Wave Equation Intuition in the case of a horizontal string

Question

I'm looking at a derivation / intuition on the wave equation
$$u''_{tt} - c^2u''_{xx} = 0.$$ where $u(x,t)$ is the displacement of a string in tension at point $x$ and time $t$.

(A) The explanation makes a leap of faith and states that the force $F$ on a small segment of the string (particle) is proportional to the concavity.

Further it recognizes that $F = ma = mu''_{tt}$ which I understand.

(B) It then states that the concavity is equal to $u''_{xx}.$ This is the part I don't get. Why is this the case? Shouldn't the concavity simply be $u(x,t)$? Isn't $u''_{xx}$ a measure of the change of the "acceleration of the slope"?

It then uses statement (A) and introduces $k=c^2$ to set up the equation $$u''_{tt} = ku''_{xx}.$$

1. What is going on in B?

2. What is the intuition behind a "inhomogeneous" wave equation? $$u''_{tt} - c^2u''_{xx} = f$$

Answer 1

Intuition for continuing your method :

(I may be wrong here) My intuition is that, from your image, the horizontal straight line is the steady state for the wave. The wave is in steady state if there is no force, while in your Image : the wave above the steady-state will have to balance (which will go down) with the one below the steady state (which will go up).

The wave above the steady state (left one) has negative concavity, while the other one has positive concavity.

Here is another logic : in linear wave equation, the right-wave (or wave with positive concavity, $u_{xx}>0$) in your Image will bounce upward..forming like the left-wave.. While the left-wave (or wave with negative concavity, $u_{xx} < 0$) will bounce downward..forming like the right-wave.

Concavity of the wave-shape is the second derivative wrt the spatial dimension $u_{xx}$. So $u_{tt}$, which is the force, will be negative or go downward when the concavity is negative like the left-wave, and vice-versa, so : $$u_{tt} = c^{2}u_{xx}$$

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I have some experiences in PDE's but never encounter this type of proof for wave equation. Do you have to understand this way? sometimes you can try to understand other proofs and then go back to this one and gain better insights.

Answer 2

Your wave's displacement also known as $u$ is dependant on both time and the location of the wave.

$u_{tt}$ is because a specific part of the wave's displacement varies due to its unsteady state.

$u_{xx}$ is because a specific part of the wave's displacement varies due to its curvature.

The equation's solution will be generally in the triangular form of a Fourier's series like $\sum {b_n sin(n \pi x/L)}$ or $\sum{a_n cos(n \pi x/L)}$, etc, depending on the boundary conditions it has.

If your system is constrained to something, a damper or friction, for instance, the PDE equation may be non-homogeneous. (e.g. $F-f_k=ma\to F=f_k+ma=ma_t \to$ the PDE will be non-homogeneous because it is dependant on another thing that cause non-homogeneity, like dampers)

For more clarification:

Imagine the wave of a rope that $u$ is displacement. For an element of the rope, we have naturally 2 forces exerted on each side of the element. Let's call them $T_1$ and $T_2$.

For the wave the displacement is only in y-direction and we have no displacement in x-direction, so:

$T_2 cos(\theta 2)-T_1 cos(\theta 1)=0 \to T_2 cos(\theta 2)=T_1 cos(\theta 1)=T$

$ \sum F=ma \to T_2 sin(\theta_2)-T_1 sin(\theta 1)=ma \to T \frac{sin(\theta 2)}{cos(\theta 2)}-T\frac{sin(\theta 1)}{cos(\theta 1)}=\rho \Delta x u_{tt}$, $(m=\rho \Delta x,a=u_{tt})$

$\to tan(\theta 2)-tan(\theta 1)=\rho \Delta x u_{tt}/T$

We know $tan$ of the angle of a tangent line is velocity. So, $tan(\theta)=u_x$

$\to tan(\theta 2)-tan(\theta 1)= u_x|_{x+\Delta x}-u_x|_{x}=\frac{\rho}{T} \Delta x u_{tt}$

$\to \frac{u_x|_{x+\Delta x}-u_x|_{x}}{\Delta x}=\frac{\rho}{T} u_{tt}=c^2 u_{tt}$

$\to u_{xx}=c^2 u_{tt}$