# Are the following PDEs linear or nonlinear?

• Thomas Moore
In summary: You might be able to figure it out if you try. In summary, the textbook says to use $L = \frac{\partial}{\partial t} - \frac{\partial^2}{\partial x^2} + 1/x$, but I'm not sure if that is the right way to do it for the first example. For the second example, I don't know how to write it so that the u^3 term will be included.
Thomas Moore
Hi. I'm a bit confused on determining whether a certain PDE is linear or non-linear.

For example, for the wave equation, we have: u_{xx} + u_{yy} = 0, where a subscript denotes a partial derivative.
So, my textbook says to write:
$L = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$

And then it is easy to deduce that $L(u+v) = L(u) + L(v)$ and $L(c u) = cL(u)$.

But, I have no idea how to do this for the following PDEs:
1. $u_{t} - u_{xx} + u/x = 0$, the $u/x$ is throwing me off.
2. $u_{tt} - u_{xx} + u^3 = 0$, the $u^3$ term is throwing me off.
I don't know how to write this as $Lu = 0$, to determine linearity. Any help would be appreciated, thanks!

Hi. That link was quite helpful. Well, this is where I got stuck though.

For the first example, I wrote:
1. $L = \frac{\partial}{\partial t} - \frac{\partial^2}{\partial x^2} + 1/x$, which seems to work if you do $L u$. But, I'm not sure if adding 1/x is right like that.

2. But for this one, I have no idea! I want to write: $L = \frac{\partial^2}{\partial t^2} - \frac{\partial}{\partial x^2}$ plus something, but I don't know how to write this so Lu will give the u^3 term at the end!

When choosing your operator, you can let ( )^3 be in your operator expression. E.g. if G is the cubic operator, then G(2)=8. The cubic operator (the operator which cubes its input) is not linear, but that doesn't mean we can't define it.

If you double up the dollar signs, the TeX will render properly.
BrianT is correct ... the naive way would have been just to write it out as if you just divided through by u.
You should not be afraid to try out stuff.

## 1. What is the definition of a linear PDE?

A linear partial differential equation (PDE) is a mathematical equation that involves partial derivatives of a function with respect to multiple independent variables, and the coefficients of these derivatives are constants or functions of the independent variables. A linear PDE can be written in the form an(x,y)un + an-1(x,y)un-1 + ... + a1(x,y)u' + a0(x,y) = f(x,y), where an, an-1, ..., a0 are constants or functions of x and y, u is the unknown function, and f(x,y) is a function of x and y.

## 2. What is the difference between a linear and a nonlinear PDE?

A linear PDE is an equation where the unknown function and its derivatives appear only in a linear form, while in a nonlinear PDE, the unknown function or its derivatives appear in a nonlinear form. In other words, the coefficients of the derivatives in a linear PDE are constants or functions of the independent variables, while in a nonlinear PDE, they can be functions of the unknown function or its derivatives.

## 3. How can I determine if a PDE is linear or nonlinear?

You can determine if a PDE is linear or nonlinear by looking at the coefficients of the derivatives. If the coefficients are constants or functions of the independent variables, then the PDE is linear. If the coefficients involve the unknown function or its derivatives, then the PDE is nonlinear.

## 4. Are there any advantages to working with linear PDEs?

Yes, there are several advantages to working with linear PDEs. They are easier to solve compared to nonlinear PDEs, and there are well-developed techniques for solving them, such as separation of variables and Fourier transforms. Additionally, the solutions to linear PDEs are usually more well-behaved and easier to interpret compared to solutions of nonlinear PDEs.

## 5. Can a nonlinear PDE be transformed into a linear PDE?

Yes, in some cases, a nonlinear PDE can be transformed into a linear PDE through a change of variables. This process is known as linearization and involves substituting a new variable for the unknown function, which transforms the nonlinear PDE into a linear one. However, this is not always possible, and it depends on the specific PDE and the chosen transformation.

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