# Homework Help: Ince's ODEs

1. May 29, 2010

### psholtz

1. The problem statement, all variables and given/known data

I'm working through the classic treatise on ODEs by Ince. I know that this is a somewhat dated text, but (imho) there are some real "gems" on the subject of ODEs here, well worthy of careful study.

I'm looking at Problem #7 at the end of Chapter 1. In it, we are given a function $$z(x,y)$$:

$$z(x,y) = 3xy - y^2 + (y^2-2x)^{3/2}$$

and we are asked to show two things. First, that:

$$\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial^2 z}{\partial y \partial x}$$

and second that:

$$\frac{\partial^2 z}{\partial x^2} \cdot \frac{\partial^2 z}{\partial y^2} = \left( \frac{\partial^2z}{\partial x \partial y} \right)^2$$

The first one is a no-brainer. It's the second one that I'm having trouble with..

2. Relevant equations

All are given above.

3. The attempt at a solution

Differentiating the above expression, we have:

$$z_x = 3y + \left( \frac{3}{2} \right)\left(y^2-2x \right)^{1/2}(-2)$$
$$z_x = 3y - 3\left(y^2- 2x\right)^{1/2}$$

and differentiating again:

$$z_{xy} = 3 - 3\left(\frac{1}{2}\right) \left(y^2-2x\right) ^ {-1/2}(2y)$$

$$z_{xy} = 3 - \frac{3y}{\sqrt{y^2-2x}}$$

Now differentiating with respect to y first, we have:

$$z_y = 3x - 2y + \left(\frac{3}{2}\right) \left(y^2-2x\right)^{1/2}(2y)$$

$$z_y = 3x - 2y + 3y\sqrt{y^2-2x}$$

and differentiating again:

$$z_{yx} = 3 +3y\left(\frac{1}{2}\right)\left(y^2-2x\right)^{-1/2}(-2)$$

$$z_{yx} = 3 - \frac{3y}{\sqrt{y^2-2x}}$$

So yes, we have $$z_{yx} = z_{xy}$$ as expected.

Now, moving on to the second part of this problem, we square this expression:

$$\left(\frac{\partial^2z}{\partial x \partial y}\right)^2 = 9 + \frac{9y^2}{y^2-2x} - \frac{18y}{\sqrt{y^2-2x}}$$

$$\left(\frac{\partial^2z}{\partial x \partial y}\right)^2 = \frac{18(y^2-x)}{y^2-2x} - \frac{18y}{\sqrt{y^2-2x}}$$

Now, going back to the expression for $$z_x$$:

$$z_x = 3y - 3\sqrt{y^2-2x}$$

and differentiating again with respect to x, we have:

$$z_{xx} = -3\left(\frac{1}{2}\right)\left(y^2-2x\right)^{-1/2}(-2)$$

$$z_{xx} = \frac{3}{\sqrt{y^2-2x}}$$

And going back to the expression for $$z_y$$:

$$z_y = 3x - 2y + 3y\sqrt{y^2-2x}$$

and differentiating with respect to y:

$$z_{yy} = -2 + 3\left[\sqrt{y^2-2x} + y \left(\frac{1}{2}\right)\left(y^2-2x\right)^{-1/2}(2y)\right]$$

$$z_{yy} = - 2 + 3 \left[\sqrt{y^2-2x} + \frac{y^2}{\sqrt{y^2-2x}}\right]$$

$$z_{yy} = - 2 + 3 \left[\frac{2(y^2-x)}{\sqrt{y^2-2x}}\right]$$

$$z_{yy} = - 2 + \frac{6(y^2-x)}{\sqrt{y^2-2x}}$$

Multiplying these together, we get:

$$z_{xx} \cdot z_{yy} = \frac{18(y^2-x)}{y^2-2x}- \frac{6}{\sqrt{y^2-2x}}$$

Comparing this to what we derived above for $$\left(z_{xy}\right)^2$$, we see it's close, but not exact. The first term is correct, but the second term is off by a factor of 3y.

What am I doing wrong?

Or is there a typo in Ince?

2. May 29, 2010

### jdwood983

I'm not sure that it is a valid equality in the first place. Imagine our function is a little bit more simple:

$$z(x,y)=3xy-y^2+x^2$$

Then

$$z_{xx}=2, z_{yy}=-2 \rightarrow z_{xx}\cdot z_{yy}=-4$$

compared to

$$z_{yx}=3\rightarrow z_{yx}^2=9$$

so these two are not equal. Unless, that is, this equality is specifically for this function only. In this latter case, I get the same results you do, that there is a missing 3y in the second term.

3. May 29, 2010

### psholtz

I think it's supposed to be for this function only..

As you showed, one can easily construct functions for which the equality does not hold.. for instance:

$$z = x^3 + y^3$$

$$z_{xx} = 6x$$

$$z_{yy} = 6y$$

and hence:

$$z_{xx} \cdot z_{yy} = 36xy$$

but

$$z_{xy} = 0$$

It is interesting, nonetheless, to ask the question: for what kinds of functions does the equality, $$z_{xx} \cdot z_{yy} = \left(z_{xy}\right)^2$$ hold? i.e., what conditions would the function have to meet, in order for the equality to hold?

