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Given

$4u_{xx} +2u_{yy} - 6u_{zz} +6u_{xy} + 10u_{xz} +4u_{yz} + 2u =0$

I need to find the type of the equation, for that i tried to get the canonical form

So, turn into lambdas

$4\lambda_1^2 + 2\lambda_2^2 - 6\lambda_3^2 + 6\lambda_1\lambda_2 + 10\lambda_1\lambda_3 + 4\lambda_2\lambda_3 = 0$

from here i can't get the form which is looks like

$\pm\xi^2 \pm \eta^2 \pm \zeta^2 =0$

any help or advice is much appreciated

Spike Bughdaryan
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    Just as with a general conic equation, you can determine the type of this differential equation by examining its discriminant and determinant of the associated $3\times3$ coefficient matrix. – amd Mar 26 '19 at 06:35
  • Yes I tried by that way, but the thing is, i would need those $\eta \zeta \xi$ after to get canonic form – Spike Bughdaryan Mar 26 '19 at 06:49
  • Ah, but you said that you needed the canonical form in order to determine they equation’s type, not that you also needed the former for some other reason. – amd Mar 26 '19 at 19:44

2 Answers2

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The equation $$\sum_{i,j=1}^na_{ij}u_{x_ix_j}+a(x,u,\nabla u)=0$$ can be reduced to canonical form by applying the non singular linear transformation $\xi=B^Tx$, where B is a matrix such that the transformation $y=B\eta$ reduces the quadratic form $\sum_{i,j=1}^na_{ij}y_iy_j$ to canonical form.

Example. Reduce to canonical form $$4u_{xx} +2u_{yy} - 6u_{zz} +6u_{xy} + 10u_{xz} +4u_{yz} + 2u =0.$$

$$4\lambda_1^2 + 2\lambda_2^2 - 6\lambda_3^2 + 6\lambda_1\lambda_2 + 10\lambda_1\lambda_3 + 4\lambda_2\lambda_3= \frac{(4\lambda_1+3\lambda_2+5\lambda_3)^2}{4}-\frac{(\lambda_2+7\lambda_3)^2}{4}$$ Let $$y_1=\frac{4\lambda_1+3\lambda_2+5\lambda_3}{2},\;y_2=\frac{\lambda_2+7\lambda_3}{2},\;y_3=\lambda_3$$ $\Rightarrow$ $$\lambda_1=\frac12y_1-\frac32y_2+4y_2,\;\lambda_2=2y_2-7y_2,\;\lambda_2=y_3$$ $\Rightarrow$ $$B^T=\begin{pmatrix}\frac{1}{2} & -\frac{3}{2} & 4\\ 0 & 2 & -7\\ 0 & 0 & 1\end{pmatrix},\quad B=\begin{pmatrix}\frac{1}{2} & 0 & 0\\ -\frac{3}{2} & 2 & 0\\ 4 & -7 & 1\end{pmatrix}$$ Transform to canonical form is: $$\xi=\frac{x}{2},\;\eta=-\frac{3}{2}x+2y,\;\zeta=4x-7 y+z$$ After reducing we get $$u_{\xi\xi}-u_{\eta\eta}+2u=0$$

Aleksas Domarkas
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  • Thank you for answer, can you please show the steps how you got there? – Spike Bughdaryan Mar 26 '19 at 07:02
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    $4\lambda_1^2 + 2\lambda_2^2 - 6\lambda_3^2 + 6\lambda_1\lambda_2 + 10\lambda_1\lambda_3 + 4\lambda_2\lambda_3=-\frac{{{\left( 23 {{\lambda }_3}-{{\lambda }_2}\right) }^{2}}}{4}+\frac{{{\left( 5 {{\lambda }_3}-3 {{\lambda }_2}+4 {{\lambda }_1}\right) }^{2}}}{4}+120 {{{{\lambda }_3}}^{2}}$ – Aleksas Domarkas Mar 26 '19 at 07:23
  • I believe that you’ve made a sign error somwhere. Expanding the R.H.S. of the above equation produces the term $-6\lambda_1\lambda_2$ instead of $+6\lambda_1\lambda_2$. – amd Mar 26 '19 at 20:41
  • @amd Sorry, I solved for $4u_{xx} +2u_{yy} - 6u_{zz} -6u_{xy} + 10u_{xz} +4u_{yz} + 2u =0$. I'll fix it. – Aleksas Domarkas Mar 27 '19 at 08:13
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Your problem is one of diagonalizing a quadratic form. There are several methods available for doing this. You might for instance compute eigenvalues and eigenvectors of the associated symmetric matrix $$Q = \begin{bmatrix}4&3&5\\3&2&2\\5&2&-6\end{bmatrix}.$$ In this case, we’re pretty lucky because the characteristic polynomial is fairly “nice:” $\lambda^3-66\lambda$, from which the eigenvalues are $0$ and $\pm\sqrt{66}$. This gives you the information you need to classify the equation, which only requires knowing the spectrum of the quadratic form. To proceed onto the canonical form, you’ll need to find a set of corresponding eigenvectors, which doesn’t take too much more work.

In practice, the eigenvalues and eigenvectors aren’t usually this nice, but there are other ways to diagonalize the quadratic form. Another common method is successively completing squares. One possible result of doing this is $$\frac14\left(4\lambda_1+3\lambda_2+5\lambda_3\right)^2-\frac14\left(\lambda_2+7\lambda_3\right)^2.$$ The spectrum of these terms matches the spectrum obtained from the eigenvalues, which we expect.

If you’re not good at completing squares, there’s an entirely mechanical algorithm described here for diagonalizing a quadratic form. Omitting the gory details of the computation, for this problem the algorithm eventually produces the augmented matrix $$\left[\begin{array}{ccc|ccc}4&0&0 & 1&0&0 \\ 0&-4&0 & -3&4&0 \\ 0&0&0 & -64&112&-16 \end{array}\right].$$ From the diagonal of the left-hand side, we get the same spectrum as with the previous two methods. To find the canonical form, we must then invert the right-hand side, which produces the diagonalization $$4\left(\lambda_1+\frac34\lambda_2+\frac54\lambda_3\right)^2-4\left(\frac14\lambda_2+\frac74\lambda_3\right)^2$$ that happens to be the same as the one obtained by completing the squares above.

amd
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