9.7 Steady-State Temperature and Laplace's Equation

Introduction to Higher Dimensions

Recall the 1-D Heat equation, where heat flows along the $x$ direction:

$$ u_t = k u_{xx} $$

In 2-Dimensions, heat can flow in both the $x$ and $y$ directions. The equation becomes:

$$ u_t = k(u_{xx} + u_{yy}) $$

The right side represents the Laplacian of $u$. For higher dimensions, this is denoted as:

$$ \nabla^2 u = \nabla \cdot \nabla u $$

Comparison of equations using the Laplacian:

Heat Equation: $ u_t = k \nabla^2 u $
Wave Equation: $ u_{tt} = a^2 \nabla^2 u $

Physical Interpretation of $\nabla^2 u$

$\nabla^2 u$ represents how the value of $u$ at a specific point compares to the average value of $u$ at all nearby points.

In the 1-D Heat equation, $\nabla^2 u = u_{xx}$.
Visually (think of a curve): If the curve is concave down (like an upside-down bowl), the value of $u$ at the peak is higher than the average of its neighbors.
If considering heat or a wave (string), $\nabla^2 u$ determines how heat flows (into or away from a point) or how the string tries to resist displacement.
Nature (heat/strings) generally "doesn't want" unevenness.

Laplace's Equation

We are looking for the Steady-State. This happens when the temperature stops changing with time, meaning $u_t = 0$.

Laplace's Equation (Steady-state of 2D Heat Equation):

$$ \nabla^2 u = 0 $$

More specifically in 2D:

$$ u_{xx} + u_{yy} = 0 $$

Boundary Value Problem Setup

Consider a rectangular region defined by $0 < x < a$ and $0 < y < b$.
We have four Boundary Conditions (BCs), one for each side of the rectangle:

Bottom: $u(x,0) = g_1(x)$
Left: $u(0,y) = f_1(y)$
Right: $u(a,y) = f_2(y)$
Top: $u(x,b) = g_2(x)$

Since $u_{xx} + u_{yy} = 0$ is a linear PDE, the principle of superposition applies.

Strategy

If a boundary condition is 0, it is homogeneous.

Solve a case with 3 zero BCs and one non-zero BC.
Change which BC is non-zero (do this for all non-zero sides).
Combine the results (sum them up).

Note: If all BCs are 0, it leads to the trivial solution $u=0$, which we generally ignore.

Example Problem

Let's solve $u_{xx} + u_{yy} = 0$ with the following boundary conditions:

Left: $u(0,y) = 0$
Bottom: $u(x,0) = 0$
Top: $u(x,b) = 0$
Right: $u(a,y) = f(y)$ (The non-homogeneous BC)

Step 1: Separation of Variables

This method is useful for linear homogeneous PDEs. We assume a solution of the form:

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

Therefore:

$$ u_{xx} = X''Y \quad \text{and} \quad u_{yy} = XY'' $$

Applying the homogeneous (zero) Boundary Conditions:

$u(0,y) = X(0)Y(y) = 0 \Rightarrow X(0) = 0$
$u(x,0) = X(x)Y(0) = 0 \Rightarrow Y(0) = 0$
$u(x,b) = X(x)Y(b) = 0 \Rightarrow Y(b) = 0$

Note: We leave the non-homogeneous BC ($u(a,y)=f(y)$) for last.

Substituting into the PDE $u_{xx} + u_{yy} = 0$:

$$ X''Y + XY'' = 0 $$

Separating the variables:

$$ \frac{X''}{X} = -\frac{Y''}{Y} = \text{constant } (C) $$

Step 2: Solving the Ordinary Differential Equations (ODEs)

We separate this into two ODEs:

$X'' - CX = 0$
$Y'' + CY = 0$

Solve the $Y$ equation first (because it has complete homogeneous BCs: $Y(0)=0$ and $Y(b)=0$).

Assuming $C > 0$ (let $C = \lambda^2$):

$$ Y'' + CY = 0 $$

The characteristic equation gives $r = \pm \sqrt{-C}$ (imaginary), leading to:

$$ Y(y) = c_1 \cos(\sqrt{C}y) + c_2 \sin(\sqrt{C}y) $$

Applying BCs:

$Y(0) = 0 \Rightarrow c_1 = 0$
$Y(b) = 0 \Rightarrow c_2 \sin(\sqrt{C}b) = 0$

Since we want a non-trivial solution ($c_2 \neq 0$), we must have $\sin(\sqrt{C}b) = 0$.

$$ \sqrt{C}b = n\pi \Rightarrow C_n = \lambda_n = \frac{n^2\pi^2}{b^2} $$

These are the eigenvalues. The associated eigenfunctions are:

$$ Y_n = \sin\left(\frac{n\pi y}{b}\right) $$

Step 3: Solving the $X$ Equation

Using the eigenvalues found above ($C = \frac{n^2\pi^2}{b^2}$):

$$ X'' - \frac{n^2\pi^2}{b^2}X = 0 $$

The characteristic roots are $r = \pm \frac{n\pi}{b}$.

The general solution can be written with exponentials, but for these boundary conditions, it is more convenient to use hyperbolic sine and cosine:

$$ X(x) = d_1 \cosh\left(\frac{n\pi x}{b}\right) + d_2 \sinh\left(\frac{n\pi x}{b}\right) $$

Recall that $\sinh(x) = \frac{1}{2}(e^x - e^{-x})$.

Applying the 3rd homogeneous BC: $X(0) = 0$:

$$ X(0) = d_1 \cosh(0) + d_2 \sinh(0) = d_1(1) + 0 \Rightarrow d_1 = 0 $$

So the eigenfunction for $X$ is:

$$ X_n = \sinh\left(\frac{n\pi x}{b}\right) $$

Step 4: Final General Solution

For each $n = 1, 2, 3, \dots$, we have a solution $u_n = X_n Y_n$:

$$ u_n = \sinh\left(\frac{n\pi x}{b}\right) \sin\left(\frac{n\pi y}{b}\right) $$

The general solution is a linear combination of all $u_n$:

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

Step 5: Applying the Non-Homogeneous BC

Finally, use the right edge condition $u(a,y) = f(y)$:

$$ f(y) = \sum_{n=1}^{\infty} \left[ c_n \sinh\left(\frac{n\pi a}{b}\right) \right] \sin\left(\frac{n\pi y}{b}\right) $$

This is a Fourier Sine Series where the coefficient (the term in brackets) depends on $f(y)$.

Let the bracketed term be the Fourier coefficient. Using the orthogonality formula for Fourier Sine Series (where length is $b$):

$$ c_n \sinh\left(\frac{n\pi a}{b}\right) = \frac{2}{b} \int_{0}^{b} f(y) \sin\left(\frac{n\pi y}{b}\right) dy $$

Solving for $c_n$:

$$ c_n = \frac{2}{b \sinh\left(\frac{n\pi a}{b}\right)} \int_{0}^{b} f(y) \sin\left(\frac{n\pi y}{b}\right) dy $$