# Reducing to the canonical form of a linear pde with constant coefficients in 4 dimensions

I have a 4-dimensional second-order PDE, which needs to be reduced to the canonical form.

$$u_{xx}+2u_{xy}-2u_{xz}-4u_{yz}+2u_{ty}+u_{zz}=0$$

I know and understand the 2-dimensional (xy space) theory, but how does one approach this case of 4 dimensions? Should one explicitly write out expressions for coefficients after variable substitutions? What form should one expect to be canonical after doing that (in order to equate proper coefficients to zero)?

I can't understand the general N-dimensional theory in the book, so i would be very grateful for a general step-by-step algorithm for a 4-dimensional 2-order linear PDE with constant coefficients.

• What's the book? – Harjeq Dec 28 '19 at 18:08
• @JeqHar a very thick russian book :) "vladimirov equations of mathematical physics". not sure you should open it :) – Nick The Dick Dec 28 '19 at 18:13
• Does it have a name? – Harjeq Dec 28 '19 at 18:16
• @JeqHar yep, check the edit – Nick The Dick Dec 28 '19 at 18:18

I've gone through pages $$39$$ - $$41$$ of your text. I'm going to restate some things in a hopefully clearer manner.

As with the usual $$2$$-D case, we seek a transformation which transforms the equation to canonical form. In the text, we want the form where there are only second derivatives of a single variable. The existence of a new set of variables related to the old set generates partial derivatives that will transform the coefficients of the second-order partial derivatives. The author handles all the chain rule steps to generate the form of the transformation.

As all your coefficients are constant, there exists a linear transformation that will take your equation to canonical form. The author notes that these coefficients transform just as coefficients of quadratic forms do under linear transformation. For quadratic forms, there always exists a transformation that reduces to only square terms. A transformation can also be chosen such that the coefficients of these squares are $$1$$, $$0$$, or $$-1$$ by Sylvester's Law of Inertia (it actually states it for a symmetric matrix, so we'll replace the matrix of coefficients by its symmetric part which will preserve the quadratic form and PDE). Accordingly, only the second derivatives of a single variable will remain in the new PDE.

Now let's look at your problem. We can produce the original $$4$$x$$4$$ coefficient matrix $$A$$ from the coefficients you have given while denoting $$x_1 = x$$, $$x_2 = y$$, $$x_3 = z$$, and $$x_4 = t$$. This matrix has the elements $$a_{ij}$$ from Equation $$1$$ in the text.

$$A =\begin{bmatrix} 1 & 2 & -2 & 0\\0 & 0 & -4 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 2 & 0 & 0 \end{bmatrix}$$

Now I'll replace $$A$$ by its symmetric part $$A_S$$.

$$A_S = \frac{A+A^\top}{2} = \left(\begin{bmatrix} 1 & 2 & -2 & 0\\0 & 0 & -4 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 2 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 & 0\\2 & 0 & 0 & 2 \\ -2 & -4 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} \right) = \begin{bmatrix} 1 & 1 & -1 & 0\\1 & 0 & -2 & 1 \\ -1 & -2 & 1 & 0\\ 0 & 1 & 0 & 0 \end{bmatrix}$$

If we apply the coefficients of this symmetric matrix with their corresponding second-order derivative, then we arrive at the following:

$$u_{xx} + u_{xy} - u_{xz} + u_{yx} - 2u_{yz} + u_{yt} - u_{zx} - 2u_{zy} + u_{zz} + u_{ty} = 0$$

Note that this equation is exactly the same as the original as the partial derivatives commute. But now we have a symmetric matrix to diagonalize to arrive at a diagonal matrix with entries of $$1$$,$$0$$, or $$-1$$.

Eigenvector Approach:

We can do this by finding the eigenvectors and dividing them by the square root of the absolute value of their nonzero eigenvalues (the square root because the diagonalization uses each eigenvector twice). The scaled eigenvector matrix (where the eigenvectors are the rows) then gives the necessary transformation to the new coordinate system. I'll call the new coordinates $$x^{\prime}$$, $$y^{\prime}$$, $$z^{\prime}$$, $$t^{\prime}$$.

