Need to check the answer of some simple PDE

[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)

Homework Statement

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

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

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?

 

[vela]

Your solution to (a) is fine.

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

 

[yungman]

Your solution to (a) is fine.

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

Thanks for checking on a).

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

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

 

[yungman]

Your solution to (a) is fine.

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

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 $$\frac{\partial x}{\partial \alpha},\frac{\partial x}{\partial \beta}?$$

 

[vela]

$$x=\frac{\alpha+\beta}{2} \Rightarrow \frac{\partial x}{\partial \alpha} = \frac{1}{2}$$

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

 

[yungman]

$$x=\frac{\alpha+\beta}{2} \Rightarrow \frac{\partial x}{\partial \alpha} = \frac{1}{2}$$

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

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?

 

[vela]

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

 

[yungman]

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

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.

 

[vela]

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 $\beta$.

 

[yungman]

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 $\beta$.

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.

 

[vela]

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.

 

[yungman]

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.

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

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

 

[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.

 

[owlpride]

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

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

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?

 

[yungman]

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?

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 traveling "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 $$\frac{\partial \alpha}{\partial \beta}=0$$?

 

[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.

 

[yungman]

Can anyone show me how to prove this part?

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

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

 

[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?

 

[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!

 

[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

 

[yungman]

Anyone please?

 

[vela]

When you take a partial derivative with respect to $\alpha$, you hold the other independent variables constant, so

$$\frac{\partial G(\alpha,\beta)}{\partial \alpha} = \lim_{h\rightarrow 0}\frac{G(\alpha+h,\beta)-G(\alpha,\beta)}{h}$$

So say $G(\alpha,\beta)=\beta$ as in your example. The numerator is 0 because regardless of the value of $\alpha$ and $\alpha+h$, G is equal to $\beta$. It will work out similarly for any function that depends only on $\beta$.

 

[yungman]

When you take a partial derivative with respect to $\alpha$, you hold the other independent variables constant, so

$$\frac{\partial G(\alpha,\beta)}{\partial \alpha} = \lim_{h\rightarrow 0}\frac{G(\alpha+h,\beta)-G(\alpha,\beta)}{h}$$

So say $G(\alpha,\beta)=\beta$ as in your example. The numerator is 0 because regardless of the value of $\alpha$ and $\alpha+h$, G is equal to $\beta$. It will work out similarly for any function that depends only on $\beta$.

I think it is supposed to be:

$$\frac{\partial G(\alpha,\beta)}{\partial \alpha} = \lim_{\Delta\alpha\rightarrow 0}\frac{G(\alpha+\Delta\alpha,\beta)-G(\alpha,\beta)}{\Delta\alpha}$$



If you don't have reason to dispute my assertion that

In the case of $$G(\beta)=-\beta = \beta - 2\beta = x-ct-2x+2ct=(x+ct)-2x=\alpha -2x$$

$$x=\frac{\alpha+\beta}{2} \Rightarrow \frac{\partial x}{\partial \alpha} = \frac{1}{2}$$

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

Then I don't think it is zero.

I really appreciate your help and I really don't mean to be difficult. I just need to be convinced like you convinced me that the $$\alpha,\beta$$ are totally independent.

I might be missing something here, but from what I understand, it still don't make sense to me.

Thanks for all your help
Alan

 

[vela]

I think it is supposed to be:

$$\frac{\partial G(\alpha,\beta)}{\partial \alpha} = \lim_{\Delta\alpha\rightarrow 0}\frac{G(\alpha+\Delta\alpha,\beta)-G(\alpha,\beta)}{\Delta\alpha}$$

I'm not sure what your point is here. I called the small increment h, and you called it $\Delta\alpha$. It doesn't make a difference what you call it.

If you don't have reason to dispute my assertion that

In the case of $$G(\beta)=-\beta = \beta - 2\beta = x-ct-2x+2ct=(x+ct)-2x=\alpha -2x$$

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

Then I don't think it is zero.

Well, you made two mistakes. First, you flipped the sign of the second term. Second, $\partial x/\partial\alpha=1/2$, so your example actually does come out to zero, but that's just an accident of how the numbers work out. Still, it's pointless to address your argument because it's based on the misconception that varying $\alpha$ while holding $\beta$ constant will lead to variation of a function of only $\beta$.

I really appreciate your help and I really don't mean to be difficult. I just need to be convinced like you convinced me that the $$\alpha,\beta$$ are totally independent.

I might be missing something here, but from what I understand, it still don't make sense to me.

Well, it's apparent you aren't convinced $\alpha$ and $\beta$ are independent since you're arguing that varying $\alpha$ will necessarily induce a change in $\beta$. The fact they are independent means you can change one while holding the other constant.

When you vary $\alpha$, you can hold $\beta$ constant if you move in the right direction in the x-ct plane. (I'm using ct instead of just t to simplify the algebra a little.) From the relationships you derived earlier, you can determine the change in x and ct caused by a change in $\alpha$ and $\beta$:

$$\delta x = 1/2 (\delta\alpha+\delta\beta)$$
$$\delta (ct) = 1/2 (\delta\alpha-\delta\beta)$$

If $\beta$ is constant, you have $\delta\beta=0$, and you get $(\delta x,\delta (ct)) = (1/2,1/2) \delta\alpha$. Recall the directional derivative

$$\nabla_{\hat{u}} f = \hat{u}\cdot \nabla f = u_x \frac{\partial f}{\partial x} + u_y \frac{\partial f}{\partial y}$$

This suggests the partial derivative with respect to $\alpha$ is

$$\frac{\partial u}{\partial \alpha} = \frac{1}{2}\frac{\partial u}{\partial x}+\frac{1}{2}\frac{\partial u}{\partial (ct)}$$

Note this is exactly the same result you get when you write

$$\frac{\partial u}{\partial \alpha} = \frac{\partial x}{\partial \alpha}\frac{\partial u}{\partial x}+\frac{\partial (ct)}{\partial \alpha}\frac{\partial u}{\partial (ct)}$$

 

[ideasrule]

beta=2x-alpha, so -beta=alpha-2x. The partial derivative you computed should be 1-1=0.

 

[yungman]

Yes, I double checked, I was wrong and is equal to 0. I did tried with $$G(\beta)=-\beta^2$$ and still come out to zero. Maybe it is really true that $$\frac{\partial G(\beta)}{\partial \alpha}=0$$. I give up.

I just not convince that $$\frac{\partial G(\beta)}{\partial \alpha}=0$$. If anyone have a more convincing way to show, I would love to see it. As I said, most likely you guys are right, I just don't see a convincing prove like I see in other theorem proves and I am not very good in buying "It's just is".

Anyway, thanks for all your time.

 

[ideasrule]

Taking the partial derivative means, by definition, "hold all other variables constant and compute the derivative of this function with respect to this variable". Since all other variables are constant, well, all other variables are constant. If a function depends only on beta and beta doesn't change, then the function can't possibly change. This is why the partial derivative of G(beta) with respect to alpha is 0: beta, by definition, must be held constant.

To reinforce the point, why is the derivative equal to the slope of the tangent? It was defined that way. Why does the "+" sign represent addition? It was defined that way. Why is i =sqrt(-1)? It was defined that way.