Question

State whether the following statements are true or false. Justify your answer with the help of a short proof or a counter example. i) Equation cos(x+y)p + sin(x+y)q = z^2 is a quasi-linear equation. ii) The solution of PDE dz/dx + dz/dy =z^2 is z = -[y+f(x-y)] Iii)(dz/dx)(dz/dy) - (dz/dy)^2 = 0 is a non-linear PDE
Please do it stepwise as I am a noob in differential equations

Answer 1

QUESTION(i)

Recall the theory.

Definition. Quasi-linear partial differential equations of first order can be written as

$$P(x,y,z)p+Q(x,y,z)q=R(x,y,z)\quad\quad (1)$$

where $P,Q,R$ are functions of $x,y$ and $z$ . They do not involve $p$ or $q$ . Here, linear means that $p$ and $q$ appear to the first degree only.

Definition taken from book : E. Rukmangadachari. Mathematical Methods

In our case,

$$\boxed{\cos(x+y)p+\sin(x+y)q=z^2-\text{s a quasi-linear equation}}$$

QUESTION(ii)

$$z=-y-f(x-y)\longrightarrow\\[0.3cm] \left\{\begin{array}{l} u=x-y\\[0.3cm] \displaystyle\frac{\partial z}{\partial x}=-\displaystyle\frac{\partial f}{\partial u}\cdot \displaystyle\frac{\partial u}{\partial x}\equiv-f' \\[0.3cm] \displaystyle\frac{\partial z}{\partial y}=-1-\displaystyle\frac{\partial f}{\partial y}\cdot \displaystyle\frac{\partial u}{\partial y}\equiv-1-f'\cdot(-1) \end{array}\right.\longrightarrow\\[0.3cm] \frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=-f' -1+f' =-1\neq z^2$$

Conclusion,

$$z=-y-f(x-y)\,\,\text{is not a solution to the differential equation}\\[0.3cm] \frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=z^2$$

QUESTION(iii)

Equation

$$\frac{\partial z}{\partial x}\cdot\frac{\partial z}{\partial y}-\left(\frac{\partial z}{\partial y}\right)^2=0$$

contains the term $\left(z'_y\right)^2$ therefore it is nonlinear.

Conclusion,

$$\boxed{\frac{\partial z}{\partial x}\cdot\frac{\partial z}{\partial y}-\left(\frac{\partial z}{\partial y}\right)^2=0-\text{is nonlinear}}$$

ANSWER

(i)

$$\cos(x+y)p+\sin(x+y)q=z^2-\text{s a quasi-linear equation}$$

(ii)

$$z=-y-f(x-y)\,\,\text{is not a solution to the differential equation}\\[0.3cm] \frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=z^2$$

(iii)

$$\frac{\partial z}{\partial x}\cdot\frac{\partial z}{\partial y}-\left(\frac{\partial z}{\partial y}\right)^2=0-\text{is nonlinear}$$