Identifying the Type of PDE and Methods for Solving It

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In summary: However, due to the ellipticity of the equation, it will not be a straightforward task. In summary, this problem seems to be very difficult to solve.
  • #1
alokgautam
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Hello dear friends,

I have this PDE



Can anyone help me in finding what type PDE is it. So that I can try to solve this .

As per my knowledge, I think this is second order and Nonlinear
Is my guess is correct?


Is it parabolic or hyperbolic, or is Elliptic and what method can I use to solve this Pde.
if it is possible please let me know about book for this problem


please see attachment for PDE
 
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  • #3


Hi Gota,
thanks for mailing again

yes this is the same problem.
because still i do not know from where shall start to solve this equation.

even i am not able to know to find what type of PDE is it.

parabolic or hyperbolic or elliptic PDEs, because i do not know what is the value of b^2-a*c.

thats why i am not able to find the method for the soluton.

looking for you reply



gato_ said:
Isn't it exactly the same problem you posted a couple of weeks ago here?
https://www.physicsforums.com/showthread.php?t=371542
 
  • #4


I don't know if the classification applies well for a non linear equation. The linear version of this sure is elliptic, and the tendency of this equation is to "smooth" the resulting surface, but no linear technique will be of help unless the curvatures involved are small. Try [tex]\Delta z(x,y)=z(x,y)/c[/tex]. Beware with boundary conditions, they are the most tricky part
 
  • #5


I think that it is very difficult (if possible) to find the general solution to your PDE. Unfortunately there is not a universal method for this.

Some methods for finding particular solutions of nonlinear PDEs see in

A. D. Polyanin and V. F. Zaitsev, Handbook of Nonlinear Partial Differential Equations, Chapman & Hall/CRC Press, Boca Raton, 2004

These particular solutions, unluckily, rarely meet the physically based boundary conditions.

The most easy way to find particular solution is to suppose that

[tex]z(x,y) = F(ax+by)[/tex]

which lead to some ODE for function F. This way gives the following particular solution to your PDE in implicit form

[tex]\pm\int_k^{z(x,y)}\frac{(a^2+b^2)(\xi^2+2C1)}{\sqrt{-(a^2+b^2)(\xi^4+4\xi^2C1+4C1^2-4c^2))}}\,d\xi -ax-by-C2 = 0[/tex]

where a, b, k, C1, C2 are arbitrary constants. Incidentally, the integral above can be evaluated.

One remark - Nevertheless, for some PDEs from your field the general solutions can be found, see, e.g. http://eqworld.ipmnet.ru/eqarchive/view.php?id=271".
 
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  • #6


Dear Gato,

Thank you once again.
i will try to apply finite element method to solve this
with best regards

ALok


gato_ said:
I don't know if the classification applies well for a non linear equation. The linear version of this sure is elliptic, and the tendency of this equation is to "smooth" the resulting surface, but no linear technique will be of help unless the curvatures involved are small. Try [tex]\Delta z(x,y)=z(x,y)/c[/tex]. Beware with boundary conditions, they are the most tricky part
 
  • #7


Dear kosovtsov,

Thank you very much.

so there is no numerical method is available to solve this PDE.

with best regards

ALOK


kosovtsov said:
I think that it is very difficult (if possible) to find the general solution to your PDE. Unfortunately there is not a universal method for this.

Some methods for finding particular solutions of nonlinear PDEs see in

A. D. Polyanin and V. F. Zaitsev, Handbook of Nonlinear Partial Differential Equations, Chapman & Hall/CRC Press, Boca Raton, 2004

These particular solutions, unluckily, rarely meet the physically based boundary conditions.

The most easy way to find particular solution is to suppose that

[tex]z(x,y) = F(ax+by)[/tex]

which lead to some ODE for function F. This way gives the following particular solution to your PDE in implicit form

[tex]\pm\int_k^{z(x,y)}\frac{(a^2+b^2)(\xi^2+2C1)}{\sqrt{-(a^2+b^2)(\xi^4+4\xi^2C1+4C1^2-4c^2))}}\,d\xi -ax-by-C2 = 0[/tex]

where a, b, k, C1, C2 are arbitrary constants. Incidentally, the integral above can be evaluated.

One remark - Nevertheless, for some PDEs from your field the general solutions can be found, see, e.g. http://eqworld.ipmnet.ru/eqarchive/view.php?id=271".
 
Last edited by a moderator:
  • #8


alokgautam said:
so there is no numerical method is available to solve this PDE.

I do not say that.

It is clear that you can turn to numerical solution of the PDE if you will.
 

1. Is a parabola and a hyperbola the same thing?

No, a parabola and a hyperbola are two different types of curves. A parabola is a U-shaped curve while a hyperbola is two separate curves that mirror each other.

2. How can I tell if a curve is parabolic or hyperbolic?

One way to tell is by looking at the equation of the curve. A parabola has an equation in the form of y = ax^2 + bx + c while a hyperbola has an equation in the form of x^2/a^2 - y^2/b^2 = 1. Another way is to look at the shape of the curve - a parabola is symmetrical while a hyperbola is not.

3. What are the real-life applications of parabolic and hyperbolic curves?

Parabolic curves are commonly seen in physics, such as the trajectory of a projectile or the shape of a satellite dish. Hyperbolic curves are often used in economics and finance to represent exponential growth or decay.

4. Are there any similarities between parabolic and hyperbolic curves?

Yes, both curves have a focus and a directrix. The focus is a fixed point within the curve and the directrix is a line that is perpendicular to the axis of symmetry. Both curves also have a vertex, which is the point where the curve changes direction.

5. Can a curve be both parabolic and hyperbolic?

No, a curve can only be classified as either parabolic or hyperbolic. However, there are some curves that have characteristics of both and are called paraboloids. These curves have a parabolic cross-section in one direction and a hyperbolic cross-section in the other direction.

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