ASSIGNED EXERCISES OF MATH544: PDE set 6 April, 2001

THE HEAT EQUATION Ronald B Guenther and John W Lee, Partial Differential Equations of Mathematical Physics and Integral Equations

1. Solve the following initial value problem

u_{t} = au_{xx}, (a > 0), t > 0, x ∈ [0, L],  (1)
u(0, t) = u(L, t) = 0, t ≥ 0,  (2)
u(x, 0) = f(x), x ∈ [0, L]  (3)

step 1. Formal Solution: Using the separation of variables we first let u(x, t) = T (t)X(x). Then find that

T′ = akT, X′′ = kX

where k is any constant. It must be a negative number k = −λ^{2}. we find that

T = e^{−aλ_{n}^{2}t}, X = sin(λ_{n}x) , n = 1, 2, ⋯

where λ_{n} = \frac{nπ}{L}. Hence

u(x, t) = ∑_{n=1}^{∞} α_{n} e^{−aλ_{n}^{2}t} sin(λ_{n}x)  (4)

Here α_{n} are constants , will be determined by the initial condition (assuming the t goes to zero exists which requires the uniform convergence of the sine series)

f(x) = ∑_{n=1}^{∞} α_{n} sin(λ_{n}x),  (5)

with

α_{n} = \frac{2}{L} ∫_{0}^{L} f(x) sin(λ_{n}x)dx, n = 1, 2, ⋯  (6)

step 2. Is the formal solution a solution of the initial and boundary value problem we started with?. For this the given function must satisfy certain conditions. Term-by term differentiation of the series representing u(x, t) in (4) can be justified at each point (x̃, t̃) with t̃> 0 and 0 ≤ x̃ ≤ L by showing that the series in (4) and the series representing u_{t}, u_{x} and u_{xx} that result from term-by-term differentiation , converge uniformly in a region D which contains the point (x̃, t̃). Such a region is

D = \{(x, t)|0 ≤ x ≤ L, t ≥ \frac{t̃}{2}\}.

We first assume that the function f(x) is bounded in [0, L]

|f(x)| ≤ M, for all x ∈ [0, L]

Then from (6) we have

|α_{n}| ≤ 2M

and series (4) satisfies

| ∑_{n=1}^{∞} α_{n} e^{−aλ_{n}^{2} t} sin(λ_{n}x) | ≤ 2M ∑_{n=1}^{∞} λ_{n}^{2} e^{−aλ_{n}^{2} t̃/2}.

The series at the right hand side converges uniformly in D. Similar reasoning applies also for the series representing u_{t}, u_{x}, and u_{xx}. Consequently (4) satisfies the equations (1) and (2).

To guarantee that (4) will also satisfy the initial condition (5), we need further assumptions on f(x). For the uniform convergence of the series in (5) we need ∑_{n=1}^{∞} |α_{n}| < ∞. If f(0) = f(L) = 0 then

α_{n} = −\frac{2}{λ_{n}^{2}L} ∫_{0}^{L} f′′ sin(λ_{n}x)dx.

Hence , if f′′ is continuous on 0 ≤ x ≤, then it is bounded say by M′′, and

|α_{n}| ≤ \frac{2M′′L^{2}}{π^{2} n^{2}}

This estimate implies that the series (4) for u(x, t) is uniformly convergent for t ≥ 0 and 0 ≤ x ≤ L. Since u(x, t) is the uniform limit of continuous functions , it is continuous. Hence we have the following:

Theorem. If f(x) is bounded and integrable, the series in (4) yields a solution to (1) and (2) for t > 0 and 0 ≤ x ≤ L. If in addition f has continuous second derivative and f(0) = f(L) = 0, then u(x, t) defined in (4) is uniformly convergent and continuous for t ≥ 0 and 0 ≤ x ≤ L and also satisfies the initial condition (3).

2. Solve the following initial and boundary value problem:

u_{t} − au_{xx} = F (x, t), t > 0, 0 < x < L,  (7)
u(0, t) = u(L, t) = 0, t ≥ 0,  (8)
u(x, 0) = f(x), 0 ≤ x ≤ L  (9)

Complete the method of solution given in class. The method of solution is similar to previous problem.

