Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the constant coefficient PDE: $$a_{11}u_{xx}+2a_{12}u_{xy}+a_{22}u_{yy}+b_{1}u_{x}+b_{2}u_{y}+cu=0$$ Show that the only ones that are unchanged under all axis-rotations (rotation invariant) have the form: $$a(u_{xx} + u_{yy}) + bu = 0$$where $a$ and $b$ are constants.

I just let $x'=x\cos A-y\sin A$ , and $y'=x\sin A+y\cos A$ ,

then get $u_{xx}\;,\;u_{xy}\;,\;u_{yy}$ in terms of $x'$ and $y'$.

But I still don't know how to get the required form since $u_{x'y'}$ doesn't vanish.

Hope someone gives me some hints or the solution.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Start with

$$\frac{\partial u}{\partial x} = \frac{\partial x'}{\partial x} \frac{\partial u}{\partial x'} + \frac{\partial y'}{\partial x} \frac{\partial u}{\partial y'} = \cos{A} \, u_{x'} + \sin{A} \, u_{y'} $$

$$\frac{\partial u}{\partial y} = \frac{\partial x'}{\partial y} \frac{\partial u}{\partial x'} + \frac{\partial y'}{\partial y} \frac{\partial u}{\partial y'} = -\sin{A} \, u_{x'} + \cos{A} \, u_{y'} $$

Then differentiate again:

$$\frac{\partial^2 u}{\partial x^2} = \cos^2{A} \, u_{x'x'} + \sin^2{A} \, u_{y'y'} + 2 \sin{A} \cos{A} u_{x'y'} $$

$$\frac{\partial^2 u}{\partial y^2} = \sin^2{A} \, u_{x'x'} + \cos^2{A} \, u_{y'y'} - 2 \sin{A} \cos{A} u_{x'y'} $$

$$\frac{\partial^2 u}{\partial x \partial y} = -\cos{A} \sin{A} \, u_{x'x'} + (\cos^2{A}-\sin^2{A}) \, u_{x'y'} + \cos{A} \sin{A} \, u_{y'y'} $$

The coefficient of $u_{x'y'}$ obtained after plugging the above into the differential expression above is set to zero:

$$(a_{11} - a_{22}) \sin{2 A} + 2 a_{12} \cos{2 A} = 0$$

which reveals the rotation angle $A$. Note that the coefficients of $u_{x'x'}$ and $u_{y'y'}$ are not equal as you state above. The quadratic terms in the rotated coordinate system become

$$\left [ 1 + \frac{a_{12}}{\sqrt{a_{12}^2 + (a_{11}-a_{22})^2}} \right ] u_{x'x'} + \left [ 1 - \frac{a_{12}}{\sqrt{a_{12}^2 + (a_{11}-a_{22})^2}} \right ] u_{y'y'}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.