Question

The partial differential equation \[ 7 \frac{\partial^2 u}{\partial x^2} + 16 \frac{\partial^2 u}{\partial x \partial y} + 4 \frac{\partial^2 u}{\partial y^2} = 0 \] is transformed to \[ A \frac{\partial^2 u}{\partial \xi^2} + B \frac{\partial^2 u}{\partial \xi \partial \eta} + C \frac{\partial^2 u}{\partial \eta^2} = 0, \] using \( \xi = y - 2x \) and \( \eta = 7y - 2x \). Then, the value of \( \frac{1}{123} (B^2 - 4AC) \) is _____

Answer 1

Correct Answer: 12

Partial Differential Equation Transformation

The given partial differential equation is:

\( 7 \frac{\partial^2 u}{\partial x^2} + 16 \frac{\partial^2 u}{\partial x \partial y} + 4 \frac{\partial^2 u}{\partial y^2} = 0 \)

The transformation to new coordinates \( \xi = y - 2x \) and \( \eta = 7y - 2x \) is required. We need to find the value of \( \frac{1}{123} (B^2 - 4AC) \) after transforming the equation.

Step 1: Compute the Jacobian Matrix

The partial derivatives of \( \xi \) and \( \eta \) with respect to \( x \) and \( y \) are:

\( \frac{\partial \xi}{\partial x} = -2, \quad \frac{\partial \xi}{\partial y} = 1 \)

\( \frac{\partial \eta}{\partial x} = -2, \quad \frac{\partial \eta}{\partial y} = 7 \)

The inverse Jacobian matrix is:

\( \frac{\partial x}{\partial \xi} = -\frac{1}{2}, \quad \frac{\partial y}{\partial \xi} = \frac{1}{2}, \quad \frac{\partial x}{\partial \eta} = -\frac{1}{2}, \quad \frac{\partial y}{\partial \eta} = \frac{7}{2} \)

Step 2: Transformation of the Partial Derivatives

We transform the second-order partial derivatives of \( u \) with respect to \( x \) and \( y \) into those with respect to \( \xi \) and \( \eta \) using the chain rule. We express the equation as:

\( A \frac{\partial^2 u}{\partial \xi^2} + B \frac{\partial^2 u}{\partial \xi \partial \eta} + C \frac{\partial^2 u}{\partial \eta^2} = 0 \)

Using the matrix form for the coefficients of the PDE:

\[ \begin{pmatrix} A & B \\ B & C \end{pmatrix} = \begin{pmatrix} 7 & 16 \\ 16 & 4 \end{pmatrix} \]

Step 3: Compute \( B^2 - 4AC \)

The determinant \( B^2 - 4AC \) is computed as follows:

\( B = 16, \quad A = 7, \quad C = 4 \)

Now calculate \( B^2 - 4AC \):

\[ B^2 - 4AC = 16^2 - 4(7)(4) = 256 - 112 = 144 \]

Step 4: Final Answer

Thus, the value of \( \frac{1}{123} (B^2 - 4AC) \) is:

\[ \frac{1}{123} (B^2 - 4AC) = \frac{1}{123} \times 144 = \frac{144}{123} = \frac{12}{1} \]

So, the value is \( \boxed{12} \).