$(\partial_{tt}-\nabla^2+\partial_t)f=g,\quad (\partial_t-\nabla^2+b)g=\partial_t f$

Question

Hi I am looking for complete solutions for $f(r,t),g(r,t)$ given in the coupled linear partial differential equations below: $$ (\partial_{tt}-a\nabla^2+b\partial_t)f(r,t)=bg(r,t) $$ $$ (\partial_t-c \nabla^2+b)g(r,t)=b\partial_t f(r,t) $$ Initial and boundary conditions are given by $$ \partial_t f(0,t)=0,\ \partial_t f(R,t)=d\cos(\omega t) $$ $$ g(0,t)=0, \ g(R,t)=d\cos (\omega t) $$ where $a,b,c,d,R,\omega>0.$ Note $$ \nabla^2\equiv \frac{1}{r}\partial_r(r\partial_r)-\frac{1}{r^2}=\partial_{rr}+\frac{1}{r}\partial_r -\frac{1}{r^2}. $$ Thank you! Some comments:

If the right hand side of both equations are zero, (equations become homogenous), then the kernel of both linear operators are known and are in terms of Bessel functions $J_1$ (we're in a cylindrical geometry hence the Laplacian like term) times oscillating functions of time: $\sin \omega t,\cos \omega t$.

Answer 1

This is an extension of the previous answers :

$(\partial_{tt}+\partial_t-\nabla^2)f(r,t)=0$

PDE $(\partial_{tt}+a\partial_t-b\nabla^2)f(r,t)=0$

In the calculus below, some results which have already been obtained are re-used without more explanation.

Answer 2

This is sketch of how to solve your problem. I still don't understand the boundary conditions/ initial conditions as stated. I have set $a,b,d$ to one, and have made generic boundary conditions.

Let $\Omega \subset \mathbb{R}^d$ be a bounded open domain. We are interested in solving the following coupled systems of PDE. This is pretty hairy.

$$ \begin{align*} 0 &= (\partial_{tt} - \nabla^2 + \partial_t) f - g&\quad \mathrm{for}\ (x,t)\in \Omega \times \mathbb{R}^+ \\ 0 &= (\partial_t-\nabla^2 + 1) g - \partial_t f&\quad \mathrm{for} (x,t) \in \Omega \times \mathbb{R}^+ \\ f &= 0 &\quad \mathrm{for} (x,t)\in \Omega\times \{0\}\\ f_t &= 0&\quad \mathrm{for} (x,t)\in \Omega\times \{0\}\\ g &= 0& \quad \mathrm{for} (x,t)\in \Omega\times \{0\}\\ f_t &= s_1 &\quad \mathrm{for} (x,t)\in \partial\Omega \times \mathbb{R}^+\\ g &= s_2 &\quad \mathrm{for} (x,t)\in \partial\Omega \times \mathbb{R}^+ \end{align*} $$ The fist two equations are the PDE. The second three are initial conditions, and the last two are boundary conditions.

Step #1 Homogenize the PDE's boundary conditions. We need to turn the boundary conditions into "forcing terms"

Let $f_{bc}$ be the solution to the following PDE

\begin{align*} -\nabla^2 f_{bc} &= 0 &\quad \mathrm{for} (x,t)\in \Omega \times \mathbb{R}^+\\ \frac{\mathrm{d}}{\mathrm{d}t} f_{bc} &= s_1 & \quad \mathrm{for} (x,t)\in \partial\Omega\times \mathbb{R}^+ \end{align*}

With simple boundary conditions and simple geometry this $f_{bc}$ might just be a constant in space and vary in time.

