Solving Partial Differential Equations

$\blacktriangleright$ To solve a partial differential equation

1.
Type the partial differential equation using standard mathematical notation.

2.
With the insertion point in the equation, choose Solve PDE.

$\blacktriangleright$ Solve PDE

${\frac{{\partial ^{2}u}}{{\partial x^{2}}}}$ - ${\frac{{\partial ^{2}u}}{{\partial y^{2}}%
}}$ = 0, Exact solution is : u(x, y) = F1(y + x) + F2(y - x)

To verify that u(x, y) = F1(y + x) + F2(y - x) is actually a solution, define F1(x) and F2(x) to be generic functions and define

u(x, y) = F1(y + x) + F2(y - x)

Then Evaluate yields

$\displaystyle {\frac{{\partial ^{2}u(x,y)}}{{\partial x^{2}}}}$ = F1′′$\displaystyle \left(\vphantom{ y+x}\right.$y + x$\displaystyle \left.\vphantom{ y+x}\right)$ + F2′′$\displaystyle \left(\vphantom{ y-x}\right.$y - x$\displaystyle \left.\vphantom{ y-x}\right)$

and

$\displaystyle {\frac{{\partial ^{2}u(x,y)}}{{\partial y^{2}}}}$ = F1′′$\displaystyle \left(\vphantom{ y+x}\right.$y + x$\displaystyle \left.\vphantom{ y+x}\right)$ + F2′′$\displaystyle \left(\vphantom{ y-x}\right.$y - x$\displaystyle \left.\vphantom{ y-x}\right)$

and hence

$\displaystyle {\frac{{\partial ^{2}u(x,y)}}{{\partial x^{2}}}}$ - $\displaystyle {\frac{{\partial ^{2}u(x,y)}}{{%
\partial y^{2}}}}$ = 0


\begin{example}
The choice
\begin{displaymath}
u(x,y)=\sin (y+x)+\cos (y-x) ...
...rac{\partial ^{2}u(x,y)}{%
\partial y^{2}}=0
\end{displaymath}
\end{example}