How would I solve this?

$\displaystyle \frac{d^{2}u}{dt^2}=4\frac{d^{2}u}{dx^2}$

with conditions:

du/dx(0,t)=0 and u(1,t)= 0

u(x,0) = x^3 +2x^2 -3

du/dt(x,0) = x^2 -1

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- Oct 26th 2012, 06:02 PMselzer9differential equation with conditions
How would I solve this?

$\displaystyle \frac{d^{2}u}{dt^2}=4\frac{d^{2}u}{dx^2}$

with conditions:

du/dx(0,t)=0 and u(1,t)= 0

u(x,0) = x^3 +2x^2 -3

du/dt(x,0) = x^2 -1 - Oct 27th 2012, 07:03 AMTheEmptySetRe: differential equation with conditions
You would use a Fourier Series expansion.

Assume $\displaystyle u(x,t) = X(x)T(t)$ this gives

$\displaystyle X\ddot{T}=4X''T \iff \frac{X''}{X}=\frac{\ddot{T}}{T}=-\lambda^2$

Now separate and solve the equation in X and use the boundary conditions to find the eigenvalues.

Post back with your work if you get stuck. - Oct 30th 2012, 01:33 AMJJacquelinRe: differential equation with conditions
Hi!

The conditions u(x,0) = x^3 +2x^2 -3 and du/dt(x,0) = x^2 -1 seem sufficient to determine an unique solution u(x,t).

But this solution is not consistent with the conditions du/dx(0,t)=0 and u(1,t)= 0

Where is the mismatch ? - Nov 3rd 2012, 03:11 PMTheEmptySetRe: differential equation with conditions
The solution needs to be found using a Fourier Series expansion as in Post #2.

If you separate and solve the system of ODE's you get

$\displaystyle X(x)=c_1\cos(\lamba x) +c_2\sin(\lambda x)$

Using the initial conditions $\displaystyle X'(0)=0$ and $\displaystyle X(1)=0$ gives

$\displaystyle 0=-\lambda c_1\sin( \lambda 0)+\lambda c_2\cos(\lambda 0) \iff c_2=0$

$\displaystyle 0=c_1\cos(\lambda) \iff \lambda = \frac{\pi}{2}+\pi n, n =0,1,2,3,...$

This gives the solution of the form

$\displaystyle u_{n}(x,t)=\cos\left(\left[\frac{\pi+2\pi n}{2}\right]x \right)\left( a_n\cos\left(\left[\frac{\pi+2\pi n}{2}\right]t \right)+b_n\sin\left(\left[\frac{\pi+2\pi n}{2}\right]t \right)\right)$

The $\displaystyle a_n,b_n$ can be determined using the initial conditions for the PDE and orthogonallity of the Eigen vectors. - Nov 6th 2012, 01:25 AMJJacquelinRe: differential equation with conditions
Not necessarily.

Any method of solving, correctly carried out, leads to the same result.

In the present case, the solution of the PDE with the third and fourth conditions leads to a unique result : the function u(x,t) in attachment.

Any other method will leads to this function and not to a different function.

But the patern of the function can be different. For example the function can be presented on the form of a Fourier series.

So the method which consists to solve the PDE thanks to Fourier series leads to the same result, but on the form of a Fourier series, which is the Fourier series corresponding to u(x,t) in attachment.

We know that the Fourier series correctly express the function on a limited range, beteween the borders, but not exactly on the border in many cases (for example the square wave function, the sawtooth wave function etc.). That is what append with the PDE considered here.

Of course, the Fourier series which is obtained satisfy the border conditions u(1,t)=0 and du/dx(0,t)=0. But it is an "artificial" agreement because the true fuction u(x,t) is deformed on the border in order to fit the patern of a Fourier series.