Partial differential equations: general solution

In summary, the conversation discusses transforming a second order homogeneous partial differential equation with mixed partial derivatives into a simpler form by making a change of variables. This is accomplished by defining new functions and using the chain rule for partial derivatives. The method of transforming the equation can be simplified by setting certain coefficients equal to zero and integrating the resulting equation. This allows for a more efficient and systematic approach to solving the partial differential equation.
  • #1
sachi
75
1
I have a 2nd order homogenous P.D.E:

(d^2)V/((dx)^2) + (d^2)V/((dy)^2) + 6(d^2)V/(dx dy) = 0

where all derivatives are partial derivatives. I need to transform this to form

a*(d^2)f/(dX^2) + b*(d^2)f/(dY^2) = 0 where a, b are constants and the derivatives are again partial, and f(X,Y) = V(x,y)

I I need to make a change of variables e.g x=cX +dY etc. but I get very messy algebraic expressions which I can't simplify. Thanks very much.
 
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  • #2
You have V = V(x,y).

Then make the change of variables,

X = X(x,y)
Y = Y(x,y)

where,

X = y - m1x
Y = y - m2x

(It doesn't need to be any more complicated)

Now use the chain rule for partial differentials involving change of variables.
 
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  • #3
You need to rotate axes to eliminate the mixed partial. This is normally done by defining a new function H(w,z) such that:

[tex]w=xCos(a)+ySin(a)\quad\text{and}\quad z=-xSin(a)+yCos(a)[/tex]

with a to be determined so that the coefficient of the mixed partial in the final results is zero.

So, start calculating all the partials and substitute them into your equation and end up with a PDE in H(w,z), then determine a so that the coefficient of the mixed partial is zero. Solve the PDE in H(w,z), then substitute back x and y:

[tex]V(x,y)=H(xCos(a)+ySin(a),-xSin(a)+yCos(a))[/tex]

I'll start off the partials:

[tex]\frac{\partial V}{\partial x}=\frac{\partial H}{\partial w}\frac{\partial w}{\partial x}+\frac{\partial H}{\partial z}\frac{\partial z}{\partial w}[/tex]

so then:

[tex]\frac{\partial^2 V}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial H}{\partial w}\frac{\partial w}{\partial x}+\frac{\partial H}{\partial z}\frac{\partial z}{\partial w}\right)[/tex]

So do that one, the mixed one, and then for y and then substitute them back into your PDE. It goes fast once you get going. Then solve for a to make the mixed one go away.:smile:

Edit: Hum . . . well maybe that too Fermat.:smile:
 
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  • #4
saltydog said:
...
Edit: Hum . . . well maybe that too Fermat.:smile:

Yeah :smile: That's the method we were shown for solving 2nd order pde's.

If the pde was of the form,

[tex]A\cdot \frac{\partial ^2V}{\partial x^2} + B\cdot \frac{\partial ^2V}{\partial x \partial y} + C\cdot \frac{\partial ^2V}{\partial y^2} = 0[/tex]

we would transform it to,

[tex]P(m_1)\cdot \frac{\partial ^2V}{\partial X^2} + Q(m_1,m_2)\cdot \frac{\partial ^2V}{\partial X \partial Y} + P(m_2)\cdot \frac{\partial ^2V}{\partial Y^2} = 0[/tex]

[tex]\mbox{However, rather than making, } Q(m_1,m_2) = 0 \mbox{ we would make } P(m_1) = 0 \mbox{ and } P(m_2) = 0[/tex]

This would give,

[tex]\frac{\partial ^2V}{\partial X \partial Y} = 0[/tex]

which is then easily integrated to give,

V = f(X) + g(Y)
============

where you would substitute for,

X = y - m1x
Y = y - m2x

You would get m1 and m2 when setting P(m1) = 0 and P(m2) = 0.

So, you could use any functions f() and g() that you liked, as long as their arguments were y - m1x and y - m2x.
 
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  • #5
Fermat said:
Yeah :smile: That's the method we were shown for solving 2nd order pde's.
If the pde was of the form,
[tex]A\cdot \frac{\partial ^2V}{\partial x^2} + B\cdot \frac{\partial ^2V}{\partial x \partial y} + C\cdot \frac{\partial ^2V}{\partial y^2} = 0[/tex]
we would transform it to,
[tex]P(m_1)\cdot \frac{\partial ^2V}{\partial X^2} + Q(m_1,m_2)\cdot \frac{\partial ^2V}{\partial X \partial Y} + P(m_2)\cdot \frac{\partial ^2V}{\partial Y^2} = 0[/tex]
[tex]\mbox{However, rather than making, } Q(m_1,m_2) = 0 \mbox{ we would make } P(m_1) = 0 \mbox{ and } P(m_2) = 0[/tex]
This would give,
[tex]\frac{\partial ^2V}{\partial X \partial Y} = 0[/tex]
which is then easily integrated to give,
V = f(X) + g(Y)
============
where you would substitute for,
X = y - m1x
Y = y - m2x
You would get m1 and m2 when setting P(m1) = 0 and P(m2) = 0.
So, you could use any functions f() and g() that you liked, as long as their arguments were y - m1x and y - m2x.

Thanks for point that out Fermat. That's a much better approach considering, after I looked into it a bit, that an arbitrary solution of the wave equation:

[tex]\frac{\partial^2 u}{\partial t^2}=a^2\frac{\partial^2}{\partial x^2}[/tex]

can be written in the form:

[tex]u(x,t)=F(x+at)+G(x-at)[/tex]

and thus I would suspect your m's above are related to the final coefficient on the x-partial term above. I'll work with it a bit.:smile:
 

1. What is a partial differential equation (PDE)?

A partial differential equation is a type of mathematical equation that involves multiple variables and their partial derivatives. These equations are commonly used to model physical phenomena in fields such as physics, engineering, and economics.

2. What is the general solution of a partial differential equation?

The general solution of a PDE is a family of solutions that satisfies the equation for all possible values of the variables. It contains arbitrary constants that can be determined by applying boundary conditions to the equation.

3. How is the general solution of a PDE different from the particular solution?

A particular solution is a specific solution that satisfies both the PDE and the given boundary conditions. On the other hand, the general solution is a more general form that includes all possible solutions to the PDE.

4. What are the methods for finding the general solution of a PDE?

There are several methods for finding the general solution of a PDE, including separation of variables, method of characteristics, and Fourier series. The choice of method depends on the type and complexity of the PDE.

5. Can the general solution of a PDE always be found?

No, the general solution of a PDE may not always be found. Some PDEs have no general solution, while others may have an infinite number of solutions. The existence and uniqueness of solutions to a PDE depend on the specific equation and boundary conditions.

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