# Is This General Solution Valid for the Wave Equation?

• tandoorichicken
In summary, the conversation discusses the problem of proving a given general solution for the wave equation. The given solution is p(x,t) = A1f1(x-c0t) + A2f2(x+c0t). The conversation also includes calculations and assumptions made along the way, such as the linearity of the wave equation, the relationship between spatial position and time, and the transformation to new variables. The nabla symbol is also mentioned and its use for more than one spatial coordinate is discussed. Finally, the conversation brings up questions about the constants A1 and A2 and the dimensionality of x.
tandoorichicken
Hello everyone.

I'm asked in a problem to prove that a given general solution is valid for the wave equation
$$\nabla^2 p - \frac{1}{c_0^2} \frac{\partial^2 p}{\partial t^2} = 0$$.

The given solution was
$$p(x, t) = A_1 f_1 (x - c_0 t) + A_2 f_2 (x + c_0 t)$$.

I just need a check of work here. I plugged in the solution and calculated out for a 1-dimensional wave equation, but I might have made some assumptions along the way that are un-kosher.

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Here is the work:

$$\nabla^2 p - \frac{1}{c_0^2}\frac{\partial^2 p}{\partial t^2} = \frac{\partial^2 p}{\partial x^2} - \frac{1}{c_0^2}\frac{\partial^2 p}{\partial t^2} = 0$$

$$\frac{\partial^2 p}{\partial x^2} = A_1 \frac{\partial^2 f_1}{\partial x^2} + A_2 \frac{\partial^2 f_2}{\partial x^2}$$
$$\frac{\partial^2 p}{\partial t^2} = A_1 c_0^2\frac{\partial^2 f_1}{\partial t^2} + A_2 c_0^2\frac{\partial^2 f_2}{\partial t^2}$$

$$\frac{\partial^2 p}{\partial x^2} - \frac{1}{c_0^2}\frac{\partial^2 p}{\partial t^2} = A_1 \frac{\partial^2 f_1}{\partial x^2} + A_2 \frac{\partial^2 f_2}{\partial x^2} - \frac{1}{c_0^2} (A_1 c_0^2\frac{\partial^2 f_1}{\partial t^2} + A_2 c_0^2\frac{\partial^2 f_2}{\partial t^2}) = 0$$

$$A_1 (\frac{\partial^2 f_1}{\partial x^2} - \frac{\partial^2 f_1}{\partial t^2}) + A_2 (\frac{\partial^2 f_2}{\partial x^2} - \frac{\partial^2 f_2}{\partial t^2}) = 0$$

At this point I assumed that since this was the LINEAR wave equation, this meant that spatial position varied linearly with time, so some equation could be written $x = kt + b$ where k and b are arbitrary constants.

$$\frac{\partial^2 f}{\partial t^2} = \frac{\partial^2 f}{\partial x^2}\frac{d^2 x}{d t^2} = 0$$

Likewise, while it doesn't make sense physically, one could make the mathematically sound statement $t = mx + d$ where m and d are arbitrary constants.

$$\frac{\partial^2 f}{\partial x^2} = \frac{\partial^2 f}{\partial t^2}\frac{d^2 t}{d x^2} = 0$$

Therefore the above equations reduce to 0.

Q.E.D?

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The assumption doesn't make sense to me. Spatial position of what? How does you assumption work for a known solution like exp[i(x - c0 t)]?

I was thinking that you might be able to do this by transforming to new variables

$\phi_\pm = x \pm c_0 t$

Sorry, completely forgot the context

This is for ultrasonic imaging. The function p(x,t) is a function of position (depth) in tissue and time.

I'm just not sure your problem is 1-D, or that you just made that assumption... Things I am just concerned about -- are your givens A1, A2 constants or vectors (like wih E-fields, they would be indicative of polarizations) and is x one-dimensional... or a vector (which would often be indicated by r, but not always)? These thoughts complicate the things, as now the spatial derivatives are more interesting...

The nabla is for more than one spatial coordinate. Let's assume for simplicity that the "x" in the solution means that the problem is essentially one-dimensional.

Then you might redo the differentiations using the chain rule properly.

E.g.

$$\frac{\partial p}{\partial x}=\frac{dp}{d\left(x-c_{0}t\right)}\frac{\partial \left(x-c_{0}t\right)}{\partial x} +\frac{dp}{d\left(x+c_{0}t\right)}\frac{\partial \left(x+c_{0}t\right)}{\partial x}$$

and using the expression the problem offers, you finally get

$$\frac{\partial p}{\partial x}=A_{1}\frac{df_{1}}{d\left(x-c_{0}t\right)}\frac{\partial \left(x-c_{0}t\right)}{\partial x}+A_{2}\frac{df_{2}}{d\left(x+c_{0}t\right)}\frac{\partial \left(x+c_{0}t\right)}{\partial x}=...$$

Daniel.

## 1. What is the wave equation and why is it important in science?

The wave equation is a mathematical formula that describes the behavior of waves in various physical systems. It is important because it allows scientists to accurately predict and analyze the properties and behavior of waves, which are fundamental to many natural phenomena, such as sound, light, and water waves.

## 2. How do you solve the wave equation?

The general solution for the wave equation involves using mathematical techniques such as separation of variables and Fourier analysis. Depending on the specific problem, the solution may also require applying boundary or initial conditions. This process can be complex and may involve using advanced mathematical tools, such as differential equations and complex numbers.

## 3. What is the role of boundary conditions in proving the solution of the wave equation?

Boundary conditions are essential in solving the wave equation because they provide information about the behavior of the wave at the boundaries of the system. By incorporating these conditions into the equation, scientists can determine the specific form of the solution and accurately model the behavior of waves in a given system.

## 4. Can the wave equation be applied to different types of waves?

Yes, the wave equation is a universal formula that can be used to describe the behavior of a wide range of waves, including mechanical, electromagnetic, and quantum waves. However, the specific parameters and boundary conditions may vary depending on the type of wave being studied.

## 5. How is the solution of the wave equation verified or tested?

The solution of the wave equation can be verified through experimental observations and measurements. Scientists can also use numerical simulations and computer modeling to test the accuracy of the solution and compare it to real-world data. Additionally, the solution can be checked using mathematical methods, such as plugging it back into the equation to see if it satisfies the original problem.

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