# PDE using D'Alembert's Solution

• Aug 16th 2010, 04:51 AM
JoernE
PDE using D'Alembert's Solution
Use D'Alembert's solution to solve the following

$\displaystyle u_{xx} - \frac{1}{c^2} u_{tt} = 4$

subject to the initial conditions $\displaystyle u(x, 0) = \sin (x)$ and $\displaystyle u_t (x, 0) = c \ \cos (x)$.

Note that $\displaystyle u_{xx} - c^{-2} u_{tt} = 4u_{\xi \eta}$ with $\displaystyle \xi = x + ct$ and $\displaystyle \eta = x - ct$
• Aug 16th 2010, 06:36 AM
Jester

$\displaystyle 4 u_{\xi \eta} = 4$ gives $\displaystyle u_{\xi \eta} = 1.$

Integrating twice gives

$\displaystyle u = \xi \eta + F(\xi) + G(\eta)$

or

$\displaystyle u = x^2 - c^2 t^2 + F(x+ct) + G(x-ct).$

Imposing the two boundary conditions gives

$\displaystyle x^2 + F(x) + G(x) = \sin x,\;\;\;(1)$
$\displaystyle cF'(x)-cG'(x) = c\cos x,\;\;\;(2)$.

Differentiating the first and then solving for F' and G' gives

$\displaystyle F'(x) = \cos x - x,\;\;\;G' = - x$.

Thus, $\displaystyle F(x) = \sin x - \frac{1}{2}x^2 + c_1,\;\;\;G(x) = - \frac{1}{2} x^2 + c_2$.

Finally, we have

$\displaystyle u = x^2 - c^2 t^2 + \sin(x+ct) - \frac{1}{2}(x+ct)^2 - \frac{1}{2}(x-ct)^2 =-2 c^2t^2 + \sin(x+ct) + c_1+c_2$

but imposing the first BC (1) gives $\displaystyle c_1+c_2 = 0$, so the final solution is

$\displaystyle u = -2 c^2t^2 + \sin(x+ct).$
• Aug 24th 2010, 04:20 AM
JoernE
Quote:

Originally Posted by Danny

$\displaystyle 4 u_{\xi \eta} = 4$ gives $\displaystyle u_{\xi \eta} = 1.$

Integrating twice gives

$\displaystyle u = \xi \eta + F(\xi) + G(\eta)$

or

$\displaystyle u = x^2 - c^2 t^2 + F(x+ct) + G(x-ct).$

Imposing the two boundary conditions gives

$\displaystyle x^2 + F(x) + G(x) = \sin x,\;\;\;(1)$
$\displaystyle cF'(x)-cG'(x) = c\cos x,\;\;\;(2)$.

Differentiating the first and then solving for F' and G' gives

$\displaystyle F'(x) = \cos x - x,\;\;\;G' = - x$.

Thus, $\displaystyle F(x) = \sin x - \frac{1}{2}x^2 + c_1,\;\;\;G(x) = - \frac{1}{2} x^2 + c_2$.

Finally, we have

$\displaystyle u = x^2 - c^2 t^2 + \sin(x+ct) - \frac{1}{2}(x+ct)^2 - \frac{1}{2}(x-ct)^2 =-2 c^2t^2 + \sin(x+ct) + c_1+c_2$

but imposing the first BC (1) gives $\displaystyle c_1+c_2 = 0$, so the final solution is

$\displaystyle u = -2 c^2t^2 + \sin(x+ct).$

Hi Danny,

I can't see where you've used d'Alembert's solution i.e.

$\displaystyle u(x,t) = \frac{1}{2} \big ( f (x + ct) + f(x - ct) \big ) + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds$

Could you please go into detail with each step? I really want to understand this.
• Aug 24th 2010, 05:34 AM
Jester
Quote:

Originally Posted by JoernE
Hi Danny,

I can't see where you've used d'Alembert's solution i.e.

$\displaystyle u(x,t) = \frac{1}{2} \big ( f (x + ct) + f(x - ct) \big ) + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds$

Could you please go into detail with each step? I really want to understand this.

I used it after making the change of variables. Let me derive my solution with the D'Alembert solution.

First the D'Alembert solution applies for

$\displaystyle u_{xx} - \frac{1}{c^2} u_{tt} = 0$

But your PDE has a nonhomogeneous term.

$\displaystyle u_{xx} - \frac{1}{c^2} u_{tt} = 4$

So we'll find a particlar solution - $\displaystyle u_p = -2c^2 t^2$ works. So let $\displaystyle u = v + u_p$ where now $\displaystyle v$ satisfies

$\displaystyle v_{xx} - \frac{1}{c^2} v_{tt} = 0$,

$\displaystyle v(x,0) = \sin x,\;\;\; v_t(x,0) = c\cos x$

D'Almbert solution

$\displaystyle v(x,t) = \frac{1}{2} \left( f (x + ct) + f(x - ct) \right) + \frac{1}{2c} \displaystyle \int_{x-ct}^{x+ct} g(s) ds$

and your IC's gives $\displaystyle f(x) = \sin x$ and $\displaystyle g(x) = c\cos x$

$\displaystyle v(x,t) = \frac{1}{2} \left( \sin (x + ct) + \sin(x - ct) \right) + \frac{1}{2c} \displaystyle \int_{x-ct}^{x+ct} c\cos(s) ds = \sin (x + ct)$

so

$\displaystyle u(x,t) = -2c^2t^2 + \sin(x+ct)$.
• Aug 25th 2010, 04:00 AM
JoernE
Quote:

so we'll find a particular solution $\displaystyle u_p = -2c^2 t^2$ works
Thanks again. I'm just curious how you arrive at this particular solution. I now understand the other parts.
• Aug 25th 2010, 04:35 AM
Jester
Quote:

Originally Posted by JoernE
Thanks again. I'm just curious how you arrive at this particular solution. I now understand the other parts.

Assume that $\displaystyle u = F(t)$ only and substitute into your PDE. You'll get an simple ODE for $\displaystyle F(t)$ from wich my particular solution will emerge. (Note - you can supress the constants of integration, it only makes things more complicated).