# Deriving the 2D Wave Equation for a Vibrating Membrane using Newton's Law of Motion and Taylor Series

This is my first time on this site so i hope i followed the guidelines well enough. I am working on deriving the two-dimensional wave equation for a vibrating membrane in my partial differential equations course. I have been given a hint to use Newton's Law of Motion and the Taylor Series. Here's the information I have:

The 2D wave equation for a vibrating membrane (the equation I'm trying to derive) is:

$$$$\frac{\partial^2 u}{\partial t^2}=c^2 \nabla^2 u=c^2\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right)$$$$

Newton's Law of Motion is given by:

$$$$F=m a$$$$

The Taylor Series is represented as: $$$$\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n !}(x-a)^n=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^2+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^3+\cdots$$$$

Here's my attempt so far:

• I considered a small rectangular segment of the highly stretched membrane between $$x$$ and $$x+\Delta x$$ and $$y$$ and $$y+\Delta y$$.
• The displacement of the membrane is given by $$z=u(x, y, t)$$.
• I assumed that all slopes $$\left(\frac{\partial u}{\partial x}\right.$$ and $$\left.\frac{\partial u}{\partial y}\right)$$ are small, so the vibrations are entirely vertical, and the tension is approximately constant.
• Let $$T_0$$ be the constant force per unit length exerted by the stretched membrane on any edge of the rectangular segment.

The mass of the small segment at $$(x, y)$$ is given by:

$$$$m=\rho_0 \Delta x \Delta y$$$$

where $$\rho_0$$ is the mass density of the membrane (mass per unit area).

Newton's second law of motion for the vertical motion of the small segment at $$(x, y)$$ is:

$$$$m \frac{\partial^2 u}{\partial t^2}(x, y, t)=F$$$$

where $$F$$ is the sum of all the vertical forces.

The vertical forces are the vertical components of the tensile forces in effect along all four sides. We can analyze the vertical components along the sides by considering the vertical component of tension (per unit length) as $$\sin \phi$$, where $$\phi$$ is the angle of inclination for fixed $$y$$. Since $$\phi$$ is assumed small, $$\sin \phi \approx \tan \phi=\partial u / \partial x$$.

Summing up the vertical forces on all four sides, we have:

$$$$F=T_0\left[\Delta y \frac{\partial u}{\partial x}(x+\Delta x, y)-\Delta y \frac{\partial u}{\partial x}(x, y)+\Delta x \frac{\partial u}{\partial y}(x, y+\Delta y)-\Delta x \frac{\partial u}{\partial y}(x, y)\right]$$$$

At this point, I am stuck and believe that I may need to use the Taylor Series. I know that I could substitute the force equation above into Newton's Law of Motion, as stated earlier in the question, which is something I know I'll have to do anyway. However, after that, I'm left with a lengthy equation that I'm unsure how to proceed with.

Can anyone help me figure this out? Thank you in advance :)

Use the Taylor series "in x-direction" for $$\frac{\partial u}{\partial x}$$ and the Taylor series "in y-direction" for $$\frac{\partial u}{\partial y}$$. $$\frac{\partial u}{\partial x}(x^{\star},y) \approx \frac{\frac{\partial u}{\partial x}(x,y)}{0!} \left(x^{\star}-x\right)^0+ \frac{\frac{\partial^2 u}{\partial x^2}(x,y)}{1!} \left(x^{\star}-x\right)^1$$ If you now set $$x^{\star} = x+\Delta x$$, you get $$\frac{\partial u}{\partial x}(x+\Delta x,y) \approx \frac{\partial u}{\partial x}(x,y) + \frac{\partial^2 u}{\partial x^2}(x,y)\; \Delta x$$ and analogously $$\frac{\partial u}{\partial y}(x,y+\Delta y) \approx \frac{\partial u}{\partial y}(x,y) + \frac{\partial^2 u}{\partial y^2}(x,y)\; \Delta y$$ This leads to $$F\approx T_0\;\Delta x\;\Delta y \left( \frac{\partial^2 u}{\partial x^2}(x,y)+ \frac{\partial^2 u}{\partial y^2}(x,y) \right)$$