0
$\begingroup$

Solve this via Fourier method:

$$u_t-u_{xx}=0 \quad\quad 0< x<\pi, \quad t >0, $$

$$u(0,t)=u_x(\pi,t)=0, \quad\quad t \ge 0$$

$$u(x,0)=2\sin\left(\frac{3x}{2}\right) \, \cos{2x},\quad\quad 0\leq x\leq\pi$$

I guess it is heat equation with insulated ends. what does it say with method of Fourier? do we use $u(x, t) =X(x)T(t)$? or

$$u(x,t)=\sum_{n=1}^{\infty}u_n(x,t)=\sum_{n=1}^{\infty}b_ne^{4n^2t}\sin(nx) $$

$u(x,0)=f(x)=\sum_{n=1}^{\infty}b_n\sin(nx)$ should I expand like $f(x)=2\sin(\frac{3x}{2})\cos2x=\sin(\frac{7x}{2})-\sin(\frac{x}2)=b_1\sin x+b_2\sin2x+b_3\sin3x....$

or finding

$$\frac2\pi\int_0^L(\sin(\frac{7x}{2})-\sin(\frac{x}2))\sin(nx)dx = $$

$$ = \frac2\pi\int_0^L\frac{\left(-\cos(x(\frac72+n))+\cos(x(\frac72-n)+\cos(x(\frac12+n))-\cos(x(\frac12-n)\right)}{2} = $$

$$= \frac2\pi \left(\frac{\frac{-\sin(x(\frac72+n))}{\frac72+n}+\frac{\sin(x(\frac72-n)}{\frac72-n}+\frac{-\sin(x(\frac12+n))}{\frac12+n}+\frac{\sin(x(\frac12-n))}{\frac12-n}}{2}\right)\Bigg|^\pi_0.$$

$\endgroup$
0
$\begingroup$

Try this:

$u(x,t) = \sum _{-\infty}^{\infty} c_k(t)e^{2\pi ikx} $. Period of $x$ is $\pi$.

Place this in the differential equation (and then manipulate), and then use the boundary methods.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.