General Solution of a PDE and Fourier Series Representations of Functionsns

Question #1:

You compare your equation with the general form of a second-order PDE,

$Au_{xx}+ Bu_{xy} +Cu_{yy} + Du_{x} + Eu_{y}+ Fu= G$

We see that A= 2, B = 6, C = 4, D= 1, E = 1, F = 0, and G= 0.

So, the characteristic equations of this PDE are given by

$\frac{dy}{dx}=\frac{1}{2A} \times(B±\sqrt{B^{2} −4AC})$



$\frac{dy}{dx}=\frac{1}{4} \times(6±\sqrt{36 −32})$

$\frac{dy}{dx}=\frac{1}{4} \times(6±2)$


Note that since $\sqrt{B^{2} −4AC}$ = 25 − 16 = 9, is greater than 0, and so the PDE is hyperbolic. And therefore the solutions to the ordinary differential equations are two real and distinct families of characteristic curves in the xy-plane.

$\frac{dy}{dx}=1$

$\frac{dy}{dx}=2$

And so the charachteristics are:

$y = x + c_1$

$y= 2x + c_2$

The linear transformation:

$\xi = y - x$

$\eta = y - 2x$

So,

$\xi_x = -1$

$\xi_{xx} = 0$

$\xi_{xy} = 0$

$\xi_y = 1$

$\xi_{yy} = 0$

$\eta_x = -2$

$\eta_{xx} = 0$

$\eta_{xy} = 0$

$\eta_y = 1$

$\eta_{yy} = 0$


So, we have:

$A^* = A\xi_x^2 + B\xi_x\xi_y + C\xi_y^2 = 0$

$B^* = 2A\xi_x\eta_x + B(\xi_x\eta_y + \xi_y\eta_x) + 2C\xi_y\eta_y = -2$

$C^* = A\eta^2_x + B\eta_x \eta_y + C\eta^2_y = 0$

$D^* = A\xi_{xx} + B\xi_{xy} + C\xi_{yy} + D\xi_x + E\xi_y = 0$

$E^* = A\eta_{xx} + B\eta_{xy} + C\eta_{yy} + D\eta_x + E\eta_y = -1$

$F^* = F = 0$

$G^* = G = 0$

Now:

$A^* u_{\xi \xi} + B^* u_{\xi \eta} + C^* u_{\eta \eta} + D^* u_{\xi} + E^* u_{\eta} + F^* u = G^*$

$u_{\xi \eta} = \frac{-1}{2} u_\eta$

By means of the substitution $v = u_\eta$, the preceding equation reduces to

$v_\xi = \frac{-1}{2} v$

Integrating with respect to $\xi$, we have

$v = e^{\left(\frac{-\xi}{2}\right)} F(\eta)$

Integrating with respect to $\eta$, we obtain

$u(\xi, \eta) = g(\eta) e^{\left(\frac{-\xi}{2}\right)} + f(\xi)$

where $f(\xi)$ and $g(\eta)$ are arbitrary functions.

The general solution of the given equation becomes

$u(x, y) = g(y-2x) e^{\frac{-1}{2}(y - x)} + f(y - x)$

----------------------------------------------
----------------------------------------------

Question #2:

The Fourier series for a function $f(x)$ defined on $-\pi < x < \pi$ is:

$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos(nx) + b_n \sin(nx)\right)$

The coefficients are:

$a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos(nx) \, dx, \quad b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(nx) \, dx$


Case 1:

$x = 2\left(\sin x - \frac{\sin 2x}{2^2} + \frac{\sin 3x}{3^2} - \dots \right)$

This corresponds to the Fourier sine series for $f(x) = x$ on $-\pi < x < \pi$. So, we have:

$b_n = \frac{1}{\pi} \int_{-\pi}^\pi x \sin(nx) \, dx$

By symmetry and integration:

$b_n = \frac{2(-1)^{n+1}}{n^2}, \quad a_n = 0$

Thus:

$x = 2\left(\sin x - \frac{\sin 2x}{2^2} + \frac{\sin 3x}{3^2} - \dots \right)$


Case 2:

$x^2 = \frac{\pi^2}{3} - 4\left(\cos x - \frac{\cos 2x}{2^2} + \frac{\cos 3x}{3^2} - \dots \right)$

This corresponds to the Fourier cosine series for $f(x) = x^2$. So, we have:

$a_n = \frac{1}{\pi} \int_{-\pi}^\pi x^2 \cos(nx) \, dx$

The coefficients are:

$a_0 = \frac{\pi^2}{3}, \quad a_n = -\frac{4}{n^2}$

Thus:

$x^2 = \frac{\pi^2}{3} - 4\left(\cos x - \frac{\cos 2x}{2^2} + \frac{\cos 3x}{3^2} - \dots \right)$


Case 3:

$x(\pi + x)(\pi - x) = 12\left(\sin x - \frac{\sin 2x}{2^2} + \frac{\sin 3x}{3^2} - \dots \right)$

The function $f(x) = x(\pi^2 - x^2)$ is odd, so it has only sine terms. So, we have:

$b_n = \frac{1}{\pi} \int_{-\pi}^\pi x(\pi^2 - x^2) \sin(nx) \, dx$

After integration:

$b_n = \frac{12(-1)^{n+1}}{n^2}$

So, we have:

$x(\pi + x)(\pi - x) = 12\left(\sin x - \frac{\sin 2x}{2^2} + \frac{\sin 3x}{3^2} - \dots \right)$