I used MATLAB for this to arrive at an approximation of the following transformation equation. Note that each element of this matrix is equal to the partial derivative of the new coordinate w.r.t. an old coordinate, as the chain rule indicates.

$$\begin{bmatrix}x^{\prime}\\ y^{\prime}\\ z^{\prime}\\ t^{\prime} \end{bmatrix} = \begin{bmatrix} 0.0677 & -0.5533 & -0.3551 & 0.2874\\-0.5774 & 0 & -0.5774 & -0.5774 \\ -0.9675 & 0.4646 & -0.0694 & 1.0369\\ 0.2618 & 0.3004 & -0.3483 & 0.0864 \end{bmatrix} \begin{bmatrix}x\\ y\\ z\\ t \end{bmatrix}$$

The resulting diagonalization is as below. The elements are the coefficients of the second derivatives of the PDE in the new coordinate system. This matrix has transformed elements $$\tilde{a}_{lk}$$ from Equation $$6$$ in the text.

$$\begin{bmatrix} -1 & 0 & 0 & 0\\0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

The new PDE is as follows. The author remarks that you may have fewer squares than the dimension of the problem. By the text, this is a parabolic PDE.

$$-u_{x^\prime x^\prime} + u_{z^\prime z^\prime} + u_{t^\prime t^\prime} = 0$$

Lagrange's Method:

The other approach to solve this problem would be to use Lagrange's method. This approach provides a more exact transformation without the need for finding the eigenvectors and eigenvalues. However, both approaches will provide the necessary canonical form. We start with the quadratic form given below.

$$x^2 + 2xy - 2xz - 4yz + 2ty + z^2 = 0$$

The reduction to all squared terms occurs as below.

$$x^2 + 2xy - 2xz - 4yz + 2ty + z^2 \\ = [x^2 + 2x(y-z) ] - 4yz + 2ty + z^2 \\ = [x^2 + 2x(y-z) + (y-z)^2] - (y-z)^2 - 4yz + 2ty + z^2 \\ = (x + y - z)^2 - (y-z)^2 - 4yz + 2ty + z^2 \\ = (x + y - z)^2 - (y^2 - 2yz + z^2) - 4yz + 2ty + z^2 \\ = (x + y - z)^2 - y^2 + 2yz - z^2 - 4yz + 2ty + z^2 \\ = (x + y - z)^2 - y^2 - 2yz + 2ty \\ = (x + y - z)^2 - [y^2 + 2y(z-t)] \\ = (x + y - z)^2 - [y^2 + 2y(z-t) + (z-t)^2] + (z-t)^2 \\ = (x + y - z)^2 - (y + z - t)^2 + (z-t)^2 \\$$

As indicated in the eigenvector approach, the quadratic reduces to the sum of three squares. Fortunately, the coefficients are already $$1$$, $$0$$, or $$-1$$. Now, this may appear to provide the necessary change in coordinates for the problem, but this isn't it. Note we also don't have a unique way to map the original $$4$$ coordinates to $$4$$ new coordinates because of the parabolic nature of the equation. We are entirely free to choose how one of these new coordinates is related to the old ones. It changes nothing about the solution. I'll establish an intermediate change in coordinates with $$x^*$$, $$y^*$$, $$z^*$$, and $$t^*$$. The transformation between coordinates is then as below with my choice of $$y^* = x$$ to ensure the transformation matrix is invertible.

$$\begin{bmatrix}x^*\\ y^*\\ z^*\\ t^* \end{bmatrix} = \begin{bmatrix} 0 & 1 & 1 & -1\\1 & 0 & 0 & 0 \\ 1 & 1 & -1 & 0\\ 0 & 0 & 1 & -1 \end{bmatrix} \begin{bmatrix}x\\ y\\ z\\ t \end{bmatrix}$$

The vector in the immediate coordinates corresponds to the quadratic form with the diagonal matrix. We want to transform the vector of original coordinates that corresponds to the original quadratic form, where the matrix is symmetric but not diagonal. To transform that vector to the intermediate coordinates, we have to invert the transformation matrix. The matrix of partial derivatives we need to use to convert to canonical form is the transpose of the inversion seen below. The primed coordinates are now used.

$$\begin{bmatrix}x^{\prime}\\ y^{\prime}\\ z^{\prime}\\ t^{\prime} \end{bmatrix} = \begin{bmatrix} 0 & 1 & 1 & 1\\1 & 0 & 1 & 1 \\ 0 & 0 & -1 & -1\\ 0 & -1 & -1 & -2 \end{bmatrix} \begin{bmatrix}x\\ y\\ z\\ t \end{bmatrix}$$

Note that this matrix and its transpose will also diagonalize the original coefficient matrix to produce the same canonical form as the eigenvector approach. This approach is probably favored because the process and results are much nicer.

The process can be summed up as follows: First, find the intermediate transformation from Lagrange's method (ensure it is invertible if you have freedom over one variable as in the parabolic case). Lagrange's method will tell you what the canonical form will look like for the second derivatives. However, you may still need the transformation to transform lower derivatives. Lastly, take the transpose of the inverse of the intermediate transformation matrix. This will provide the change in coordinates for that canonical form. That's it.

