For $u:\mathbb{R}\times [0,1]$ with boundary conditions $u(0,x)=\cos (2\pi x)$ and $u(t,0)=u(t,1)$.
Solve $u_t=u_{xx}+u_x$.
I had this on an exam and tried to write $u$ as a product of two single variate functions and convert to an ode using the usual methods but things got messy and I couldn't finish. I also tried to write $u$ as a Fourier series in $x$ with coefficients $a_k(t)$ depending on $t$, but this also didn't seem to yield anything.
Is there a more clever way to approach this, possibly with Fourier series? I heard from a friend there may be a Fourier transform method since the equation was "homogeneous in momentum space."
Thanks!
Best Answer
The Fourier series method should work! (I would definitely go for Fourier series rather than Fourier transforms, since your solution is meant to be periodic in $x$.)
Having said that, your initial condition only involves $\cos(2\pi x)$, so I would be tempted to try an ansatz of the form, $$ u(t,x) = a(t) \cos(2\pi x) + b(t) \sin(2\pi x).$$ The PDE gives \begin{multline} \dot a(t) \cos (2\pi x) + \dot b(t)\sin(2\pi x)\\ = (-4\pi^2 a(t)+2\pi b(t))\cos(2\pi x) +(-2\pi a(t)-4\pi^2b(t))\sin(2\pi x).\end{multline} So we need to solve the ODEs $$ \frac{d}{dt}\left[ \begin{array}{c} a(t) \\ b(t) \end{array}\right] = \left[\begin{array}{cc} -4\pi^2 & 2\pi \\ - 2\pi & -4\pi^2 \end{array} \right]\left[ \begin{array}{c} a(t) \\ b(t) \end{array}\right]$$ and the initial conditions (coming from the initial conditions for the PDE) are $$ a(0) = 1, \ \ \ \ \ b(0) = 0.$$ We can solve this ODE by finding the eigenvalues and eigenvectors of the matrix. The result is $$ \left[ \begin{array}{c} a(t) \\ b(t) \end{array}\right] = \frac 1 2 \left[ \begin{array}{c} 1 \\ i \end{array}\right] \exp((-4\pi^2 + 2\pi i)t) + \frac 1 2 \left[ \begin{array}{c} 1 \\ -i \end{array}\right] \exp((-4\pi^2 - 2\pi i)t)$$ i.e. $$ a(t) = \exp(-4\pi^2 t) \cos(2\pi t), \ \ \ b(t) = -\exp(-4\pi^2 t) \sin(2\pi t).$$