# Solving a simple PDE

A linear equation has the form:

$$a(x,y)u_x + b(x,y)u_y = c_0(x,y)u+c_1(x,y)$$

where $$a,b,c_0,c_1$$ are given functions of $$(x,y)$$.

I now want to do a simple example from my book. I think what we want to do here is just solve it directly. Example:

$$u_x=c_0u+c_1\tag{1}$$

In This example we set $$a=1, b=0, c_0$$ is constant, and $$c_1=c_1(x,y)$$.

Comment 1: If we look at the general form of a linear equation, we can see that $$a=1, b=0$$. But how do we know that $$c_0$$ is constant? Did we just choose it like that? If so, why did we do that? And why did we do it at this stage?

Since (1) contains no derivative with respect to the $$y$$ variable, we can regard this variable as a parameter.

Comment 2: I mean sure, why not - but why should we? What's the point?

Recall from the theory of ODEs that in order t obtain a unique solution we must supply an additional condition. We saw in Chapter 1 that there are many ways to supply additional conditions to a PDE. The natural condition for a first-order PDE is a curve laying on the surface.

Comment: I don't really get exactly what they meant here with natural condition but I do get the "we need more conditions to get a specific solution." What surface are they exactly talking about?

We shall refer to such a condition as a initial condition, and the problem will be called a initial value problem. For example, we can supplement (1) with the initial condition

$$u(0,y)=y$$

Comment: We did just this right? Probably has to do about $$y$$ not ocuring in the problem.

Since we are actually dealing with an ODE, the solution is immediate:

$$u(x,y)=e^{e_0 x}\big[\int_0^x e^{-c_0 \chi} c_1(\chi, <)d\chi + y\big]$$

Comment: If we have a PDE, the above form of the solution makes sense to me. We probably used Euler to get it. What I can't yet 100% see is the PDE. Can someone write down the PDE fully?

## 1 Answer

1. We don't know that $$c_0$$ is a constant, the author chose that arbitrarily, for the sake of exposition. Equations with constant coefficients are easier.

2. Solving ODEs is million times easier than PDEs. Looking at $$u_x(x,y)=c_0u(x,y)+c_1(x,y),$$ the derivative $$u_y$$ does not appear explicitly and we can see a family of ODEs $$u'(x)=c_0u(x)+c_1(x).$$

3. The solution of the PDE is a bivariate function $$u(x,y)$$, and the locus of $$z=u(x,y)$$ is a surface in 3D space. Referring to the previous topic, we understand that the ODE requires an initial condition such as $$u(x_0)=u_0$$, and this condition is also a function of $$y$$, $$u(x_0,y)=u_0(y)$$. This actually describes a curve that belongs to the surface. For more general PDEs, the "initial" curve condition remains.

• 1. ok, good. 2. Okay, so that's a bit more radical than what I tried to think about. Thought we can actually kind of reduce it to a single partial derivative, couldn't figure that out. But I think that's clear now. - thanks – xotix Oct 8 '18 at 12:56