PDE in cylindrical coordinate system

Question

Starting from here: $$ \frac{\partial}{\partial r}u\left(r,t\right)+\frac{1}{r}u\left(r,t\right)=f\left(t\right)$$

(this is a partial differential equation in a cylindrical coordinate system $\left(r,\theta,z\right)$)

Show that

$$u=\frac{1}{2}f\left(t\right)r+\frac{g\left(t\right)}{r}$$

where $f$ and $g$ are arbitrary functions of time.

Answer 1

let $y(x)=u(r,t)$

let $x=r$

let $f=f(t)$

hence we can re-write the expression as:

$$ \frac{dy}{dx} + \frac{1}{x} y = f$$

we have a linear first order ODE:

$$ \frac{dy}{dx} + p(x)y(x) = q(x)$$

using Integrating Factor method...

i.e. we multiply across by a function $\mu(x)$ that has the property that:

$$\frac{d}{dx}\left[\mu(x) y(x)\right]=\mu(x) \left(\frac{dy}{dx} + p(x)y(x)\right) $$

this allows us to write: $$\frac{d}{dx}\left[\mu(x) y(x)\right] = \mu (x) q(x)$$

$$\mu(x) y(x) = \int \mu (x) q(x) dx$$ $$ y(x) = \mu(x)^{-1}\int \mu (x) q(x) dx$$

In order for integrating factor $\mu$ to have the property seen above it must satisfy the ODE:

$$\mu(x)p(x)=\frac{d\mu}{d x}$$

hence:

$$\mu(x)=e^{\int p(x) dx}$$

---

In our situation:

$$p(x)=\frac{1}{x}$$

$$\mu(x)=e^{\int p(x) dx}$$

$$\mu(x)=e^{\int \frac{1}{x} dx}=e^{\log(x)}=x$$

---

$$ y(x) = \mu(x)^{-1}\int \mu (x) q(x) dx$$

$$q(x)= f$$

$$ y(x) = x^{-1}\int x q(x) dx$$

$$ y(x) = x^{-1}\int x f dx$$

$$ y(x) = x^{-1} \left(\frac{1}{2}x^{2}f + g\right)$$

$$ y(x) = \frac{1}{2}f x + \frac{1}{x}g$$

---

$$ u = \frac{1}{2}f\left(t\right) r + \frac{1}{r}g\left(t\right)$$

Answer 2

Note that the PDE depends only on the variable $r$ wheras $t$ is a parameter. Thus, use standard methods for ODEs: The Guessing Method.

Solve the homogeneous equation $r \partial_r u + u=0$. Solution is obvious: $u_0 = \frac{A_0}{r}$ for some constant $A_0$. Here constant should be understood w.r.t. the variable $r$.

To solve the inhomogenous equation we vary the constant $A_0 \to A=A(r)$ and insert it into the ODE \begin{align*} f &= r \partial_r u + u = r \partial_r \left( \frac{A(r)}{r} \right) + \frac{A(r)}{r} \\ &= r \left( \frac{A'}{r} - \frac{A}{r^2} \right) + \frac{A(r)}{r} = A' \\ \implies A(r) &= f r + g \end{align*} where $g=\text{const}$ is another constant.

This gives the solution as $u(r) = \frac{A(r)}{r} = f + \frac{g}{r}$.

To include the parameter $t$ one can think of the constants $A_0, g, f$ depending on $t$ but not on $r$ which gives the solution as $u(r, t) = f(t) + \frac{g(t)}{r}$.