Partial Differential Equation : Partial Differential Equation is a relation between partial derivatives of independent variable , dependent variable and their partial derivatives.
A general form of a partial differential equation can be written as
\[A\frac{{{\partial ^2}u}}{{\partial {x^2}}} + B\frac{{{\partial ^2}u}}{{\partial x\partial y}} + C\frac{{{\partial ^2}u}}{{\partial {y^2}}} + D\frac{{\partial u}}{{\partial x}} + E\frac{{\partial u}}{{\partial x}} + F\frac{{\partial u}}{{\partial y}} = G\]
If the coefficients of this equation satisfies condition
\[{B^2} – 4AC \gt 0\]
then the pde is called Elliptic.
Example (1)
\[({x^2} + {y^2} – 1)\frac{{{\partial ^2}u}}{{\partial {x^2}}} + 2\frac{{{\partial ^2}u}}{{\partial x\partial y}} + ({x^2} + {y^2} – 1)\frac{{{\partial ^2}u}}{{\partial {y^2}}} = 0\] is elliptic parabolic in the region \({x^2} + {y^2} \gt 2 \)
Solution:
Given partial differential equation
\[({x^2} + {y^2} – 1)\frac{{{\partial ^2}u}}{{\partial {x^2}}} + 2\frac{{{\partial ^2}u}}{{\partial x\partial y}} + ({x^2} + {y^2} – 1)\frac{{{\partial ^2}u}}{{\partial {y^2}}} = 0 \hspace{5 cm} (1)\]
A general form of second order partial differential equation
\[A\frac{{{\partial ^2}u}}{{\partial {x^2}}} + B\frac{{{\partial ^2}u}}{{\partial x\partial y}} + C\frac{{{\partial ^2}u}}{{\partial {y^2}}} + other lower order derivative terms = 0\hspace{5 cm} (2) \]
is elliptic if \( {B^2} – 4AC \lt 0 \)
Since for eqn (1), we have
\[ A = ({x^2} + {y^2} – 1) , B = 2 , C = ({x^2} + {y^2} – 1)\]
Hence \[{B^2} – 4AC = 2^2 – 4 ({x^2} + {y^2} – 1)({x^2} + {y^2} – 1) \]
\[ = 4 – 4 ({x^2} + {y^2} – 1)^2 \]
\[ = 4 \left[1 – ({x^2} + {y^2} – 1)^2 \right] \]
If \( ({x^2} + {y^2} \gt 2 \) then \(({x^2} + {y^2} – 1)^2 \gt 1\) this means \( \left[1 – ({x^2} + {y^2} – 1)^2 \right] \lt 0\)
\[\Rightarrow {B^2} – 4AC = 4 \left[1 – ({x^2} + {y^2} – 1)^2 \right] \lt 0\]
Thus \(\Rightarrow {B^2} – 4AC \lt 0 \) on the region \({x^2} + {y^2} \gt 2 \)
Therefore pde (1) is elliptic in the region \({x^2} + {y^2} \gt 2 \)
hyperbolic pde
Example (1)
Prove that pde \[({x^2} + {y^2} – 1)\frac{{{\partial ^2}u}}{{\partial {x^2}}} + 2\frac{{{\partial ^2}u}}{{\partial x\partial y}} + ({x^2} + {y^2} – 1)\frac{{{\partial ^2}u}}{{\partial {y^2}}} = 0\] is hyperbolic in the region \(0 \lt {x^2} + {y^2} \lt 2 \)
Solution:
Given partial differential equation
\[({x^2} + {y^2} – 1)\frac{{{\partial ^2}u}}{{\partial {x^2}}} + 2\frac{{{\partial ^2}u}}{{\partial x\partial y}} + ({x^2} + {y^2} – 1)\frac{{{\partial ^2}u}}{{\partial {y^2}}} = 0 \hspace{5 cm} (1)\]
We know general form of second order pde
\[A\frac{{{\partial ^2}u}}{{\partial {x^2}}} + B\frac{{{\partial ^2}u}}{{\partial x\partial y}} + C\frac{{{\partial ^2}u}}{{\partial {y^2}}} + other\hspace{0.1cm} lower \hspace{0.1cm} order \hspace{0.1cm} partial \hspace{0.1cm}derivative \hspace{0.1cm} terms = 0\hspace{5 cm} (2) \]
is hyperbolic if \( {B^2} – 4AC \gt 0 \)
Since for eqn (1), we have
\[ A = ({x^2} + {y^2} – 1) , B = 2 , C = ({x^2} + {y^2} – 1)\]
Hence \[{B^2} – 4AC = 2^2 – 4 ({x^2} + {y^2} – 1)({x^2} + {y^2} – 1) \]
\[ = 4 – 4 ({x^2} + {y^2} – 1)^2 \]
\[ = 4 \left[1 – ({x^2} + {y^2} – 1)^2 \right] \]
If \( 0\lt {x^2} + {y^2} \lt 2 \) then \(({x^2} + {y^2} – 1)^2 \lt 1\) this means \( \left[1 – ({x^2} + {y^2} – 1)^2 \right] \gt 0\)
\[\Rightarrow {B^2} – 4AC = 4 \left[1 – ({x^2} + {y^2} – 1)^2 \right] \gt 0\]
Thus \(\Rightarrow {B^2} – 4AC \gt 0 \) in the region \(0\lt {x^2} + {y^2} \lt 2 \)
Therefore eqn (1) is hyperbolic in the region \(0\lt {x^2} + {y^2} \lt 2 \)