A problem on maximum principle

 

Problem. Let \(D = \{ x \in \mathbb{R}, y \in \mathbb{R}: x^2 + y^2<1 \}\) be the open unit disk, and \(u\) be a twice continuously differentiable function on \(\bar{D} = \{ x \in \mathbb{R}, y \in \mathbb{R}: x^2 + y^2 ≤ 1 \}\).

(a) Suppose \(u\) satisfies the equation

\[u_{xx} + u_{yy} = f \text{ in } D.\]

State and prove the Maximum Principle Theorem for this equation.

Theorem. If \(f ≥ 0\) on \(D\) then \(u\) attains a maximum on \(∂D\).

Proof. \(u \in C^2\) and \(\overline{D}\) is compact\(⇒ \max_{\overline{D}} u\) exists.

If \(f > 0\) suppose \(\max u\) is attained at \(p∈D\), then \(u_x(p)= u_y(p)= 0, u_{x x}(p)≤ 0, u_{y y}(p)≤ 0⇒f(p)≤0\), contradiction.

If \(f ≥ 0\), let \(v = u + ε (x^2 + y^2), ε > 0\). Since \(ℒv = f + 4 ε > 0\), the maximum of \(v\) must occur on \(∂D\).

\[\max_{\overline{D}} u<\max_{\overline{D}} v = \max_{∂D} v = \max_{∂D} u + ε\]

\(\max_{\overline{D}} u≤\lim_{ε→0} \max_{\overline{D}} v=\max_{∂D} u\).

(b) Suppose that \(u\) satisfies the equation

\[u_{xx} + u_{yy} + r (x, y) u_x + s (x, y) u_y = f \text{ in } D\]

where \(r\) and \(s\) are two continuous functions on \(\bar{D}\).

(i) Show that if \(f > 0\) in \(D\), then \(u\) attains its maximum on \(∂D\).

(ii) Show that if \(f ≥ 0\) in \(D\), then \(u\) attains its maximum on \(∂D\). [You may wish to consider \(v = u + ε e^{ax}\) for suitable \(a\).]

Proof. (i) Assume \(u\) attains maximum at \(p∈D\), then \(u_x(p)= u_y(p)= 0, u_{x x} (p) ≤ 0, u_{y y}(p) ≤ 0 ⇒ f(p)≤ 0\) but \(f > 0\) in $D$, contradiction. So $u$ attains its maximum on $∂D$.

(ii) Let \(v = u + εe^{ax}\) for suitable \(a\)

\(ℒv =ℒu + ε (a + r) ae^{ax}\)

choose \(a > \max_{\overline{D}}{|r(x, y)|}\) then \((a + r) a > 0\)

\[\max_{\overline{D}} u<\max_{\overline{D}} v = \max_{∂D} v≤\max_{∂D} u + ε \max_{∂D} (a + r) ae^{ax}\]

\(\max_{\overline{D}} u≤\lim_{ε→0} \max_{\overline{D}} v≤\max_{∂D} u\).

(c) Suppose that \(u\) is a solution of the boundary value problem

\begin{align*} u_{xx} + u_{yy} + r (x, y) u_x + s (x, y) u_y & = f \text{ in } D\\ u & = g \text{ on } ∂D \end{align*}

where \(r, s, f\) and \(g\) are given continuous functions. Deduce from parts (a) and (b) that \(u\) is unique. [You may assume without proof that \(u\) exists.]

Proof. Assume \(u_1, u_2\) are 2 solutions of \(u\), let \(w = u_1 - u_2\).

\(ℒw = 0\) in \(D\) and \(w = 0\) on \(∂D\). By the maximum principle:

\(ℒw ≥ 0 ⇒ \max_{\overline{D}} w = \max_{∂D} w = 0\)

\(ℒw ≤ 0 ⇒ \min_{\overline{D}} w = \min_{∂D} w = 0\)

So \(w≡0\). So \(u\) is unique.

(d) Find all non-negative solutions \(u \in C^2 (\bar{D})\) of

\begin{align*} u_{xx} + u_{yy} & = u^4 \text{ in } D\\ u & = 0 \text{ on } ∂D \end{align*}
Solution. \(u^4≥0\), by Maximum Principle, \(\max_D u=\max_{∂D} u=0 ⇒\)only non-negative solution is \(u = 0\).