[Partial differential equation Tutorial Exercise Reference Solution The derivation of]1.1 equation and the formulation of definite solution problem

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