = 0$$ (hyperbolic) into $$U_{xx} + U_{yy} = 0$$ (elliptic). The former has two characteristics while the latter has none. This is related to the theory of quadratic forms in the following way: if your 2nd order PDE is given by $$a_{ij} U_{x_i x_j}$$ then there is an associated quadratic form $$a_{ij} \xi^i \xi^j = 0$$ (summation over repeated indices understood).

Now, there are two classical invariants for quadratic forms, the rank (being the rank of the matrix $$a_{ij}$$) and the signature. Recall that the signature is found by diagonalizing $$a_{ij}$$ and reducing it to a form with only +1 and -1 on the diagonal; the number of -1s is the signature. Linear transformations leave rank and signature invariant. For instance, the hyperbolic equation I mentioned earlier has signature 0 and the elliptic equation has signature 1 and are hence inequivalent.

So the answer to your question is no.

 

[Clausius2]

I think that if the equation is originally elliptic, I think it should be some way of transforming the equation to canonical form (i.e. lap(U)=0)

 

[Feynman]

If we call $$U_{xx}= \partial _{x} \partial _{x} U$$ the second partial differential derivative so we have for the Laplace operator:

$$\nabla ^{2} U = U_{xx}+U_{yy}+U_{zz}$$ then let,s suppose we have the differential equation:

$$aU_{xx}+bU_{yy}+cU_{zz}+dU_{xy}+eU_{xz}+fU_{yz}=0$$

then my question is if we can use a linear transform to choose another coordinate system so the equation read: $$\nabla^{2} U=0$$

thanks.

You must presise your space (sobolev, holder, besov,...) and your boundary conditions

 

[HallsofIvy]

I think eljose's question was really about getting rid of the first derivative forms which you can (almost) always do.

Tantoblin's objection was that you cannot, in general, convert specifically to the elliptic form U_xx+ U_yy+ U_zz= 0, Laplace's equation.

It might, rather, be the case than you can convert to the hyperbolic forms U_xx+ U_yy- U_zz= 0 or U_xx- U_yy- U_zz= 0, the wave equation, or
the parabolic forms U_xx+ U_yy+ U_z= 0 or U_xx+ U_y+ V_z= 0, the heat(diffusion) equation.