Find the solution of the form u=f(x)cos y of (*): 2x (∂…

Find the solution of the form $u=f(x)\cos y$ of $(*)\!: \qquad 2x\,\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}\cot y + x^2\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = 0$, satisfying $\left.\dfrac{\partial u}{\partial x}\right|_{x=a} = -\cos y$ ($a$ a constant) and $\displaystyle\lim_{x\to\infty}\frac{\partial u}{\partial x} = 0$.

Step 1: Substituting the separated ansatz

With $u=f(x)\cos y$, the required partial derivatives are

\[ \frac{\partial u}{\partial x}=f'\cos y,\qquad \frac{\partial^2u}{\partial x^2}=f''\cos y,\qquad \frac{\partial u}{\partial y}=-f\sin y,\qquad \frac{\partial^2u}{\partial y^2}=-f\cos y. \]

Substituting into $(*)$: $2xf'\cos y + (-f\sin y)\cot y + x^2f''\cos y + (-f\cos y) = 0$.

Step 2: Reducing to an ODE in $f$

Since $\sin y\cot y = \cos y$, the second term equals $-f\cos y$, so every term carries a common factor $\cos y$: \[\cos y\,\Bigl[\,2xf' - f + x^2f'' - f\,\Bigr] = 0 \;\Longrightarrow\; x^2f'' + 2xf' - 2f = 0,\]

after dividing by $\cos y$ (valid away from the isolated values where $\cos y=0$, by continuity elsewhere). This is a homogeneous Euler equation in $f$.

Step 3: Solving the Euler equation

Try $f=x^k$: $k(k-1)+2k-2=0$, i.e. $k^2+k-2=0=(k+2)(k-1)$, so $k=1$ or $k=-2$. Hence

\[ f(x) = C_1x + C_2x^{-2}, \qquad f'(x) = C_1 - 2C_2x^{-3}. \]

Step 4: Applying the boundary conditions

As $x\to\infty$, $f'(x)\to C_1$; the condition $\lim_{x\to\infty}\partial u/\partial x = 0$ (i.e. $f'(x)\cos y\to0$ for every $y$) forces $C_1=0$. With $C_1=0$, $f(x) = \frac{C_2}{x^2}, \qquad f'(x) = -\frac{2C_2}{x^3}$.

The condition at $x=a$, $\partial u/\partial x|_{x=a}=f'(a)\cos y=-\cos y$, requires $f'(a)=-1$: \[-\frac{2C_2}{a^3} = -1 \;\Longrightarrow\; C_2 = \frac{a^3}{2}.\]

Step 5: The solution

\[ f(x) = \frac{a^3}{2x^2}, \qquad\qquad u(x,y) = \frac{a^3}{2x^2}\cos y. \]

Step 6: Verification

With $f(x)=a^3/(2x^2)$: $f'(x)=-a^3/x^3$ and $f''(x)=3a^3/x^4$. Substituting into the Euler equation, \[x^2f''+2xf'-2f = \frac{3a^3}{x^2} - \frac{2a^3}{x^2} - \frac{a^3}{x^2} = 0.\ \checkmark\]

At $x=a$: $f'(a)=-a^3/a^3=-1$, so $\partial u/\partial x|_{x=a}=-\cos y\ \checkmark$. As $x\to\infty$: $f'(x)=-a^3/x^3\to0$, so $\partial u/\partial x\to0\ \checkmark$. Since every step above was reversible (the reduction to the Euler equation, its two-root solution, and the two linear conditions on $C_1,C_2$), $u$ is confirmed to solve the full PDE $(*)$ together with both boundary conditions.

\[u(x,y) = \frac{a^3}{2x^2}\cos y\]