Solve the heat equation using a transform methodk∂2U∂x2=∂U∂tsubject toU(0,t)=1,t&gt;0U(x,0)=e−x,x&gt;0

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Solve the heat equation using a transform methodk∂2U∂x2=∂U∂tsubject toU(0,t)=1,t&amp;gt;0U(x,0)=e−x,x&amp;gt;0

diocedss33 · Accepted Answer

Step 1If we were to take the Laplace Transform of this expression, assumingL(U(x,t))(x,s)=V(x,s)we havek∂2V∂x2=sV−U(x,0)=sV−e−xFrom here we can solve for V in terms of constants that may depend on s. The homogeneous solution isVh(x,s)=C1(s)exsk+C2(s)e−xskThe particular solution must take the form of&amp;nbsp;A(s)e−x, sokA(s)e−x=sA(s)e−x−e−x⇒A(s)=1s−kThus, the general solution isV(x,s)=Vh(x,s)=C1(s)exsk+C2(s)e−xsk+e−xs−kWe have to employ an assumption that for any fixed value of t,&amp;nbsp;U(x,t)→0&amp;nbsp;as x gets large, meaning that the total heat density falls to zero. This translates to the fact that as x gets large, V(x,s) should remain bounded (vague issues regarding convergence here), and so any growth in x should be ignored. As such, we can assume that&amp;nbsp;C1(s)=0Next, using the boundary condition, we transform U to getV(0,s)=1sThus,C2(s)+1s−k=1s⇒C2(s)=−ks(s−k)and soV(x,s)=(1s−1s−k)e−xsk+e−xs−kFrom here, we need to leverage the well-known and famous Laplace Transform ofL12πt32e-14t=e-s(this can be shown with a variety of clever calculus tricks) The modification to get our version with&amp;nbsp;xsk&amp;nbsp;is to include a parameter in the exponent and we can show (through u-substitutions) thatLx2kπt32e-x24kt=e-xskNow we can throw everything together using the convolution theorem. Thus,U(x,t)=∫0t(1−ek(t−u))x2kπu32e−x24ku,du+ekt−xor a little more cleanly:U(x,t)=x2kπ∫0t1u3/2(1-ek(t-u))e-x24ku du+ekt-xFrom here, the integration is nasty, but doable, albeit in terms of non-elementary functions.

Makenzie Hart · Accepted Answer

Step 1Applying the Laplace transform we obtainsU(x,s)=kU×(x,s)+e−x, U(0,s)=1sand after solving for x we obtainU(x,s)=2C0(s)sinh(sxk)+ke−sxk−se−xs(k−s)here C0(s) appears due to the fact of incomplete boundary conditions. The problem needs one more condition.

Solve the heat equation using a transform method

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