# Need to check the answer of some simple PDE

1. Feb 6, 2010

### yungman I don't have the answer of these question. Can someone take a look at a) and tell me am I correct? I don't even know how to solve b)

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

a) Show $$u(x,t)=F(x+ct) + G(x-ct)$$ is solution of $$\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}$$

3. The attempt at a solution

let $$\alpha = x+ct,\beta = x-ct \Rightarrow \frac{\partial \alpha}{\partial x}= 1, \frac{\partial \alpha}{\partial t}= c,$$ and also $$\frac{\partial \beta}{\partial x}= 1, \frac{\partial \beta}{\partial t}= -c$$

$$\frac{\partial u}{\partial x} = \frac{\partial F(\alpha)}{\partial \alpha} \frac{\partial \alpha}{\partial x} + \frac{\partial G(\beta)}{\partial \beta}\frac{\partial \beta}{\partial x} = \frac{\partial F(\alpha)}{\partial \alpha} + \frac{\partial G(\beta)}{\partial \beta}$$

$$\frac{\partial^2 u}{\partial x^2} = \frac {\partial (\frac{\partial F(\alpha)} {\partial \alpha}) }{\partial \alpha} \frac{\partial \alpha}{\partial x} + \frac {\partial (\frac{\partial G(\beta)}{\partial \beta })} {\partial \beta} \frac{\partial \beta}{\partial x} = \frac{\partial^2 F(\alpha)}{\partial \alpha^2} + \frac{\partial^2 G(\beta)}{\partial \beta^2}$$

$$\frac{\partial u}{\partial t} = \frac{\partial F(\alpha)}{\partial \alpha}\frac{\partial \alpha}{\partial t} + \frac{\partial G(\beta)}{\partial \beta}\frac{\partial \beta}{\partial t} = c[\frac{\partial F(\alpha)}{\partial \alpha} - \frac{\partial G(\beta)}{\partial \beta}]$$

$$\frac{\partial^2 u}{\partial t^2} = c\frac {\partial (\frac{\partial F(\alpha)} {\partial \alpha}) }{\partial \alpha} \frac{\partial \alpha}{\partial t} - c\frac {\partial (\frac{\partial G(\beta)}{\partial \beta })} {\partial \beta} \frac{\partial \beta}{\partial t} = c^2[\frac{\partial^2 F(\alpha)}{\partial \alpha^2} + \frac{\partial^2 G(\beta)}{\partial \beta^2}]$$

$$\Rightarrow \frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}$$

Therefore:
$$u(x,t)=F(x+ct) + G(x-ct)$$ is solution of $$\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}$$

b) Transform $$\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}$$ in to $$\frac{\partial^2 u}{\partial \alpha \partial \beta}=0$$

where $$u(x,t)=F(x+ct) + G(x-ct)$$

and $$\alpha = x+ct,\beta = x-ct$$

Can someone give me a hint how to go by this?

Last edited: Feb 6, 2010
2. 3. Feb 6, 2010

### vela Staff Emeritus
Your solution to (a) is fine.

For part (b), try start by solving for x and t in terms of $\alpha$ and $\beta$.

4. Feb 6, 2010

### yungman Thanks for checking on a).

$$\alpha = x+ct,\beta = x-ct \\Rightarrow x=\frac{\alpha + \beta}{2}, t=\frac{\alpha - \beta}{2c}$$

$$\frac{\partial u}{\partial \alpha}= \frac{\partial u}{\partial x} \frac{\partial x}{\partial \alpha}=\frac{\partial F(\alpha}{\partial \alpha}+ \frac{\partial G(\beta)}{\partial x}\frac{\partial x}{\partial \beta}$$

5. Feb 6, 2010

### yungman Thanks for checking on a).

$$\alpha = x+ct,\beta = x-ct \Rightarrow x=\frac{\alpha + \beta}{2}, t=\frac{\alpha - \beta}{2c}$$

$$\frac{\partial u}{\partial \alpha}= \frac{\partial u}{\partial x} \frac{\partial x}{\partial \alpha} + \frac{\partial u}{\partial t} \frac{\partial t}{\partial \alpha} = \frac{\partial F(\alpha)}{\partial \alpha}+ \frac{\partial G(\beta)}{\partial x}\frac{\partial x}{\partial \beta} + \frac{\partial G(\beta)}{\partial t}\frac{\partial t}{\partial \beta}$$

But how do I find $$\frac{\partial x}{\partial \alpha},\frac{\partial x}{\partial \beta}?$$

6. Feb 6, 2010

### vela Staff Emeritus
$$x=\frac{\alpha+\beta}{2} \Rightarrow \frac{\partial x}{\partial \alpha} = \frac{1}{2}$$

and so on. You do pretty much the same thing you did in part (a) just with different variables.

