Find the expression of $\frac{1}{Dx^2-Dy^2} \sin (x-y)$

Question

Find the expression of $\frac{1}{Dx^2-Dy^2} \sin (x-y)$

May I ask if anybody could, please, tell me how to do this? I am not getting anything on this.

$$ \frac{1}{Dx^2-Dy^2} \sin (x-y) = \frac{1}{D(x^2-y^2)} \sin (x-y) = \int \frac{\sin (x-y)}{x^2-y^2} $$

Not sure which one to tag. So including differential and partial differential tags

I do not know what details I need to add. I am preparing for a competitive exam and the question is asked in previous year model paper. I just posted it as it is. After seeing the explanation I got to see it belongs to partial differential equations, but I thought it is from differential equations. Please correct me if I am wrong somewhere.

Answer 1

The result of that expression actually gives a particular solution to the PDE $D_x^2u-D_y^2u=u_{xx}-u_{yy}=\sin(x-y).$ This is called an operator method to find particular solutions.

$u_p= \frac{1}{Dx^2-Dy^2} \sin (x-y)=\frac{1}{-1-1}\sin (x-y)=-\frac{1}{2}\sin(x-y),$

where we have used $D_x^2=-1$ and $D_y^2=-1$ coming from the general formula $$\frac{1}{P(Dx^2, Dy^2)} \sin (ax+by)=\frac{1}{P(-a^2, -b^2)} \sin (ax+by)$$ and in your case the partial differential operator $P$ is $P(Dx^2, Dy^2)=Dx^2-Dy^2$.

Edit: I have made a mistake because the above formula does not work as @achille hui commented because $P(-a^2, -b^2)=0$. To resolve this after factorizing the operator we use the formula $$\frac{1}{P(D_x, D_y)}e^{ax+by}=\frac{1}{P(a, b)}e^{ax+by}$$ provided that $P(a, b)\neq 0$. $$u_p= \frac{1}{Dx^2-Dy^2} \sin (x-y)=\frac{1}{Dx^2-Dy^2} \text{Im} e^{i(x-y)}=\frac{1}{(D_x+D_y)(D_x-D_y)} \text{Im} e^{i(x-y)}$$

$$=\frac{1}{(D_x+D_y)(D_x-D_y)} \text{Im} e^{i(x-y)}=\frac{1}{(D_x+D_y)} \text{Im}\frac{e^{i(x-y)}}{2i}=\frac{1}{2}(x+y)\text{Im} \frac{e^{i(x-y)}}{2i}=-\frac{1}{4}(x+y)\cos(x-y).$$

For the last two steps first we have to calculate $\frac{1}{bD_x-aD_y} e^{ax+by}$. Notice that in this case when $D_x=a$ and $D_y=b$, the denominator is zero. To do so, I have used this formula $$\frac{1}{bD_x-aD_y} e^{ax+by}=\frac{ay-bx}{2ab}e^{ax+by}.\qquad (1) $$

Applying (1) to $\frac{1}{D_x+D_y} e^{ix-iy}$ with $a=i$ and $b=-i$ we get $$\frac{1}{D_x+D_y} e^{ix-iy}=\frac{(-iy-ix)}{2(i)(-i)}e^{ix-iy}=-\frac{i}{2}(x+y)e^{ix-iy}=-\frac{i}{2}(x+y)e^{i(x-y)}.$$ Thus using this and Euler's formula $e^{it}=\cos t+i\sin t$ and taking the imaginary part of the resulting expression with $t=x-y$ we find $-\frac{1}{4}(x+y)\cos(x-y).$ Note: Of course for some operators it may be better to use another way to find a particular solutions. I have used this because your question about operator method.

Answer 2

This answer assumes that $\frac{1}{Dx^2 - Dy^2}$ is meant to represent the inverse of the differential operator $$ {\left(\frac{\partial}{\partial x}\right)}^2 - {\left(\frac{\partial}{\partial y}\right)}^2 = \partial_x^2 - \partial_y^2\, . $$

We are trying to find a function $f(x,y)$ such that $$ \left(\partial_x^2 - \partial_y^2\right)f(x,y) \;=\; \sin(x - y)\, . $$

Define new variables $u = x - y$ and $v = x+y$. Then $$ \partial_x^2 - \partial_y^2 \;=\; \left(\partial_x - \partial_y\right)\left(\partial_x + \partial_y\right)\;=\; 4\, \partial_u \partial_v\, $$ and our problem simplifies to finding a function $g(u, v)$ such that $$ 4\, \partial_u \partial_v\, g(u, v) \;=\; \sin u\, . $$

Either by inspection or by indefinitely integrating both sides once each with respect to both $u$ and $v$, we see that the general solution to this is $$ g(u, v) \;=\; -\frac{v}{4}\, \cos u \;+\; A(u) \;+\; B(v)\, , $$ where $A(u)$ and $B(v)$ are entirely arbitrary functions of $u$ and $v$, respectively. Substituting in our original change of variables for $u$ and $v$, we then find that the (non-unique) answer to the original problem is $$ f(x, y) \;=\; -\frac{x+y}{4} \cos(x - y) \;+\; A(x-y) \;+\; B(x+y)\, . $$