Classification of a second order partial differential equation

[sylent33]

Homework Statement

classify the partial differential equation

Relevant Equations

DE solving

Hello! Consider this partial differential equation

$$ zu_{xx}+x^2u_{yy}+zu_{zz}+2(y-z)u_{xz}+y^3u_x-sin(xyz)u=0 $$

Now I've got the solution and I have a few questions regarding how we get there. Now we've always done it like this.We built the matrix and then find the eigenvalues.

And here is where my first question arrives. I am not sure how they arive at this matrix.

$$ \left[ \begin{matrix} z & 0 & y-z \\\ 0 & x^2 & 0 \\\ y-z & 0 & z \end{matrix} \right] $$

Now I am guessing that the xx,yy,zz are the diagonal,and that is okay,but I am not sure how get the y-z at the bottom left corner? Its "coordinates" are xz meaning top right,why is it in the bottom left as well? Also what happened to the 2 infront of it?

Now after finding the eigenvalues (I know how to that) we get that the eigenvalues are;

$\lambda_1 = z$
$\lambda_2 = x^2$
$\lambda_3 = z$

And now comes the second part I don't understand.According to the book we are using for this class the classification should be done like this (and I quote)

elliptic if all EV of A(x) are non-zero and have the same sign
(ii) hyperbolic if all EV of A(x) are non-zero and have different signs.
(iii) parabolic if one or more EW of A(x) is zero.

Okay,and now the solutions are given like this;

Elliptic when : $x\neq 0 \wedge z>0$
Hyperbolic when : $x\neq 0 \wedge z<0$
Parabolic when : x = 0 or z = 0 or x and z = 0

Now the parabolic one,at least to me makes sence.Because it covers all the "cases" if you will.But the first 2 just don't click with me,

The fact that ALL of the EV should be non zero shouldn't we also say z unlike 0? and that they have the same sign,at least to me we are saying that all EV are positive,but nothing says what the sign of x is,it just states that is nonzero it might as well be negative.

Same for the hyperbolic, we are only stating that z is negative,but how do we know that x is positive?

Thanks and sorry for the long post.

 

[Orodruin]

Homework Statement:: classify the partial differential equation
Relevant Equations:: DE solving

Also what happened to the 2 infront of it?

For the off-diagonals, there are a factor of two because the matrix is symmetric and $u_{xz} = u_{zx}$.

Homework Statement:: classify the partial differential equation
Relevant Equations:: DE solving

The fact that ALL of the EV should be non zero shouldn't we also say z unlike 0?

If $z>0$ (as explicitly stated in case 1) then it cannot be zero.

Homework Statement:: classify the partial differential equation
Relevant Equations:: DE solving

but nothing says what the sign of x is,it just states that is nonzero it might as well be negative.

Because the eigenvalue that depends on $x$ is positive regardless of the sign of $x$ (it is $x^2$).

 

[anuttarasammyak]

I observe the equation is written as
Tr[\begin{bmatrix}<br /> z &amp; 0 &amp; y-z \\<br /> 0 &amp; x^2 &amp; 0 \\<br /> y-z &amp; 0 &amp; z \\<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> u_{xx} &amp; u_{xy} &amp; u_{xz} \\<br /> u_{yx} &amp; u_{yy} &amp; u_{yz} \\<br /> u_{zz} &amp; u_{zy} &amp; u_{zz} \\<br /> \end{bmatrix}<br /> -<br /> \begin{bmatrix}<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 \\<br /> \end{bmatrix}<br /> \frac{sin(xyz)}{3}u]=0
Is it helpful ?

 

[arhzz]

Oh its x squared...well that makes sense now,thanks for pointing it out.

But I didnt get the part with the off-diagonal as such, what do you mean by "For the off-diagonals, there are a factor of two because the matrix is symmetric",I know that the matrix,and if I understood you correctly (not sure I did) if we had it given as 3(y-z) would $u_xz = u_zx$ still hold true,since the matrix is symmetric? Or is it because of the 2 infront (in the example i just gave 3) of the apprentacies.

 

[Orodruin]

No. There is a factor of 2 between the differential equation and the entries in the matrix. Technically the matrix tells you that you have $(y-z)(u_{xz} + u_{zx})$ but this is equal to $2(y-z)u_{xz}$. If you had $3(y-x)u_{xz}$ in the equation then the matrix entries would be $3(y-z)/2$.

 

[sylent33]

Makes sence,also I see the x squared as well now.Thanks for the help as always