Consider The Wave Equation On The Finite Domain [0, 1] With Both Ends Fixed And A Linear Restoring Force.

To solve, this PDE using the Fourier series we have to follow the below process

We know that u(x,t) can be represented in the Fourier series as:

u(x, t) = Σn=1∞ {An sin( nπx) + Bn cos( nπx)}\( \times \left ( Cn cos(n\pi t)+Dn sin(n\pi t)\right )\)

Where An, Bn, Cn, and Dn are constants that depend on the given function f(x) and g(x).

Next, we differentiate u(x, t) twice with respect to t and once with respect to x, and then we plug it into the given wave equation to get:

∑n=1∞ (Cn sin(nπx) – Dn cos(nπx))[(nπ)^2 + 1]cos(nπt) = ∑n=1∞ An sin(nπx) (nπ)^2 [Cn cos(nπt) + Dn sin(nπt)]

To find An, we multiply both sides of the above equation by sin(nπx) and integrate it with respect to x.

∫[0,1] sin(nπx)[∑n=1∞ (Cn sin(nπx) – Dn cos(nπx))[(nπ)^2 + 1]cos(nπt) dx] = ∫[0,1] sin(nπx) [∑n=1∞ An sin(nπx) (nπ)^2 [Cn cos(nπt) + Dn sin(nπt)] dx]

Using the orthogonality property of sine and cosine,

we get An (nπ)^2 [Cn cos(nπt) + Dn sin(nπt)] = ∫[0,1] sin(nπx) [∑n=1∞ (Cn sin(nπx) – Dn cos(nπx))[(nπ)^2 + 1]cos(nπt) dx]

We can find the constants Cn and Dn by using the initial conditions:u(x, 0) = f(x), x ∈ [0, 1], and ut(x, 0) = g(x), x ∈ [0, 1].

Applying Fourier series to initial conditions:

u(x, 0) = f(x) = ∑n=1∞ An sin(nπx) Cn cos(nπt) + Dn sin(nπt)ut(x, 0) = g(x) = ∑n=1∞ An sin(nπx) nπ [Cn cos(nπt) + Dn sin(nπt)]

Therefore, we have to solve the four equations given by:

An (nπ)^2 [Cn cos(nπt) + Dn sin(nπt)] = ∫[0,1] sin(nπx) [∑n=1∞ (Cn sin(nπx) – Dn cos(nπx))[(nπ)^2 + 1]cos(nπt) dx]f(x)

= ∑n=1∞ An sin(nπx) Cn cos(nπt) + Dn sin(nπt)g(x) = ∑n=1∞ An sin(nπx) nπ [Cn cos(nπt) + Dn sin(nπt)]u(0, t) = 0u(1, t) = 0

To know more about PDE: brainly.com/question/31129985

#SPJ11