Verified answer

Particular solution:

Assume that u(x,y) = A e^(2y) sin x.

Substituting this into the PDE yields

-A e^(2y) sin x + 4A e^(2y) sin x = e^(2y) sin x

==> A = 1/3.

--------------------

Now, let u(x,y) = v(x,y) + (1/3) e^(2y) sin x.

So, the BVP reduces to v_xx + v_yy = 0

with v(0, y) = 0, v(π,y) = 0, v(x,0) = (-1/3) sin x, v(x, L) = [1 - (1/3)e^(2L)] sin x.

We solve this by adding the solutions to v_xx + v_yy = 0 to the individual BVP's

(i) v₁(0, y) = 0, v₁(π,y) = 0, v₁(x,0) = (-1/3) sin x, v₁(x, L) = 0.

(ii) v₂(0, y) = 0, v₂(π,y) = 0, v₂(x,0) = 0, v₂(x, L) = [1 - (1/3)e^(2L)] sin x.

---

Summarizing the separation of variables procedure...

Letting v(x,y) = X(x) Y(y) yields X'' Y + XY'' = 0 ==> X''/X = -Y''/Y = -k^2 (for nonzero solutions).

So, we have

X'' + k^2 X = 0 with X(0) = X(π) = 0

Y'' - k^2 Y = 0

Solving the first ODE yields X_n = sin(nx) with n = 1, 2, 3, ... (and k = n).

--------

For (i), solving Y'' - n^2 Y = 0 with Y(L) = 0 yields Y_n = A cosh(n(y-L)) + B sinh(n(y-L))

(the -L's rearrange the arbitrary constants to make it easier to apply the B.C.)

Since Y(L) = 0, we have A = 0.

==> Y_n = B_n sinh(n(y-L)).

Hence,

v₁(x,y) = Σ(n = 1 to ∞) v_n(x, y), since the PDE is linear and homogeneous

.........= Σ(n = 1 to ∞) B_n sinh(n(y-L)) sin(nx).

Since v₁(x,0) = (-1/3) sin x, we find that

(-1/3) sin x = Σ(n = 1 to ∞) -B_n sinh(nL) sin(nx).

==> -1/3 = -B₁ sinh L ==> B₁ = 1/(3 sinh L) and all other B_n's =0.

Thus, v₁(x,y) = (1/(3 sinh L)) sinh(y-L) sin x.

-------

For (ii), solving Y'' - n^2 Y = 0 with Y(L) = 0 yields Y_n = A cosh(ny) + B sinh(ny)

Since Y(L) = 0, we have A = 0.

==> Y_n = B_n sinh(ny).

Hence,

v₂(x,y) = Σ(n = 1 to ∞) v_n(x, y), since the PDE is linear and homogeneous

.........= Σ(n = 1 to ∞) B_n sinh(ny) sin(nx).

Since v₂(x, L) = [1 - (1/3)e^(2L)] sin x, we find that

[1 - (1/3)e^(2L)] sin x = Σ(n = 1 to ∞) B_n sinh(nL) sin(nx).

==> [1 - (1/3)e^(2L)] = B₁ sinh L ==> B₁ = (1 - (1/3)e^(2L))/sinh L and all other B_n = 0.

Thus, v₂(x,y) = ((1 - (1/3)e^(2L))/sinh L) sinh y sin x.

----------------

So, v(x, y) = v₁(x,y) + v₂(x,y)

.................= [(1/(3 sinh L)) sinh(y-L) + ((1 - (1/3)e^(2L))/sinh L) sinh y] sin x

Finally,

u(x, y) = [(1/(3 sinh L)) sinh(y-L) + ((1 - (1/3)e^(2L))/sinh L) sinh y] sin x + (1/3) e^(2y) sin x.

----------------

I hope this helps!

First, lets try to find a particular solution to the inhomogeneous equation. The rhs of the equation strongly suggests we try something of the form:

U(x,y) = αe^2y sin(x).

I will use the notation Ux to indicate ∂U/∂x, etc.

Uxx = -αe^2y sin(x)

Uyy = 4αe^2y sin(x)

The differential equation tells us

Uxx + Uyy = e^2y sin(x)

-α e^2y sin(x) + 4α e^2y sin(x) = e^2y sin(x)

Solve for α to get

3α = 1

α = 1/3

So our particular solution is

U(x,y) = (1/3) e^2y sin(x)

Next, check to see if this matches the boundary conditions. If it does, then we are all done.

U(0,y) = (1/3)e^2y sin(0) = 0 .............. so the u(0,y) boundary condition is satisfied.

Similarly, the u(π,y) boundary condition is satisfied.

BUT

u(x,0) = (1/3)sin(x) ≠ 0 so the boundary conditions at y = 0 is not satisfied.

The same is true for the y=L boundary condition except for the very special case where

(1/3)e^2L = 1

Since the particular solution does satisfy the y boundary conditions we need to find the general solution to the homogeneous equation. However, because the particular solution does fit the x boundary conditions we don't actually need the general solution for the x partial derivatives.

Now the general solution to

∇^2u(x,y) = 0 is found by separation of variables. Assume u(x,y) = X(x)Y(y). Then

∇^2u = X_xx Y + X Y_yy =0

divide by XY to get

X_xx/X + Y_yy/Y = 0

this can only be true for all x and y if

X_xx/X = -k^2 = - Y_yy/Y

I have chosen to set the constant to k^2 because then the general solutions are sin(kx),cos(kx) and e^ky, e^-ky

Now we know that the boundary conditions at x=0 restrict us to the sin(kx) solution. The BC at x=π means that

sin(kπ) = 0. This is satisfied as long as k is an integer.

So we now have general solutions of the form:

sin(kx)[Ae^-ky + Be^ky]

The BC at y = 0 tells us A+B = 0

So our general solution now looks like

Asin(kx)[e^ky - e^-ky]

I can rewrite this as

Asin(kx) sinh(ky) ....... I've redefined the constant A but that's OK because it still hadn't been determined.

Finally the BC at y = L is

Asin(kx) sinh(kL) = sin(x).

This tells us that k=1 and

A sinh(2L) = 1

A = 1/sinh(2L)