Let $g_{bc}$ be the solution of a similar PDE

\begin{align*} -\nabla^2 g_{bc} &= 0&\quad \mathrm{for} (x,t)\in \Omega \times \mathbb{R}^+\\ g_{bc} &= s_2& \quad \mathrm{for} (x,t)\in \partial\Omega \times \mathbb{R}^+. \end{align*}

Now we have a set of new PDE's.

$$ \begin{align*} -(\partial_{tt} + \partial_t )f_{bc} + g_{bc}&= (\partial_{tt} - \nabla^2 + \partial_t) (f-f_{bc}) -(g-g_{bc})&\quad \mathrm{for}\ (x,t)\in \Omega \times \mathbb{R}^+ \\ -(\partial_t +1 )g_{bc} + (f_{bc})_t&= (\partial_t-\nabla^2 + 1) (g-g_{bc}) - \partial_t (f-f_{bc})&\quad \mathrm{for} (x,t) \in \Omega \times \mathbb{R}^+ \\ f-f_{bc} &= -f_{bc} &\quad \mathrm{for} (x,t)\in \Omega\times \{0\}\\ f_t-(f_{bc})_t &= -(f_{bc})_t&\quad \mathrm{for} (x,t)\in \Omega\times \{0\}\\ g-g_{bc} &= -g_{bc}& \quad \mathrm{for} (x,t)\in \Omega\times \{0\}\\ f_t-(f_{bc})_t &= 0 &\quad \mathrm{for} (x,t)\in \partial\Omega \times \mathbb{R}^+\\ g-g_{bc} &= 0 &\quad \mathrm{for} (x,t)\in \partial\Omega \times \mathbb{R}^+ \end{align*} $$

Notice what we have done, we have changed a PDE with with inhomogeneous boundary conditions and no "forcing functions" into one with homogeneous boundary conditions with forcing functions.

Step #2.

Denote $\{\Phi_i\}_{i=1}^\infty$ as a sequence of eigenfunctions for the associated homogeneous Poisson equation, ordered by increasing eigenvalue. We mean to say that $\int_{\Omega} \Phi_i \Phi_j \mathrm{d} x = \delta_{i,j}$ and $\int_{\Omega} (\nabla \Phi_i) \cdot (\nabla \Phi_j) \mathrm{d}x = \delta_{i,j} \lambda_i^2$. (For the cylindrical case we have $\Phi_i$ are the Bessel functions.)

Now, these eigenfunctions are complete so we can expand $(f-f_{bc})$ as

$$ (f-f_{bc})(x,t) = \sum_{i=1}^\infty \tilde{f}_i(t) \Phi_i(x) $$

and $$(g-g_bc)(x,t) = \sum_{i=1}^\infty \tilde{g}_i(t) \Phi_i(x)$$

Now multiplying the PDE by $\Phi_j$ integrating by parts the Laplacian term we get $$ \underbrace{\int_{\Omega} \Phi_i ((-\partial_{tt} - \partial_t )f_{bc} + g_{bc} )\, \mathrm{d} x}_{\alpha_i(t)} = (\tilde{f}_i)_{tt} + (\tilde{f}_i)_t + \lambda_i^2 \tilde{f}_i - \tilde{g}_i $$

and $$ \underbrace{\int_{\Omega} \Phi_i ((-\partial_t+1) g_{bc} + \partial_t f_{bc})\mathrm{d}x }_{\beta_i(t)} = (\tilde{g}_i)_t+\lambda_i^2 \tilde{g}_i + \tilde{g}_i - (\tilde{f}_i)_t $$

The beauty is that, at this point $\alpha_i$ and $\beta_i$ are all known, and $\tilde{f}_i$ only couples with $\tilde{g}_i$.

Step #3. Solve ODE.

We write a state-space representation of the ODE's.
$$ \begin{align*} \frac{\mathrm{d}}{\mathrm{d} t}\left\lbrack \begin{array}{c} \tilde{f}_i\\ (\tilde{f}_i)_t\\ \tilde{g}_i \end{array}\right\rbrack + \left\lbrack \begin{array}{ccc} 0&-1&0\\ \lambda_i^2& 1& -1\\ 0& -1 & \lambda_i^2+1 \end{array} \right\rbrack \left\lbrack \begin{array}{c} \tilde{f}_i\\ (\tilde{f}_i)_t\\ \tilde{g}_i \end{array} \right\rbrack = \left\lbrack \begin{array}{c} 0\\ \alpha_i\\ \beta_i\\ \end{array} \right\rbrack \end{align*} $$

Solve this system how ever you want. This solves $f-f_{bc}$ and $g-g_{bc}$ so to get $f$ and $g$ you just add the boundary functions. Do you see how to handle the initial conditions? I can explain any steps that I skipped or ones that are unclear.