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

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

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

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


1 Answer 1


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.


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