7. Feb 6, 2010

### yungman I try this part and I got a different answer:

$$\frac{\partial x}{\partial \alpha} = \frac{\partial (\alpha + \beta)}{2} = \frac{1}{2} [\frac{\partial \alpha}{\partial \alpha} + \frac{\partial \beta}{\partial x} \frac{\partial x}{\partial \alpha}] = \frac{1}{2}[ 1 + \frac{\partial x}{\partial \alpha}]$$

$$\Rightarrow \frac{\partial x}{\partial \alpha} = 1$$

Which is the same as $$\frac {1}{(\frac{\partial \alpha}{\partial x})} = 1$$

What did I do wrong?

8. Feb 6, 2010

### vela Staff Emeritus
$x=x(\alpha,\beta)$ is a function of $\alpha$ and $\beta$. When you take the partial derivative wrt $\alpha$, you hold $\beta$ constant.

9. Feb 6, 2010

### yungman The thing that confuse me is both $$\alpha, \beta$$ have x and t. $$\alpha, \beta$$ are not independent variables like x and y in the normal case. $$\alpha = 2x-\beta$$.

That is the thing that really throw me off all this time. That is the reason I did it in post #6. Why it is wrong in post #6.

10. Feb 6, 2010

### vela Staff Emeritus
You can say the same thing about x and t. They both depend on $\alpha$ and $\beta$. It's like a change of basis from x and t to $\alpha$ and $\beta$.

11. Feb 6, 2010

### yungman I still don't get it. Let say x is position, t is time. We equate $$\alpha = x+ct, \beta=x-ct$$. This imply only $$\alpha, \beta$$ is depending on both position and time. That cannot reverse to imply x and t have a relation. x and t are completely indenpend variable. I just cannot turn this around in my mind.

Also in post #6, I don't assume anything there, I just step by step derive the formula. I look at it again and again, I just don't see what I did wrong. Can you comment about post #6 where exactly I did wrong that it won't work even if $$\alpha, \beta$$ are being treated as independent variables.

Thanks for your patient. I feel I am stuck in a spot.

12. Feb 6, 2010

### vela Staff Emeritus
Just as x and t are independent when you write $\alpha=\alpha(x,t)$ and $\beta=\beta(x,t)$, you consider $\alpha$ and $\beta$ to be independent when you write $x=x(\alpha,\beta)$ and $y=y(\alpha,\beta)$.

In #6, you're taking the partial derivative with respect to $\alpha$. That means $\beta$ is held constant so any derivative of $\beta$ will equal 0.

13. Feb 6, 2010

### yungman Please bear with me, I am still stuck!!! What is wrong with this:

$$\frac{\partial \beta}{\partial \alpha} = \frac{\partial \beta}{\partial x} \frac{\partial x}{\partial \alpha} = \frac{\partial \beta}{\partial x}$$

14. Feb 6, 2010

### owlpride Think about it in terms of linear algebra: (1,0) and (0,1) are two independent vectors that span R^2. Spanning R^2 means that given any point p in R^2, you can find coordinates x and y such that p = x (1,0) + y (0,1). (1,0) and (1,1) are also independent and form a basis for R^2, but they are not orthogonal.

The same is true for alpha and beta. They are two independent coordinates (in the linear algebra sense) and form a basis for R^2, but they are not orthogonal.

15. Feb 6, 2010

### owlpride Consider another example: take standard x-y coordinates and define a function t(x,y) = x+y. By the same argument,

$$\frac{\partial y}{\partial x} = \frac{\partial y}{\partial t} \frac{\partial t}{\partial x} = 1 \cdot 1 = 1$$

Do you see what happened?

