Chain rule for multivariable calculus.

Question

I can't seem to figure out this problem.

We let s and p be real functions that are twice continuously differentiable. And we define r and f such that: $$ r =r(x,y,z) = \sqrt{x^2+y^2+z^2} $$ and $$ f(x,y,z,t) = \frac{1}{r}( \ p (r-ct) + s(r+ct) \ ) $$ We need to show that $ \frac{\partial^2f}{\partial t^2} = c^2(\frac{\partial^2f}{\partial x^2} + \frac{\partial^2f}{\partial y^2} + \frac{\partial^2f}{\partial z^2})$

I know that we have to use the chain rule, so I tried finding functions g and h so that $ f = g \circ h$ but I can't seem to do so. Any tips to how I can understand this problem?

Thanks for your time, K.

Answer 1

Here is a hint on how to proceed:

The first term of $f$ is:

$$f(x,y,z,t) = \frac{p(r(x,y,z) - c t )}{r(x,y,z)}$$

We differentiate both sides with respect to t first:

$$\frac{\partial f}{\partial t} = \frac{\partial}{\partial t}\frac{p(r(x,y,z) - c t )}{r(x,y,z)} = \frac{r \frac{\partial p}{\partial t} - p \frac{\partial r}{\partial t}}{r^2}$$

Here is where the chain rule comes into play:

$$\frac{\partial p(r(x,y,z)-ct))}{\partial t} = \frac{\partial p(r(x,y,z)-ct)}{\partial (r(x,y,z)-ct)}\frac{\partial (r(x,y,z)-ct)}{\partial t}$$

This is in spirit of the chain rule since $$\frac{\partial}{\partial x} f (g(x,y,z)) = \frac{\partial f}{\partial g}\frac{\partial g}{\partial x}$$

The spatial derivatives are computed similarly, you just need to know if the spatial variables are dependent on t, if not the derivative of r is zero with respect to t: $\dfrac{\partial r}{\partial t} = 0$.

For a hint on the spatial part (we need the second derivatives w.r.t. each variable):

$$\frac{\partial f}{ \partial x} = \frac{r \frac{\partial p}{\partial x} - p \frac{\partial r}{\partial x}}{r^2}$$

Where

$$\frac{\partial p(r- ct)}{\partial x} = \frac{\partial p(r-ct)}{\partial (r-ct)}\frac{\partial (r-ct)}{\partial x}$$

Once you are done with all derivatives you should see that the equation orginaly given is satisfied.

Answer 2

As Triatticus said, all you need is a straightforward application of the chain rule without needing to introduce any extraneous functions to define $f$.

HINT: I would define $u=r+ct$ and $v=r-ct$ so that you have terms like $\frac{\partial p}{\partial t}=c\frac{\partial p}{\partial u}$ and $\frac{\partial p}{\partial x}=\frac{\partial p}{\partial u}\frac{\partial r}{\partial x}$, etc etc and then you can collect terms with the partial derivatives of $p$ and $s$ with respect to $u$ and $v$ to simplify your algebra.

You're being asked in essence to derive the wave equation in 3 spatial dimensions.