Solution of the differential equation: $E^2(E^2\psi)=0$

How do we solve the differential equation $E^2(E^2\psi)=0$?

We have, in an earlier article, defined the Stokes stream function for an incompressible fluid at the boundaries of r = a and r = ∞ where:

$\psi=0\; \; \; \; \; if\; r=a$

$\psi=\frac{\textbf{\textit{u}}}{2}r^2sin^2\theta\; \; \; \; \; if\; r=\infty$

To develop a Stokes stream function that satisfies eq36 for a ≤ r ≤ ∞ , let’s consider a solution of the form:

$\psi=fsin^2\theta\; \; \; \; \; \; \; (37)$

where $f$ is a function of r.

Substitute eq37 in eq36 and noting that $\frac{\partial }{\partial \theta}\left ( \frac{1}{sin\theta}\frac{\partial }{\partial \theta} \right )sin^2\theta=-2sin\theta$ , we have:

$\left ( \frac{\partial^2 }{\partial r^2} -\frac{2}{r^2}\right )^2f=0\; \; \; \; \; \; \; (38)$

Letting $f=r^n$ and solving eq38, we get n = ±1, 2, 4. Therefore, the function $f$ takes the form:

$f=\frac{A}{r}+Br+Cr^2+Dr^4\; \; \; \; \; \; \; (39)$

Substitute eq39 in eq37,

$\psi=\left ( \frac{A}{r}+Br+Cr^2+Dr^4 \right )sin^2\theta\; \; \; \; \; \; \; (40)$

To satisfy the function at the top of the page at r → ∞, we let D = 0 and C = u/2. Eq40 becomes:

$\psi=\left ( \frac{A}{r}+Br+\frac{\textbf{\textit{u}}}{2}r^2 \right )sin^2\theta\; \; \; \; \; \; \; (41)$

Substitute eq41 in eq15 and eq16, we have:

$u_r=2cos\theta\left ( \frac{A}{r^3}+\frac{B}{r}+\frac{\textbf{\textit{u}}}{2} \right )\; \; \; \; \; \; \; (42)$

$u_\theta=sin\theta\left (- \frac{A}{r^3}+\frac{B}{r}+\textbf{\textit{u}} \right )\; \; \; \; \; \; \; (43)$

To satisfy the function at the top of the page at r = a, u_r = 0 and u_θ = 0, i.e. velocity of the fluid on the surface of the object is zero relative to the velocity of the object. This is because the force of attraction between the fluid particles and solid particles of the object is greater than that between the fluid particles. Eq42 and eq43 become:

$\frac{A}{a^3}+\frac{B}{a}+\frac{\textbf{\textit{u}}}{2}=0\; \; \; \; and\; \; \; \;-\frac{A}{a^3}+\frac{B}{a}+\textbf{\textit{u}}=0$

Solving the simultaneous equations gives:

$A=\frac{\textbf{\textit{u}}}{4}a^3\; \; \; \; and\; \; \; \; B=-\frac{3\textbf{\textit{u}}}{4}a$

Substitute A and B in eq41,

$\psi=sin^2\theta\frac{\textbf{\textit{u}}}{4}\left ( \frac{a^3}{r}-3ar+2r^2 \right )\; \; \; \; \; \; \; (44)$

Therefore, we have a Stokes stream function that satisfies the differential equation eq36 for a ≤ r ≤ ∞ .

Substitute eq44 in eq15 and eq16, we have:

$u_r=\textbf{\textit{u}}cos\theta\left ( \frac{a^3}{2r^3}-\frac{3a}{2r}+1 \right )\; \; \; \; \; \; \; (45)$

$u_\theta=-\textbf{\textit{u}}sin\theta\left ( -\frac{a^3}{4r^3}-\frac{3a}{4r}+1 \right )\; \; \; \; \; \; \; (46)$

Eq45 and eq46 are the component flow velocities of the viscous incompressible fluid.

Solution of the differential equation: \(E^2(E^2\psi)=0\)

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