1. Dec 8, 2009

### yungman

I am a self studier, there might be things seems obvious to you guys that are not obvious for me. I have a few questions here, I write down the equation first and present the question in the last part. Please bare with me. Thanks.

Eulers equation of 2nd order: $$L[y]=x^{2} y''+ \alpha xy'+ \beta y = 0$$ (3)

Let $$y = x^{r}$$

=> $$L[x^{r}]=x^{r}[r(r-1)+\alpha x(x^{r})'+ \beta (x^{r})] = 0$$

If r is a root of the quadratic equation: $$F(r)=r(r-1)+\alpha r + \beta = 0$$

$$r_{1} , r_{2} =\frac{-(\alpha - 1)^{+}_{-}\sqrt{(\alpha-1)^{2}-4\beta}}{2}$$

and

$$F(r)=(r-r_{1})(r-r_{2})$$

For $$r_{1} = r_{2}$$, $$F(r)=(r-r_{1})^{2}$$, Both F(r) and F'(r) =0.

We only find $$y_{1}=x^{r}$$. We need to find $$y_{2}.$$

We differentiating(3) $$\frac{d}{dr}[L(x^{r}]=\frac{d}{dr}{x^{r}F(r)]$$(4)

$$\frac{dx^{r}}{dr}=x^{r}ln(x)$$

$$L[x^{r}ln(x)]=(r-r_{1})^{2}x^{r}ln(x)+2(r-r_{1})x^{r}$$(5)

For $$r=r_{1}$$, right side of (5) equal to zero.

Therefore $$y_{2}(x)=x^{r_{1}}ln(x).$$

My questions are:

1) How is $$\frac{d}{dr}[L(x^{r}]=\frac{d[x^{r}ln(x)]}{dr}$$ from (4) to (5)?

2) Why For $$r=r_{1}$$, right side of (5) equal to zero. And therefore $$y_{2}(x)=x^{r_{1}}ln(x)?$$

Last edited: Dec 8, 2009
2. Dec 8, 2009

### yungman

I have updated the post. Thanks.

Alan

3. Dec 8, 2009

### yungman

I want to clarify question 1): Why is

$$\frac{d}{dr}[L(x^{r}]=L[\frac{d(x^{r})}{dr}]=L[x^{r}ln(x)]$$

This is an example that I follow the logic and put in my own function. If you let $$y=x^{2}+2x, L[y]=4y+3$$

$$\frac{d}{dx}[L(y)]=L[\frac{d(x^{2}+2x)}{dx}]=L[2x+2]=4(2x+2)+3=8x+11$$

Where if you substitude $$y=x^{2}+2x$$ into 4y+3 then differentiate will give $$\frac{d[4(x^{2}+2x)+3]}{dx}=8x+8$$

Obvious the two are not equal!!! This is how I under what the book try to use.

I have one more question:

3) I fail to see how $$\frac{dx^{r}}{dr}=x^{r}ln(x)$$and if this equal to zero, then $$x^{r}ln(x)$$ is the second solution. Why differentiate respect to r give the second solution to the original equation which is a differential equation respect to x.

4. Dec 8, 2009

### LCKurtz

What you have is a statement about order of differentiation. Your "counterexample" is not two differentiation operators. Look at it this way. xris a function of x and r. Let's say R is an operator that differentiates with respect to r and L differentiates with respect to x. Now what happens to f(x,r) when operated on by these operators?

$$R(L(f(x,r)) = R(f_x(x,r)) = f_{xr}(x,r)$$

$$L(R(f(x,r)) = L(f_r(x,r)) = f_{rx}(x,r)$$

This is the result of Clairaut's theorem, which says if the mixed partials are continuous as functions of two variables, they are equal. This combined with linearity gives the result for general linear L.

5. Dec 8, 2009

### matematikawan

For this one, I think we should be thinking along partial derivatives.
$$\frac{\partial}{\partial r} ( \frac{\partial y}{\partial x}) = \frac{\partial}{\partial x} ( \frac{\partial y}{\partial r})$$

$$\frac{d}{dr}[L(y)]= \frac{d}{dr}( x^{2} y\prime\prime + \alpha xy\prime+ \beta y )= L[\frac{dy}{dr}]$$

ps. Help!! How do we write y" and y' in latex?

6. Dec 8, 2009

### LCKurtz

$$y',\ y''$$

I just use the single apostrophe two keys to the left of the L on the keyboard with no ^.

7. Dec 8, 2009

### yungman

I read in other section on finding second solution in regular singular series with indicial root r=r1=r2. the method used is exactly the same:

Since differentiate resp to x only get one solution. differentiate resp to r and if equal to zero, then that would be the second solution!!!

That sounds very thin!!!! What is the reason behind this?

Last edited: Dec 9, 2009
8. Dec 9, 2009

### yungman

I think I understand what you mean. I just want to verify with you:

1) For a function f(x,r) where it's dirivatives resp to x and r are both continuous on [a,b], then $$f_{xr}=f_{rx}$$.

