Solution to homogeneous wave equation

[kiwi101]

Homework Statement

Prove by direct substitution that any twice differentiable function of (t-R$\sqrt{με}$) or of (t+R$\sqrt{με}$) is a solution of the homogeneous wave equation.

Homework Equations

Homogeneous wave equation = ∂²U/ ∂R² - με ∂²U/∂t² = 0

The Attempt at a Solution

Could you guys give me a direction or a way I should think. I have no idea on how I should approach this problem.

 

[PeroK]

Why not start by differentiating a general function and see what you get?

 

[kiwi101]

By differentiating a general function you mean differentiating t-R√με in terms of t?

 

[PeroK]

No, that's not a general function. What do you think this means:

Any twice differentiable function of (t-R√με) or of (t+R√με)

 

[kiwi101]

I think it means f"(t-R√με)

 

[PeroK]

Not quite. That's the second derivative of a general function (f) of t - R√με.

You also need to relate partial derivatives (wrt t and R) to ordinary derivatives.

 

[kiwi101]

Well isn't a twice differentiable function one that has a second derivative?
Could you elaborate or demonstrate another example about linking partial derivatives to the ordinary ones

 

[PeroK]

Well isn't a twice differentiable function one that has a second derivative?

There's a difference between a twice differentiable function and the second derivative of that function. E.g:

$$f(x) = x^3$$
Is a twice differentiable function. And:
$$f''(x) = 6x$$ is its second derivative.

To get you started, you would say something like:

$$Let \ U(t, R) = f(t - R\sqrt{μ ε})$$ where f is a twice differentiable function.

Now you need to calculate the partial derivatives of U. Do you know how to do that?

 

[kiwi101]

∂U/∂t = 1

∂U/∂R = √με

So that is the partial derivative of U(t,R)

I can't take the second derivative of this because it is constants.

 

[PeroK]

∂U/∂t = 1

∂U/∂R = √με

So that is the partial derivative of U(t,R)

That's not right. You will need to use the chain rule for the partial derivatives of U.

 

[kiwi101]

I know how to solve via chain rule
but there is no given value of t & R so...

∂U/∂f = ∂U/∂t * dt/dU + ∂U/∂R * dR/dU
= (1) * (?) + (√με) * (?)

 

[PeroK]

The chain rule works like this:

$$\frac{\partial{U}}{\partial t} = f'(t - R\sqrt{με}).1 \ \ and \ \ \frac{\partial{U}}{\partial R} = f'(t - R\sqrt{με}).(-\sqrt{με})$$

What you actually have here is:

$$U(t, R) = f(g(t,R)) \ where \ g(t, R) = t - R\sqrt{με}$$So:

$$\frac{\partial{U(t, R)}}{\partial t} = f'(g(t, R)).\frac{\partial{g(t, R)}}{\partial t}$$

 

[kiwi101]

What I am confused about is what are u deriving f'(g(t,R) in terms of?
So for example the answer to
∂U(t,R)/∂t = f'g(t,R) * 1
I don't understand where we are going with this.
I am sorry for testing your patience but I am confused and I want to learn this

 

[PeroK]

Okay. I find it really important with multi-variable calculus to key an eye on what everything is is function of. Functions can be functions of one variable, two variables or vectors and be real or vector valued. So:

In this case, U is real-valued function of two variables (R and t). So, it has partial derivatives wrt R and t.

f, on the other hand, is an ordinary real valued function of a single variable. So, f' and f'' are also ordinary real-valued functions. These are well-defined functions (if f is twice differentiable).

And, g is the specific function of two variables: g(t, R) = t - R√με

Now, one form of the chain rule is:

$$\frac{\partial{U(t, R)}}{\partial t} = f'(g(t, R)).\frac{\partial{g(t, R)}}{\partial t}$$

That's the one that's relevant here. It's probably worth taking time to fully understand this.

For example, if f(x) = sin(x), then f'(x) = cos(x) etc.

And, if $$U(r, R) = sin(t^2 - R^2) \ with \ g(t, R) = t^2 - R^2$$, then:

$$\frac{\partial{U(t, R)}}{\partial t} = f'(g(t, R)).\frac{\partial{g(t, R)}}{\partial t} = cos(t^2 - R^2).(2t)$$

 

[PeroK]

... alternatively, you could let x = t - R√με, then:

U = f(x) and

$$\frac{\partial{U}}{\partial{t}} = \frac{df}{dx}\frac{\partial{x}}{\partial{t}}$$

Maybe you prefer to look at it that way?

You can think of these things in terms of functions or variables.

 

[kiwi101]

Thanks the example helped
So according to this ∂U(t,R)/∂t = t -R√με * (1)

 

[PeroK]

Thanks the example helped
So according to this ∂U(t,R)/∂t = t -R√με * (1)

What happened to f?

 

[kiwi101]

Actually if x = t-R√με
& U = f(x)

∂U/ ∂t = df/dx * ∂x/∂t
= 1 * 1
= 1

 

[kiwi101]

This is how I interpreted your example
x=t² + R²
∂U(t,R)/∂t = f'(x) * ∂x/∂t
= cos(x) * (2t)
= cos(t² + R²) * 2t

 

[PeroK]

So, the derivative of any function is 1?

What if f(x) = sin(x)? Then df/dx = cos(x). It's certainly not 1.

 

[PeroK]

The chain rule works like this:

$$\frac{\partial{U}}{\partial t} = f'(t - R\sqrt{με}).1 \ \ and \ \ \frac{\partial{U}}{\partial R} = f'(t - R\sqrt{με}).(-\sqrt{με})$$

Start from here! I've given you the first pd's!

 

[kiwi101]

Can you tell me what the answer for U(r,t) = t² + R² would be?
Unless I see a visual I am going to remain confused because I know the answer is very obvious. I am just confused.

 

[PeroK]

Can you tell me what the answer for U(r,t) = t² + R² would be?
Unless I see a visual I am going to remain confused because I know the answer is very obvious. I am just confused.

That was just an example to explain the chain rule. See my last post. You've got the first partial derivatives. You need the second partials. Do you want to have a go at doing those?

 

[kiwi101]

Is the second partial derivative going to be:

∂²U/∂t = f"(t-R√με) *(1)

∂²U/∂R = f"(t-R√με) * (με)

 

[PeroK]

Yes! Nearly there. Just plug those into the wave equation.

 

[kiwi101]

I get it!
Thank you so much for bearing with me and explaining this to me step by step!