I have the following exercise:
Verify that Ex=f(z-ct)+g(z+ct) is a solution of the one dimensional wave equation.
But I don't get what I should do, I mean, isn't Ex the formula of the one dimensional wave equation itself? U.U
How can I verify it?
Thank you.
review section 5.2 of this: https://web.stanford.edu/class/math220a/handouts/waveequation1.pdf
Bobpursley coudl I get an email from you? I need to ask you something.
No emails. Our web safety issues don't allow that.
:( I could really use some assistance better than the provided from a input field of HTML xD, and my questions usually have symbols that require more precision in order for me to understand them.
Is there nothing we can do?
Not now, we are working on a more flexible input field. Or a white board.
The equation is:
d^2y/dx^2 = (1/c^2 )[d^2y/dt^2]
here:
d^2x/dz^2 = (1/c^2 )[d^2x/dt^2] with the strange letters for displacements
I will just do the half that is moving right (z+ct)= constant or f(z-ct)
write as f(w) where w = z-ct
df/dz = df/dw * dw/dz = df/dw
d^2f/dz^2 = d^2f/dw^2 dw/dz = d^2f/dw^2
and then
df/dt = df/dw * dw/dt = df/dw (-c) ah hah !
d^2f/dt^2 =d^2f/dw^2 (-c)(-c) = c^2 d^2f/dw^2
so in the end
d^2f/dt^2 /c^2 = d^2f/dw^2 = d^2f/dz^2
which is what we had to show.
note that due to the (-c)(-c) = c*c
it will come out the same for x+ct as for x-ct
To verify that `Ex = f(z-ct) + g(z+ct)` is a solution of the one-dimensional wave equation, we need to substitute `Ex` into the wave equation and confirm that it satisfies the equation.
The one-dimensional wave equation is given by:
```
∂²Ex / ∂t² - c² ∂²Ex / ∂z² = 0
```
where `c` is the wave propagation speed.
To verify the solution, we need to calculate the second partial derivatives of `Ex` with respect to both `t` and `z` and substitute them into the wave equation to check if it equals zero.
Let's start by calculating the second partial derivatives:
1. Calculate the first partial derivative with respect to `t`:
```
∂Ex / ∂t = -cf'(z-ct) + cg'(z+ct)
```
2. Calculate the second partial derivative with respect to `t`:
```
∂²Ex / ∂t² = c²f''(z-ct) + c²g''(z+ct)
```
3. Calculate the first partial derivative with respect to `z`:
```
∂Ex / ∂z = f'(z-ct) + g'(z+ct)
```
4. Calculate the second partial derivative with respect to `z`:
```
∂²Ex / ∂z² = f''(z-ct) + g''(z+ct)
```
Now, substitute these derivatives back into the wave equation:
```
c²f''(z-ct) + c²g''(z+ct) - c²(f''(z-ct) + g''(z+ct)) = 0
```
Simplifying the equation:
```
c²f''(z-ct) + c²g''(z+ct) - c²f''(z-ct) - c²g''(z+ct) = 0
```
Notice that the terms with first and second derivatives cancel each other, so we are left with:
```
0 = 0
```
Since the equation is identically satisfied as `0 = 0`, we can conclude that `Ex = f(z-ct) + g(z+ct)` is indeed a solution of the one-dimensional wave equation.
In summary, to verify a solution of the wave equation, substitute the proposed solution into the equation and check if it satisfies the equation identically.