Question

Solve the IBVP using Fourier series method utt −c2uxx = 100, 0 < x < l, t > 0 u$x, 0$ = x$1 −x$, 0 < x < l, ∂u ∂t $x, 0$ = x, u$0, t$ = u$l, t$ = 0, t > 0. Show all steps even integration

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Answer to a math question Solve the IBVP using Fourier series method utt −c2uxx = 100, 0 < x < l, t > 0 u$x, 0$ = x$1 −x$, 0 < x < l, ∂u ∂t $x, 0$ = x, u$0, t$ = u$l, t$ = 0, t > 0. Show all steps even integration

Hermann
4.6
To solve this IBVP using the method of Fourier series, we first express the solution as an infinite series:

u$x, t$ = \sum_{n=1}^{\infty}$A_n\cos(\omega_nt$ + B_n\sin$\omega_nt$)\sin$\frac{n\pi x}{l}$

where
\omega_n = \frac{nc\pi}{l}

and the coefficients A_n, B_n are to be determined.

Step 1:
Substitute the given initial conditions into the Fourier series solution:
u$x, 0$ = \sum_{n=1}^{\infty} A_n \sin$\frac{n\pi x}{l}$ = x$1-x$
\frac{\partial u}{\partial t}$x, 0$ = \sum_{n=1}^{\infty} B_n\omega_n \sin$\frac{n\pi x}{l}$ = x

Step 2:
Solve for A_n, B_n by using the orthogonality property of sine functions, we multiply both sides by \sin$\frac{m\pi x}{l}$ and integrate over [0, l] :
For A_n :
\int_{0}^{l} x$1-x$ \sin$\frac{m\pi x}{l}$ dx = \int_{0}^{l} $\sum_{n=1}^{\infty} A_n \sin(\frac{n\pi x}{l}$) \sin$\frac{m\pi x}{l}$ dx
A_m = \frac{2}{l} \int_{0}^{l} x$1-x$\sin$\frac{m\pi x}{l}$ dx

For B_n :
\int_{0}^{l} x \sin$\frac{m\pi x}{l}$ dx = \int_{0}^{l} $\sum_{n=1}^{\infty} B_n \omega_n \sin(\frac{n\pi x}{l}$) \sin$\frac{m\pi x}{l}$ dx
B_m\omega_m = \frac{2}{l} \int_{0}^{l} x \sin$\frac{m\pi x}{l}$ dx

Step 3:
Solve for A_n and B_n using the integrals defined above.

Step 4:
Substitute the obtained coefficients back into the Fourier series solution and calculate the solution for u$x, t$ .

Answer: u$x, t$ = \ldots

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