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Classifying PDEs
Problem 1. Consider first order equations and determine if they are linear homogeneous, linear inhomogeneous, or nonlinear; for nonlinear equations, indicate if they are also semilinear, or quasilinear:
ut + xux − u = 0, (1) ut + ux − u^2 = 0, (2) ut + uux + x = 0. (3)
Solution 1.
(1) Since the coefficients in front of ut, ux, and u are functions of x and t only, the equation is linear. There is also no term that depends only on x or t, so it is homogeneous. To prove that the equation is linear, notice that
L[au + bv] = (au + bv)t + x(au + bv)x − (au + bv) = a(ut + xux − u) + b(vt + xvx − v) = aL[u] + bL[v].
(2) There is a u^2 term, so the function is nonlinear. However, the coefficients of the highest order terms are functions of x and t, so the function is semilinear. To prove that the operator is non-linear, we show that the scaling property fails,
L[2x] = (2x)t + (2x)x − (2x)^2 = 2 − 4 x^2 6 = 2 − 2 x^2 = 2L[x].
(3) There is a uux term, so the function is nonlinear. However, the coefficients of the highest order terms are functions of x, t and u, so the function is quasilinear.
Solving Basic PDEs
PDEs with only one term
Problem 2. Find the general solutions to the following equations:
uxxy = 0, (1) uxyz = sin(x) + sin(y) sin(z). (2)
Solution 2.
(1) We integrate out each of the partial derivatives and introduce an integration constant in each step,
uxxy = 0 ⇒ uxx = f (x) ⇒ ux = f˜ (x) + g(y) f˜x = f ⇒ u = f˜˜ (x) + xg(y) + h(y) f˜˜xx = f
where f˜˜ is a twice differentiable function.
(2) We integrate out each of the partial derivatives and introduce an integration constant in each step,
uxyz = sin(x) + sin(y) sin(z) ⇒ uxy = z sin(x) − sin(y) cos(z) + f (x, y) ⇒ ux = yz sin(x) + cos(y) cos(z) + f˜ (x, y) + g(x, z) f˜y = f ⇒ u = −yz cos(x) + x cos(y) cos(z) + f˜˜ (x, y) + ˜g(x, z) + h(y, z) ˜gx = g, f˜˜yx = f ,
where f˜˜ is differentiable in each of its coordinates, and ˜g is differentiable in its first coordinate.
Semilinear First Order PDEs
Problem 3. Find the general solutions to the following equations
ut − 4 ux + u = 0, (1) − 2 ux + 4uy = ex+3y^ − 5 u. (2)
Solution 3.
(1) We have the system of equations dt 1 = dx − 4 = du −u
Characteristic Curve: We start by solving the equation involving the first and second term,
dt 1 =^
dx − 4 ⇒^
dx dt =^ −^4 ⇒^ C^ =^ x^ + 4t.
General Solution: We now solve the equation involving the first and third term,
dt 1 =^
du −u ⇒^
du dt =^ −u.
This is a separable ODE, which has solution
log |u| = −t + f (C) ⇒ u = ±ef^ (C)e−t
Since f (C) is an arbitrary function, we might can redefine ±ef^ (C)^ =: g(C). Since C = x + 4t, we have our general solution is u(t, x) = g(x + 4t)e−t.
(2) We have the system of equations
dx − 2
dy 4
du ex+3y^ − 5 u
Characteristic Curve: We start by solving the equation involving the first and second term,
dx − 2
dy 4
dy dx = − 2 ⇒ C = y + 2x.
General Solution: We now solve the equation involving the first and second term,
dx − 2
du ex+3y^ − 5 u
du dx
(ex+3y^ − 5 u) ⇒ du dx
u = −
ex+3y^.
Semilinear First Order PDEs in Higher Dimensions
Problem 5. Find the general solution to the equation
ux + 3uy − 2 uz = u.
Solution 5. We have the system of equations,
dx 1 =^
dy 3 =^
dz − 2 =^
du u.
Characteristic Curves Part 1: We start by solving the equation involving the first and second term,
dx 1
dy 3 =⇒ C = 3x − y.
Characteristic Curves Part 2: We now solve the equation involving the first and third term,
dx 1
dz − 2 =⇒ D = 2x + z.
General Solution: We now solve the equation involving the first and fourth term,
dx 1
du u =⇒ x = log |u| + f (C, D) =⇒ u(x, y, z) = f (C, D)ex^ = f (3x − y, 2 x + z)ex,
for some function f of two variables.
Problem 6. Find the general solution to the equation
ut + yux + xuy = 0.
Find the particular solution when u(0, x, y) = f (x, y).
Solution 6. We have the system of equations,
dt 1 =^
dx y =^
dy x =^
du
Characteristic Curves Part 1: We start by solving the equation involving the second and third term,
dx y
dy x =⇒ x^2 = y^2 + C =⇒ C = x^2 − y^2.
Characteristic Curves Part 2: We now solve the equation involving the first and second term using the fact y =
x^2 − C, dt 1 = dx y = √ dx x^2 − C
=⇒ t = log |
x^2 − C + x| + D = log |y + x| + D =⇒ D = (x^ +^ y) et^
General Solution: We now solve the equation involving the first and fourth term,
dt 1
du 0 =⇒ u(t, x, y) = g(C, D) = g
x^2 − y^2 , (x + y) et
Particular Solution: Plugging in our initial conditions, we have
u(0, x, y) = g(x^2 − y^2 , x + y) = f (x, y).
We set u = x^2 − y^2 and v = x + y. Our goal is to write x and y as some functions of u and v. We see that u = x^2 − y^2 = (x − y)(x + y) = (x − y)v =⇒ x − y = u v
Since x + y = v and x − y = uv , we can add and subtract our answers to conclude
x =
v + u v
y =
v − u v
so u(0, x, y) = g(u, v) = f
v + u v
v − u v
Therefore, our particular solution is given by
u(t, x, y) = g
x^2 − y^2 , (x + y) et
= f
( (^) x + y et^
x^2 − y^2 (x+y) et
( (^) x + y et^
x^2 − y^2 (x+y) et
= f
( (^) x + y et^
( (^) x + y et^ − x^ −^ y e−t
Remark: We were a bit sloppy with the constants and domains of our functions above. The constants C and D changed each line and writing x^2 − y^2 = C implicitly in terms of y depends on the value of C. We should check our general solution to ensure that it is a solution to our PDE by differentiating.
