# Solving forced PDE with method of Characteristics

• A
• nickthequick
In summary, to solve the given PDE, we can use the method of characteristics to find the characteristic curves and then use those curves to solve for u.

#### nickthequick

I'm trying to solve the following PDE:
$$u_t+yu_x=-y-\mathbb{H}(y_x)$$
where y satisfies the inviscid Burgers equation
$$y_t+yy_x=0$$
and the Hilbert Transform is defined as
$$\mathbb{H}(f) = PV \int_{-\infty}^{\infty} \frac{f(x')}{x-x'} \ dx',$$
where PV means principal value.

The solutions for y are determined by the initial conditions, so that if y(t=0,x)=F(x) then y=F(x-yt).

Now, I know how to solve the homogeneous part of the equation, as well as the particular solution for the term -y on the right hand side of the governing equation. However, I cannot find a solution to

$$u_t+yu_x=-\mathbb{H}(y_x).$$

Any tips on how to find solutions for a system like this are appreciated.

As a scientist, I would approach this problem by first understanding the physical meaning of the equations and the variables involved. The equations represent a system of two partial differential equations, where u is the dependent variable and y is the independent variable. The first equation, known as the inviscid Burgers equation, describes the evolution of a fluid flow with time and space. The second equation, which includes the Hilbert Transform, represents the effect of viscosity on the fluid flow.

To solve this system, we need to find a solution for both equations simultaneously. This can be done by using a method known as the method of characteristics. This method involves finding a set of curves in the x-t plane along which the solution to the system is constant. These curves are known as the characteristic curves.

To find the characteristic curves, we can use the initial condition given for y, which is y(t=0,x)=F(x). This means that at t=0, the value of y is known at all points in the x-axis. We can then use this information to find the characteristic curves by solving the following differential equation:

$$\frac{dy}{dt}=y$$

This equation can be solved using separation of variables, and the solution is given by:

$$y=Ce^t$$

where C is a constant. This means that the characteristic curves are given by the equation x-yt=C, where C is the constant determined by the initial condition.

Now, to find a solution for u, we can use the method of characteristics again. We can write the governing equation as:

$$\frac{du}{dt}=-\mathbb{H}(y_x)$$

Using the definition of the Hilbert Transform, we can rewrite this equation as:

$$\frac{du}{dt}=-PV \int_{-\infty}^{\infty} \frac{y_x(x')}{x-x'} \ dx'$$

We can then use the characteristic curves to transform the integral into a single variable integral. The solution for u can then be found by solving the following differential equation:

$$\frac{du}{dt}=-PV \int_{C}^{x-yt} \frac{y_x(x')}{x-x'} \ dx'$$

This equation can be solved using standard methods, and the solution can be expressed in terms of the characteristic curves and the initial conditions.

In conclusion, to solve the given system of equations, we