I would like to find the solution $(U,V)\equiv (U(x,t),V(x,t))$ of the following system.
\begin{equation} \displaystyle\left\{\begin{array}{l} \frac{\partial U}{\partial t}+(a+b x)\frac{\partial U}{\partial x}-(c+k_1)U+k_1V=0, \\ \frac{\partial V}{\partial t}+(a+b x)\frac{\partial V}{\partial x}-(c+k_2)V+k_2 U=0, \\ \end{array}\right. t\in [s,T] \end{equation}
with the boundary conditions (where $W$ could be $U$ and $V$)
\begin{equation} \left\{\begin{array}{l} W(x,t)=0 \ \ \text{as} \ \ x\to-\infty,\\ W(x,t)\to e^x \ \ \text{as} \ \ x\to\infty,\\ W(x,T)=\max \{ e^x-100,0\} \\ \end{array}\right. \end{equation}
Assuming that above system has a unique solution, $(U,V)$, how can I find that solution. I would be satisfied to given a reference to a paper from which I can learn a method to solve it.
Can Mathematica solve this system?
Added after I got the answer from the above system: when I modified the codes of bbgodfrey to solve the followin system :
\begin{equation} \displaystyle\left\{\begin{array}{l} \frac{\partial U}{\partial t}+(\textbf{a}_1+b x)\frac{\partial U}{\partial x}-(c+k_1)U+k_1V=0, \\ \frac{\partial V}{\partial t}+(\textbf{a}_2+b x)\frac{\partial V}{\partial x}-(c+k_2)V+k_2 U=0, \\ \end{array}\right. t\in [s,T] \end{equation}
when I run the program, Mathematican seems runs out of memory. Is there any way to fix this issues or this is because Mathematica cannot solve the new system ?
Answer
The constant c can be eliminated from the equations by a standard transformation.
eq1 = (Unevaluated[D[u[x, t], t] + (a + b x) D[u[x, t], x] - (c + k1) u[x, t] + k1 v[x, t]]
/. {u[x, t] -> uu[x, t] E^(c t), v[x, t] -> vv[x, t] E^(c t)})/E^(c t) // Simplify
(* -(k1*uu[x, t]) + k1*vv[x, t] + Derivative[0, 1][uu][x, t] +
a*Derivative[1, 0][uu][x, t] + b*x*Derivative[1, 0][uu][x, t] *)
and similarly for the second expression, designated eq2. Taking the difference between these two expressions yields
Collect[(eq1 - eq2) // Simplify, {a + b x, k1 + k2}, Simplify]
(* (k1 + k2)*(-uu[x, t] + vv[x, t]) + Derivative[0, 1][uu][x, t] - Derivative[0, 1][vv][x, t]
+ (a + b*x)*(Derivative[1, 0][uu][x, t] - Derivative[1, 0][vv][x, t]) *)
which can be rewritten as
-((k1 + k2)*zz[x, t]) + Derivative[0, 1][zz][x, t] + (a + b*x)*Derivative[1, 0][zz][x, t]
where zz == uu - vv. And, this equation can be solved
First@DSolve[% == 0, zz[x, t], {x, t}]
(* {zz[x, t] -> (a + b*x)^(k1/b + k2/b) C[1][(b t - Log[a + b x])/b]} *)
Undoing the original transformation then yields
First@Solve[% /. Rule -> Equal /. zz[x, t] -> z[x, t] E^(-c t), z[x, t]]
{* {z[x, t] -> E^(c t) (a + b*x)^(k1/b + k2/b) C[1][(b t - Log[a + b x])/b]} *)
Note that C[1] is an arbitrary function of (b t - Log[a + b x])/b, which can be chosen to satisfy the boundary conditions in x.
Alternative, Simpler Approach
DSolve cannot integrate the two equations as written in the question. However, the equations can be separated, after which DSolve can integrate them without difficulty.
r0 = First@Solve[{z[x, t] == u[x, t] - v[x, t],
y[x, t] == u[x, t] + v[x, t] k1/k2}, {u[x, t], v[x, t]}];
eq1 = (Unevaluated[D[u[x, t], t] + (a + b x) D[u[x, t], x] - (c + k1) u[x, t] +
k1 v[x, t]] /. %) // Simplify;
eq2 = (Unevaluated[D[v[x, t], t] + (a + b x) D[v[x, t], x] - (c + k2) v[x, t] +
k2 u[x, t]] /. %%) // Simplify;
Simplify[eq1 - eq2];
r1 = Simplify[#] & /@ DSolve[% == 0, z[x, t], {x, t}][[1, 1]]
(* z[x, t] -> (a + b x)^((c + k1 + k2)/b) C[1][t - Log[a + b x]/b] *)
Simplify[eq1 + eq2 k1/k2];
r2 = Simplify[#] & /@ DSolve[% == 0, y[x, t], {x, t}][[1, 1]] /. C[1] -> C[2]
(* y[x, t] -> (a + b x)^(c/b) C[2][t - Log[a + b x]/b] *)
r0 /. {r1, r2}
(* {u[x, t] -> -((-k1 (a + b x)^((c + k1 + k2)/b)
C[1][t - Log[a + b x]/b] - k2 (a + b x)^(c/b) C[2][t - Log[a + b x]/b])/(k1 + k2)),
v[x, t] -> -((k2 (a + b x)^((c + k1 + k2)/b)
C[1][t - Log[a + b x]/b] - k2 (a + b x)^(c/b) C[2][t - Log[a + b x]/b])/(k1 + k2))} *)
Note that the solution contains two arbitrary functions C[1] and C[2] of (b t - Log[a + b x])/b, as it must, because the original equations form a second order system of advective equations.
Note also that the earlier, incomplete solution would eventually have reached the same point.
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