Chapter 12, Problem 1RE

Use separation of variables to find product solutions of

$\frac{{\partial }^{2}u}{\partial x\partial y}=u.$

To determine

The product solutions of the given equation.

Answer to Problem 1RE

When $\lambda =0$ the product solution is $\underset{\_}{u\left(x,y\right)=0}$ and when $\lambda \ne 0$ the product solution is $\underset{\_}{u\left(x,y\right)=A{e}^{-\lambda x-\frac{y}{\lambda }}}$.

Explanation of Solution

Given:

The equation is $\frac{{\partial }^{2}u}{\partial x\partial y}=u$.

Calculation:

The given equation is as follows.

$\frac{{\partial }^{2}u}{\partial x\partial y}=u \dots \dots \left(1\right)$

Consider the solution of the equation as given below.

$u\left(x,y\right)=X\left(x\right)+Y\left(y\right) \dots \dots \left(2\right)$

Separate the variables of the above equation.

$\frac{X^{\prime }}{X}=-\lambda \text{and}Y{Y}^{\prime }=-\lambda$

The equation of the variable $X$ is given below.

$X^{\prime }+\lambda X=0 \dots \dots \left(3\right)$

The equation of the variable $Y$ is given below.

$Y^{\prime }+\frac{Y}{\lambda }=0 \dots \dots \left(4\right)$

The solution of the equations can be obtained for various cases considering the value of $\lambda$ to be strictly zero and non zero.

When the value of $\lambda$ is zero, substitute $\lambda =0$ in equation (3).

$X^{\prime }=0$

The general solution of the above equation can be calculated as follows.

$X\left(x\right)=C \dots \dots \left(5\right)$

Substitute $\lambda =0$ in equation (4).

$Y^{\prime }=0$

The general solution of the above equation is as follows.

$Y\left(y\right)=D \dots \dots \left(6\right)$

Substitute the value of $X\left(x\right)$ from equation (5) and the value of $Y\left(y\right)$ from equation (6) in equation (2).

$\begin{array}{c}u\left(x,y\right)=C\cdot D\\ =A \dots \dots \left(7\right)\end{array}$

When the value of $\lambda$ is not equal to zero.

The general solution of the equation (3) is as follows.

$X\left(x\right)=c_1{e}^{-\lambda x} \dots \dots \left(8\right)$

The general solution of the equation (4) is as follows.

$Y\left(y\right)=c_2{e}^{-\frac{y}{\lambda }} \dots \dots \left(9\right)$

Substitute the value of $X\left(x\right)$ from equation (8) and the value of $Y\left(y\right)$ from equation (9) in equation (2).

$u\left(x,y\right)=c_1{c}_2{e}^{-\frac{y}{\lambda }}{e}^{-\lambda x}$

Rewrite the term $c_1{c}_2=A$ in the above equation to obtain the solution of the given equation.

$u\left(x,y\right)=A{e}^{-\lambda x-\frac{y}{\lambda }}$

Thus, when $\lambda =0$ the product solution is $\underset{\_}{u\left(x,y\right)=0}$ and when $\lambda \ne 0$, the product solution is $\underset{\_}{u\left(x,y\right)=A{e}^{-\lambda x-\frac{y}{\lambda }}}$.