### Problem StatementFind the steady-state temperature on the plate shown below. Your solution \( u(x, y) \) should depend on the constants \( a \) and \( b \), and the function \( f(y) \).### Description of DiagramThe diagram depicts a rectangular plate positioned in a coordinate system with axes labeled \( x \) and \( y \). The plate boundaries are defined by certain temperature conditions:- **Left Boundary**: The temperature equation is given by \( u = b - y \).- **Top Boundary**: The temperature is constant at \( u = 0 \).- **Right Boundary**: The partial derivative of temperature with respect to \( x \) is \( u_x = f(y) \). The top right corner of the plate is at the coordinates \( (a, b) \).- **Bottom Boundary**: The temperature is constant at \( u = b \).These conditions will influence the solution for the temperature distribution \( u(x, y) \) across the plate.

The temperature function u(x,y) will be given by the solution of partial differential equation uxx+uyy=0, with the boundary conditions u(x,0)=b, u(x,b)=0,u(0,y)=b-y,ux(a,y)=f(y) Using the superposition position principle, the solution of this equation can be written as the sum of three functions u1(x,y), u2(x,y) and u3(x,y), where u1, u2 and u3 are solutions of the following three problems. Problem (1): u1xx+u1yy=0, with the boundary conditions u1(x,0)=b ...(1a) u1(x,b)=0 ...(1b) u1(0,y)=0 ...(1c) u1x(a,y)=0 ...(1d) Problem (2): u2xx+u2yy=0, with the boundary conditions u2(x,0)=0 ...(2a) u2(x,b)=0 ...(2b) u2(0,y)=b-y ...(2c) u2x(a,y)=0 ...(2d) Problem (3): u3xx+u3yy=0, with the boundary conditions u3(x,0)=0 ...(3a) u3(x,b)=0 ...(3b) u3(0,y)=0 ...(3c) u3x(a,y)=f(y) ...(3d) So, we find the solution of these 3 problems separately and then use superposition principle to write the final solution. Consider the solution u1(x,y) as a product of two functions X(x) and Y(y), where X(x) is a function of x and Y(y) is a function of y. u1(x,y)=X(x)Y(y) Differentiating u1 twice with respect to x and with respect to y individually, we get u1xx=X”(x)Y(y)u1yy=X(x)Y”(y) Substitute it in the differential equation and simplify. X"(x)Y(y)=-X(x)Y"(y) X”X=-Y”y Now, since the left-side of this equation is a function of x and the right side is a function of y the equation can be equated to constant k. This gives us the following two differential equations X”-kX=0 …(4a) Y”+kY=0 …(4b)Consider 3 cases based on the value of constant k. Case k=0: For k=0 we have X”=0. The solution of which is X=c1x+c2, where c1 and c2 are constants. Now, the boundary condition (1c) and (1d) suggests X(0)=0 and X’(a)=0. Use these conditions to find the values of constants. X(0)=0 ⇒ c2=0 X’(a)=0⇒ c1=0 So, we get the trivial solution X(x)=0 in this case, which further gives u(x,y)=X(x)Y(y)=0. Case k>0: Write k=l2, then the differential equation of X becomes X”-l2X=0. The solution of which is of the form X=c1cosh(lx)+c2sinh(lx), where c1 and c2 are constants. Now, use the conditions X(0)=0 and X’(a)=0 to find the values of constants. X(0)=0 ⇒ c1=0 X’(a)=0⇒ c2=0 In this case too we get the trivial solution.Case k<0: Write k=-l2, then the differential equation of X becomes X”+l2X=0. The solution of which is of the form X=c1cos(lx)+c2sin(lx), where c1 and c2 are constants. Now, use the conditions X(0)=0 and X’(a)=0 to find the values of constants. X(0)=0 ⇒ c1=0 X’(a)=0⇒ c2cos(la)=0 The equation c2cos(la)=0 implies either c2=0 or cos(la)=0. For a non-trivial solution we take c2≠0, so