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Sahithyan's S3
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Sahithyan's S3 — Differential Equations

Solving Partial Differential Equations

A PDE of order nn has at most nn arbitrary functions in its general solution. The general solution is written as linear combination of the nn functions.

An additional requirement imposed on the solution of a partial differential equation. Defines a condition fo the boundary of the domain in which the PDE is posed.

Suppose a PDE is defined on a region DD with boundary D\partial D. A boundary condition prescribes a relation of the form:

B(x,u(x),u(x))=0,xDB\big(x, u(x), \nabla u(x)\big) = 0, \qquad x \in \partial D

A constraint that specifies the value of the solution at the initial time on the entire spatial domain for a time-dependent PDE.

Used when variables can be integrated directly with respect to one variable.

This method assumes a product solution u=X(x)Y(y)u = X(x)Y(y). A PDE is converted into two ODEs.

Used in Laplace, heat, and wave equations.

Assumptions when formulating a mathematical model:

  • Thermal conductivity of the metal is uniform.
  • The plate is sufficiently thin (no heat flow in z-axis).
  • The temperature distribution is in a steady state.
    If not, eigenvalues and eigenfunctions are used to model transient states. And they are not covered in this module.

The model in Cartesian coordinates:

2Tx2+2Ty2=0\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0

The final solution in Cartesian coordinates:

T(x,y)=n=1Ansin(nπxa)sinh(nπya).T(x,y)=\sum_{n=1}^\infty A_n \sin\left(\frac{n\pi x}{a}\right) \sinh\left(\frac{n\pi y}{a}\right).

With BCθ(x,b)=100BC\theta(x,b)=100, Fourier sine series gives constants.

The model in polar coordinates:

r22Tr2+rTr+2Tθ2=0r^2 \frac{\partial^2 T}{\partial r^2} + r\frac{\partial T}{\partial r} + \frac{\partial^2 T}{\partial \theta^2} = 0

Here:

  • TT - temperature
  • x,yx,y - spatial coordinates
  • r,θr, \theta - polar coordinates

Temperatures of the edges of the plate are the boundary conditions.

Laws of heat flows:

  • The amount of heat in a body is proportional to its mass and temperature.
  • The rate of heat flow through a plane surface is proportional to the area and the rate of change of temperature with respect to the perpendicular distance.
2Tx2=1kTt\frac{\partial^2 T}{\partial x^2}=\frac1k\frac{\partial T}{\partial t}

Here k>0k \gt 0.

Boundary conditions:

  • T(0,t)=0T(0,t)=0
  • T(l,t)=0T(l,t)=0
  • T(x,0)=f(x)T(x,0)=f(x)

General solution:

T(x,t)=r=1Brsin(rπxl)exp(r2π2ktl2)T(x,t)=\sum_{r=1}^\infty B_r \sin\left(\frac{r\pi x}{l}\right) \exp \left(\frac{r^2\pi^2 k t}{l^2} \right) Br=2l0lf(x)sin(rπxl)dxB_r=\frac{2}{l}\int_0^l f(x)\sin\left(\frac{r\pi x}{l}\right)\,\text{d}x 2yt2=c22yx2\frac{\partial^2 y}{\partial t^2}=c^2\frac{\partial^2 y}{\partial x^2}

Boundary conditions:

  • y(0,t)=0y(0,t)=0
  • y(L,t)=0y(L,t)=0
  • y(x,0)=f(x)y(x,0)=f(x)
  • yt(x,0)=0y_t(x,0)=0
y(x,t)=n=1Ansin(nπxL)cos(nπctL)y(x,t)=\sum_{n=1}^\infty A_n \sin\left(\frac{n\pi x}{L}\right) \cos\left(\frac{n\pi c t}{L}\right) An=2L0Lf(x)sin(nπxL)dxA_n=\frac{2}{L}\int_0^L f(x)\sin\left(\frac{n\pi x}{L}\right)\,dx