2.1.18 (Haberman 12.2.5 (d)) ωt+3tωx=ω(x,t) and ω(x,0)=f(x)

problem number 18

Added Nov 25, 2018.

Problem 12.2.5 (d) from Richard Haberman applied partial differential equations book, 5th edition

Solve for u(x,t) ωt+3tωx=ω(x,t)

with ω(x,0)=f(x).

See my HW 12, Math 322, UW Madison.

Mathematica

ClearAll["Global`*"]; 
pde = D[w[x, t], t] + 3*t*D[w[x, t], x] == w[x, t]; 
ic  = w[x, 0] == f[x]; 
sol =  AbsoluteTiming[TimeConstrained[Simplify[DSolve[{pde, ic}, w[x, t], {x, t}]], 60*10]];
 

{w(x,t)et2f(x3t22)}{w(x,t)et2f(x3t22)}

Maple

restart; 
pde := diff(w(x,t),t)+3*t*diff(w(x,t),x)=w(x,t); 
ic:=w(x,0)=f(x); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],w(x,t))),output='realtime'));
 

w(x,t)=f(3t22+x)et

Hand solution

Solve (1)wt+3twx=w(x,t)

With initial conditions w(x,0)=f(x)

Solution

Let ww(x(t),t) then (2)dwdt=wxdxdt+wt

Comparing (1,2) shows that

(3)dwdt=w(4)dxdt=3t

Solving (3) gives

w=Cet

From initial conditions at t=0, the above becomes f(x(0))=C. Hence the above becomes

(5)w(x,t)=f(x(0))et

From (4)

x=32t2+x(0)x(0)=x32t2

Substituting the above in (5) gives

w(x(t),t)=f(x32t2)et

Alternative solution

Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)(1)dtds=1(2)dxds=3t(3)dwds=w

With initial conditions at s=0t(0)=t1,x(0)=t2,w(0)=t3 And w(x,0)=f(x) becomes (4)t3=f(t2),t1=0 Equation (1) givest=s+t1(5)=s

Equation (2) gives, after replacing t by s from (5)dxds=3s(6)x=32s2+t2

Solving for t2 gives(7)t2=x32s2 Equation (3) giveslnw=s+t3w=t3es=f(t2)es

Using (7,5) in the above gives the solutionw(x,t)=f(x32t2)et

____________________________________________________________________________________