Question

Which of the following statements is/are true for the PDE: \((1+x^2)u_{xx}+(1+y^2)u_{yy}+xu_x+yu_y=0\)?
(I) It is classified as parabolic PDE.
(II) The canonical equation is \(u_{\xi\xi}+u_{\eta\eta}=0\).

• A. Only (I)
• B. Only (II)
• C. Both (I) and (II)
• D. Neither (I) nor (II)

Hint:

Compute B^2 - 4AC to classify the PDE, then use xi = sinh^-1(x), eta = sinh^-1(y) to find the canonical form.

Answer 1

The Correct Option is B

Step 1: For the general second-order PDE \(Au_{xx}+2Bu_{xy}+Cu_{yy}+\dots=0\), the type is determined by the discriminant \(B^2-4AC\). Here \(A=1+x^2\), \(B=0\), \(C=1+y^2\).

Step 2: \[B^2-4AC=0-4(1+x^2)(1+y^2)=-4(1+x^2)(1+y^2)\] Since \(1+x^2>0\) and \(1+y^2>0\) for all real \(x,y\), this discriminant is always negative. So the PDE is elliptic everywhere, never parabolic. Statement (I) is false.

Step 3: To find the canonical form, use \(\xi=\sinh^{-1}x\), \(\eta=\sinh^{-1}y\), so \(\xi_x=\dfrac{1}{\sqrt{1+x^2}}\), \(\eta_y=\dfrac{1}{\sqrt{1+y^2}}\), \(\xi_y=\eta_x=0\).

Step 4: Compute \(u_x=u_\xi\xi_x\) and \[u_{xx}=u_{\xi\xi}\xi_x^2+u_\xi\xi_{xx}=\frac{u_{\xi\xi}}{1+x^2}-\frac{x\,u_\xi}{(1+x^2)^{3/2}}\] so \[(1+x^2)u_{xx}+xu_x=u_{\xi\xi}-\frac{x\,u_\xi}{\sqrt{1+x^2}}+\frac{x\,u_\xi}{\sqrt{1+x^2}}=u_{\xi\xi}\] The lower-order terms cancel exactly.

Step 5: By the same computation in \(y\), \((1+y^2)u_{yy}+yu_y=u_{\eta\eta}\).

Step 6: Adding, the PDE becomes exactly \[u_{\xi\xi}+u_{\eta\eta}=0\] which matches statement (II), so (II) is true.

Step 7: Hence only statement (II) is true. \[\boxed{\text{Only (II)}}\]