Solve the characteristic of the PDE $z=p^2-q^2$, and find the integral surface which passes through the parabola $4z+x^2=0,y=0$

I used Clairaut's methods and found a relation between $p,q$ as $\frac{q}{p}=c(constant),q=pc$

Thus $z=p^2(1-c^2)\;$, and $\;\;dz=pdx+qdy=p(dx+cdy)$. Substituting the value of $p$, we get $$dz=\frac{\sqrt {z}}{\sqrt {1-c^2}}(dx+cdy)$$

Using this I solved the differential equation and got the solution as $$2\sqrt z=\frac{x}{\sqrt {1-c^2}}+\frac{c}{\sqrt {1-c^2}}y+c_1$$, where $c,c_1$ are constants. Now I am not able to find the integral surface which passes through the given parabola. Please help solve the problem


$2\sqrt z=\frac{x}{\sqrt {1-c^2}}+\frac{c}{\sqrt {1-c^2}}y+c_1$ is correct. $$z=\frac14\left(\frac{x+cy}{\sqrt {1-c^2}}+c_1\right)^2\tag 1$$

$$z(x,0)=\frac14\left(\frac{x}{\sqrt {1-c^2}}+c_1\right)^2$$ $$4z(x,0)+x^2=\left(\frac{x}{\sqrt {1-c^2}}+c_1 \right)^2+x^2=0$$ $$\left(\frac{1}{(1-c^2)}+1 \right)x^2+\frac{c_1}{\sqrt{1-c^2}}x+c_1^2=0$$ $$\begin{cases} c_1=0\\ \frac{1}{(1-c^2)}+1=0 \quad\implies\quad c=\pm\sqrt{2} \end{cases}$$ We put $c_1=0$ and $c=\pm\sqrt{2}$ into Eq.$(1)$ : $$z=\frac14\left(\frac{x\pm\sqrt{2}\:y}{i}\right)^2$$ $$z(x,y)=-\frac14\left(x\pm\sqrt{2}\:y\right)^2$$ One can check that this solution satisfies both the PDE and the boundary condition.

  • $\begingroup$ Thx for the solution , one small question, if I want to find an integral surface through the surface . I must substitute the equations of the surface in the characteristic solution and find a relation between the constants and eliminate them right ? Is this procedure right ? $\endgroup$ Mar 21 '19 at 7:59
  • $\begingroup$ I am not sure to well understand your question about ''an integral surface through the surface''. I think that $z(x,y)=-\frac14\left(x\pm\sqrt{2}\:y\right)^2$ answers to the question without further calculus. $\endgroup$
    – JJacquelin
    Mar 21 '19 at 8:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.