The Ince text, while deep and powerful, has typos in it that can be frustrating. In Question #2 in Chapter 1, he asks you to prove that given a (different) y (I won't go through the whole problem statement here), that:

$$(y-x)y'' = 2y(1+y')$$

In fact, this is impossible (given the problem statement), and the correct answer is:

$$(y-x)y'' = 2y'(1+y')$$

It took me about a week of staring at that problem to figure out that it was a typo.

I'm stating this only in case other people are reading this text as well.

4. May 29, 2010

### jdwood983

I started thinking that it might be for this function only after I typed my response. I would think, then, that there must be a typo in the problem. What the error is, I can't be sure; perhaps the power should be something else?

5. May 30, 2010

### psholtz

I tried using a power of 5/2, and it doesn't seem to get anywhere useful.

In terms of the kinds of functions that do satisfy the general relation, $$z_{xx} \cdot z_{yy} = \left(z_{xy}\right)^2$$, there are the following observations:

$$z = x + y$$

$$z_{xx} = z_{yy} = z_{xy} = 0$$

But that's a sort of "trivial" case.

I also thought you might be able to use something like:

$$z = kx^4y^4$$

$$z_{xx} = 12kx^2y^4$$

$$z_{yy} = 12kx^4y^2$$

and hence that:

$$z_{xx} \cdot z_{yy} = 144k^2x^6y^6$$

and likewise that:

$$z_{xy} = 16kx^3y^3$$

so that:

$$\left(z_{xy}\right)^2 = 256k^2x^6y^6$$

so the degrees of the polynomials come out correctly, but the trouble is that you can't get the coefficients to come out right.

6. May 30, 2010

### psholtz

The typo is in one digit. Rather than using:

$$z = 3xy - y^2 + \left(y^2-2x\right)^{3/2}$$

we must use:

$$z = 3xy - y^3 + \left(y^2-2x\right)^{3/2}$$

In other words, the second term is raised to the third power, not the second power.

Using this, we have:

$$z_x = 3y + \left(\frac{3}{2}\right)\left(y^2-2x\right)^{1/2}(-2)$$

$$z_x = 3y - 3\left(y^2-2x\right)^{1/2}$$

and taking the second derivatives:

$$z_{xx} = -3\left(\frac{1}{2}\right)\left(y^2-2x\right)^{-1/2}(-2)$$

$$z_{xx} = \frac{3}{\sqrt{y^2-x}}$$

and

$$z_{xy} = 3 - 3 \left(\frac{1}{2}\right)\left(y^2-2x\right)^{-1/2}(2y)$$

$$z_{xy} = 3 - \frac{3y}{\sqrt{y^2-2x}}$$

Taking the y-derivatives:

$$z_y = 3x - 3y^2 + \left(\frac{3}{2}\right)\left(y^2-2x\right)^{1/2}(2y)$$

$$z_y = 3x - 3y^2 + 3y\left(y^2-2x\right)^{1/2}$$

and its second derivatives:

$$z_{yx} = 3 + 3y\left(\frac{1}{2}\right)\left(y^2-2x\right)^{-1/2}(-2)$$

$$z_{yx} = 3 - \frac{3y}{\sqrt{y^2-2x}}$$

and

$$z_{yy} = -6y + 3 \left[\sqrt{y^2-2x} + y \left(\frac{1}{2}\right)\left(y^2-2x\right)^{-1/2}(2y)\right]$$

$$z_{yy} = -6y + 3 \left[\sqrt{y^2-2x} + \frac{y^2}{\sqrt{y^2-2x}}\right]$$

$$z_{yy} = -6y +3\left[\frac{2(y^2-x)}{\sqrt{y^2-2x}}\right]$$

$$z_{yy} = -6y + \frac{6(y^2-x)}{\sqrt{y^2-2x}}$$

So clearly we have $$z_{xy} = z_{yx}$$, as expected, and we also have:

$$\left(z_{xy}\right)^2 = 9 + \frac{9y^2}{y^2-2x} - \frac{18y}{\sqrt{y^2-2x}}$$

$$\left(z_{xy}\right)^2 = \frac{18(y^2-x)}{y^2-2x} - \frac{18y}{\sqrt{y^2-2x}}$$

and we also have:

$$z_{xx} \cdot z_{yy} = \frac{3}{\sqrt{y^2-2x}} \cdot \left(-6y + \frac{6(y^2-x)}{\sqrt{y^2-2x}}\right)$$

$$z_{xx} \cdot z_{yy} = \frac{18(y^2-x)}{y^2-2x} - \frac{18y}{\sqrt{y^2-2x}}$$

and so $$z_{xx} \cdot z_{yy} = \left(z_{xy}\right)^2$$ as desired.

Ince can be a very frustrating text. It's filled w/ great information, but this is the second (serious) typo in just the first problem set...

7. May 30, 2010

### jdwood983

Woohoo, I was right! It just wasn't the power I was thinking of (you already pointed out the 3/2 power should be as is due to the 5/2 not working). I'm glad you were able to figure the solution out. I imagine that there's plenty more typos in that text, and am very surprised that the text made it through 4 editions with zero (it seems) corrections to such egregious errors. I did a quick google search for errata, but found none.