1. Point out the order of the following equation and determine whether it is linear, or nonlinear, if it is linear, indicating whether it is homogeneous or non-homogeneous.
(1). $u _t-(U_{xx}+u_{yy}) +1=0.$
Solution: This is the second order linear non-homogeneous equation.
(2). $u _t-u_{xx}+xu=0$.
Answer: This is the second order linear homogeneous equation.
(3). $u _t-u_{xxt}+uu_x=0$.
Answer: This is a third-order semilinear equation.
(4). $u _x^2+uu_y=0$.
Solution: This is a first order complete nonlinear equation.
(5). $u _{tt}-u_{xx}+t^2+x^2=0$.
Answer: This is the second order linear homogeneous equation.
(6). $u _x+e^y u_y=0$.
Answer: This is the first order linear homogeneous equation.
(7). $u _t+u_{xxxx}+\sqrt{1+u}=0$.
Answer: This is a four-order quasilinear linear equation.
(8). $u _x (1+u_x^2) ^{-\frac{1}{2}}+u_y (1+u_y^2) ^{-\frac{1}{2}}=0$.
Solution: This is the first order nonlinear equation.
(9). $\dps{\frac{\p^4u}{\p x^4} +2\frac{\p^4u}{\p x^2\p y^2} +\frac{\p^4u}{\p y^4}=0}$.
Answer: This is a four-order linear homogeneous equation.
(Ten). $\dps{\frac{\p^3u}{\p x^3} +\frac{\p^3u}{\p x\p y^2}+\ln u=0}$.
Answer: This is a third order quasilinear linear equation.
2. Set the $f (x) $ and $g (y) $ is any of the two consecutive micro functions that verify the function $u =f (x) g (Y) to satisfy the equation $$\bex uu_{xy}-u_xu_y=0. \eex$$
Proof: by $$\bex u_x=f ' (x) g (y), \quad u_y=f (x) g ' (y), \quad u_{xy}=f ' (x) g ' (y) \eex$$ is known as the conclusion.
3. Verify that the function $\dps{u (x, y) =\frac{1}{6}x^3y^2+x^2+\sin x+\cos y-\frac{1}{3}y^2+4}$ is the equation $$\bex \frac{\p^2u}{\p x\p y}=x^2y \eex$ $ of solution.
Answer: Note that $$\bex \frac{\p^2}{\p x\p y}\sex{\frac{1}{6}x^3y^2} =x^2y \eex$$ is known to the conclusion.
4. Verify that the function $u (x, y) =x^2-y^2$ and $u (x, y) =e^x\sin y$ are the solutions of the equation $u _{xx}+u_{yy}=0$.
Answer: There is a direct generation of authentication.
5. Verify function $$\bex u (x,y,z,t) =\frac{1}{\sqrt{a^2t^2-(x-\xi) ^2-(Y-\eta) ^2}} \eex$$ in zone $\omega=\sed{(x,y,z,t);(x-\xi) ^2+ ( Y-\eta) ^2<a^2t^2}$ satisfies the equation $$\bex u_{tt}=a^2 (u_{xx}+u_{yy}), \eex$$ where $a $ is normal and $\xi,\eta$ is any real number.
Proof: Without losing generality, can be set $a =\xi=\eta=0$. and $$\bex u=\frac{1}{\sqrt{t^2-x^2-y^2}}. The direct calculation of \eex$$ is there.
[Partial differential equation Tutorial Exercise Reference Solution The derivation of]1.1 equation and the formulation of definite solution problem