• thanks! i was researching this problem also and found another source where it became clear to me what to do. i am going to diagonalise the quadratic form by Lagrange's method though :) in 20 hours i can award the bounty to you - thank you for your efforts ! – Nick The Dick Dec 28 '19 at 21:09
• the source i was looking in is "tikhonov samarskii equations of mathematical physics", in case you are interested – Nick The Dick Dec 28 '19 at 21:13
• Thanks. I was wondering what it was. – Harjeq Dec 28 '19 at 21:18
• I've added the result from Lagrange's method as well because that approach is so nice. – Harjeq Dec 29 '19 at 18:35
• thank you. that's great! – Nick The Dick Dec 29 '19 at 18:37

$$u_{xx}+2u_{xy}-2u_{xz}-4u_{yz}+2u_{ty}+u_{zz}=0$$

is a linear PDE with constant coefficients and can be represented as

$$\left(\partial_X\cdot M\cdot\partial_X^T\right)u(X) = 0$$

with

$$\partial_X = (\partial_x,\partial_y,\partial_z,\partial_t)\\ M = \left( \begin{array}{cccc} 1 & 1 & -1 & 0 \\ 1 & 0 & -2 & 1 \\ -1 & -2 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array} \right)$$

where $$\left(\partial_X\cdot M\cdot\partial_X^T\right)$$ represents the differential operator.

Considering now a change of variables on the differential operator as

$$\partial_X = \partial_W\cdot Q$$

with

$$\partial_W = (\partial_{w_1},\partial_{w_2},\partial_{w_3},\partial_{w_4})$$

we have on $$W$$ coordinates:

$$\left(\partial_W\cdot Q\cdot M\cdot Q^T\cdot\partial_W^T\right)u(W) = 0$$

now as $$M$$ is hermitian we can proceed calculating the $$M$$ eigenvectors and eigenvalues, giving a set of real eigenvalues. Extracting the eigenvectors $$Q$$ and eigenvalues $$\Lambda$$ we have

$$Q\cdot M\cdot Q^T = \Lambda$$

then choosing $$Q$$ as the normalized eigenvectors matrix associated to $$M$$ we obtain

$$\left(\partial_W\cdot \Lambda\cdot\partial_W^T\right)u(W) = 0$$

Follows a MATHEMATICA script which calculates the variable change

ClearAll[DChange]
DChange[expr_, transformations_List, oldVars_List, newVars_List, functions_List] := Module[{pos, functionsReplacements, variablesReplacements, arguments, heads, newVarsSolved},
pos = Flatten[Outer[Position, functions, oldVars], {{1}, {2}, {3, 4}}];
arguments = List @@@ functions;
newVarsSolved = newVars /. Solve[transformations, newVars][[1]];
functionsReplacements = Map[Function[i, heads[[i]] -> (Function[#, #2] &[arguments[[i]],
ReplacePart[functions[[i]],
variablesReplacements = Solve[transformations, oldVars][[1]];
expr /. functionsReplacements /. variablesReplacements // Simplify // Normal];
DChange[expr_, functions : {(_[___] == _) ..}] := expr /. Replace[ functions, (f_[vars__] == body_) :> (f -> Function[{vars}, body]), {1}]
DChange[expr_, x___] := DChange[expr, ##] & @@ Replace[{x}, var : Except[_List] :> {var}, {1}];


The previous code is a module responsible for the variables change from $$X$$ to $$W$$

M = {{1, 1, -1, 0}, {1, 0, -2, 1}, {-1, -2, 1, 0}, {0, 1, 0, 0}};


Follows the eigenvalues and eigenvectors calculation

esys = Eigensystem[M] // N;
Lambda = First[esys];
Q = Last[esys];


Normalizing $$Q$$

Qn = Table[Q[[k]]/Norm[Q[[k]]], {k, 1, 4}]


Proposed change of variables

transf = Table[Subscript[w , k] == Qn[[k]].{x, y, z, t}, {k, 1, 4}];


and finally the use of DChange

Clear[u]
DChange[D[u[x, y, z, t], x, x] + 2 D[u[x, y, z, t], x, y] - 2 D[u[x, y, z, t], x, z] - 4 D[u[x, y, z, t], y, z] + 2 D[u[x, y, z, t], t, y] + D[u[x, y, z, t], z, z] == 0, transf, {x, y, z, t}, {Subscript[w, 1], Subscript[w, 2], Subscript[w, 3], Subscript[w, 4]}, u[x, y, z, t]] // Chop


giving

$$7.76072 u_{w_1^2}(W)-4.29714 u_{w_2^2}(W)+u_{w_3^2}(W) = 0$$

Follows the representation of the conic

$$7.76072 w_1^2-4.29714w_2^2+w_3^2 = 0$$