# Transforming a parabolic pde to normal form

• saching
In summary, the conversation discusses transforming a parabolic PDE to normal form. The characteristic equation and characteristic curve are used to find new variables, v and w, and the normal form is determined to be U_{ww}=F_{2}. The process involves using the chain rule and integration, with the final solution being a function of the new variables v and w. Similar techniques can be used for hyperbolic PDEs, as shown in the example provided.
saching
[SOLVED] transforming a parabolic pde to normal form

## Homework Statement

The problem is to transform the PDE to normal form.

The PDE in question is parabolic: U$$_{xx}$$ - 2U$$_{xy}$$ + U$$_{yy}$$ = 0 but I also need to solve other problems for hyperbolic pde's so general advice would be appreciated.

## Homework Equations

The characteristic equation is: Ay'$$^{2}$$ - 2By' + C = 0

The new variables should be v=x, w=psi, and the normal form is U$$_{ww}$$=F$$_{2}$$

## The Attempt at a Solution

The solutions manual provides:

I get lost right after we solve the characteristic equation. I don't understand how the variable substitution works or what is going on after that. My textbook only offers 1 example similar to this problem with no explanation of how it goes from step to step...so I'm completely lost. I looked online for information but the limited amount of stuff I did find is too technical(I read through all of them).

saching said:

## Homework Statement

The problem is to transform the PDE to normal form.

The PDE in question is parabolic: U$$_{xx}$$ - 2U$$_{xy}$$ + U$$_{yy}$$ = 0 but I also need to solve other problems for hyperbolic pde's so general advice would be appreciated.

## Homework Equations

The characteristic equation is: Ay'$$^{2}$$ - 2By' + C = 0

The new variables should be v=x, w=psi, and the normal form is U$$_{ww}$$=F$$_{2}$$

## The Attempt at a Solution

The solutions manual provides:

I get lost right after we solve the characteristic equation. I don't understand how the variable substitution works or what is going on after that. My textbook only offers 1 example similar to this problem with no explanation of how it goes from step to step...so I'm completely lost. I looked online for information but the limited amount of stuff I did find is too technical(I read through all of them).
Here's how I would do it (finding the characteristic equation slightly differently):
From $U_{xx}- 2U_{xy}+ U_{yy}= 0$ we get x2- 2xy+ y2= (x- y)2= 0 which has only x-y= 0 or x= y as solution. Perpendicular to that line is the "characteristic curve" (line in this case) x+y= 0. We change coordinate systems so that line is one of the axes: let v= x+ y. Since that is the only characteristic line, just let u= x be the other. In these new variables, by the chain rule,
$$U_x= U_u\frac{\partial u}{\partial x}+ U_v\frac{\partial v}{\partial x}$$
Since $\partial u/\partial x= 1$ and $\partial v/\partial x= 1$ that is
$$U_x= U_u+ U_v$$

$$U_y= U_u\frac{\partial u}{\partial y}+ U_v\frac{\partial v}{\partial y}$$
Since $\partial u/\partial y= 0$ and $\partial v/\partial y= 1$ that is
$$U_y= U_v[/itex] Continuing to use the chain rule, [tex]U_{xx}= (U_x)_x= (U_u+ U_v)_x= (U_{uu}\frac{\partial u}{\partial x}+ U_{uv}\frac{\partial v}{\partial x})+ (U_{vu}\frac{\partial u}{\partial x}+ U_{vv}\frac{\partial v}{\partial x})$$
$$= U_{uu}+ 2U_{uv}+ U_{vv}$$

$$U_{xy}= (U_x)_y= (U_u+ U_v)_y= (U_{uu}\frac{\partial u}{\partial y}+ U_{uv}\frac{\partial v}{\partial y})+ (U_{vu}\frac{\partial u}{\partial y}+ U_{vv}\frac{\partial v}{\partial y})$$
$$= U_{uv}+ U_{vv}$$

$$U_{yy}= (U_y)_y= (U_u+ U_v)_y= (U_{uu}\frac{\partial u}{\partial y}+ U_{uv}\frac{\partial v}{\partial y})+ (U_{vu}\frac{\partial u}{\partial y}+ U_{vv}\frac{\partial v}{\partial y})$$
$$= U_{vv}$$

Putting those into the equation $U_{xx}- 2U_{xy}+ U_{yy}= 0$ we get
$(U_{uu}+ 2U_{uv}+ U_{vv})-2(U_{uv}+ U_{vv})+ U_{vv}= U_{uu}= 0$
the "normal" form.