3. Solve the problem:

u_{t} − au_{xx} = F (x, t), 0 < x < L, t > 0,  (10)
u_{x}(0, t) = u_{x}(L, t) = 0, t ≥ 0,  (11)
u(x, 0) = f(x), 0 ≤ x ≤ L.  (12)

Explain also how you can prove the uniqueness of the solution of this problem.

4. Let u(x, t) be a solution to the problem

u_{t} − au_{xx} = 0, t > 0, 0 ≤ x ≤ L,  (13)
u(0, t) = u(L, t) = 0, t ≥ 0  (14)
u(x, 0) = f(x), 0 ≤ x ≤ L  (15)

let ũ be a solution of the above problem with f replaced by f̃. Denote the respective Fourier coefficients by α_{n} nad α̃_{n}. Show |α_{n}−α̃_{n}| ≤ 2||f′′−f̃′′||/λ_{n}^{2} and hence

||u − ũ|| ≤ 2||f′′ − f̃′′|| (∑_{n=1}^{∞} \frac{1}{λ_{n}^{2}} )

where ||h|| = max|h(x)| for 0 ≤ x ≤ L. Deduce continuous dependence and hence the well-posedness of the problem.

5. For T > 0 let

Q_{T} = {(x, t)|0 < x < L, 0 < t ≤ T }

(this is exactly the region D defined in the lecture), be a rectangle with its bottom and two vertical sides removed. Let

Q̄_{T} = {(x, t)|0 ≤ x ≤ L, 0 ≤ t ≤ T }.

Then B_{T} = Q̄_{T} − Q_{T} consists of the three missing sides of the original rectangle Q_{T}. Suppose that w(x, t) satisfies the differential inequality

w_{t} − a(x, t)w_{xx} − b(x, t)w_{x} − c(x, t)w < 0 (x, t) ∈ Q_{T}

where a(x, t) ≥ 0 and c(x, t) ≤ 0 in Q_{T}. Then w(x, t) can not achieve a positive local maximum in Q_{T}

6. Let a, b and c be as in the previous problem. Suppose that u(x, t) is continuous on Q̄_{T} and satisfies

u_{t} − a(x, t)u_{xx} − b(x, t)u_{x} − c(x, t)u ≤ 0, (x, t) ∈ Q_{T} ,  (16)
u(x, t) ≤ 0, (x, t) ∈ B_{T}  (17)

Then u(x, t) ≤ 0 on Q̄_{T}.

7. Let a, c and c be as in Prob.4. Suppose that u(x, t) is continuous on Q̄_{T} and satisfies

u_{t} − a(x, t)u_{xx} − b(x, t)u_{x} − c(x, t)u = 0

in Q_{T}. If u has a positive maximum , it is assumed on the boundary B_{T} and if u has a negative minimum , it is assumed on B_{T}.

8. Let a, b and c be as in Prob. 4. Suppose that F (x, t) is defined in Q_{T} and bounded by the constant N there. Suppose that u(x, t) is continuous on Q̄_{T}, M = max_{B_{T}} |u(x, t)|, and u(x, t) satisfies

u_{t} − a(x, t)u_{xx} − b(x, t)u_{x} − c(x, t)u = F (x, t)

in Q_{T}. The |u(x, t)| ≤ M + T N in Q_{T}.

9. Let a(x, t) > 0 and c(x, t) ≤ 0 for all 0 < x < L and t >). Show that the initial boundary value problem

u_{t} − a(x, t)u_{xx} − b(x, t)u_{x} − c(x, t)u = F (x, t), 0 < x, L, t > 0, (18)
u(0, t) = g(t), u(L, t) = h(t), t ≤ 0, (19)
u(x, 0) = f(x), 0 ≤ x ≤ L (20)

has at most one solution which is continuous on 0 ≤ x ≤ L and t ≥ 0 [Notice that if there is , in fact a continuous solution, then the boundary and initial data must be compatible. That is ,g(0) = f(0), h(0) = f(L)].