2) I follow what you say that the example that I gave is not correct because L[y] in the example is not a differential operator.

Therefore $$\frac{d}{dr}[L(x^{r}]=L[\frac{d(x^{r})}{dr}]=L[x^{r}ln(x)]$$ is true and I can move d/dr inside the L[] as shown in the equation.

Thanks
Alan

I still don't get the reasoning how the author get the second solution by differentiating by r and if the result is zero, then the value inside L[] is the second solution.

Thanks

9. Dec 9, 2009

### LCKurtz

When you substitute y = xr into your Euler DE you saw that the xr factored out and left it multiplied by the indicial polynomial which I will call p(r):

(*) L(xr) = xrp(r)

If r is a root of the indicial polynomial p(r), then p(r) = 0 and the right side of that equation is zero so you know L(xr)=0, so xr is a solution to your DE L(y) = 0. So you have your first solution. So far so good.

Now suppose r is a double root of p(r). That means both p(r) and p'(r) are zero. Watch what happens if we differentiate equation (*) above with respect to r. On the left side, as we have seen, we can differentiate inside the L operator, and on the right side we can use the product rule, so we get:

L(xrln(x)) = xrln(x)p(r) + xrp'(r) = 0

This shows you that xrln(x) is a solution to L(y) = 0 when r is a double root.

[Edit: Corrected a term in the second line above]

Last edited: Dec 9, 2009
10. Dec 9, 2009

### yungman

So what you are saying is $$L[y]=x^{2} y''+ \alpha xy'+ \beta y = 0$$

which give $$L[y]=x^{r}[r^{2}+\alpha r+\beta]=0$$

If r is the root, $$[r^{2}+\alpha r+\beta]=0$$

$$L[x^{r}ln(x)]=x^{2} [x^{r}ln(x)]''+ \alpha x[x^{r}ln(x)]'+ \beta [x^{r}ln(x)]$$

From what you said:$$L[x^{r}ln(x)]=p(r)x^{r}ln(x)+p'(r)x^{r}$$ and both p(r) and p'(r) = 0.

Therefore $$L[x^{r}ln(x)]=x^{2} [x^{r}ln(x)]''+ \alpha x[x^{r}ln(x)]'+ \beta [x^{r}ln(x)]=x^{r}( .............) =0$$

This prove $$x^{r}ln(x)$$ is a solution and W[]not equal to zero means this is the second independent solution.

Am I getting it correct?

Alan

Last edited: Dec 9, 2009
11. Dec 9, 2009

### LCKurtz

Once you have established the equation

(*) L(xr) = xrp(r)

the argument I gave you settles it. While it is correct to put the alphas and betas back in and write out L(y), I think that just obscures what is going on.

12. Dec 9, 2009

### yungman

Alan

13. Dec 9, 2009

### LCKurtz

Alan, I just noticed and error which I edited at the bottom of post #9 in this thread. It doesn't change the thrust of the argument.

14. Dec 9, 2009

### yungman

I notice that, as you can see from my reply. I am not going to tell the teacher his mistake, that would be like a smart alec!!!!

Alan

15. Dec 9, 2009

### matematikawan

What good is this method?
http://komplexify.com/epsilon/category/quotes/" (Ron Getoor)

The method doesn't works on linear DE with constant coefficient
$$L[y]= ay''+ by' + c y = 0$$
because erx and rerx are not linearly independent.

ps. Now I know how to write y" in latex. Thanks.

Last edited by a moderator: Apr 24, 2017
16. Dec 9, 2009

### yungman

This kind of x^r only work with Euler, Polynomial coef regular singular series type differential equation and actually in my case, I am studying Bessel equation with r1=r2. They all use the same method. This is the reason I try so hard to understand this step by step instead of just use the equation to solve the problem.

Regarding to typing the equation. I actually went through some of the posts here, when I found equation that I want to write out, I just pretend to respond to the post by hitting the "quote" tap. The text form of the equation will show up. I just copy the tex with the equation onto the word file. Then later when I want to write similar equation, I just copy from the word file and start changing to my own equation!!!

Actually, after I do this for half an hour, I start to remember them and I start typing the tex myself. YOu notice I type in a lot of equation on #1, half of it is to pratice with writing the equation!!! Since I am studying PDE on my own, I have a feeling that I'll be on here quite often, might as well learn the "language"!!

Won't take long before you can pretty much type it in tex on most equations.

Last edited by a moderator: Apr 24, 2017
17. Dec 10, 2009

### elibj123

The method doesn't work for constant coefficients
+Differentiating erx wrt to r gives you xerx which is not linearly dependent on erx

Last edited by a moderator: Apr 24, 2017
18. Dec 10, 2009

### LCKurtz

Yes, it is linearly independent of erx. To be dependent it would have to be a constant times erx.

19. Dec 10, 2009

### elibj123

You've just repeated what I said....

20. Dec 10, 2009

### LCKurtz

Yep. I must have read too fast. When you said the method doesn't work on constant coefficient equations I didn't read carefully from there. Now I'm getting mixed up on who said what.