# Thread: PDE using D'Alembert's Solution

1. ## PDE using D'Alembert's Solution

Use D'Alembert's solution to solve the following

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

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

Note that $u_{xx} - c^{-2} u_{tt} = 4u_{\xi \eta}$ with $\xi = x + ct$ and $\eta = x - ct$

2. With your change of variables

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

Integrating twice gives

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

or

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

Imposing the two boundary conditions gives

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

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

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

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

Finally, we have

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

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

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

3. Originally Posted by Danny

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

Integrating twice gives

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

or

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

Imposing the two boundary conditions gives

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

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

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

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

Finally, we have

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

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

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

Hi Danny,

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

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

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

4. Originally Posted by JoernE
Hi Danny,

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

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

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

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

But your PDE has a nonhomogeneous term.

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

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

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

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

D'Almbert solution

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

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

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

so

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

5. so we'll find a particular solution $u_p = -2c^2 t^2$ works
Thanks again. I'm just curious how you arrive at this particular solution. I now understand the other parts.

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