we should have cos(la)=0. Thus, we get the values of l as l=(2n-1)π2a, where n=1,2,3,.... Therefore, the solutions of X”+l2X=0 are Xn=C2n-1sin(2n-1)π2a x. Now, after substitution of k=-l2 in Y”+kY=0 we get Y”-l2Y=0, the solution of which is of form Y=c1cosh(ly)+c2sinh(ly). Plugging the values of l, we get the solutions Yn as Yn= C12n-1cosh(2n-1)pi2ay+C22n-1sinh(2n-1)pi2ay Multiplying Xn and Yn and using superposition principle, we get the solution u1 as u1(x,y)=∑n=1∞ A2n-1cosh(2n-1)π2ay+B2n-1sinh(2n-1)π2aysin(2n-1)π2ax Now, applying the conditions (1a), we get ∑n=1∞ A2n-1sin(2n-1)π2ax=b So, from Fourier sine series the constants A2n-1 are given as A2n-1=2ba∫0asin(2n-1)π2axdx And applying condition (1b), we get ∑n=1∞ A2n-1cosh(2n-1)π2ab+B2n-1sinh(2n-1)π2absin(2n-1)π2ax=0B2n-1=-A2n-1coth(2n-1)π2ab Thus, the solution u1(x,y)=∑n=1∞ A2n-1cosh(2n-1)π2ay+B2n-1sinh(2n-1)π2aysin(2n-1)π2ax, Where, A2n-1=2ba∫0asin(2n-1)π2axdx and B2n-1=-A2n-1coth(2n-1)π2ab.Considering the solution u2(x,y)=X(x)Y(y) and following the similar procedure as in problem(1) we get the same set of differential equations (4a) and (4b). In this case, the boundary conditions (2a) and (2b) suggest Y(0)=0 and Y(b)=0. So, we solve Y”+kY=0 first. Considering the 3 cases based on the value of k as did in problem 1, we will find that for k=0 and k<0 the equation will have the trivial solution. So, only consider the case k>0. Write k=l2, then the differential equation of Y becomes Y”+l2Y=0. The solution of which is of the form Y=c1cos(ly)+c2sin(ly), where c1 and c2 are constants. Now, use the conditions Y(0)=0 and Y(b)=0 to find the values of constants. Y(0)=0 ⇒ c1=0 Y(b)=0⇒ c2sin(lb)=0 The equation c2sin(lb)=0 implies either c2=0 or sin(lb)=0. For a non-trivial solution we take c2≠0, so we should have sin(lb)=0. Thus, we get the values of l as ll=nπb, where n=1,2,3,... Therefore, the solutions are Yn=Cnsinnπb y. Now, after substitution of k=l2 in X”-kX=0 we get the solution of which is of form X=c1cosh(lx)+c2sinh(lx). Plugging the values of l, we get the solutions Xn as Xn=C1ncoshnπbx+C2nsinhnπbx. Multiplying Xn and Yn and using superposition principle, we get the solution u2 as u2(x,y)=∑n=1∞ Ancoshnπbx+Bnsinhnπbxsinnπby Now, applying the conditions (2c), we get ∑n=1∞ Ansinnπby=b-y So, from Fourier sine series the constants An are given as An=2b∫0b(b-y)sinnπbydy And applying condition (2d), we get ∑n=1∞ nπbAnsinhnπba+nπbBncoshnπbasinnπby=0Bn=-An tanhnπba Thus, the solution is u2(x,y)=∑n=1∞ Ancoshnπbx+Bnsinhnπbxsinnπby, where An=2b∫0b(b-y)sinnπbydy and Bn=-An tanhnπba. In this case also, if we let u3(x,y)=X(x)Y(y) we get the same set of differential equations (4a) and (4b). And, the boundary conditions (3a) and (3b) also suggest Y(0)=0 and Y(b)=0. Therefore, following the same process as in problem 2, we get the solution u3(x,y)=∑n=1∞ Ancoshnπbx+Bnsinhnπbxsinnπby Now, applying the conditions (3c), we get ∑n=1∞ Ansinnπby=0 For non-trivial solution sinnπby≠0, therefore we have An=0. And applying condition (3d), we get ∑n=1∞ nπbBncoshnπbasinnπby=f(y) Consider nπbBncoshnπba=Dn. As, the term is constant for each n. Then, from Fourier sine series the constants Dn are given as Dn=2b∫0bf(y)sinnπbydy. This implies, Bn=2nπcoshnπba∫0bf(y)sinnπbydy Thus, the solution is u3(x,y)=∑n=1∞ Bnsinhnπbxsinnπby, where Bn=2nπcoshnπba∫0bf(y)sinnπbydy.Thus, the final solution is u(x,y)=u1(x,y)+u2(x,y)+u3(x,y), where u1(x,y), u2(x,y) and u3(x,y) are as found above.