Saying that $U_{uu}= (U_u)_u= 0$ means that $U_u$ is a "constant" with respect to u. It must depend only on v: $U_u= F(v)$. Integrating with respect to u again, U(u,v)= F(v)u+ G(v) where G(v) is the "constant" of itegration- since we are only integrating with respect to u, G may depend upon v. That is the general solution to the differential equation in terms of u and v. Since u= x and v= x+y, that is the same as U(x,y)= F(x+y)x+ G(x+y). Since F and G, above, were arbitrary functions of one variable, I can choose any such functions and replace the one variable with x+ y. For example, suppose F(x)= x2 and G(x)= sin(x) (chosen pretty much arbitrarily). Then U(x,y)= (x+y)2x+ sin(x+ y)= x3+ 2x2y+ xy2+ sin(x+y). I'll leave it to you to show that this function does, in fact, satisfy the original differential equation.

Last edited by a moderator:
Take deep breath, here we go again for hyperbolic equations!

An an example, I will use $U_{xx}- 5U_{xy}+ 6U_{yy}= 0$.

Convert that to the algebraic equation x2- 5xy+ 6y2= (x- 3y)(x- 2y)= 0. The fact that there are two distinct factors tells us that there are two distinct characteristic curves (lines) and so this equation is hyperbolic. The line, through the origin, perpendicular to x- 3y= 0 is y+ 3x= 0. The line, through the origin, perpendicular to x- 2y= 0 is y+ 2x= 0.

Let u= 3x+ y and v= y+ 2x. so that
$$\frac{\partial u}{\partial x}= 3[/itex] [tex]\frac{\partial v}{\partial x}= 2[/itex] [tex]\frac{\partial u}{\partial y}= 1[/itex] [tex]\frac{\partial v}{\partial y}= 1[/itex] Then [tex]U_x= 3U_u+ 2U_v$$
$$U_y= U_u+ U_v$$

$$U_{xx}= 3(3U_{uu}+2U_{vu})+ 2(3U_{uv}+ 2U_{vv})$$
$$= 9U_{uu}+ 12U_{uv}+ 4U{vv}$$

$$U_{yy}= (U_{uu}+ U_{vu})+ (U_{vu}+ U_{vv})$$
$$= U_{uu}+ 2U_{uv}+ U_{vv}$$

$$U_{xy}= (U_x)_y= (3U_{uu}+ 2U_{uv})+ (3U_{uv}+ 2U{vv})$$
$$= 3U_{uu}+ 5U_{uv}+ 2U_{vv}$$

$$U_{xx}= 9U_{uu}+ 12U_{uv}+ 4U{vv}$$
$$-5U_{xy}= -15U_{uu}- 25U_{uv}- 10U{vv}$$
$$6U_{yy}= 6U_{uu}+ 12U+{uv}+ 6U{vv}$$
$$U_{xx}- 5U_{xy}+ 6U{yy}= -U_{uv}= 0$$

That tells us the derivative of $U_u$, with respect to v, is 0 and so $U_u$ is a function of U only: $U_u= f(u)$. Integrating, with respect to u now, U(u,v)= F(u)+ G(v) where F is the anti-derivative of f with respect to u and G is the "constant of integration" which, since the derivative is with respect to u, may be a function of v. That is the general solution to the differential equation in terms of u and v. Putting it back into x and y, U(x,y)= F(y+ 3x)+ G(y+ 2x).

Again, I will leave it to you to show that, taking F(x)= x2 and G(x)= sin(x), U(x,y)= (y+ 3x)2+ sin(y+ 2x) satisfies the original differential equation.

## 1. What is a parabolic PDE?

A parabolic PDE (partial differential equation) is a type of mathematical equation that involves multiple variables and their partial derivatives with respect to time. It is commonly used to model physical processes that involve diffusion, such as heat transfer or diffusion of particles.

## 2. Why is it important to transform a parabolic PDE to normal form?

Transforming a parabolic PDE to normal form allows for easier analysis and solution of the equation. Normal form is a standard format that simplifies the equation and makes it easier to identify important features and properties. It also allows for comparison with other equations and helps in the development of analytical and numerical methods for solving the equation.

## 3. What is the process for transforming a parabolic PDE to normal form?

The process for transforming a parabolic PDE to normal form involves using a change of variables to eliminate terms involving the highest order derivatives. This results in a simpler equation with only first and second order derivatives. The transformed equation can then be analyzed and solved using various techniques.

## 4. What are some common techniques used to transform a parabolic PDE to normal form?

Some common techniques for transforming a parabolic PDE to normal form include the method of characteristics, separation of variables, and similarity transformations. Each technique has its own advantages and is suited for different types of PDEs. Choosing the right technique depends on the specific equation and its properties.

## 5. Are there any limitations to transforming a parabolic PDE to normal form?

Yes, there are limitations to transforming a parabolic PDE to normal form. Some PDEs may not have a simple transformation that results in a normal form, making it difficult to analyze and solve. Additionally, the transformed equation may lose some physical meaning or properties, so it is important to carefully consider the implications of the transformation.

• Calculus and Beyond Homework Help
Replies
11
Views
738
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
571
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
19
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
2K