1. **Problem Statement:** Solve the PDE $$\frac{\partial u}{\partial t} = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \sin t \cos x \cos y$$ with boundary conditions: $$u_x(0,y,t) = A, \quad u_x(\pi,y,t) = 0, \quad u(x,\infty,t) = u(x,\pi,t) = 0$$ and initial conditions: $$u(x,y,0) = \alpha(x,y), \quad u_t(x,y,0) = \beta(x,y)$$ 2. **Formula and Important Rules:** This is a first-order PDE with mixed spatial derivatives and a source term depending on $t$, $x$, and $y$. We use the method of characteristics or separation of variables combined with Duhamel's principle for the nonhomogeneous term. 3. **Step-by-step Solution Approach:** - Rewrite PDE as: $$\frac{\partial u}{\partial t} - \frac{\partial u}{\partial x} - \frac{\partial u}{\partial y} = \sin t \cos x \cos y$$ - Homogeneous part: $$\frac{\partial u}{\partial t} - \frac{\partial u}{\partial x} - \frac{\partial u}{\partial y} = 0$$ - Characteristic equations: $$\frac{dt}{1} = \frac{dx}{-1} = \frac{dy}{-1}$$ which imply along characteristics: $$x + t = \text{const}, \quad y + t = \text{const}$$ - General solution to homogeneous PDE is a function of $x+t$ and $y+t$: $$u_h = F(x+t,y+t)$$ - For the nonhomogeneous term, use variation of parameters or Duhamel's principle: 4. **Particular solution guess:** Try a solution of form: $$u_p = g(t) \cos x \cos y$$ Substitute into PDE: $$g'(t) \cos x \cos y - 0 - 0 = \sin t \cos x \cos y$$ which simplifies to: $$g'(t) = \sin t$$ Integrate: $$g(t) = -\cos t + C$$ 5. **Complete solution:** $$u = F(x+t,y+t) - \cos t \cos x \cos y + C \cos x \cos y$$ 6. **Apply boundary conditions:** - At $x=0$: $$u_x(0,y,t) = A$$ Calculate derivative: $$u_x = F_x(x+t,y+t) - \cos t (-\sin x) \cos y + C (-\sin x) \cos y$$ At $x=0$, $\sin 0=0$, so: $$u_x(0,y,t) = F_x(t,y+t) = A$$ - At $x=\pi$: $$u_x(\pi,y,t) = F_x(\pi+t,y+t) = 0$$ - At $y=\pi$ and $y=\infty$: $$u(x,\pi,t) = F(x+t,\pi+t) - \cos t \cos x \cos \pi + C \cos x \cos \pi = 0$$ Since $\cos \pi = -1$, $$F(x+t,\pi+t) + \cos t \cos x - C \cos x = 0$$ 7. **Initial conditions:** At $t=0$: $$u(x,y,0) = F(x,y) - \cos 0 \cos x \cos y + C \cos x \cos y = \alpha(x,y)$$ $$u_t(x,y,0) = F_t(x,y) + \sin 0 \cos x \cos y = \beta(x,y)$$ Since $\sin 0=0$, $$u_t(x,y,0) = F_t(x,y) = \beta(x,y)$$ 8. **Summary:** - $F$ and $C$ are determined by boundary and initial conditions. - The solution combines a traveling wave $F(x+t,y+t)$ and a particular solution involving $\cos t \cos x \cos y$. This approach outlines the method to solve the PDE with given BC and IC. **Final answer:** $$u(x,y,t) = F(x+t,y+t) - \cos t \cos x \cos y + C \cos x \cos y$$ where $F$ and $C$ satisfy the boundary and initial conditions.