Question

solve d^2u/dx^2 +d^2u/dy^2 = 0 which satisfies u (0,y) = u (l,y) = u (x, 0) =0 & u(x,a)= sin nπx/L

Answer 1

Answer on Question #82430 – Math – Differential Equations

Question

solve $d^2 u/dx^2 + d^2 u/dy^2 = 0$ which satisfies $u(0,y) = u(l,y) = u(x,0) = 0$ & $u(x,a) = \sin n\pi x / L$

Solution

DE

$$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0;$$

Boundary conditions

$$u(0,y) = u(L,y) = u(x,0) = 0; \quad u(x,L) = \sin \frac{n\pi x}{L}.$$

This is Laplace's equation, which one can solve by the separation of variables.

Denote

$$u(x,y) = X(x)Y(y).$$

Then

$$\frac{\partial^2 (XY)}{\partial x^2} + \frac{\partial^2 (XY)}{\partial y^2} = 0;$$ $$Y \frac{\partial^2 X}{\partial x^2} + X \frac{\partial^2 Y}{\partial y^2} = 0;$$

Divide by $XY$ to get

$$\frac{1}{X} \frac{\partial^2 X}{\partial x^2} + \frac{1}{Y} \frac{\partial^2 Y}{\partial y^2} = 0;$$

We can see that these two parts are functions of only $x$ and only $y$. Because of $(x,y)$ are independent variables, the first term must be a constant, and the second term must be a constant. We call them $-k^2$ and $+k^2$ (their sum must be null).

The new equations are

$$\frac{1}{X} \frac{\partial^2 X}{\partial x^2} = -k^2; \qquad \text{and} \qquad \frac{1}{Y} \frac{\partial^2 Y}{\partial y^2} = k^2.$$

a) Solution for $X$:

$$\frac{\partial^2 X}{\partial x^2} = -k^2 X;$$ $$X'' + k^2 X = 0.$$

Solution is

$$X(x) = A \sin kx + B \cos kx.$$

Boundary conditions: $X(0) = 0$; $X(L) = 0$.

$$\left\{ \begin{array}{l} A \sin k \cdot 0 + B \cos k \cdot 0 = 0; \\ A \sin k L + B \cos k l = 0; \end{array} \right. \Rightarrow \left\{ \begin{array}{c} B = 0; \\ k L = n \pi, \end{array} \right. \quad n \in \mathbb{Z}.$$ $$u(x, L) = \sin \frac{n \pi x}{L} \Rightarrow A = 1.$$ $$X_n(x) = \sin \frac{n \pi}{L} x.$$

b) Solution for $Y$:

$$\frac{\partial^2 Y}{\partial y^2} = k^2 Y;$$ $$Y'' - k^2 Y = 0.$$

Solution is

$$Y(y) = C e^{-k y} + D e^{k y}.$$

Boundary conditions: $Y(0) = 0$; $Y(L) = 1$.

$$0 = C + D \Rightarrow C = -D;$$ $$Y(y) = C (e^{-k y} - e^{k y}) = C_1 \sinh k y; \quad \text{where} \quad C_1 = 2C.$$ $$1 = C_1 \sinh k L; \quad \Rightarrow \quad C_{1n} = \frac{1}{\sinh k L} = \frac{1}{\sinh n \pi}.$$ $$Y_n(y) = \frac{1}{\sinh n \pi} \sinh \frac{n \pi}{L} y.$$

So we can write the general solution. It must be combination of all solutions for $X(x)$ and $Y(y)$:

$$u(x, y) = \sum_{n=1}^{\infty} \frac{1}{\sinh n \pi} \sin \left( \frac{n \pi}{L} x \right) \sinh \left( \frac{n \pi}{L} y \right).$$

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