16. Feb 7, 2010

### yungman I got out the multi variable book and read up on the chain rule, I am still reading and thinking about it. I thing I see wrong with my assumption is $$\frac{\partial x}{\partial \alpha}$$is not legal in the chain rule because it is travelling "up" the diagram.

I can see

$$\frac{\partial y}{\partial x} = \frac{\partial y}{\partial t} \frac{\partial t}{\partial x} = 1 \cdot 1 = 1$$

Is wrong because t is function of y, y is not a function of t.

$$\frac{\partial y}{\partial t}$$ is not allowed.

Am I correct?

Also is $$\alpha,\beta$$ independent variable that $$\frac{\partial \alpha}{\partial \beta}=0$$?

Last edited: Feb 7, 2010
17. Feb 7, 2010

### owlpride The diagrams are a good way to take partial derivatives. When you try to differentiate alpha with respect to beta, you are either moving up in the diagram or you need to build a circular diagram (beta is a function of x and t, x is a function of alpha and beta, etc).

Have you tried drawing a picture of what the alpha and beta coordinate curves look like? Once you draw a picture, you can see that you can freely vary alpha without affecting beta. You can also do this algebraically: Suppose beta = x-ct = const or t = (x - const)/c. If you move along the line t = (x - const)/c, you will vary alpha while keeping beta fixed.

It's the analog of moving parallel to the x axis (keeping the y coordinate fixed) in the x-y world. Alpha and beta are two independent coordinates, and the equations alpha = x + ct, beta = x - ct tell you how to convert between alpha/beta and x/y coordinates.

18. Feb 8, 2010

### yungman Can anyone show me how to prove this part?

Transform $$\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}$$ in to $$\frac{\partial^2 u}{\partial \alpha \partial \beta}=0$$

where $$u(x,t)=F(x+ct) + G(x-ct)$$

and $$\alpha = x+ct,\beta = x-ct$$

This is not a school work and I am still struggling on the Chain rule. I am hopping if I get the steps to solve this problem, I can reverse it and try to understand this. I think I spent enough time on this already.

Thanks

19. Feb 8, 2010

### owlpride I think you and vela have established that

du/d(alpha) = du/dx dx/d(alpha) + du/dt dt/d(alpha) = 1/2 du/dx + 1/2 du/dt
du/d(beta) = du/dx dx/d(beta) + du/dt dt/d(beta) = 1/(2c) du/dx - 1/(2c) du/dt

Can you say anything about the second partial derivatives of u with respect to alpha and beta?

20. Feb 8, 2010

### owlpride Or if you already know this part:
$$u(x,t)=F(x+ct) + G(x-ct) = F(\alpha) + G(\beta)$$
it's even easier:

The derivative of F(alpha) with respect to beta is zero. Differentiating G'(beta) with respect to alpha makes that zero. QED. Alpha and beta are independent variables!

21. Feb 8, 2010

### yungman I gave this a lot of thoughts. I still cannot accept $$\frac{\partial G(\beta)}{\partial \alpha}=0$$

Please bear with me. Let's take a look at this example:

Let $$G(\beta)=-\beta = \beta - 2\beta = x-ct-2x+2ct=(x+ct)-2x=\alpha -2x$$

$$\Rightarrow\frac{\partial G(\beta)}{\partial \alpha}= \frac{\partial \alpha}{\partial \alpha} + \frac{\partial 2x}{\partial \alpha}= 1+0=1$$

Yes I know I am playing around and the argument is very thin, but never the less, it is valid. My whole point is just because $$\alpha$$ and $$\beta$$ are independent variable, $$\frac{\partial G(\beta)}{\partial \alpha}$$ not necessary equal 0.

$$\frac{\partial \beta}{\partial \alpha}=0$$ do not imply at all $$\frac{\partial G(\beta)}{\partial \alpha}=0$$

I don't think the question is good to say $$\frac{\partial^2 u}{\partial \alpha \partial \beta}}=0$$ I have been struggling on this very point for two days!!!!

Please tell me if I am correct.

Thanks a million.

Alan

Last edited: Feb 9, 2010