6 An Introduction to Partial Differential Equations with MATLAB©R
c) Isu=x+ya solution?
d) Find all solutions of the formu=ax+by,whereaandbare
constants.- Consider the simple first-order PDEux=0,whereu=u(x, y).
a) Find all solutions. (Compare this problem with that of finding all
solutions of theODEdydx= 0.) Describe them (compare to Exercise
3).
b) Describe the set of solutions which satisfy the additional require-
ment thatu(0,0) = 0. How many are there?
c) Do the same, but for the requirementu(0,y)=y^2 −cosy.
d) Do the same, but foru(x,0) =x^3.1.2 PDEsWeCanAlreadySolve
Let’s go back and look at Exercise 6 of the previous section. However, first,
remember how we would solve the ODE
dy
dx=0.
We integrated both sides to gety= constant (after having proved in calculus
thaty= constant is the only function whose derivative is identically zero).
We can do the same with PDEs—except that we must remember that the
derivatives arepartial derivatives, so any antiderivatives we take will be, in
a sense, “partial antiderivatives” or “partial integrals.” That is, we “anti-
differentiate” with respect to one variable while treating the other variables
as constants.
So for the PDEux= 0, any function which is independent ofxwill be a
solution. (Further, similarly to above, these will be theonlysolutions.) To
be more precise, in order to find all functionsu=u(x, y) which solve
ux=0, (1.1)we get
u=∫
0 dx=f(y) (1.2)wherefis any∗arbitrary function ofy(and where
∫
...dxis, as we’ve men-
tioned, any antiderivative with respect toxwhile treatingyas a constant).
∗Again, from our definition of solution,f′must be continuous.