Why do not we solve two dimensional wave equation directly without using method of descent ? Why do not we solve two dimensional wave equation directly without using method of descent ?  Is there any problem with two dimensional wave equation ? Two solve two dimensional wave equation, we use three dimensional solution to wave equation. Why do we do this ?  Thank you for your help. . 
 A: Indeed I don't understand why we don't solve it by Fourier analysis.
We can.
I prefer to answer you through an example: consider a thin elastic membrane stretched tightly over a rectangular frame. Suppose the dimensions of the frame are $a \times b$ and that we keep the edges of the membrane fixed to the frame.
1) Perturbing the membrane from equilibrium results in some sort of vibration of the surface.
2) Our goal is to mathematically model the vibrations of the membrane surface.
We let $u(x, y, t)$ as the deflection of membrane from equilibrium at position $x, y$ and at time $t$.
For a fixed $t$, the surface $z = u(x,y,t)$ gives the shape of the
membrane at time $t$.
Under ideal assumptions (e.g. uniform membrane density, uniform tension, no resistance to motion, small deflection, etc.) one can show that $u$ satisfies the two dimensional wave equation:
$$\frac{\partial^2 u}{\partial t^2} = c^2\nabla^2 u$$
for $0<x<a$ and $0<y<b$
As in the one dimensional situation, the constant c has the units of velocity. It is given by
$$c^2 = \frac{\tau}{\rho}$$
where $\tau$ is the tension per unit length, and $\rho$ is mass density.
The fact that we are keeping the edges of the membrane fixed is expressed by the boundary conditions:
$$u(0, y, t) = u(a, y, t) = 0$$
$$u(x, 0, t) = u(x, b, t) = 0$$
We must also specify how the membrane is initially deformed and set into motion. This is done via the initial conditions
$$u(x,y,0) = f (x,y)$$
$$u_t(x,y,0) = g(x,y)$$
Where $u_t$ is the derivative of $u$ with respect to $t$.
The goal is now to solve the equation, and we will use the separation of variables and the superposition principle.
Let's tart with
$$u(x,y,t) = X(x)Y(y)T(t)$$
Plugging this into the wave equation we get
$$XYT′′ =c2 X′′YT +XY′′T $$
If we divide both sides by $c^2XYT$ this becomes
$$\frac{T''}{c^2 T} = \frac{X''}{X} + \frac{Y''}{Y}$$
Because the two sides are functions of different independent
variables, they must be constant
$$\frac{T''}{c^2 T} = \frac{X''}{X} + \frac{Y''}{Y} = A$$
That is for the first equality:
$$T'' - c^2AT = 0$$
and for the second
$$\frac{X''}{X} = -\frac{Y''}{Y} + A$$
Once again, the two sides involve unrelated variables, so both are
  constant:
$$\frac{X''}{X} = -\frac{Y''}{Y} + A = B$$
If we now let $C = A-B$ we get 
$$X'' - BX = 0$$
$$Y'' - CY = 0$$
By the first boundary condition we notice that since we want nontrivial solutions only, we can cancel $Y$ and $T$, yielding
$$X(0) = 0$$
When we perform similar computations with the other three
boundary conditions we also get
$$X(a) = 0$$
$$Y(0) = Y(b) = 0$$
And there are no boundary conditions on $T$.
You can easily solve the two boundary conditions for $X$ and $Y$ so you can easily get
$$X_m (x) = \sin\mu x ~~~~~~~~~~~ \mu = \frac{m\pi}{a}$$
$$Y_n(y) = \sin\nu y ~~~~~~~~~~~ \nu = \frac{n\pi}{b}$$
For $n$ and $m$ natural numbers.
The separation constants are then $B = -\mu^2$ and $C = -\nu^2$.
Recalling that $T$ must satisfy $T''  - c^2 AT = 0$ we get, with $A + B = C = (\mu^2 + \nu^2) < 0$, then for any choice of $n$ and $m$ we have
$$T_{m n} (t) = B_{nm}\cos\lambda_{nm}t  + B^*_{nm}\sin\lambda_{mn} t$$
Where
$$\lambda_{nm} = c\sqrt{\mu^2 + \nu^2} = c\pi \sqrt{\frac{m^2}{a^2} + \frac{n^2}{b^2}}$$
These are the characteristic frequencies of the membrane.
Remarks:
Note that the normal modes:
1) oscillate spatially with frequency $\mu$ in the $x$-direction
2) oscillate spatially with frequency $\nu$ in the $y$-direction
3) oscillate in time with frequency $\lambda_{nm}$
Eventually, According to the principle of superposition, we may add them to obtain the general solution:
$$u(x, y, t) = \sum_{n = 1}^{+\infty}\sum_{m = 1}^{+\infty} \sin\mu x \sin\nu y (B_{nm}\cos\lambda_{mn}t + B^*_{nm}\sin\lambda_{nm} t)$$
P.s.
We must use a double series since the indices $m$ and $n$ vary independently throughout the Natural numbers set.
Finally, we must determine the values of the coefficients $B_{mn}$ and $B^∗$ that are required so that our solution also satisfies the initial
conditions.
We easily get
$$f(x, y) = u(x, y. 0) = \sum_{n = 1}^{+\infty}\sum_{m = 1}^{+\infty} B_{nm}\sin\frac{m\pi}{a}x \sin\frac{n\pi}{b} y$$
and
$$g(x, y) = u_t(x, y, 0) = \sum_{n = 1}^{+\infty}\sum_{m = 1}^{+\infty}\lambda_{nm}B^*_{nm}\sin\frac{m\pi}{a}x \sin\frac{n\pi}{b} y$$
And by the way these are examples of double Fourier series.
There are ways to determine the Fourier Coefficients, and its something a bit long to do. If you know Fourier analysis you can try by yourself considering the function 
$$Z_{nm}(x, y) = B_{nm}\sin\frac{m\pi}{a}x \sin\frac{n\pi}{b} y$$
are pairwise orthogonal relative to the inner product $\langle f, g\rangle$.
The calculation is not difficult but quite tedious. Eventually you'll get the solution:
$$u(x, y, t) = \frac{576}{\pi^6} \sum_{n = 1}^{+\infty}\sum_{m = 1}^{+\infty} \left(\frac{(1 + (-1)^{m+1})(1 + (-1)^{n+1})}{m^3n^3}\sin\frac{m\pi}{2}x \sin\frac{n\pi}{3}y\cos \pi\sqrt{9m^2 + 4n^2t}\right)$$
Hence no steepest descent or other methods have been used.
