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Mathematics LibreTexts

5.6: Inhomogeneous Equations

( \newcommand{\kernel}{\mathrm{null}\,}\)

Consider a rod of length 2m, laterally insulated (heat only flows inside the rod). Initially the temperature u is

1ksin(πx2)+500 K.

The left and right ends are both attached to a thermostat, and the temperature at the left side is fixed at a temperature of 500 K and the right end at 100 K. There is also a heater attached to the rod that adds a constant heat of sin(πx2) to the rod. The differential equation describing this is inhomogeneous

tu=k2x2u+sin(πx2),u(0,t)=500,u(2,t)=100,u(x,0)=1ksin(πx2)+500.

ince the inhomogeneity is time-independent we write

u(x,t)=v(x,t)+h(x),

where h will be determined so as to make v satisfy a homogeneous equation. Substituting this form, we find

tv=k2x2v+kh+sin(πx2).

To make the equation for v homogeneous we require h(x)=1ksin(πx2),

which has the solution

h(x)=C1x+C2+4kπ2sin(πx2).

At the same time we let h carry the boundary conditions, h(0)=500, h(2)=100, and thus h(x)=200x+500+4kπ2sin(πx2). The function v satisfies

tv=k2x2v,v(0,t)=v(π,t)=0,v(x,0)=u(x,0)h(x)=200x.

This is a problem of a type that we have seen before. By separation of variables we find

v(x,t)=n=1bnexp(n2π24kt)sinnπ2x.

The initial condition gives

n=1bnsinnx=200x.

from which we find

bn=(1)n+1800nπ.

And thus

u(x,t)=200x+500+4π2ksin(πx2)+800πn=1(1)nn+1sin(πnx2)ek(nπ/2)2t.

Note: as t, u(x,t)400πx+500+sinπ2xk. As can be seen in Fig. 5.6.1 this approach is quite rapid – we have chosen k=1/500 in that figure, and summed over the first 60 solutions.

Inhomgen2.png

Figure 5.6.1: Time dependence of the solution to the inhomogeneous Equation ???.


This page titled 5.6: Inhomogeneous Equations is shared under a CC BY-NC-SA 2.0 license and was authored, remixed, and/or curated by Niels Walet via source content that was edited to the style and standards of the LibreTexts platform.

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